Slide 1

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 2

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 3

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 4

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 5

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 6

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 7

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 8

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 9

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 10

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 11

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 12

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 13

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 14

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 15

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 16

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 17

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 18

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 19

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 20

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 21

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 22

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 23

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 24

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 25

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING


Slide 26

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY

KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING

ORGANIC REACTION SEQUENCES
INTRODUCTION
This Powerpoint show is one of several produced to help students understand
selected topics at AS and A2 level Chemistry. It is based on the requirements of
the AQA and OCR specifications but is suitable for other examination boards.
Individual students may use the material at home for revision purposes or it may
be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available
from the KNOCKHARDY SCIENCE WEBSITE at...

www.argonet.co.uk/users/hoptonj/sci.htm
Navigation is achieved by...
either

clicking on the grey arrows at the foot of each page

or

using the left and right arrow keys on the keyboard

CONVERSIONS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

ALKANES

DIBROMOALKANES

ALKENES

KETONES

ALCOHOLS

ALDEHYDES

HALOGENOALKANES

AMINES

ESTERS

NITRILES

CONVERSIONS

CARBOXYLIC ACIDS

REACTIONS OF ORGANIC COMPOUNDS

POLYMERS

DIBROMOALKANES

KETONES
P

F

C

S

D
ALKANES

ALKENES

E

ALCOHOLS
M
N

A

B

G

L

T

U

R

V

HALOGENOALKANES

ALDEHYDES

ESTERS
U

H

O

Q

I
T

AMINES

J

NITRILES

K

CARBOXYLIC ACIDS

A

Initiation

CHLORINATION OF METHANE
Cl2

Propagation

Cl• + CH4
Cl2 + CH3•

Termination

Cl• + Cl•
Cl• + CH3•
CH3• + CH3•

——> 2Cl•

radicals created

——> CH3• + HCl
——> CH3Cl + Cl•

radicals used and
then re-generated

——>
——>
——>

radicals removed

Cl2
CH3Cl
C2H6

Summary
Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to
react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you will get further substitution and a mixture of chlorinated products

CONVERSIONS

ELECTROPHILIC ADDITION OF HBr

B

Reagent

Hydrogen bromide... it is electrophilic as the H is slightly positive

Condition

Room temperature.

Equation

C2H4(g) + HBr(g) ———> C2H5Br(l)

bromoethane

Mechanism

Step 1

As the HBr nears the alkene, one of the carbon-carbon bonds breaks
The pair of electrons attaches to the slightly positive H end of H-Br.
The HBr bond breaks to form a bromide ion.
A carbocation (positively charged carbon species) is formed.

Step 2

The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

C

ELECTROPHILIC ADDITION OF BROMINE

Reagent

Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 )

Conditions

Room temperature. No catalyst or UV light required!

Equation

C2H4(g) + Br2(l)

——>

CH2BrCH2Br(l)

1,2 - dibromoethane

Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.

CONVERSIONS

D

Reagent

DIRECT HYDRATION OF ALKENES
steam

Conditions high pressure
Catalyst

phosphoric acid

Product

alcohol

Equation

C2H4(g) +

Use

ethanol manufacture

Comments

It may be surprising that water needs such vigorous conditions to
react with ethene. It is a highly polar molecule and you would expect
it to be a good electrophile.

H2O(g)

C2H5OH(g)

ethanol

However, the O-H bonds are very strong so require a great deal of
energy to be broken. This necessitates the need for a catalyst.

CONVERSIONS

HYDROGENATION

E

Reagent
Conditions

hydrogen
nickel catalyst - finely divided

Product

alkanes

Equation

C2H4(g) + H2(g)

Use

margarine manufacture

———>

C2H6(g)

CONVERSIONS

ethane

POLYMERISATION OF ALKENES

F

EXAMPLES OF ADDITION POLYMERISATION
ETHENE

POLY(ETHENE)

PROPENE

POLY(PROPENE)

CHLOROETHENE

POLY(CHLOROETHENE)
POLYVINYLCHLORIDE

TETRAFLUOROETHENE

PVC

POLY(TETRAFLUOROETHENE)
PTFE

CONVERSIONS

“Teflon”

G

NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE

Reagent
Conditions
Product
Nucleophile
Equation

Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l)

+

NaBr(aq)

Mechanism

* WARNING

It is important to quote the solvent when answering questions.
Elimination takes place when ethanol is the solvent
The reaction (and the one with water) is known as HYDROLYSIS

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Reagent
Conditions
Product
Nucleophile

Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

Equation

(i)
(ii)

C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes

The equation shows two ammonia molecules.
The second one ensures that a salt is not formed.
Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

H

AMMONIA
Why excess ammonia?
The second ammonia molecule ensures the removal of HBr which would lead to the
formation of a salt. A large excess ammonia ensures that further substitution doesn’t
take place - see below
Problem
The amine produced is also a nucleophile (lone pair on N) and can attack another
molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and
reacts further producing a 3° amine and, eventually a quarternary ammonium salt.

C2H5Br

——>

HBr

+

(C2H5)2NH

diethylamine, a 2° amine

(C2H5)2NH + C2H5Br

——>

HBr

+

(C2H5)3N

triethylamine, a 3° amine

(C2H5)3N

——>

(C2H5)4N+ Br¯

C2H5NH2

+

+

C2H5Br

tetraethylammonium bromide, a 4° salt

CONVERSIONS

NUCLEOPHILIC SUBSTITUTION

I

POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation

Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +

KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance

J
K

it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.

Hydrolysis
Reduction

C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS

NH3

ELIMINATION

L

Reagent

Alcoholic sodium (or potassium) hydroxide

Conditions

Reflux in alcoholic solution

Product

Alkene

Mechanism

Elimination

Equation

C3H7Br

+

NaOH(alc)

———>

C3H6

+

H2O

+

NaBr

Mechanism

the OH¯ ion acts as a base and picks up a proton
the proton comes from a C atom next to the one bonded to the halogen
the electron pair moves to form a second bond between the carbon atoms
the halogen is displaced;
overall there is ELIMINATION of HBr.
With unsymmetrical halogenoalkanes, a mixture of products may be formed.
CONVERSIONS

ELIMINATION OF WATER (DEHYDRATION)

L

Reagent/catalyst

conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)

Conditions

reflux at 180°C

Product

alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Equation
Mechanism

Step 1
Step 2
Step 3
Note

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene

Alcohols with the OH in the middle of a chain can have
two ways of losing water. In Step 3 of the mechanism,
a proton can be lost from either side of the carbocation.
This gives a mixture of alkenes from unsymmetrical alcohols...

CONVERSIONS

OXIDATION OF PRIMARY ALCOHOLS

N

Primary alcohols are easily oxidised to aldehydes
e.g.

———>

CH3CH2OH(l) + [O]

CH3CHO(l) + H2O(l)

it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]

———>

OXIDATION TO
ALDEHYDES
DISTILLATION

CH3COOH(l)

OXIDATION TO
CARBOXYLIC ACIDS
REFLUX

Aldehyde has a lower boiling point so
distils off before being oxidised further

Aldehyde condenses back into the
mixture and gets oxidised to the acid
CONVERSIONS

O

OXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O]

•
•
•
•

———>

CH3COOH(l)

one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation
ALDEHYES are EASILY OXIDISED
KETONES are RESISTANT TO MILD OXIDATION
reagents include TOLLENS’ REAGENT
and FEHLING’S SOLUTION

TOLLENS’ REAGENT
Reagent
ammoniacal silver nitrate solution
Observation
a silver mirror is formed on the inside of the test tube
Products
silver + carboxylic acid
Equation
Ag+ + e- ——> Ag
FEHLING’S SOLUTION
Reagent
a solution of a copper(II) complex
Observation
a red precipitate forms in the blue solution
Products
copper(I) oxide + carboxylic acid
Equation
Cu2+ + e- ——> Cu+
CONVERSIONS

OXIDATION OF SECONDARY ALCOHOLS

P

Secondary alcohols are easily oxidised to ketones
e.g.

CH3CHOHCH3(l) + [O]

———> CH3COCH3(l) + H2O(l)

The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment
with a powerful oxidising agent they can be further oxidised to a mixture of acids
with fewer carbon atoms than the original alcohol.

CONVERSIONS

Q

REDUCTION OF CARBOXYLIC ACIDS

Reagent/catalyst

lithium tetrahydridoaluminate(III) LiAlH4

Conditions

reflux in ethoxyethane

Product

aldehyde

Equation

e.g.

CH3COOH(l)

+ 2[H]

———>

CONVERSIONS

CH3CHO(l) + H2O(l)

REDUCTION OF ALDEHYDES

R

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

primary alcohol

Equation

e.g.

C2H5CHO(l) + 2[H]

———>

CONVERSIONS

C3H7OH(l)

S

REDUCTION OF KETONES

Reagent

sodium tetrahydridoborate(III) NaBH4

Conditions

warm in water or ethanol

Product

secondary alcohol

Equation

e.g. CH3COCH3(l) + 2[H]

———>

CONVERSIONS

CH3CH(OH)CH3(l)

ESTERIFICATION

T

Reagent(s)

carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions

reflux

Product

ester

Equation

e.g. CH3CH2OH(l) + CH3COOH(l)

CH3COOC2H5(l) + H2O(l)

Notes

Concentrated H2SO4 is also a dehydrating agent, it removes
water as it is formed causing the equilibrium to move to the right
and thus increasing the yield of ester.

Uses of esters

Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings

Naming esters

Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH
from ethanoic acid

CH3COOCH3 + H2O

CH3COOCH3
METHYL ETHANOATE
CONVERSIONS

from methanol

U

HYDROLYSIS OF ESTERS

Reagent(s)

dilute acid or dilute alkali

Conditions

reflux

Product

carboxylic acid and an alcohol

Equation

Notes

e.g. CH3COOC2H5(l) + H2O(l)

CH3CH2OH(l) + CH3COOH(l)

If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)

CONVERSIONS

BROMINATION OF ALCOHOLS

V

Reagent(s)

conc. hydrobromic acid HBr(aq) or
sodium (or potassium) bromide and concentrated sulphuric acid

Conditions

reflux

Product

haloalkane

Equation

C2H5OH(l) + conc. HBr(aq)

Mechanism

The mechanism starts off in a similar way to dehydration
(protonation of the alcohol and loss of water) but the carbocation
(carbonium ion) is attacked by a nucleophilic bromide ion in step

———>

C2H5Br(l) + H2O(l)

3.

Step 1
Step 2
Step 3

protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation

CONVERSIONS

AN INTRODUCTION TO

SYNTHETIC ORGANIC
CHEMISTRY
THE END

© 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING