Transcript www.hfcrd.ab.ca

Organic Chemistry

Carbon is a tramp - video
Organic Chemistry
A. Carbon Compounds
 organic compounds are those in which carbon atoms
are almost always bonded to each other, to hydrogen
atoms and a few other atoms (O, N, S, P)
inorganic exceptions are the oxides of carbon,
carbonates, cyanides and carbides
(no C-C bonds or C-H bonds)
eg) CO2, CaCO3, NaCN, SiC
 there are millions of organic compounds and
all contain
covalent bonds
carbon is unique for two reasons:
a) it can bond with other carbon atoms to
form long chains, rings, spheres,
tubes, sheets etc.
b) it can form combinations of single, double
and triple bonds
(no other element does this!!!!)
Practice question
 recall polarity and intermolecular forces from the
chemical bonding unit
 polar bonds are formed when there is an uneven
pull on e
 polar compounds are formed when the polar bonds
within a molecule do not cancel each other out
 the presence of dipole-dipole forces and hydrogen
bonding will allow polar compounds to dissolve in
water, since it is also polar
 non-polar compounds only have LD forces between
molecules and will not dissolve in water
B. Structural Isomers
isomers are compounds with the same molecular
formula but a different structure
Example 1
Draw the three structural isomers for C5H12.
1.
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
3.
H
2.
H
H
H
CH3 H
C
C
H
C H3 H
C
H
H
H
H
C
C
C
C
H
CH3 H H
H
H
Example 2
Draw three structural isomers for C4H8F2 .
1.
H
H
H
H
F
C
C
C
C
H
H
2.
H
H
H F
3.
H
H
H
H
F
C
C
C
C
H
F
H H
H
H
H
F
C
C
C
C
H
H
F
H
H
H
different structures result in different properties
 the arrangement of the atoms determines the types of
intermolecular attractions which then determines
properties such as boiling point and solubility
in water
Example
Draw two isomers of C3H8O. Which one would have a
boiling point of 7.4C and which would have a boiling
point of 82.5C? Explain why the boiling points are so
different.
H
H
H
H
C
C
C
H
OH H
H
H
82.5C – has HB
between molecules
which makes the boiling
point quite high
H
C
H
O
H
H
C
C
H
H
H
7.4C – does not have HB
between molecules
therefore the boiling point
is significantly lower
Practice Question
C. Formulas and Structural Diagrams
 organic molecules can be drawn in three different
ways:
1. complete structural diagram – shows all bonds
eg) C3H8
H
H
H
H
eg)
C
C
C
H
H
H
C3H7F
H
H
H
H
H
C
C
C
H
H
H
F
2. condensed structural diagrams – shows carbon to
carbon bonds but includes the hydrogens etc.
attached to each carbon
eg)
C3H8
eg)
C3H7F
CH3
CH2
CH3
CH3
CH2
CH2F
3. line structural diagrams – shows only
carbon
bonds
eg)
C3H8
eg)
C5H12
eg)
C4H8
carbon to
D. Prefixes
1. Number of Functional Groups
6 = hexa
2 = di
7 = hepta
3 = tri
8 = octa
4 = tetra
9 = nona
5 = penta
10 = deca
2. Number of Carbons
1 = meth
6 = hex
2 = eth
7 = hept
3 = prop
8 = oct
4 = but
9 = non
5 = pent
10 = dec
E. Alkanes
CnH2n+2
eg) C5H12, C20H42, etc.
 hydrocarbons containing only single bonds
ie) they are
 can be long
ring
SATURATED
continuous chains, branched chains,
structures (cycloalkanes)
1. Properties
 nonpolar not soluble
in water
 can be solid, liquid or gas
depending on
number of carbon atoms
 relatively unreactive because the single bonds
are very stable
2. Uses
 natural gas, BBQ’s, lighter fluid, gasoline etc
 good for making plastics, lubricants
3. Naming
 IUPAC = International Union of Pure and Applied
Chemistry
i. Continuous Chains
 prefix + “ANE” (suffix)
eg)
H
H
H
H
H
C
C
C
C
H
H
H
H
H
butane
ii. Branched Chains
 branches are called alkyl
functional group
 1 C = methyl ; 2C = ethyl ; 3 C = propyl etc
 find the longest carbon chain and number it
so the branches get the lowest possible numbers
 to name: name the groups first (in alphabetical
order), including the number of the carbon where
each group is found, then name the longest chain
(parent name)
H
eg)
H
H
H
H
1 2 3 4
5
H C C C C C
H
H
CH3 H
H
3-methylpentane
H
methyl
H
H
1
C
H
H
2
C
H
3
C
H
4
C
H
5
C
CH3 CH2 H
H
methyl CH3
ethyl
H
3-ethyl-2-methylpentane
Step #1 – find the longest parent chain
Step #2 – draw a square around the branches
Step #3 – give numbers to the chain: make sure
the branches get the smallest number possible
H
H
H
H
H
H
1 2 3 4 5
C C C C C
H
H
CH3 H
H
H
3 – methyl pentane
eg)
H
H
H
H
H
5 4 3 2
1
H C C C C C
H
H
H
CH3 CH3 H
methyl methyl
2,3-dimethylpentane
iii. Cycloalkanes
 use the
ring structure as the “parent”
 “cyclo”+ prefix + ANE
eg)
cyclobutane
cyclopropane
name
 if there are branches, number the carbons in the ring
so the branches get the
sequence
lowest possible number
eg)
5
4
3
1
2
CH2 CH3
ethyl
CH3
methyl
1-ethyl-3-methylcyclopentane
Practice Questions
F. Alkenes
CnH2n
eg) C5H10, C20H40, etc.
 hydrocarbons containing one or more double
bonds
ie) they are UNSATURATED
 can be long
continuous chains, branched chains,
ring structures (cycloalkenes)
1. Properties
 nonpolar not soluble in water
 lower boiling point
than corresponding alkane
because they have fewer e which makes the LD
forces of attraction weaker
eg) ethane BP = 88.6C
 more reactive
ethene BP = 103.8C
than alkanes
 double bond has more e-
in the same area
 greater repulsion and bond less stable
 diagnostic test: use KMnO4(aq) or Br2(l)
Show video
***alkenes (double bond) will react with these
substances causing a noticeable colour change,
alkanes will not
Br2(l)
A
alkene
B
KMnO4(l)
alkane
A
alkane
B
alkene
*** the alkenes will react causing the colour to
disappear as the coloured substance is used up in
the reaction
2. Uses
 plastics (PVC)
 steroids
 welding torches
3. Naming
i. Continuous Chains
 prefix + “ENE” (suffix)
 number carbons to give the double bond the
lowest number
 the number where the double bond starts is to be
given as a “ # ” between the prefix and the suffix
eg)
H H H
H H
H C C C‗ C C C H
H H
hex-3-ene
H H H
H H H H H
‗
H C C C C C H
H
H H
H
H H H
C
H
pent-2-ene
‗
C
C
C
H H
H
but-1-ene
ii. Branched Chains
 find the longest carbon chain and number it so the
double bond
gets the lowest possible number
 to name: name the groups first (in alphabetical
order), including the number of the carbon
where each is found, then the parent name
including the number of the carbon where
the double bond starts
eg)
CH2‗ C
CH2 CH2 CH2 CH2 CH3
CH2
2-ethylhept-1-ene
CH3
CH3 CH
‗
CH
CH
CH
CH
CH3
CH3 CH3 CH2
CH2
CH3
4,5,6-trimethylnon-2-ene
iii. Cycloalkenes
 double bond is always numbered
1, 2
 branches get the lowest numbering sequence
after the double bond
 “cyclo”+ prefix + ENE
 list branches in alphabetical order with the
number of the carbon they are on
cyclohexene
eg)
cyclopropene
2
CH2 CH3
3
3-ethylcyclobutene
1
4
CH3
1
4
2
3
CH3
CH2 CH3
3-ethyl-1,3-dimethylcyclobutene
G. Alkynes
CnH2n-2
eg) C5H8, C20H38, etc.
 hydrocarbons containing one or more triple bonds
ie) they are also UNSATURATED
 can be long continuous chains, branched chains
 not plentiful in nature
1. Properties
 nonpolar
 very reactive (more than alkanes and alkenes)
 triple bond has 6 e- in the same area  high force
of repulsion
 boiling points are higher
than corresponding
alkanes and alkenes because of their linear structure
and the nature of triple bonds
2. Uses
 welding torches
3. Naming
i. Continuous Chains
 prefix + “YNE”
 number carbons to give the triple bond
the
lowest number
 the number where the triple bond starts
is to
be given as a “ # ” between the prefix and the
suffix
eg)
H H
H C
C
H H
H
H H
C≡ C
C
C
hex-3-yne
H H
H H
H C C≡ C C C H
H
H
H H
pent-2-yne
ii. Branched Chains
 find the longest carbon chain and number it so the
triple bond
 to name:
gets the lowest number
name the groups first (in alphabetical order),
including the number of the carbon where each is
found, then the parent name including the
number of the carbon where the triple bond starts
eg)
CH≡ C
CH
CH2 CH2 CH2 CH2 CH3
CH2 CH2 CH3
3-propyloct-1-yne
CH3 CH
CH3
C≡ C
CH
CH
CH3
CH3 CH2
CH3
2,5,6-trimethyloct-3-yne
Review
 alkanes – branches, rings – least reactive
 alkenes – branches, rings
 alkynes – branches – most reactive
 all called aliphatics
 all nonpolar and not soluble in water
 major intermolecular forces are LD
boiling points are low
Video
Practice Question
Practice Question
H. Aromatics
 hydrocarbons containing one or more
Video
benzene rings C6H6
OR
 all bonds are the
same length and strength
we draw benzene like this:
1. Properties
 nonpolar
 the benzene ring structure is very stable
 aromatics are characterized by strong aromas
2. Uses
 ASA, amphetamines, adrenaline, benzocaine
(anesthetic)
 moth balls, TNT
 wintergreen, menthol, vanilla, cinnamon
 SPF in sunscreen
3. Naming
i. Benzene as a Branch
 if you have a really long carbon chain, it is easier
to call the benzene ring a “phenyl” group
eg)
4-phenylheptane
CH3 CH2 CH2 CH CH2 CH2 CH3
‗
CH2 CH
CH
CH2 C
CH2 CH2 CH3
CH3
3-methyl-5,5-diphenyloct-1-ene
ii. Benzene as a the Main Compound
 if only one group is attached, give the alkyl name
attached to “benzene”
(no number is necessary)
eg)
CH3
methylbenzene
 if there is more than one branch, number them so
they get the lowest sequence
alphabetically
and name
CH3
eg)
1-ethyl-3-methylbenzene
CH2 CH3
CH3
1,3-dimethyl-5-propylbenzene
CH3
CH3 CH2 CH2
C2H5
CH3
1-ethyl-3-methylcyclohexane
***Watch out for this!!!
http://www.bing.com/videos/search?q=benzene&mid=013F6F84DA6614E381A6013F6F84DA6614
E381A6&FORM=LKVR4#
Practice Question
I. Alcohols
R - OH
 organic compounds with one or more
OH (hydroxyl) groups
 ** OH, its an Alcohol!!!
1. Properties
 have much higher boiling points
corresponding aliphatics because of
bonding!
eg) methane (CH4) BP = -162C
methanol (CH3OH) BP = 65C
than
hydrogen
 polar
 the –OH
end of the alcohol is polar while the
carbon chain end is not  small alcohols are
soluble in water and large alcohols are not
2. Uses
 antifreeze, rubbing alcohol, beverages, moistening
agent
3. Naming
 number the longest carbon chain containing the
hydroxyl group so the –OH group gets the
lowest number
 aliphatic name (without “e” at end) + “OL”
 give the number for the carbon
where the
–OH group is found between the parent name and
the suffix
 if there is more than one
hydroxyl group, use a
prefix ( di, tri, tetra ) to indicate the number
of OH groups and place the numbers between the
parent name and the suffix
***Note, if the suffix starts with a vowel, drop the “e”
on the parent name; if the suffix starts with a
consonant, keep the “e” on the parent name
eg)
H
H
H
H
H
H
H
C
C
C
C
H
H
OH H
CH3 H
H
C
C
C
C
H
H
OH H
H
H
butan-2-ol
3-methylbutan-2-ol
H
H
CH3 H
H
C
C
C
H
OH OH H
C
H
2-methylbutane-2,3-diol
 an unusual case:
OH
phenol
J. Organic Halides
R-X
 where R is carbon chain or ring
and X is a
halogen
 organic compounds where hydrogen
has been
replaced by one or more halogens (F, Cl, Br, I)
 do not readily occur in nature
1. Properties
 can be polar
or nonpolar , depending on
the placement of the halogen groups
 many are
toxic and dangerous
2. Uses
 manufactured for human use eg) DDT, PCB, CFC
 anesthetics
 dry cleaning fluid
 plastics, polymers (Teflon)
3. Naming
 same rules as before… name branches ( halogens
included now) alphabetically
F = fluoro Cl = chloro
eg
)
H
H
Cl
Cl
H
C
C
C
C
H
H
H
H
H
Br = bromo
I = iodo
2,3-dichlorobutane
I
2-fluoro-4-iodo-1-methylbenzene
CH3
F
K. Carboxylic Acids
R
O
║
C
OH
 where R is carbon chain or ring
 organic compounds containing the carboxyl
functional group (-COOH)
** o – oh, its an acid!!! – mnemonic device
1. Properties
 polar  dissolve in water
 high
boiling points due to hydrogen bonding
 weakly acidic
 diagnostic test: use litmus paper (will turn red ),
readily react with metals, neutralize bases
2. Uses
 recycling rubber – methanoic acid
 vinegar – ethanoic (acetic) acid
 rust remover – oxalic acid
 fruits – citric acid
3. Naming
 count the longest carbon chain including the
carbon in the carboxyl group
 the carbon in the carboxyl group is always number 1
 drop “e” and add
“OIC ACID”
eg)
O
methanoic acid
║
H
C
H
H
C
H
OH
O
║
C
OH
ethanoic acid
O
║
C
I
OH
H
H
H
C
C
C
H
H
H
benzoic acid
O
║
C
OH
4-iodobutanoic acid
Read pg 438 - 441

Read pg 438 – 441 on esters
L. Esters
R
O
║
C
O
R’
 where R
can be a carbon chain or hydrogen and R’
can be a carbon chain
 combination of a carboxylic acid
and an Alcohol
 ** Ester is a cow that likes to eat OATES
1. Properties
 polar
 small esters dissolve in water, large esters
do not
 boiling points slightly lower than corresponding
carboxylic acids and alcohols due to lack of hydrogen
bonding
 very volatile
which allows them to generate
aromas
2. Uses
 flavouring agents
3. Naming
 identify the alcohol
used to make the ester
 change the alcohol name to the corresponding alkyl
name
eg) methanol would become methyl
 identify the
made from
carboxylic acid
the ester was
 drop the “oic acid” and replace with “oate”
eg) butanoic acid would become “butanoate”
 put the
two names
together with a space
in
between
eg) methyl butanoate
 you can have branches
on esters…they follow
the alphabetical rule, numbering begins at the
O end of the alcohol and the C=O end of the
carboxylic acid
eg)
H H H
H C C C
H H H
H H
H C C
H H
O
H H
║
C O C C H
H H
ethyl butanoate
O
H CH3
║
C O C C H
H H
propyl propanoate
eg)
Video
O
║
H H
H C
C
C
CH3CH3
H H
O C
C
H
H H
ethyl 2-methylbutanoate
Practice Question
Practice Questions
Practice Questions
M. Boiling Points and Solubility
 we can compare the boiling points of various organic
compounds using their
polarity
and the intermolecular attractions
between the molecules
Example 1
Put the following organic compounds in order from
highest boiling point to lowest boiling point.
alcohol, alkane, alkene, aromatic, carboxylic acid
carboxylic acid alcohol aromatic alkane alkene
highest
lowest
Example 2
Put the following homologous series of organic
compounds in order from highest boiling point to
lowest boiling point.
C2H6, C2H5OH, CH3COOH, C2H4
CH3COOH C2H5OH
highest
C2H6
C2H4
lowest
 we can also compare the solubility of various
organic compounds using their polarity
Insoluble Organic
Compounds
Soluble Organic
Compounds
aliphatics – alkanes,
alkenes, alkynes
carboxylic acids
aromatics
alcohols – large (7 or
more carbons)
alcohols – small (less
than 7 carbons)
esters – large
esters – small
organic halides
organic halides
Practice Question
N. Organic Reactions
1. Combustion Reactions
 occurs when a hydrocarbon reacts with oxygen
 products are always
carbon dioxide and water
 these are economically important reactions for they are
the major reactions that produce thermal energy
required for fuelling our vehicles, heating our homes,
and producing electricity
eg) 1 C5H12(l) + 8 O2(g)  5 CO2(g)
+ 6H O
2 (g)
eg) 1 C5H12(l) + 8 O2(g)  5 CO2(g) + 6 H2O(g)
H
H
H
H
H
C
C
C
C
H
H
H
H
H
+ 6.5 O2(g)  4 CO2(g) + 5 H2O(g)
***balance these reactions
2. Addition Reactions
 a double or triple bond in an alkene or alkyne
is broken and a group or element is added
(a catalyst is present)
eg)
H
H
C
H
‗
C
+
H
Cl Cl

Cl
H
H
C
C Cl
H H
H
H
C
‗
C
+
H
H
H
H
C
H
‗
C
OH

H
H
C
C OH
H
+
H
H
H
H H

H
H
H
H
C
C H
H H
H
C
H
‗
C
+
H Br

H
H
H
H
C
C Br
H
H
H
H
H C≡ C H +
1 Cl2

H
C
‗
C
Cl
Cl
Cl Cl
H C≡ C H +
2 Cl2

H
C
C
Cl Cl
H
3. Substitution Reactions
 the replacement ( substitution ) of a hydrogen on
an alkane or aromatic with another atom (eg. F, Cl
etc)
 commonly used to make organic halides
Examples
1.
H
H
C
H
H
H +
cat
Cl – Cl 
H
C
Cl
H
+
H - Cl
2.
+
Br – Br
3.
+
Br
cat

+
cat
I–I 
+
H - Br
H-I
I
4.
H
H
H
H H
C
C
C
C
H
H
H H
H + Br2
cat

H
H
Br
H
H
C
C
C
C
H
H
H
H
H +
H - Br
Practice Question
4. Esterification Reactions
 the reaction of a carboxylic acid with an alcohol
 the catalyst is sulphuric acid
Catalyst video
Examples
1.
H
O
║
C
H
OH +
HO C
H
H2SO4(aq)

H
H
O
║
C
H
O
C
H
H
+
H – OH
2.
H
H
H
C
C
H
H
O
║
C
H
H
OH + HO C
C
H
H
H
H
H
C
C
H
H
O
║
C
H2SO4(aq)
H
O

H
H
C
C
H
H
H +
H – OH
5. Elimination Reactions
 an alcohol has water removed, forming an alkene
plus the water
 organic halides can react with a
base
(hydroxide) to produce an alkene , a
halide ion and water
Examples
1.
H
H
H
C
C
H
H
OH
cat

H
H
H
C‗ C
H
+
H – OH
2.
H
3.
H
H
H
H
H
C
C
C
C
OH H
H
H
H
H
C
C
H
H
Cl
cat

H
+ OH
H
C
‗
H
H
H
C
C
C
H
H
H

cat

H
H
H +
H – OH
H – OH
+ Cl 
H
C‗ C
H
+
6. Polymerization Reactions
 a monomer is a simple molecule
the base unit for a polymer
that forms
 a polymer is a very, very long molecule formed
by the covalent bonding of
monomers
 depending on the polymer,
the monomers that make
it up can be the same or
different
bazillions of
 polymers can be natural
eg) carbohydrates, proteins, DNA
 polymers can be synthetic
eg) nylon, PVC, Teflon, polyester
 polymers that can be heated and molded into specific
shapes are commonly called plastics
 plastics are one type of synthetic compound that has
been of great benefit to society (although there are also
problems associated with then)
 the names of polymers are the monomer name with
“poly” in front
 many have classical names instead of IUPAC names
Addition Polymers
 formed when the electrons in double or triple bonds in
the monomer units are rearranged
 the polymer is the only product
formed
Examples
1.
H
H
‗
C C
H
H
‗
+
H
H
C C
H
H
cat

…
H
H
H
H
C
C
C
C
H
H
H
H
polyethene
…
n
2.
F
F
‗
C C
F
F
‗
+
F
F
C C
F
cat

…
F
F
F
F
F
C
C
C
C
F
F
F
F
…
n
Teflon
3.
H
Cl
‗
C C
H
H
‗
C C
+
H
Cl
H
cat
…
H
H
Cl H
Cl
C
C
C
C
H
H
H
H
…
n
polyvinyl chloride (PVC)
4.
H
H
‗
C C
H
‗
+
H
cat
…
C C
H
H
H
H
C
C
C
C
H
H
H
H
…
n
polystyrene (styrofoam)
Practice Question
Condensation Polymers
 polymerization reactions that involve the formation of
a small molecule (commonly water ) as well as
the polymer
 each monomer must have two functional groups
 two common linkages formed:
1. ester linkage – between carboxyl group
(COOH) and hydroxyl group (OH)
2. amide linkage – between amino group
(NH2) and carboxyl group (COOH)
Examples
1.
O
║
HO C
O
║
C
H
H
OH + HO C
C
H
H
∙∙∙
O
║
O C
OH
cat

ester
linkage
O
H H
║
C O C C
H H
∙∙∙ + H2O
n
polyethylene terephthalate
PET
2.
HO
O H H
║
C C N H + HO
H
O H H
║
C C N H
H
∙∙∙
cat

amide
linkage
O H H O H H
║
║
C C N C C N
H
H
protein
∙∙∙ + H2O
n
Organic Chemistry

Song
O. Petroleum Refining
 Alberta has vast reserves of petroleum in the form of
natural gas, crude oil and oil sand deposits
 most of this petroleum is refined and then burned as
fuel
 petrochemicals are also used in the production of
plastics
 refining of petroleum separates the crude mixture into
purified components
 fractional distillation is used to separate the
components
 distillation works because of the different
boiling points of the components of crude oil
the molecule and the lower the
boiling point, the higher it rises in the tower
(asphalt, fuel oil, wax at bottom; gasoline at top)
 the lighter
distillation tower:
Video
Fractional Distillation




1.
http://www.learnalberta.ca/content/t4tes/courses/senior/chemistry_30/mm/
m5/Fractional/main.html
http://resources.schoolscience.co.uk/spe/knowl/4/flash/cracking.htm
(animation of fractioning tower and cracking)
2. Animated tour http://sciencenetlinks.com/interactives/energy/interactive/api_treat_01281
0.swf
3. Animated tour - http://irvingoil.com/resource/flash/RefineryTour.swf
Fractioning Tower
 two types of reactions in petroleum refining:
1. Cracking
 breaks long chain hydrocarbons into smaller units
 cracking requires heat and pressure
 there are many different types of cracking reactions,
forming different products (alkanes, alkenes)
eg) catalytic cracking, steam cracking, hydrocracking
 hydrocracking requires H2(g)
Examples
1. C17H36 +
cat
H2(g) 
C9H20 + C 8H18
2.
CH3 CH2 CH2 CH2
cat

CH3 CH2 CH2 + H2(g)
CH3 CH2 CH2 CH3
+
CH3 CH2 CH3
2. Reforming
 small hydrocarbons are joined to make
larger molecules
 requires heat and pressure
 there are several types of reforming reactions
eg) alkylation to produce “high octane” gasoline
 all reforming reactions produce H2(g)
Example
C7H16 + C12H26
 C 19H 40
+
H2
Examples
1.
C7H16 + C12H26
 C 19H 40
+
H2
2.
ethane + octane
decane + hydrogen gas
Extraction of Bitumen in Oil sands
lab

http://www.learnalberta.ca/content/t4tes/cours
es/senior/chemistry_30/mm/m5/c30_m5_l05_
mm006/c30_m5_l05_mm006.html
Best song in the world

Family song