TMK CHAPTER 12.0 HYDROCARBONS

TUTORIAL 12A: ALKANES CHEMISTRY SK025
CHAPTER 12:
1. Classify c)
a) saturated d) HYDROCARBONS
b) unsaturated
unsaturated
2. IUPAC name saturated

a) 2-methylbutane d) 3-isopropyl-1-methyl-1-propylcyclopentane
5-ethyl-1,2,3-trimethylcyclohexane
b) 3-ethyl-2-methylpentane e)
c) Cl
c) 4-isopropyl-3,6-dimethyloctane Cl

3. Draw structural formula Br
a) Cl

b)

4. increasing order of boiling points
a) methane < propane < pentane < hexane
Reason:
 Size of hexane > pentane > propane > methane.
 Strength of Van der Waals forces between hexane molecules > pentane
molecules > propane molecules > methane molecules.
 Hence, energy needed to separate hexane molecules > pentane molecules >
propane molecules > methane molecules.
 Boiling point of hexane > pentane > propane > methane.

b) 2,2-dimethylpropane < 2-methylbutane < pentane
 Number of branch in 2,2-dimethylpropane > 2-methylbutane > pentane.
 Surface area of 2,2-dimethylpropane < 2-methylbutane < pentane.
 Strength of Van der Waals forces between 2,2-dimethylpropane molecules <
2-methylbutane molecules < pentane molecules.
 Therefore, energy needed to separate 2,2-dimethylpropane molecules < 2-
methylbutane molecules < pentane molecules.
 Boiling point 2,2-dimethylpropane < 2-methylbutane < pentane.

5. Butane soluble in benzene but insoluble in water,
Reason:
 Benzene is a non-polar solvent. Butane is soluble in benzene because butane and
benzene have the same type of intermolecular forces which is Van der Waals forces,
so they attract each other.
 Water is a polar solvent. Butane is insoluble in water because it is a non-polar
compound and cannot form hydrogen bond with water

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6. Chemical equation
a) In excess oxygen
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
b) In limited oxygen
2C3H8(g) + 7O2(g)  6CO(g) + 8H2O(g)

7. a) Structural formulae of monosubstituted major and minor product formed.

i. CH3CH2Br iv. Br
ii. major product : CH3CHBrCH3

minor product : CH3CH2CH2Br

iii. major product:
CH3

H3C C CH3

Br
minor product:

CH3

CH3 CHCH2Br

b) i. Equation

CH3 Br2, CH2Cl2 CH3 CH3
H3C C CH3 uv
+H3C C CH3 H3C C CH 2Br
H
Br H
major minor

ii. Reaction mechanism

Initiation step

UV Br + Br
Br Br

Propagation step CH3
CH3
+H Br H3C C CH3
+Br H3C C CH3

H

CH3 CH3

+H3C C CH3 Br Br +H3C C CH3 Br

Br

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Termination step

Br + Br Br Br
CH3 CH3
+ H3C C CH3 H3C CH3 CH3
H3C C CH3 C C CH3
CH3 CH3

CH3 Br CH3
H3C C CH3
+H3C C CH3

Br
iii. Observation of the reaction

The reddish-brown colour of bromine is decolourised.

iv. Observation in the dark
The reddish-brown colour of bromine is remain unchanged.

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TUTORIAL 12B: ALKENES

1. IUPAC name d) 4-ethyl-2,7-nonadiene
a) methylpropene e) 1-ethyl-2-isopropylcyclopentene
b) cis-3-heptene
c) 3,5-dimethylcyclohexene

2. Structural formulae

a) H3C CH3 c) Cl Cl
CC

H3C CH3

b) CH 3CH 2CH 2 CH 2CH 2CH 3 d) H3C H

CC CC

HH H CH 2Cl

3. trans-3-pentene

a) H3C H b) Correct name: trans-2-pentene
CC Reason:

H CH2CH3  The chain is numbered from the end nearer to the double
bond.

 The position of the double bond is designated using the

lower number of the two doubly bonded carbon atoms.

2-methylcyclohexene b) Correct name: 1-methylcyclohexene
a) Reason:
 The position of the double bond is designated using the
CH3
lower number of the two doubly bonded carbon atoms

cis-2,3-dimethyl-2-pentene b) Correct name: 2,3-dimethyl-2-pentene
a) Reason:
 Does not show any cis-trans isomerism because one of C

at restricted rotation site attached to same group of atoms.

4. a) State Saytzeff’s rule
Sayzteff’s rule states that an elimination reaction will give an alkene with the most

highly substituted double bond as the major product.

b) Structural formulae of major and minor product

CH3 H3C CH3
CH3CH C CH2C

CH3 CH2

Major product Minor product

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Mechanism for the formation of major product:
Step 1:

CH3 CH3 + - O SO3H
H3C C CH2 CH3
H3C C CH2 CH3 + H O SO3H
+
OH
OH

H

Step 2: CH3 H
CH3 OH
H3C C CH2CH3 +
H3C C CH2 CH3
+
+

OH

H

Step 3: CH3 + H CH3H H
CH3 H OH
H3C C C CH3 + +
H3C C C
OH
+
H
H

5. a) State Markovnikov’s rule
Markovnikov’s rule states that in the addition of HX to an alkene or alkyne, the
hydrogen atom adds to the carbon – carbon double bond that already has the greater

number of hydrogens atoms.

b) i. Equation

CH3 HBr CH3
H3C C CH3
H3C C CH2

Br

ii. Type of reaction
Electrophilic addition reaction.

iii. Reaction mechanism slow CH3
Step 1:
H Br H3C C CH3
CH3 +

+H3C C CH2

Step 2:

CH3 - CH3
H3C C CH3
H3C C CH3 Br
+ fast Br

6. Complete reaction scheme

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a) Alkene to vicinal dihalide and halohydrine

Br

Br2, CH2Cl2

Br

Br2 (aq) Br

OH

b) Alkene to haloalkane (Markovnikov and anti-Markovnikov) and alcohol.

Br

CH3CH2C CH2 HBr CH3CH2C CH3

CH3 CH3

HBr CH3 CH2 CH CH2Br
CH3CH2OOCH2CH3 CH3

H2O, H2SO4 OH

CH3CH2 C CH3
CH3

c) Oxidation of alkene

CH3 KMnO4, OH- CH3
CH3CH2C CH CH3 cold CH3CH2C CH CH3

KMnO4, H+ OH OH C CH3

CH3

CH3CH2C O + O

i. O3 CH3 OH
ii. Zn, H2O
CH3CH2C O + O C CH3

H

d) Alkene to oxidize and haloalkane

CH2 KMnO4, H+ + +O CO2 H2O


i. O3 +O CH2O
ii. Zn, H2O
Br
HBr
CH3

e) Alkene to oxidize and alcohol
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HOOC O

Br CH3 i. KMnO4, H+, Br CH3
ii. H2O, H+

i. O3 O O
ii. Zn, H2O Br
CH3

H2O, H+ OH

Br CH3

7. State reagents and condition c) R6: HBr, H2O2
a) R1: Cl2, CH2Cl2, uv R7: H2O, H+
R2: KOH, ethanol, reflux R8: i. KMnO4, OH-, ∆
R3: Br2, CH2Cl2 ii. H2O, H+
R9: KMnO4, OH-, cold
b) R4: conc. H2SO4, ∆
R5: i. O3 Br2, CCl4 Br
ii. Zn, H2O HCl Br
Cl
8. Outline the synthesis
a)
Br KOH, EtOH
reflux

b)
OH conc. H2SO4



9. Order of reactivity.
The order reflects the relative ease with which these alkenes accept a proton and form a
carbocation. 2-methylpropene, (CH3)2C=CH2, reacts fastest because it leads to a 3
carbocation which it’s more stable than 1o carbocation. Ethene CH2=CH2 reacts slowest
because it leads to a 1 carbocation.

10. Chemical test to distinguish between 2-butene and n-butane
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Test Unsaturation test / bromine test
Reagent &
Conditions Reagent: Br2 in CCl4
Compound Conditions: In the dark

Observation butane 1-butene

The reddish brown colour of Decolorize the reddish brown

bromine solution remains colour of bromine solution.

unchanged.

Equation H2C CHCH2CH3 Br2, CCl4 CH2BrCHBr CH2CH3
dark places

11. Suggest structural formula and write chemical equation.

A KOH, Ethanol B
reflux
CH3CHCH3 CH3CH CH2

Br i) O3

ii) Zn/ H2O

H3C C H + HC H
O O

C D

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MEKA 12 d) 1-butene
e) trans-2-hexene
1. naming f) 1-chloro-1,3,5-trimethylcyclohexane
a) 2-methylbutane
b) 1-cyclopentyl-3-methylcyclohexane e) CH3
c) 3-bromo-1-methylcyclobutene

2. draw
a) H3C

CH3 f) H2C=CHCH2CHBrCHBrCH2CH3
g)
b)

c) h)

d) Cl

H

H

Cl

3. a) Free radical substitution mechanism

CH4 + Cl2 uv +CH3 Cl HCl

Initiation Step:

Cl Cl uv Cl + Cl

Propagation Step:

H H + Cl HCl + H
HC HC

H H

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H H
H C + Cl Cl
+H C Cl Cl
H
H
Termination Step:

Cl + Cl Cl2

H H
H C + Cl H C Cl

H H

HH HH
HC + CH H C CH

HH HH

b) electrophilic addition mechanism

Br

CH3CH2C CH2 HBr CH3CH2C CH3

CH3 CH3

Step 1:

+CH3CH2C CH2 H Br slow +
CH3CH2C CH3
CH3
CH3

Step 2:

++ Br - fast Br
CH3CH2C CH3
CH3CH2C CH3
CH3
CH3

4. a) Sayzteff’s rule states that the major product in elimination is the most stable alkenes
which has the most highly substituted double bond.

b) A: B:

CH3 CH2

Mechanism:

Step 1:

H H

+CH3 H O+ H fast CH3

OH + H +H O

O H

H H

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Step 2: slow H

H C+ +HO
CH3 CH3
O+ H H
H

Step 3:

H fast +CH3 +
H OH
+C+ H O H

CH3 H

5. chemical test to differentiate cyclopentane(alkane) and cyclopentene(alkene).
Test: Bromine test
Reagent and conditions: Br2, CH2Cl2
dark place
Observation and equation:
cyclopentane
Reddish brown colour of bromine remain unchanged.

cyclopentene

Reddish brown colour of bromine decolourised.

Br2, CH2Cl2 Br
dark place

Br

6. Structure of A to H. B: C:
A: OH O
OH
H3C C CH3 H3C C CH2Br + +H3C C
CH3 CH3 CO 2 H2O

D: E: CH3
OH H
F: G:
H3C C CH2OH H3C C CH2Br H OH
CH3 CH3
H3C C CH3 H3C C CH3
H: CH3 CH3
H

H3C C CH2Br
CH3

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KUMBE 12

1. naming
a) 2-chloro-5-methylheptane
b) 3-methylpentane
c) 5-isopropyl-5,6-dimethyl-2-heptene
d) 2-isopropyl-1-pentene
e) 1-bromo-3-(2-chloroethyl)-1-methylcyclopentane
f) 4-bromo-4-chloro-5-isopropylcyclohexene

2. a) free radical substitution reaction Cl

+ h
Cl 2

Initiation Step:

Cl Cl uv Cl + Cl

Propagation step:

H H
CH C

+ Cl HCl +

H H
C Cl

C

+ Cl Cl + Cl

Termination Step:

Cl + Cl Cl Cl

H H SK 12
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H H
+ Cl Cl

b) i. X is cycloalkane because it undergoes free radical substitution with bromine

when heated with bromine at very high temperature.

ii. X:

iii.
Br

+ CH2Cl2
Br 2 high
temperature

3. 3-methylhexane < heptane < octane
 Octane has the highest boiling point because it has the highest relative molecular
mass. Van der Waals forces between molecules strongest.
 3-methylhexane has lower boiling point than heptane because it has one branch.
Hence, the molecular surface area of 3-methylhexane < heptane. The strength of Van
der Waals forces between molecules 3-methylhexane is weaker.

4. Alkane is generally unreactive
Reason:
Alkane is a saturated hydrocarbon where all C atoms are bonded with four others C atoms.

5. a) A: c) C:
H3C C CHCH3
CH3

b) B: CH2

CH3

or

6. dehydrohalogenation of 2-bromo-3-methylbutane
CH3

CH3CHCHCH3
Br

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Major product Minor product CHEMISTRY SK025
CH3 C CH CH3 CH3CHCH CH2 CHAPTER 12:

CH3 CH3 HYDROCARBONS

Minor product

H2C C CH2CH3

CH3

7. Product of cyclobutene d)
Br
a)
H

I Br
b)
e)
c) OO
H
C CH2 CH2C
Br HH

f)

OO
C CH2 CH2C

HO OH

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