Transcript Chemistry 30 – Organic Chemistry – Part 1 To accompany Inquiry into Chemistry

Chemistry 30 – Organic
Chemistry – Part 1
To accompany
Inquiry into Chemistry
PowerPoint Presentation
prepared by Robert Schultz
[email protected]
Organic Chemistry – Preparation – VSEPR
• Recall VSEPR Theory (valence shell electron
pair repulson theory) from Chemistry 20
• Organic chemistry will involve 3 particular
groupings:
• 0 lone pairs, 4 bonding pairs - tetrahedral
••
••
H
••
H C H
••
H
Organic Chemistry – Preparation - VSEPR
• 0 lone pairs, 3 bonding pairs – trigonal planar
••
•• ••
••
O
••
••
H C H
• 0 lone pairs, 2 bonding pairs - linear
••
••
••
••
••
••
O C O
••
••
Organic Chemistry - Preparation
• Recall polarity of covalent bonds from
Chemistry 20:
2 particular polar bonds important in organic
chemistry
O—H
• C – H bonds are virtually non-polar
C=O
Organic Chemistry – Preparation – Intermolecular
Forces
• London Dispersion Forces – all
moleculars – temporary dipoles –
affected by total # of e- and shape
• Dipole-dipole Forces – polar
moleculars
• Hydrogen Bonding (H covalently
bonded to F, O, or N)
affect melting
point, boiling point,
and solubility
Organic Chemistry – 14.1 - Introduction
• Organic compounds – originally defined to be
compounds from living or once-living organisms
• Wohler, 1828, synthesized urea (an organic
compound) from inorganic chemicals
• Today organic compounds defined to be
molecular compounds of carbon – exception:
oxides of carbon – CO, CO2
Organic Chemistry – 14.1 - Introduction
• Most existing compounds are organic!
• Special things about carbon that allow it to
form so many different compounds:
• 4 bonding electrons
• ability to form single, double, triple bonds
with itself
• ability to bond with itself in many different
configurations
Organic Chemistry – 14.1 - Introduction
• Classification:
organic compounds
hydrocarbon derivatives
C and H along with O, N,
and/or halogen atoms
hydrocarbons
C and H only
aliphatics
without
alkanes – all
single bonds –
CnH2n+2
aromatics
with
alkenes – 1 double
bond between C’s –
CnH2n
alkynes – 1 triple
bond between C’s –
CnH2n-2
Organic Chemistry – 14.2 - Hydrocarbons
• Alkanes - saturated hydrocarbons
• Term saturated used because alkanes have the
maximum number of hydrogens
• General formula: CnH2n+2
first 4
alkanes
methane
ethane
propane
butane
Organic Chemistry – 14.2 - Hydrocarbons
• The unbranched alkanes are a homologous
series because they differ by the number of CH2
units in each
• Alkanes are tetrahedral around each carbon
Organic Chemistry – 14.2 - Hydrocarbons
• Since carbons and
hydrogens can join up in
so many ways, structural
formulas are used
• Different types
of structural formulas:
we won’t use
this type
3
3
2
3
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
• Nomenclature of alkanes:
• You must learn the following
prefixes:
# of C’s
prefix
1
meth
2
eth
3
prop
4
but
5
pent
6
hex
7
hept
8
oct
9
non
10
dec
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
• Start naming by finding the longest continuous
chain of carbon atoms. Name the long chain
using its prefix with an ane ending.
• Identify branches, and name using their prefix
with a yl ending.
• Number the longest continuous chain from the
end closest to the branching and use the
numbers like addresses for the branches.
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
• These rules will be introduced by the following
examples
• Several additional rules will be presented with
the examples
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
Example 1:
CH3
CH3 – CH – CH – CH2 – CH2 – CH3
CH2 - CH3
CH3
CH3 – CH – CH – CH2 – CH2 – CH3
CH2 - CH3
Root name: hexane
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
Example 1:
CH3
CH3 – CH – CH – CH2 – CH2 – CH3
CH2 - CH3
CH3 methyl
1
2
3
4
5
6
CH3 – CH – CH – CH2 – CH2 – CH3
Root name: hexane
CH2 - CH3 ethyl
Identify side groups
number carbon chain to locate branches
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
• Compound name:
3-ethyl-2-methylhexane
side group
side group
long chain
position on
long chain
Additional rule: list side groups in alphabetical order
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
Example:
CH3
CH3 – CH – CH – CH – CH3
CH3
CH3
CH3
CH3
CH3
CH3 – CH – CH – CH – CH3
CH3 – CH – CH – CH – CH3
CH3 – CH – CH – CH – CH3
CH3
CH3
CH3
CH3 – CH – CH – CH – CH3
CH3
CH3
CH3
CH3
CH3
CH3
No matter how the long chain is
selected, the name is the same:
2, 3, 4 - trimethylpentane
Note the tri; use di, tri, tetra, etc, but don’t use
them for alphabetical order
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
• Example:
CH2 – CH3
CH3 – CH2 – C – CH3
CH – CH3
CH2 – CH3
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
CH2 – CH3
CH3 – CH2 – C – CH3
CH – CH3
CH2 – CH3
3 – ethyl – 3, 4 – dimethylhexane
or
4 – ethyl – 3, 4 - dimethylhexane
Which one???
lowest set of numbers
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
• Doing the reverse process is actually easier –
draw your long chain and attach the groups in
the addressed spots
• Start by drawing the long chain without any
hydrogens – don’t worry about orientation
• Add side groups in their addressed spots
• Add hydrogens (each C gets 4 bonds)
• Do alkane nomenclature worksheet
Organic Chemistry – 14.2 – Hydrocarbons: Alkanes
• Physical Properties of Alkanes:
• All alkanes are non-polar,

only intermolecular forces = London Dispersion
Forces – boiling point and melting point
increase with number of carbons (see chart
page 551) KNOW
all alkanes are insoluble in water
Organic Chemistry – 14.2 – Hydrocarbons: Alkenes
• Alkenes are hydrocarbons with 1 double bond
• Note dienes and trienes also exist – we’ll focus
on compounds with 1 double bond
• Alkenes with 1 double bond have the general
formula, CnH2n
• Since they have 2 less hydrogens than
corresponding alkanes, they’re called
unsaturated hydrocarbons
Organic Chemistry – 14.2 – Hydrocarbons: Alkenes
• Alkene formulas:
we won’t use
this type
3
3
3
• Alkenes are trigonal planar around the doubly
bonded C’s and tetrahedral around the others
Organic Chemistry – 14.2 – Hydrocarbons: Alkenes
• Nomenclature of alkenes:
find longest continuous chain of carbons that
contains the double bond – same prefixes as for
alkanes
add ene to the prefix along with a number to
indicate the position of the double bond (for
ethene and propene a position number is not
needed)
number the long chain from the end closest to
the double bond (not the branching)
Organic Chemistry – 14.2 – Hydrocarbons: Alkenes
• Example:
CH3 – CH2 – CH2 – C = CH2
CH2
CH3
sidegroup
CH3 – CH2 – CH2 – C = CH2
CH2
CH3
position of
double bond
2 – ethylpent-1-ene
position
of sidegroup
length of
long chain
containing
double
bond
Organic Chemistry – 14.2 – Hydrocarbons: Alkenes
• Do questions 10 – 14 on pages 554-5
Organic Chemistry – 14.2 – Hydrocarbons: Alkenes
• Physical properties of alkenes:
• Like alkanes, alkenes are non-polar and are
insoluble in water
• Boiling points are slightly lower than those for
alkanes with the same number of carbons
Why?
Smaller # of electrons, weaker LDF
boiling point
lower
Organic Chemistry – 14.2 – Hydrocarbons: Alkynes
• Alkynes are unsaturated hydrocarbons with 1
triple bond
• General formula CnH2n-2
• Alkynes are linear around the triply bonded
carbons and tetrahedral around other carbons
Organic Chemistry – 14.2 – Hydrocarbons: Alkynes
• Alkynes are non-polar aliphatic hydrocarbons
like alkanes and alkenes
• They are insoluble in water
Organic Chemistry – 14.2 – Hydrocarbons: Alkynes
• Note that alkynes have higher boiling points
than alkanes or alkenes
• Obviously the explanation used for alkenes
being lower than alkanes doesn’t apply here
Table 14.5, page 557
Organic Chemistry – 14.2 – Hydrocarbons: Alkynes
• Accepted explanation is that for short chain
alkynes, the linear structure around triple bond
allows them to come closer together than
alkanes or alkenes with same number of
carbons, causing stronger London Dispersion
Forces
Organic Chemistry – 14.2 – Hydrocarbons: Alkynes
• Nomenclature of alkynes is identical to that of
alkenes, the only exception is the ending:
yne, not ene
• Do Practice Problems 16 and 17 on pages 556
and 557
Organic Chemistry – 14.2 – Hydrocarbons: Cyclics
• Cyclic analogues exist for alkanes, alkenes, and
alkynes
• General formulas will contain 2 less hydrogens
than the open chain hydrocarbons:
cycloalkanes CnH2n, cycloalkenes CnH2n-2,
cycloalkynes CnH2n-4
• Small cycloalkynes don’t exist because of the
large bond strain that would exist around the
linear triple bond
Organic Chemistry – 14.2 – Hydrocarbons: Cyclics
• Line structures are commonly used for the ring
part of cyclic hydrocarbons
• Always draw them this way
• Examples:
cyclopropane:
cyclobutene:
CH2
not
CH2
CH2
CH2
CH
CH2
CH
not
Organic Chemistry – 14.2 – Hydrocarbons: Cyclics
• Cyclics will always have names ending with
cyclo_____ane or cyclo_____ene
• Don’t worry about cyclo_____ynes, you will not
encounter them, except my favourite one,
STOP
Name?
stopsyne!
• Consider the following examples to learn how
to do the nomenclature for substituted cyclics
Organic Chemistry – 14.2 – Hydrocarbons: Cyclics
CH2 – CH3
ethylcyclopentane
CH2 – CH3
3-ethylcyclopentene
CH2 – CH3
CH3
4-ethyl-3-methlycyclopentene
No numbers needed. Why?
Always start at far side of
double bond and number
clockwise or counterclockwise towards group
As above. This one must be
numbered counter-clockwise
to give lowest set of
numbers, even though 1st
group gets a higher number
Organic Chemistry – 14.2 – Hydrocarbons: Cyclics
CH2 – CH3
CH3 1-ethyl-2-methylcyclopentane
This time the numbering is
clockwise since double
bond isn’t a factor and
when possible lowest
number goes on first group
Do Practice Problems 18 – 23 page 559 and 560
Do Aliphatics Review WS
Quiz coming up!
Organic Chemistry – 14.1 - Introduction
• Classification:
organic compounds
hydrocarbon derivatives
C and H along with O, N,
and/or halogen atoms
hydrocarbons
C and H only
aliphatics
without
alkanes – all
single bonds –
CnH2n+2
aromatics
with
alkenes – 1 double
bond between C’s –
CnH2n
finished with aliphatics;
aromatics today
alkynes – 1 triple
bond between C’s –
CnH2n-2
Organic Chemistry – 14.2 – Hydrocarbons:
Aromatics
• Aromatics: all contain the grouping C6H6
• Originally this grouping thought to be:
or
• Problems: • all bonds found to be equal length
• this compound should be very
reactive but is actually very stable
Organic Chemistry – 14.2 – Hydrocarbons:
Aromatics
• Today we believe it to be made up of bonds
that are neither single nor double but a hybrid
of both
• We draw the structure
• Its name is benzene
• Benzene is the root common to all aromatics
Organic Chemistry – 14.2 – Hydrocarbons:
Aromatics
• Nomenclature of Aromatics:
page 561
Where numbering starts
Organic Chemistry – 14.2 – Hydrocarbons:
Aromatics
• Examples:
CH2 – CH2 – CH3
propylbenzene
CH3
1-ethyl-3-methylbenzene
CH3 – CH – CH3
CH2 – CH3
2-phenylpropane
Organic Chemistry – 14.2 – Hydrocarbons:
Aromatics
• Do Practice Problems 24 – 27, page 562
• Aromatics WS
Organic Chemistry – 14.2 – Hydrocarbons:
Aromatics
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives
organic compounds
hydrocarbons C
and H only
alcohols
R-OH
esters
akyl halides
R-X
alkanes – all
single bonds –
CnH2n+2
alkenes – 1 double
bond between C’s –
CnH2n
alkynes – 1 triple
bond between C’s –
CnH2n-2
O
=
aromatics
with
R1 – C – O – R2
carboxylic acids
O
=
aliphatics
without
hydrocarbon derivatives
C and H along with O, N,
and/or halogen atoms
R-C-OH
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives
• Hydrocarbon derivatives contain other elements
besides C and H; most commonly O, N, or
halogen atom
• Functional group: group of atoms that gives the
compound its characteristic properties
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alcohols
• Alcohols – functional group:
“-OH” hydroxyl group
• Common alcohols: table 14.7, page 566
3
3
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alcohols
• Nomenclature of alcohols
• Key points – long chain must have “–OH”
attached to it
• Numbering of the long chain starts from the
end closest to “-OH”
• Ending of root is ol
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alcohols
Example
CH3 – CH2 – CH2
CH3 – CH – CH2 – CH2 – OH
CH3 – CH2 – CH2
CH3 – CH – CH2 – CH2 – OH
position of
OH
side group
3-methylhexan-1-ol
position of
side group
length of long
chain containing
OH*
* don’t count OH
in length of chain
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alcohols
• Example
OH
CH2 – CH – CH2
OH
OH
position of
OH’s
OH
CH2 – CH – CH2
OH
OH
propane - 1, 2, 3 - triol
length of long
chain containing
OH’s
common name of this compound: glycerol
number
of OH’s
Advantages to above name??
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alcohols
• Do Practice Problems 28 – 30 on page 567
• Omit 28d, 29c, 30a
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alcohols
• Physical properties of alcohols
• Because of the hydrogen bonding between OH
groups in adjacent molecules,
• alcohols have much higher boiling points than
hydrocarbons (1-12 C’s are liquids at SATP)
• small alcohols are totally miscible with
water, but ……………
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alkyl Halides
• Alkyl halides contain at least 1 halogen atom,
(F, Cl, Br, I)
• Alkyl halides are all synthetic compounds
• CFC’s (chlorofluorocarbons) are alkyl halides
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alkyl Halides
• Nomenclature of alkyl halides:
long chain must be attached to halogen atom(s)
identical to nomenclature of hydrocarbons
side groups end in o, not yl – fluoro, chloro,
bromo, iodo
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alkyl Halides
• Example:
Cl
CH3 – CH2 – CH – CH – CH – CH3
Br
Br
Cl
CH3 – CH2 – CH – CH – CH – CH3
Br
Br
2, 4 – dibromo – 3 - chlorohexane
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Alkyl Halides
Br
1,4 – dibromo – 2 - chlorocyclohexane
Cl
Br
• Do Practice Problems 31, 32, page 569
• Do Alcohols/Alkyl Halides Nomenclature WS
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Carboxylic Acids
• Carboxylic acids are weak organic acids
containing the carboxyl functional group,
=
O
- C – OH ,
O
=
often written –COOH
R - C – OH(aq) 
=
=
• When carboxylic acids, R - C – OH , ionize,
the process is:
O
O
H+(aq) + R - C – O-(aq)
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Carboxylic Acids
• Common carboxylic acids, acetic acid (active
ingredient of vinegar) and citric acid
• Nomenclature of carboxylic acids:
In all carboxylic acids the carboxyl group is at
one end of the molecule
It is always carbon #1 in the chain
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Carboxylic Acids
• Example:
O
=
CH3
CH3 – C – CH2 – CH2 – C – OH
CH2
CH3
O
=
CH3
CH3 – C – CH2 – CH2 – C – OH
4, 4 – dimethylhexanoic acid
CH2
CH3
note that the carboxyl carbon does
get counted in the long chain – it is
carbon #1
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Carboxylic Acids
• Do Practice Problems 33 – 35, page 570
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Carboxylic Acids
• Physical properties of carboxylic acids:
• Like alcohols they have hydrogen bonding, but
hydrogen bonding at 2 sites, -C=O and –OH
• This leads to higher boiling points and greater
solubility than alcohols with same number of C’s
• Carboxylic acids with 1-4 C’s are completely
miscible in water
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Esters
• Esters have the general formula:
=
O
R(or H) - C – O – R′
often written RCOOR′
• Esters are formed from the reaction of an
alcohol and a carboxylic acid; the formation or
esterification reaction is the key to naming
them
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Esters
=
R-C–O-H +
carboxylic
acid
H - O - R′
alcohol
=
O
O
R - C – O - R′ + HOH
ester
It’s important that when you look at ester, that
you’re able to recognize part that came from
alcohol and part that came from acid
=
O
Acid part contains C; alcohol part is bonded
directly to O
water
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Esters
• General form of name:
_______yl _________oate
from
alcohol
from
acid
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Esters
• Examples:
=
O
CH3 – CH2 – C – O – CH3
acid part:
propanoate
alcohol part:
methyl
methyl propanoate
=
O
CH3 – CH2 – CH2 – CH2 – O – C – H
alcohol part:
butyl
acid part:
methanoate
butyl methanoate
Organic Chemistry – 14.4 – Refining and Using
Organic Compounds
• Do questions 37 and 38 page 572
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives: Esters
• Physical properties of esters:
• fruity odour in some cases
• polar but lack of OH bond means no hydrogen
bonding, so lower boiling points than alcohols
and carboxylic acids
• esters with few carbons are polar enough to be
soluble in water
Organic Chemistry – 14.3 – Hydrocarbon
Derivatives
• Structural isomers: compounds with same
molecular formula but different structural
formulas
Organic Chemistry – 14.4 – Refining and Using
Organic Compounds
• Petroleum: mixture of hydrocarbons (primarily
alkanes and alkenes) found in natural gas,
crude oil, and bitumen (from tar sands)
• Petrochemicals: hydrocarbon materials from
petroleum used to produce plastics and other
synthetic materials
Organic Chemistry – 14.4 – Refining and Using
Organic Compounds
• Fractional
distillation: a
means of
separating
petroleum
components
based on
differing boiling
points
Organic Chemistry – 14.4 – Refining and Using
Organic Compounds
• Read and discuss page 578 regarding
fractional distillation
• Fractional distillation is a physical
process; mixture is separated into
fragments with a small range of boiling
points – there is no chemical change in the
fractions
Organic Chemistry – 14.4 – Refining and Using
Organic Compounds
• Next stages of petroleum refining are chemical
processes:
• cracking – breaks carbon-carbon bonds
• reforming – forms carbon-carbon bonds
alkylation (special case of reforming) forms
2,2,4-trimethylpentane from smaller
hydrocarbons
• Both of these can be divided into many
subgroups
• Read page 579-80 and page 581
Organic Chemistry – 14.4 – Refining and Using
Organic Compounds
Organic Chemistry – 14.4 – Refining and Using
Organic Compounds