NCERT Solutions for Class 11 Chemistry Chapter 13 Hydrocarbons

Ans : The reaction begins with the homolytic cleavage of Cl — Cl bond as:
Step 1: Initiation:
\( \mathrm{Cl}-\mathrm{Cl} \stackrel{\mathrm{hv}}{\longrightarrow}
\dot{\mathrm{Cl}}+\dot{\mathrm{C}}\)
Step 2: Propagation:
In the second step, chlorine free radicals attack methane molecules and break down the bond to generate methyl radicals as:
\( \mathrm{CH}_{4}+\dot{\mathrm{C}} 1 \stackrel{\mathrm{h} v}{\longrightarrow} \dot{\mathrm{C}}
\mathrm{H}_{3}+\mathrm{H}-\mathrm{Cl}\)
These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.
\( \dot{\mathrm{C}} \mathrm{H}_{3}+\mathrm{Cl}-\mathrm{Cl} \longrightarrow
\mathrm{CH}_{3}-\mathrm{Cl}+\dot{\mathrm{C} 1}\)
Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCI and \( \mathrm{CH}_{3} \mathrm{C}\) are the major products formed, other higher halogenated compounds are a so formed as:
\( \mathrm{CH}_{3} \mathrm{Cl}+\dot{\mathrm{C}} 1 \longrightarrow \dot{\mathrm{C}}
\mathrm{H}_{2} \mathrm{Cl}+\mathrm{HCl}\)
\( \dot{\mathrm{C}} \mathrm{H}_{2} \mathrm{Cl}+\mathrm{Cl}-\mathrm{Cl} \longrightarrow
\mathrm{CH}_{2} \mathrm{Cl}_{2}+\dot{\mathrm{Cl}}\)
Step 3: Termination:
Formation of ethane is a result of the termination of chain reactions taking place as a result of the consumption of reactants as:
\( \dot{\mathrm{Cl}}+\dot{\mathrm{C}} \mathrm{l} \longrightarrow \mathrm{Cl}-\mathrm{Cl}\)
\( \mathrm{H}_{3} \dot{\mathrm{C}}+\dot{\mathrm{C}} \mathrm{H}_{3} \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}\)
Hence, by this process, ethane is obtained as a by-product of chlorination of methane. 

Ans : (a)
 
IUPAC name: 2-Methylbut-2-ene
(b)
IUPAC name: Pen-1 -ene-3-yne
(c)
IUPAC name: 1, 3-Butadiene or Buta-1,3-diene
(d)
IUPAC name: 4-Phenyl but-1-ene
(e)
IUPAC name: 2-Methyl phenol
(f)
IUPAC name: 5-(2-Methylpropyl)-decane
(g)
IUPAC name: 4-Ethyldeca-1, 5, 8 triene 

Ans : (a) The following structural isomers are possible for \( \mathrm{C}_{4} \mathrm{H}_{8}\) with one double bond:

The IUPAC name of
Compound (I) is But-1 -ene,
Compound (II) is But-2-ene, and
Compound (Ill) is 2-Methylprop-1 -ene.


(b) The following structural isomers are possible for \( \mathrm{C}_{5} \mathrm{C}_{8}\) with one triple bond .

The IUPAC name of
Compound (I) is Pent-1 -yne,
Compound (II) is Pent-2-yne, and
Compound (Ill) is 3-Methylbut-1-ene. 

Ans : (i ) Pent-2-ene undergoes ozonolysis as:

The IUPAC name of Product (1) is ethanol and Product (II)is propanel.


(ii) 3, 4-Dimethylhept-3-ene undergoes ozonolysis as:

The IUPAC næ•ne of Product (I)is butan-2-one and Product (II) is Pentan-2-one.


(iii) 2-Ethylbut-1-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is pentan-3-one and Product (II)is methanal.


(iv) 1 -Phenylbut-l -ene undergoes ozonolysis as:

The IUPAC name of Product (I)is benzaldehyde and Product (II)is propenel, 

Ans : 
During ozonolysis, an ozonide having a cyclic structure is formed as an intermediate which undergoes cleavage to give the final products.Ethanal and pentan-3-one are obtained from the intermediate ozonide. Hence, the expected structure of the ozonide is:

This ozonide is formed as addition of ozone to The desired structure of 'A' can be obtained by the removal of ozone from the ozonide. Hence. the structural formula of 'A'



The IUPAC name of 'A' is 3-Ethylpent-2-ene.