Syed Arshad Mushtaq

 Identify the following different classes: alkanes, alkenes, alkynes, halogenoalkanes, alcohols, ethers, aldehydes, ketones, esters, carboxylic acids, amines, amides, nitriles and arenes
 Identify typical functional groups in molecules e.g. phenyl, hydroxyl, carbonyl, carboxyl, carboxamide, aldehyde, ester, ether, amine, nitrile, alkyl, alkenyl and alkynyl.
It’s Family thing

Picture Source https://www.pinterest.com/pin/199073245998510273
/

Example
Butanenitrile

Identify functional groups
Alcohol
Ether

Identify functional groups in
Acetylsalicylic acid (Aspirin )
Carboxylic acid group
Ester Group

A
B
D
C

System of Naming

Should be able to name
 Non-cyclic alkanes and halogenoalkanes containing up to six carbon atoms.
 Alkenes up to hexene and alkynes up to hexyne.
 Compounds up to six carbon atoms (in the basic chain for nomenclature purposes) containing only one of the following functional groups: hydroxyl, ether, aldehyde, halogenoalkane, carbonyl, ester and carboxyl.

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Originally compounds were named based on their source or use
Many organic compounds were given common names which are still in use
However many ambiguities resulted
With the large number of organic compounds, a method for systematically naming them is very important
10

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The International Union of Pure and Applied
Chemists (IUPAC) developed a system for naming organic compounds.
This system eliminated many of the ambiguities that plagued earlier naming systems
Common names for many substances are still widely used
11

A series of prefixes are used to designate the number of carbon atoms in a carbon chain meth 1 C eth 2 C prop 3 C but 4 C pent 5 C hex 6 C hept 7 C oct 8 C non 9 C dec 10 C
12

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For straight chain hydrocarbons. The prefix indicates the number of carbon atoms.
The suffix ane is added to designate that the compound is an alkane
13

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For branched chain hydrocarbons, identify the longest consecutive (straight) chain first. Then name the side chains or branches.
The name of the branches end in “yl” and go before the name of the straight chain
-methylpropane dimethyl propane methyl butane
Video
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2methyl
H
C
H
H
H
H H
C C C C
H
H H H
H
C
H
H
C
H
H
C
H
H
H
2methyl
Hexane
2methyl, 2methyl hexane
Sounds redundant
di

H
C
H
H
C
H
H
H
C C C
H
H
H
H
H
C
H
H
C
H
H
H
2-methyl hexane
2,3-Dimethyl hexane

• Are you sick to death of writing all those carbons?
• Even worse, are you sick of writing all those Hydrogens?
• How about this…

H H H H
H C C C C H
H H H H
Keep in mind that we have been ignoring the hydrogens for a long time.
Our names have been based entirely on the positioning of the carbons.
So lets now ignore the hydrogens completely!

H
H H H
C C C C
H
H H H
H
C
H
H
C
H
H
C
H
H
C
H
H
H
H
H
C
H
H
H H
C C C C
H
H H H H
C
H
H
H
C C
H
H
H
C
H
H
C
H
H
H

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Alkenes have one (or more) carbon to carbon double bonds
When there are 4 or more carbon atoms in a chain, the location of the double bond is indicated by a number.
Numbering the location of the double bond(s) takes precedence over the location of side chains but-1-ene but-2-ene 2-methylpropene
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H
H H H
C C C C
H
H H
C
H
H
H
C
H
H hex-2-ene
H
C
H
H
C
H
H
C
H
C
H
H
H
H
C
H
H
H H
C C C C
H
H H H
C
H
H but-1-ene
3 methyl-1-pentene

3-methylpent-1-ene
,
Notice the two lines means the double bond is there!
pent2 -ene
2,3-dimethylbut-2-ene
,

Methyl propene
2,4-dimethylpent-2-ene
3-ethyl-2,4,4-trimethylpent-1-ene

 A cis isomer is one in which the substituents are on the same side of the C=C cis-but-2-ene
 A trans isomer is one in which the substituents are on the opposite sides of the C=C trans-but-2-ene
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trans but-2-ene
2,3-dimethylbut-2-ene does not have geometrical isomers because there are two identical groups attached to the same carbon of the double bond.

 Fluorides, Chlorides,
Bromides, and
Iodides
 Simply name the molecule as normal but add the prefix
Fluoro, Chloro,
Bromo, or Iodo as necessary

1-bromopropane
2 chlorobutane
1,2 diiodoethane cis 1,2difluroethene
Trans 1,2 difluoroethene
1,1,2 trifluorothene prefixes = “fluoro, chloro, bromo, iodo”
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I
I
Cl
Cl
2, 3 di chloro hexane
3, 3 di iodo pent-1-ene

 Halogenoalkanes
 Alcohols
 Ethers
 Aldehydes
 Ketones
 Carboxylic Acids
 Esters
 Amines
 R-F, R-Cl, R-Br, R-I
 R-OH
 R-O-R
 R-COH
 R-CO-R
 R-COOH
 R-COO-R
 R-NH
2

Various functional groups have unique suffixes that designate the functional group.
The functional group takes precedence in numbering the carbon chain.
Branches to the carbon chain are named in the usual manner.
alcohols “ol”
Aldehydes “al”
Ketones
Acids
“one”
“oic”
Amides
Amines
“amide”
“amine” or amino as a prefix
Ethers Ethoxy as prefix halohydrocarbons Fluoro, bromo, chloro or iodo
Esters “oate”
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• R-OH
• Name like normal except add an
–ol suffix

Propan-1ol
Propan-2ol
Suffix = “ol”
2-methyl propan-2ol
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OH OH
Propan-2-ol
H
H H
C C OH
H H ethanol cyclobutanol

 R-O-R
 Name two “R” groups with –yl endings
 End name in ether

Suffix = “oxy”on first branch
Ethoxyethane
(diethylether)
Ethoxybutane
(ethylbutyl ether)
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O
O
Dimethyl ether
(Methoxymethane)
Ethyl methyl ether
(Methoxyethane)

 R-COH
 This is a carbon to oxygen double bond with a hydrogen at the end.
 Name as normal except use a “-al” suffix

Suffix = “al”
Propan al
Note that the aldeyhde group is always on an end carbon or carbon 1
41

H H H O
H C C C C H
H H H
butan al
H H Cl H O
H C C C C C H
H H Cl H
3,3 dichloro pentan al

Suffix = “one”
Propan one
(also known as acetone)
Butan one
( also known as methyl ethyl ketone)
Pentan-2one
(note the number is necessary
Because the C=O could be on carbon
2 or carbon 3)
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H O H
H C C C
H H
H H
C C
H H
H
C H
H pentan-2one
H H H H O H
H C C C C C C H
H H H H H hexan-2one

 R-COOH or R-CO
2
H
 This is a carbon to oxygn double bond with the same carbon single-bonded to an
OH group.
 Name as normal except give it the suffix “-anoic acid”.

Suffix = “oic”
Butan oic acid
Note that the acid group
(called a “carboxyl”) is always on an end carbon or carbon 1
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H H H O
H C C C C OH
H H H
But anoic acid
HO
O H H
C C C F
H H
3-Fluoro prop anoic acid

 R-COO-R
 This is a carbon to oxygen double bond with a carbon to oxygen single bonded to another single bonded carbon
 Name by given secondary branch “-yl” suffix and main branch “-anoate” suffix.

Suffix = “oate”
Ethyl butan oate
Butyl ethan oate
There are two branches. The branch with the carbonyl gets the suffix
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Main Branch pentanoate
H H H H O
H C C C C C O
H H H H
H
C H
H
Secondary
Branch methyl
Methyl Pentanoate

H H O
H C C C O
H H
H H H H
C C C C
H H H H
H
Butyl prop anoate
H O H
H C O C C
H H
H H H H
C C C C
H H H H
H
Methyl hexanoate

 R-NH
2
 Name the “R” group or groups with “-yl” endings
 Add the word
“amine”

Suffix = “amine”
Or prefix = “amino”
Propyl amine or 1amino propane
2-propyl amine or 2amino propane
2-methyl-2-propyl amine or
2-methyl -2amino propane
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H
H C N
H
H
H
H
H C N
H
H
H
C
H
H
Methyl amine
Dimethyl amine

Suffix = “amide” butan amide
Note that the amide group is always on an end carbon or carbon 1
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R Cl
Halide
O
R C H
Aldehyde
O
R C
Ester
O R
R OH
Alcohol
O
R C
Ketone
R
R NH
2
Amine
R O
Ether
R
O
R C OH
Carboxylic Acid

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Alkanes
Alkenes
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Alkynes
Halides
Alcohols
 Ethers
 Aldehydes
 Ketones
Carboxylic Acids
 Esters
 Amines
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-
=
R-X
R-OH
R-O-R
“-ane”
“-ene”
“-yne”
“-o”
“-ol”
“-yl ether”
R-COH “-al”
R-CO-R “-one”
R-COOH “-anoic acid”
R-COO-R “-yl” “-anoate”
R-NH
2
“-yl amine”

 YES!
 It takes:
 Memorization
 Practice
 Practice
 Practice
 Practice
 And, oh yes…
 Practice!

Halogenoalkanes , Alcohols and Amines

 Distinguish between primary secondary and tertiary
Alcohols.
 Distinguish between primary secondary and tertiary halogenoalkanes.
 Distinguish between primary secondary and tertiary
Amines.

Learning Objectives
Distinguish between primary secondary and tertiary Alcohols

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Primary = a carbon attached to only ONE other carbon
Secondary = a carbon attached to only TWO other carbons
Tertiary = a carbon attached to THREE other carbons

A primary alcohol has only one carbon atom attached

A secondary alcohol has 2 carbon chains attached to the group on which the
–OH resides

 Is cholesterol a primary, secondary or tertiary alcohol?
 Secondary

Learning Objectives
Distinguish between primary secondary and tertiary
Halogenoalkanes

 In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group.
 Some examples of primary halogenoalkanes include:

 In a secondary (2°) halogenoalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different.
 Examples:

 In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different.
 Examples:

Learning Objectives
Distinguish between primary secondary and tertiary
Amines

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In primary amines, only one of the hydrogen atoms in the ammonia molecule has been replaced. That means that the formula of the primary amine will be RNH
2 where "R" is an alkyl group.
Examples include:

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In a secondary amine, two of the hydrogens in an ammonia molecule have been replaced by hydrocarbon groups. At this level, you are only likely to come across simple ones where both of the hydrocarbon groups are alkyl groups and both are the same.
For example:

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In a tertiary amine, all of the hydrogens in an ammonia molecule have been replaced by hydrocarbon groups. Again, you are only likely to come across simple ones where all three of the hydrocarbon groups are alkyl groups and all three are the same.
The naming is similar to secondary amines. For example:

Learning outcome
Benzene is an aromatic, unsaturated hydrocarbon represented by .
Use physical and chemical evidence to discuss the structure of benzene.

The Kekulé structure for benzene, C
6
H
6
 Kekulé was the first to suggest a sensible structure for benzene.
The carbons are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has a hydrogen attached to it.

Kekule’dream

 This diagram is often simplified by leaving out all the carbon and hydrogen atoms!


Problems with the Kekulé structure
 Problems with the chemistry
 Because of the three double bonds, you might expect benzene to have reactions like ethene.
 Ethene undergoes addition reactions in which one of the two bonds joining the carbon atoms breaks, and the electrons are used to bond with additional atoms.
 Benzene rarely does this. Instead, it usually undergoes substitution reactions in which one of the hydrogen atoms is replaced by something new

Problems with the Kekulé structure
Problems with the shape
 Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekulé structure. The problem is that C-C single and double bonds are different lengths.
 C-C 0.154 nm C=C 0.134 nm
 That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides.
 In real benzene all the bonds are exactly the same intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon

Problems with the Kekulé structure
Problems with the stability of benzene
 Real benzene is a lot more stable than the
Kekulé structure would give it credit for. Every time you do a thermochemistry calculation based on the
Kekulé structure, you get an answer which is wrong by about 150 kJ mol -1 is most easily shown
. This using enthalpy changes of hydrogenation.

 Although you will still come across the Kekulé structure for benzene, for most purposes we use the structure on the right.
 The hexagon shows the ring of six carbon atoms, each of which has one hydrogen attached.
 The circle represents the delocalised electrons. It is essential that you include the circle. If you miss it out, you are drawing cyclohexane and not benzene.

 http://www.chemguide.co.uk