complete modul sem 1 2021 2022

SCI 1054 CHEMISTRY SEMESTER 1

Examples: Monocyclic compounds.

; C3H6 ; cyclopropane
cyclobutane
; C4H8 ;

; C5H10 ; cyclopentane

2. If only one substituent is present, it is not necessary to designate its position.
Examples:

CH3

Cl

Chlorocyclopropane methylcyclohexane

3. When two substituents are present, number carbon in the ring beginning with the
substituent according to the alphabetical order and number in the direction that gives
the next substituent the lowest number possible.
Examples:

CH3

CH2CH3

1-ethyl-2-methylcyclohexane
(NOT 1-ethyl-6-methylcyclohexane)

Cl Cl

1,3-dichlorocyclopentane
(NOT 1,5-dichlorocyclopentane)

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4. When three or more substituents are present, begin at the carbon with substituent that
leads to the lowest set of locants.
Example:

CH3 4- chloro-2-ethyl-1-
methylcyclohexane
(NOT 1-chloro-3-ethyl-4-
Cl CH2CH3 methylcyclohexane)

CH2CH3 1-ethyl-1,3-dimethylcyclopentane
CH3 (NOT 3-ethyl-1,3-
H3C dimethylcyclopentane)

5. When a single ring system is attached to a single chain with a greater number of carbon
atoms, or when more than one ring system is attached to a single chain, then it is
appropriate to name the compounds as cycloalkylalkane.
Examples:

CH2CH2CH2CH2CH3 1,3-
dicyclohexylpropane
1-cyclobutylpentane

Practice Exercises 8.1.1

1. Give the systematic (IUPAC) names for the alkanes with the following formulae

a. CH3CH2CH(CH3)2 b. (CH3)3CC2H5
c. CH3(CH2)8CH3
d. CH3CHCH2CHCH3
e. CH3
CH3 CH2CH3
CH3CHCHCH3
CH2CH2CH3

2. Write the structural formulae for the following hydrocarbon.
a. 3-methylpentane b. 1,4-dichlorocyclohexane
c. 3-ethyl-2,2,4-trimethylhexane d. ethylcyclopentane
e. 1,2-dichloro-3-methylbutane

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8.1.4 SYNTHESIS OF ALKANES

1. Hydrogenation of alkenes
 Alkenes and alkynes react with hydrogen in the presence of metal catalysts such
as nikel, palladium and platinum to produce alkanes.

General reaction:

CC + Pt or Pd or Ni CC
H2 HH

Examples: +CH CH2 H2 Ni H3C CH CH2
i. HH

H3C

ii.

CH3 CH3
C CH2
+H3C C CH2 H2 Pt H3C HH

iii.

+ H2 Pd

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2. Hydrolysis of Grignard reagents
 Grignard reagents are usually prepared by the reaction of an alkyl halide and
magnesium metal in ether solvent.

RX + Mg ether RMgX alkylmagnesium halide

Examples:
CH3CH2CH2Cl + Mg ether CH3CH2CH2MgCl

 Hydrolysis of Grignard reagent gives an alkane.
RMgX + H2O H  RH + Mg(OH)X

Examples:
i. CH3CH2CH2MgCl + H2O H  CH3CH2CH3 + Mg(OH)Cl

ii.

H3O+

CH3CHCH3 CH3CHCH3 + Mg(OH)Cl

MgCl H

3. Wurtz reaction
 A process utilising an active reducing agent (such as Na) to synthesize longer
alkanes from:

a) identical alkyl halides

2RX + 2Na  RR + 2NaX

Example:
2CH3CH2Br + 2Na  CH3CH2CH2CH3 + 2NaBr
b) more than one types of alkyl halide; a mixture of products will
form.
RX + R‟X + 6Na  RR + RR‟ + R‟R‟ + 6NaX

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Example :

CH3CH2Br + CH3Br + Na  CH3CH2CH2CH3 + CH3CH2CH3 + CH3CH3 + NaBr
 The reaction has its limitation as it gives a low yield when reacted with haloalkanes

of smaller molecular mass.

4. Decarboxylation of alkanoate salt.
 Alkanes can be produced by heating the alkanoate salt with soda lime.
RCOONa + NaOH  RH + Na2CO3
Only sodium etanoate can react easily, while the others give a mixture of product.
Examples:
i. CH3COONa + NaOH  CH4 + Na2CO3
(90%)
ii. CH3CH2COONa + NaOH  CH3CH3 + CH4 + H2
(major product)
+ other unsaturated compounds

8.1.5 CHEMICAL REACTION OF ALKANES
A. Non-reactivity of alkanes
 Alkanes are generally inert towards many chemical reagents. Carbon and hydrogen have
nearly the same electronegativity, the carbon-hydrogen bonds of alkanes are only
slightly polarised. As a consequence, they are generally unaffected by most bases.
 The low reactivity of alkanes toward many reagents accounts for the fact that alkanes
were originally called paraffins. (Latin : parum affinis= low affinity).
B. Halogenation: a free radical substitution reaction.
 Alkenes react with halogen such as chlorine and bromine to produce haloalkanes in the
presence of light or temperature greater than 100 oC.
R–H + X2 hv R–X + HX
 With methane, the reaction produces a mixture of halomethanes and a hydrogen halide.

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Examples:
i. CH4 + Cl2 hv CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl

ii. CH3CH3 + Cl2 hv CH3CH2Cl + HCl
iii.

CH3 hv CH3

H3C C CH3 + Cl2 H3C C CH2Cl

CH3 hv CH3
CH3CH2CH2Cl + CH3CHCH3 + HCl
iv.

CH3CH2CH3 + Cl2

Cl

minor major

Note: (ii) & (iii) will produce 1 product only because all the hydrogen atoms are identical.

Practice Exercises 8.1.2

1. A hydrocarbon has the structural formula

CH3 CH3

CH3CCH2CH

CH3 CH3

a. Give the systematic name of this hydrocarbon
b. Describe and explain what you would observe when a small amount of this

hydrocarbon is shaken with water and allowed to stand.

2. Outline how you would carry out the preparation of the following alkanes by using the
method specified.
a. 2,3-dimethylbutane by Wurtz reaction.
b. 2-methylbutane using Grignard reagent.
c. 3-methylpentane by decarboxylation

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8.2 ALKENES

8.2.1 INTRODUCTION

 General formula CnH2n , n  2.
 Functional group  double bond C=C
 Unsaturated hydrocorbon
 Each carbon atom ( C=C ) is sp2 hybridised.
 Restricted rotation of carbon-carbon double bond causes cis-trans isomerism.

8.2.2 IUPAC NOMENCLATURE

1. Determine the parent name by selecting the longest chain that contains the double
bond and change the ending of the name of the alkane of identical length from –ane to
–ene.

2. Number the chain to include both carbon atom of the double bond and begin numbering
at the end of the chain nearer the double bond.

123 45 6 7 8

CH3CH2CH CHCH2CH2CH2CH3
3-octene (not 5-octene)

3. Indicate the location of the substituent groups by the numbers of the carbon atoms to
which they are attached.

CH3 123456

H3C1 C CH C4 H3 H3C C CH CH2CH CH3

2 3 CH3 CH3
2,5-dimethyl-2-hexene
2-methyl-2-butene (not 2,5-dimethyl-4-hexene)
(not 3-methyl-2-butene)

4. The ending of the alkenes with more than one double bond should be change from -
ene to :
diene – if there are two double bonds
triene – if there are three double bonds

1 23 4 1 23 45 67

H2C CHCH CH2 H2C CHCH CHCH CHCH3

1,3-butadiene 1,3,5-heptatriene

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5. Number substituted cycloalkenes in the way that gives the carbon atoms of the double
bond the 1 and 2 positions and that also gives the substituent groups the lower
numbers at the first point of difference.

CH3 1
62
1
52 53

43 H3C 4 CH3

1-methylcyclopentene 3,5-dimethylcyclohexene
(not 2-methylcyclopentene) (not 4,6-dimethylcyclohexene)

6. Two frequently encountered alkenyl groups are vinyl group and allyl group.

CH2=CH- CH2=CHCH2-
vinyl group allyl group

7. If two identical groups are on the same side of the double bond, the compound can be
designated cis, if they are on opposite sides it can be designated trans.

Cl Cl Cl H

CC CC

HH H Cl

cis-1,2-dichloroethene trans-1,2-dichloroethene

8. In the case that the atoms (or groups) that attached to carbon double bond are not
identical,designation cis-trans isomer can be made based on it relative Mr (or total Mr).
 cis-isomer : higher relative Mr (or total Mr) at the same side of the double bond.
 Trans-isomer: higher relative Mr (or total Mr) at the opposite side of the double
bond.

Cl ClBr Cl CH3
CC CC

H CH3 H ClBr
trans-2-bromo-1,2-
cis-2-bromo-1,2-dichloroethene
dichloroethene
Note :Mr=molar mass ( Mr Cl >H and Br>CH3 )

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Practice Exercises 8.2.1
1. Give IUPAC names for the following alkenes

a. b.

c. d. Br
e. CH3CH=CHCH2C(CH3)2CH3
CH2CH3

f. (CH3)2C=CHCl

2. Write the structural formula of the following alkenes:

a. cis-2-pentene b. trans-3-methyl-2-pentene
c. 2,3-dimethyl-2-butene d. 2-methyl-1,3-butadiene
e. 1,2-dibromocyclopentene f. 2-methylpropene
g. 1,3-pentadiene h. cyclohexene
i. cis-1,2-dimethylcyclopentane j. 1,4-cyclohexadiene

8.2.3 SYNTHESIS OF ALKENES

A. Dehydration of alcohols
 Alcohols react with strong acids in the presence of heat to form alkenes and water.

CC H2SO 4 (conc.) +C C H2O
H OH 

 Concentrated sulphuric acid or phosphoric acid are often used as reagents for
dehydration because these acids act both as acidic catalysts and as dehydrating agents.

 Saytzeff’s Rule:
The major product is the most stable alkenes. The most stable is the alkene that has
greater number of alkyl group.

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Examples:
+ H2O
1) H2SO4 (conc.)

CH3CH2OH  H2C CH2 + H2O

2) H3C CHCH3 H2SO4 (conc.) H2C CHCH3 + H2O
OH


3) OH H2SO4 (conc.)

+



(major) (minor)

B. Dehydrohalogenation of Alkyl Halides

The elimination of ONE hydrogen and ONE halogen from an alkyl halide to form an
alkene.
General equation:

CC + KOH alcohol C C + HX
HX reflux

Saytzeff’s rule is used to determine the major product.

Examples:

1) Cl alcohol +
reflux
+ KOH

(major) (minor)

+ HCl

2) alcohol H2C CHCH3 + HBr
reflux
H3C CHCH3 + KOH
Br

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3) alcohol + + HBr
reflux
+ KOH
Br

(major) (minor)

8.2.4 CHEMICAL REACTION OF ALKENES

I. Comparison of Reactivity Between Alkanes and Alkenes

Alkenes are more reactive compared to alkanes.

Alkanes have carbon-carbon single bonds (σ bonds) while alkenes have carbon-carbon double
bonds (π bonds).

The double bond is a site of high electron density (nucleophilic).Therefore most alkenes
reactions are electrophilic additions (susceptible to attack by electrophiles).

Electrophiles are species that have low electron density. They are attracted to centres of high
electron density. Examples of electrophiles are H+, Br+ and NO2+. They are Lewis acids.

II. Reactions of alkenes

Addition Reaction of Alkenes

① Hydrogenation

 The reaction of an alkene with hydrogen in the presence of catalyst such as
platinum, nickel and palladium to form alkane.

+C C H2 Pt or Ni or Pd CC
HH

Examples:

(1)

+H3C C CH2 H2 Pt or Ni or Pd H3C CH CH3
CH3
CH3

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(2)

+ H2 Pt or Ni or Pd

② Halogenation of Alkenes

a) In inert solvent (CCl4)

Alkenes react rapidly with chlorine or bromine in CCl4 at room temperature to form
vicinal dihalides.
General equation:

C C + X2 CH2Cl2 CC
XX

Example:

H3C CH3 CH3 + Cl2 CCl 4 H3C Cl Cl
CC C C CH3
CH3H
H

IDENTIFICATION TEST OF ALKENE : Unsaturation test

CC CH2Cl2 CC
(colorless) Br Br
+ Br2

room temp.

(reddish (colorless compound)
brown)

When bromine is used for this reaction, it can serve as a test for the presence of
carbon-carbon double bonds.
If bromine is added to alkene, the reddish brown color of the bromine disappears
almost instantly as long as the alkene is present in excess.

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b) In aqueous (Halohydrin Formation)

If the halogenation of an alkene is carried out in aqueous solution, the major product of the
overall reaction is a haloalcohol called a halohydrin.

+ +C C
X2 H2O CC
X OH
X2 = Cl or Br halohydrin

Example:

HH + Br2 + H2O HH
CC HCCH

HH Br OH
2-bromobutanol

 If the alkenes is unsymmetrical, the halogen ends up on the carbon atom with the
greater number of hydrogen atoms.

Example:

H CH3 H CH3
H C C CH3
CC + Br2 + H2O
Br OH
H CH3
1-bromo-2-methyl-2-propanol

Practice Exercise 8.2.2
Predict the structure of A.

+ Br2 + H2O A

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③ Hydrohalogenation : Markovnikov’s rule

Hydrogen halides (HI, HBr and HCl) add to the double bond of alkenes to form haloalkanes.
General equation:

C C + HX CC
HX

The reactivity decreases in the order HI > HBr > HCl due to the increasing H-X bond
energy from HI to HCl.

The addition of HX to an unsymmetrical alkenes, follows Markovnikov’s rule
(produced two products).

Markovnikov’s (Markownikoff) Rule:

Markovnikov’s states that in the reaction between an unsymmetrical alkene and HX, the
more electropositive atom (usually H atom) or group will attach itself to the unsaturated

H H (major product)
CH3C CH
H
X

HH 

CH3C C H + H X H H
CH3C CH
X (minor product)
H

In the addition of HX to an alkenes, the hydrogen atom adds to the carbon atom of the
double bond that already has the greater number of hydrogen atoms.

Examples:

i) The addition of HBr to propene, could conceivably lead to either 1-
bromopropane or 2-bromopropane. The main product, however, is 2-
bromopropane.

H2C CHCH3 + HBr HH HH
+ H C C CH3
H C C CH3
H Br Br H
1-bromopropane
2-bromopropane
(major) (minor)

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ii) When 2-methylpropane reacts with HBr, the main product is tert-butyl bromide, not
isobutyl bromide.

Br H H Br

H3C C CH2 + HBr H3C C CH2 + H3C C CH2
CH3 CH3 CH3

ter-butyl bromide isobutyl bromide
(major) (minor)

Consideration of the example above, shown that the addition of HX to an
unsymmetrical alkenes, yield the main product according to the Markovnikov’s
rule.

The secondary carbanium ion (major product) is more stable than the primary
carbonium ion(minor product), because the positive charge in secondary carbanium ion
can be stabilized by two alkyl groups which are electron-donating groups.

H3C C CH3 > H
CH3 H3C C CH2

tertiary carbocation CH3

primary carbocation

Note: Stability of carbocation is 3 > 2 > 1.

④ Addition of HBr to Alkenes in The Presence of Peroxides (Anti
Markovnikov’s rule)
 When alkenes are treated with HBr in the presence of peroxides, ROOR (eg: H2O2)

the addition occurs in an anti-Markovnikov manner whereby that the hydrogen
atom of HBr attached to the doubly bonded carbon with fewer hydrogen atoms.

Example:

+H2C CH CH3 HBr ROOR Br CH2CH2CH3

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Practice Exercises 8.2.3

Complete the following reactions:

(1)

+H3C CH2C CH2 HBr H2O2

CH3

(2)

+CH2 H2O2
HBr

⑤ Hydration of Alkene

The acid-catalyzed addition of water (steam under high temperature) to the double
bond of an alkene (hydration of an alkene) is a method for the preparation of low
molecular weight alcohols.

The acid most commonly used to catalyze the hydration of alkenes are dilute solution of
sulphuric acid and phosphoric (V) acid.

The addition of water to the double bond follows Markovnikov’s rule.

General equation:

+C C H2O H3O+ CC
H OH

Example:

+H3C C CH2 H2O H3O+ OH
H3C C CH3
CH3
CH3

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Practice Exercises 8.2.4

Complete the following reactions:

(1)

+CH2 H2O H3O+

(2)

+H2C CH CH3 H2O H3O+

⑥ Addition of Sulphuric (VI) Acid to Alkenes

 Alkenes dissolve in concentrated sulphuric (VI) acid to form alkyl hydrogen
sulphates(VI) which is soluble in water.

 Alkyl hydrogen sulfates can be easily hydrolyzed to alcohols by heating them with
water.

 The overall result of the addition of sulphuric acid to alkenes followed by hydrolysis
is the Markovnikov addition of -H and -OH.

H3C C CH3 conc. H2SO 4 CH3CH3 H2O CH3CH3
C CH3 H3C C C CH3 heat H3C C C CH3

H3C H OSO 3H H OH
alkyl hydrogen sulphate

Example:

H2C CH CH3 conc.H2SO 4 H3C OSO 3H H2O OH
heat
CH CH2 H3C CH CH2
H H

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⑦ Oxidation of Alkenes

Alkenes undergo a number of reactions in which the carbon-carbon double bond is oxidized.

a) With cold and dilute potassium permanganate, KMnO4
Potassium permanganate in base can be used to oxidize
C alkenes to 1,2-diols (glycols).
General equation:

+C KMnO4 OH-,cold CC + MnO2
(purple) OH OH (brown precipitate)

This reaction is called Baeyer’s test. It can serve as a test for the presence of carbon-
carbon double bonds where the purple colour of the KMnO4 decolourised, and brown
precipitate of MnO2 is formed.

Example: OH-,H 2O HH
H2C cold
+CH2 KMnO4 +H C C H MnO2

OH OH

Practice Exercises 8.2.5

Complete the following reactions:

+H2C CH CH2CH3 KMnO4 OH-,H2O ?
cold

b) With hot potassium permanganate solutions to alkenes (oxidative
cleavage of alkenes)

 When oxidation of the alkene is carried out with KMnO4, in acidic rather than basic
solution, cleavage of the double bond occurs and carbonyl-containing products are
obtained.

 If the double bond is tetrasubstituted, the two carbonyl-containing products are
ketones; if a hydrogen is present on the double bond, one of the carbonyl-containing
products is a carboxylic acid; and if two hydrogens are present on one carbon, CO2
is formed.

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Example:

(1)

CH3 (i) OH-,heat OO
(ii) H3O+
+H3C C C CH3 KMnO4 C+C

CH3 H3C CH3 H3C CH3
(2)

+H2C CH CH3 KMnO4 (i) OH-,heat O
(ii) H3O+
+ +C CO2 H2O

HO CH3

The oxidative cleavage of alkenes can be used to establish the location of the double bond in an
unknown alkene.

Example:
An unknown alkene with the formula C7H14 was found on oxidation with hot basic
permanganate, to form a three-carbon carboxylic acid (propanoic acid) and four-carbon
carboxylic acid (butanoic acid). What is the structure of this alkene?

C7H14 + KMnO4 i) OH-, heat O O
ii) H3O+ C +C
Solution: HO CH2CH3
HO CH2CH2CH3

O O
C C
HO CH2CH3 HO CH2CH2CH3

propanoic acid butanoic acid

HH
H3CH2C C C CH2CH2CH3

3-heptene

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c) Ozonolysis of Alkenes (Oxidative Cleavage of Alkenes)

 A more widely used method for locating the double bond of an alkene involves the use
of ozone (O3).

 Ozone reacts vigorously with alkenes to form unstable compounds called molozonides,
which rearrange spontaneously to form compounds as ozonides.

O

CC + O3 CC
OO

ozonide

 Ozonides, themselves, are very unstable compounds and can easily explode violently.
Because of that they are not usually isolated but are reduced directly by treatment with
water and in the presence of zinc and acid (normally acetic acid) to give carbonyl
compounds (either aldehydes or ketones).

Example:

Practice Exercise 8.2.6
1. Write the structure of alkene that would produce the following products when treated with

ozone followed by water, zinc and acid.
CH3COCH3 and CH3CH(CH3)CHO
2. Draw the structural formula of the organic products formed when cyclohexene reacts with
cold dilute potassium manganate (VII)
3. Predict the major products in the reactions between hydrogen iodide and
a. cyclohexene
b. 1-methylcyclohexene
c. 2-methyl-but-1-ene

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4. Give the structure of A ,B, C,D and E.

B

Br2 H2O

Br2, CH2Cl2 CH2CH3 C

A H2O
dilute H2SO4
in the dark,
room temp.

concentrated
H2SO4, cold

D

H3O+ heat

E

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8.3 ARENES (AROMATIC COMPOUNDS)

8.3.1 INTRODUCTION
 Benzene is a liquid hydrocarbon with the molecular formula of C6H6. It is an example
of an aromatic hydrocarbon or arene.

Benzene as Aromatic Compound
 Aromaticity refers to a level of stability for arenes. It consists of aromatic compound

or others which resembles benzene and its chemical behaviour. It undergoes
electrophilic substitution reaction.

Kekulé structure
 In 1865, August Kekulé, proposed the first definite structure for benzene. He

suggested that the carbon atoms of benzene ring are in the ring, that they are
bonded to each other by alternating single and double bonds. One hydrogen is
attached to each carbon atoms.

 This structure satisfied the requirements of the structural theory that carbon atoms
form four bonds and that all the hydrogen atoms of benzene are equivalent.

H
HH

or

HH
H

The Kekule Structure for benzene

 However, Kekulé structure can‟t explain why benzene molecules are not undergoes
addition reaction.

 The Kekulé structure was not accepted because of the following reasons.
o Benzene does not decolourise acidified potassium manganate (VII), showing that
it does not have C=C bond.
o Benzene does not decolourise bromine dissolved in tetrachloromethane, hence it
cannot contain C=C.
o All the carbon-carbon bond length in benzene molecule are the same (0.139
nm), and is about half-way between the length of a single bond (0.154 nm) and
a double bond (0.134 nm)

 The Kekulé structure was superseded by the delocalized structure. The delocalized
ring of electrons makes the „benzene ring‟ very stable. The term aromacity is used to
describe a system which is stabilized by a ring of delocalized electrons.

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 The actual structure of benzene is shown below:
HH

H H Or

HH

Delocalized electron

Resonance Structure
 In 1930s, Linus Pauling introduced the concept of hybrid orbital and resonance. The

bond angles of the carbon atoms in the benzene ring are 120. The carbon atoms
are sp2 hybridized and the six overlapping p orbitals form a set of three delocalised 
bonds.

 As a result, the resonance structure of benzene is a hybrid resonance from two
Kekulé structure as shown below:



hybrid resonance structure

4N + 2 = 4 pi electrons

N = 42
4

= ½ or 0.5 (not integer number so they are
Antiaromatic compound

8.3.2 NOMENCLATURE
A) MONOSUBSTITUTED BENZENE

2 systems are used in naming monosubstituted benzene
1) Benzene as the parent name & the substituent is indicated by a prefix.

Examples:

F Cl Br NO2

fluorobenzene chlorobenzene bromobenzene nitrobenzene

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2) The benzene ring and the substituent combine to form a parent name (common name).

Examples:

CH3 OH NH2

toluene phenol aniline
O CH3
SO3H COOH

Benzenesulphonic acid Benzoic acid acetophenone

B) DISUBSTITUTED BENZENE

 When 2 substituents are present, their position is indicated by prefixes ortho (o-),
meta (m-) and para (p-) or by the use of numbers.

Examples:

Br COOH CH3

Br H3C
1,4-dimethylbenzene
1,2-dibromobenzene NO2
(o- dibromobenzene) 1,3-nitrobenzoic acid (p-xylene)

8.3.3 REACTION OF BENZENE
8.3.3.1 NITRATION OF BENZENE
Benzene reacts with concentrated nitric acid at 323-333k in the presence of
concentrated sulphuric acid to form nitrobenzene. This reaction is known as nitration of
benzene.

Nitration of nitrobenzene

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The mechanism for nitration of benzene:
Step 1: Nitric acid accepts a proton from sulphuric acid and then dissociates to form nitronium
ion.

Nitric acid and nitronium ion
Step 2: The nitronium ion acts as an electrophile in the process which further reacts with
benzene to form an arenium ion.

Formation of arenium ion
Step 3: The arenium ion then loses its proton to Lewis base forming nitrobenzene.

“Don‟t Talk, Just Act.
Don‟t Say, Just Show.
Don‟t Promise, Just Prove.”

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8.3.3.2 HALOGENATION OF BENZENE
Benzene reacts with halogens in the presence of Lewis acid like FeCl3, FeBr3 to form aryl
halides. This reaction is termed as halogenation of benzene.

Halogenation of benzene
8.3.3.3 FRIEDEL CRAFT ALKYLATION
Friedel-Crafts Alkylation refers to the replacement of an aromatic proton with an alkyl group.
This is done through an electrophilic attack on the aromatic ring with the help of a carbocation.
The Friedel-Crafts alkylation reaction is a method of generating alkylbenzenes by using alkyl
halides as reactants.
The Friedel-Crafts alkylation reaction of benzene is illustrated below.

A Lewis acid catalyst such as FeCl3 or AlCl3 is employed in this reaction in order to form a
carbocation by facilitating the removal of the halide. The resulting carbocation undergoes a
rearrangement before proceeding with the alkylation reaction.

8.3.3.4 FRIEDEL CRAFT ACYLATION
The Friedel-Crafts Acylation, also known as Friedel-Crafts Alkanoylation. The goal of the
reaction is the following:

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8.4 CLASSIFICATION OF HALOALKANES

 Haloalkanes are classified as 1°(primary) , 2°(secondary) and 3°(tertiary) haloalkane.
HR

CH3 X R CH2 X RC X RCX
methyl halide 1° halide R R

2° halide 3° halide

 Examples:

CH3 Br CH3CH2CH2F Cl Cl
CH3CHCH3
CH3CCH3
Bromomethane 1-fluoropropane 2-chloropropane CH3
(methyl halide) (1°) (2°)
2-chloro-2-
methylpropane

(3°)

8.4.1 NOMENCLATURE OF HALOALKANE.
i. Common or trivial names.
Haloalkane can be named as alkyl halide for example:

CH3CHCH3

CH3CH2 F CH3CH2CH2CH2 Cl Br

isopropyl bromide
ethyl fluoride n-butyl chloride (2-bromopropane)
ii. IUPAC names
Haloalkane are named as alkanes with halogen as substituents.

Example:
Br

5-bromo-2,4- Br
dimethylheptane 2-bromo-4,5-
dimethylheptane
Cl
Cl Cl
2,3-dichloro-4-methylhexane Br
1-bromo-3-chloro-4-
methylpentane

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SCI 1054 CHEMISTRY SEMESTER 1

Cl I

Br Br 1-iodo-3-methylbutane
3,5-dibromo-4-chlorooctane

8.5 REACTIONS OF HALOALKANES.

8.5.1 HYDROLYSIS
Nucleophilic substitution with aqueous hydroxide ions Change in functional group:
haloalkane to alcohol Reagent: potassium (or sodium) hydroxide Conditions: In aqueous
solution; Heat under reflux

Br OH
H3C
H3C + Br-
OH- /aq

CH3 CH3

8.5.2 FORMATION OF NITRILES AND PRIMARY AMINES

Haloalkanes can form nitriles compound when treated with potassium cyanide or sodium
cyanide

Br H3C

H3C N

CN- / aq

This reaction increases the length of the carbon chain (which is reflected in the name)
In the above example butanenitrile includes the C in the nitrile group

Haloalkanes can form primary amines upon treatment with Ammonia

Br NH2
H3C H3C + HBr

NH3

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8.5.3 ELIMINATION REACTIONS

Haloalkane is used in the synthesis of Grignard reagents.

ether R Mg X
R X + Mg

Grignard reagents are examples of organometallic compounds because they contain a
carbon-metal bond. In general Grignard reagents are called organomagnesium halides,
of empirical formula R-Mg-X. Grignard reagents results from the reaction of haloalkanes
with magnesium metal. This reaction is always carried out in an ether solvent.

The carbon-magnesium bond is polarized, making the carbon atom both nucleophilic
and basic

 Basic and nucleophilic site

MgX



C

Grignard reagents are used in the synthesis of:

a) Alkanes

H H H2O / H+ HH
H C C H + Mg(OH)X
HC C MgX +
HH
H H
ethane

b) Alcohol R
R
R O (1) ether R' C O MgX (2) H3O+ R' C O H + Mg(OH)X
R' MgX + C
R R
R alkylmagnesium halide alcohol

c) Carboxylic acid

HH HH O

H C C MgBr + O C O H CC C

HH OMgBr
HH

H2O /H+

HH O

H CC C + MgBr

OH

HH

Propanoic acid

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ANSWER:

Practice Exercises 8.1.1

1. Give the systematic (IUPAC) names for the alkanes with the following formulae

a. CH3CH2CH(CH3)2 b. (CH3)3CC2H5
2-methylbutane 2,2-dimethylbutane

c. CH3(CH2)8CH3 d. CH3CHCH2CHCH3
decane
CH3 CH2CH3

2,4-dimethylhexane

e. CH3

CH3CHCHCH3

CH2CH2CH3

2,3-dimethylhexane

2. Write the structural formulae for the following hydrocarbon.
a. 3-methylpentane b. 1,4-dichlorocyclohexane

Cl

c. 3-ethyl-2,2,4-trimethylhexane Cl

d. ethylcyclopentane

e. 1,2-dichloro-3-methylbutane
Cl

Cl

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Practice Exercises 8.1.2
1. A hydrocarbon has the structural formula

CH3 CH3

CH3CCH2CH

CH3 CH3

c. Give the systematic name of this hydrocarbon
2,2,4-trimethylpentane

d. Describe and explain what you would observe when a small amount of this
hydrocarbon is shaken with water and allowed to stand.
The two layers solution formed because hydrocarbon is insoluble in
water

2. Outline how you would carry out the preparation of the following alkanes by using the
method specified.
a. 2,3-dimethylbutane by Wurtz reaction.

CH3CHCH3 + 2 Na CH3 + 2 NaCl
Cl CH3CH CHCH3

CH3

b. 2-methylbutane using Grignard reagent.

CH3 ether CH3 H2O / H+ CH3
CH3CCH2CH3 + Mg CH3CCH2CH3 CH3CCH2CH3

Cl MgCl H

c. 3-methylpentane by decarboxylation + CH4 + H2
+ other unsaturated products
COONa + NaOH

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SCI 1054 CHEMISTRY SEMESTER 1

Practice Exercises 8.2.1
1. Give IUPAC names for the following alkenes

a. b.
3-hexene
1,3-
dimethylcycloheptane

c. d. Br
4-ethyl-2-methylhexene
CH2CH3

3-bromo-5-
ethylcyclopentene

e. CH3CH=CHCH2C(CH3)2CH3 f. (CH3)2C=CHCl
5,5-diethyl-2-methylhexene 1-chloro-2-methylpropene

2. Write the structural formula of the following alkenes:

a. cis-2-pentene b. trans-3-methyl-2-pentene

CH3 CH2CH3 H CH3
H3C CH2CH3
CC CC

HH

c. 2,3-dimethyl-2-butene d. 2-methyl-1,3-butadiene

e. 1,2-dibromocyclopentene f. 2-methylpropene

Br
Br

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SCI 1054 CHEMISTRY SEMESTER 1

g. 1,3-pentadiene h. cyclohexene

i. cis-1,2-dimethylcyclopentane j. 1,4-cyclohexadiene

Practice Exercise 8.2.2
Predict the structure of A.

+ Br2 + H2O A Br

OH

Practice Exercises 8.2.3
Complete the following reactions:
(1)

+H3C CH2C CH2 HBr H2O2 Br
CH3CH2CCH3
CH3
CH3

(2)

+CH2 H2O2
HBr

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SCI 1054 CHEMISTRY SEMESTER 1
Practice Exercises 8.2.4

Complete the following reactions:

(1)

+CH2 H2O H3O+

OH

(2)

+H2C CH CH3 H2O H3O+ CH3CH(OH)CH3

Practice Exercises 8.2.5

Complete the following reactions:

+H2C CH CH2CH3 KMnO4 OH-,H2O CH2CHCH2CH3
cold OH OH

Practice Exercise 8.2.6
2. Write the structure of alkene that would produce the following products when treated with

ozone followed by water, zinc and acid.
CH3COCH3 and CH3CH(CH3)CHO

+ i) O3 O
ii)Zn, H2O/H+
O+

2. Draw the structural formula of the organic products formed when cyclohexene reacts with
cold dilute potassium manganate (VII)

OH

OH

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3. Predict the major products in the reactions between hydrogen iodide and
a. cyclohexene

H

I

b. 1-methylcyclohexene

I

c. 2-methyl-but-1-ene

4. Give the structure of A ,B, C,D and E. OH
CH2CH3
Br

Cl Cl OH
CH2CH3

OSO3H
CH2CH3

OH
CH2CH3

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