INTRODUCTION - Chapter 13 - Hydrocarbons - Ncert Solutions class 11 - Chemistry

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Question 13.1:

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

Answer:

Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps.

Step 1: Initiation:

The reaction begins with the homolytic cleavage of Cl – Cl bond as:

Step 2: Propagation:

In the second step, chlorine free radicals attack methane molecules and break down the C–H bond to generate methyl radicals as:

These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical.

Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and CH₃Cl are the major products formed, other higher halogenated compounds are also formed as:

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:

Hence, by this process, ethane is obtained as a by-product of chlorination of methane.

Question 13.2:

Write IUPAC names of the following compounds:

a. 

b. 

c. 

d. 

e. 

f.

g.

Answer:

(a)

IUPAC name: 2-Methylbut-2-ene

(b) 

IUPAC name: Pen-1-ene-3-yne

(c)   can be written as:

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

Question 13.3:

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)

Answer:

(a) The following structural isomers are possible for C₄H₈ with one double bond:

The IUPAC name of

Compound (I) is But-1-ene,

Compound (II) is But-2-ene, and

Compound (III) is 2-Methylprop-1-ene.

(b) The following structural isomers are possible for C₅C₈ with one triple bond:

The IUPAC name of

Compound (I) is Pent-1-yne,

Compound (II) is Pent-2-yne, and

Compound (III) is 3-Methylbut-1-yne.

Question 13.4:

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

Answer:

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

The IUPAC name of Product (I) is ethanal and Product (II)is propanal.

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

The IUPAC name 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-1-ene undergoes ozonolysis as:

The IUPAC name of Product (I)is benzaldehyde and Product (II)is propanal.