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Question-1

How do you account for the formation of ethane during chlorination of methane?

Solution:
The formation of ethane is due to the side reaction in termination step by the combination of two CH3 free radicals.

Question-2

Write IUPAC names of the following compounds:

Solution:
(a) 2-Methyl - but-2-ene

(b) Pent-1-ene-3-yne

(c) Buta 1, 3-diene.

(d) 4-Phenyl-but-1-ene

(e) 2-Methyl phenol

(f) 5-(2-Methyl propyl)-decane

(g) 4-Ethyl-deca-1, 5, 8-triene.

Question-3

For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated :
(a) C4H8 (one double bond)
(b) C5H8 (one triple bond)

Solution:
(a) (i) CH2 = CH - CH2 - CH3 But - 1-ene

(ii) CH3 - CH2 = CH - CH3 But - 2-ene

(iii) 2-Methyl propene

(b) (i) HC C - CH2 - CH2 - CH2 Pent - 1-yne

(ii) CH3 - C C - CH2 - CH2 Pent - 2-yne

(iii) 3-Methyl-but- 1-yne.

Question-4

Write IUPAC names of the products obtained by the ozonolysis of the following compounds :

(i) Pent-2-ene

(ii) 3,4-Dimethyl-hept-3-ene

(iii) 2-Ethylbut-1-ene

(iv) 1-Phenylbut-1-ene

Solution:
(i) Ethanal and propanal.

(ii) Butan-2-one and pentan-2-one.

(iii) Methanal and pentan-3-one.

(iv) Propanal and benzaldehyde.

Question-5

An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the and IUPAC name of ‘A’.

Solution:
The IUPAC name of A is 3-Ethyl-pent-2-ene.

Question-6

An alkene ‘A’ contains three C – C, eight C – H bonds and one C – C p bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.

Solution:
The IUPAC name of A is But – 2 –ene

Question-7

Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

Solution:
4-Ethyl-hex-3-ene

Question-8

Write chemical equations for combustion reaction of the following hydrocarbons:
(i) Butane

(ii) Pentene

(iii) Hexyne

(iv) Toluene

Solution:
(a) C4H10(g) + 4CO2(g) +5H2O(g)

(b) C5H10(g) + 5CO2(g) +5H2O(g)

(c) C6H10(g) + 6CO2(g) +5H2O(g)

(d) C7H8(g) +5O2(g) 7CO2(g) +4H2O(g).

Question-9

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Solution:
The cis and trans structures of hex-2-ene

The cis form will have higher boiling point . This is due to more polar nature leading to strong intermolecular dipole-dipole interaction, thus requiring more heat energy to separate them.

Question-10

Why is benzene extra ordinarily stable though it contains three double bonds?

Solution:
Benzene is extra ordinarily stable due to resonance.

Question-11

What are the necessary conditions for any system to be aromatic?

Solution:
The essential conditions for any system to be aromatic are Planar, conjugated ring system with delocalisation of (4π + 2) p electrons, where n is an integer.

Question-12

Explain why the following systems are not aromatic?


Solution:
The compounds are not aromatic due to lack of cyclic cloud p-electrons having (4n + 2) p electrons.

Question-13

How will you convert benzene into
(i) p-nitrobromobenzene
(ii) m- nitrochlorobenzene
(iii) p - nitrotoluene
(iv) acetophenone?

Solution:


Question-14

In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these.

Solution:


6H attached to 10 carbons
4H attached to 20 carbons
1H attached to 30 carbons

Question-15

What effect does branching of an alkane chain has on its boiling point?

Solution:
The boiling point will be lower if there are more branching in alkane.

Question-16

Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

Solution:



The result tells us that all the three products cannot be obtained by any one of the Kekule’s structures showing that benzene is a resonance hybrid of the two resonating structure.

Question-17

Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.

Solution:
H – C – C – H > C6H6 > C6H14

The acidic property is maximum in ethyne due to maximum s orbital character in ethyne (50 percent) as compared to 33 percent in benzene and 25 percent in n-hexane.

Question-18

Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

Solution:
Due to the presence of p -electrons in benzene it can undergo electrophilic substitution reactions easily. In Nucleophilic substitutions the nucleophile will be repelled by p -electrons. So benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty.

Question-19

How would you convert the following compounds into benzene?
(i) Ethyne (ii) Ethene (iii) Hexane

Solution:

 

Question-20

Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Solution:

Question-21

Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?

Solution:
Due to electron releasing nature of the methyl group toluene undergoes nitration most easily.

Question-22

Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.

Solution:
Another Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene is FeCl3 .

Question-23

Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

Solution:
Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms because of the formation of side products. For example, by starting with 1-bromo propane and 1-bromo butane, hexane and octane are the side products besides heptane.




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