Question

1) - Among the isomeric alkanes of molecular formula $C_5{H}_{12},$ identify the one that on photochemical chlorination yields
A) A single monochloride.

2) - Among the isomeric alkanes of molecular formula $C_5{H}_{12},$ identify the one that on photochemical chlorination yields
B) Three isomeric monochlorides.

3) - Among the isomeric alkanes of molecular formula $C_5{H}_{12},$ identify the one that on photochemical chlorination yields
C) Four isomeric monochlorides.

Answer 1

1) - Alkane halogenation is a type of substitution reaction.
To get a single monochloride, there should be only one type of $H-$ atom in the isomer of the alkane of the molecular formula $C_5{H}_{12}.$

This is because, replacement of any $H-\text{atom}$ leads to the formation of the same product.

Therefore, that isomer is $2,2-\text{dimethylpropane.}$


All the structures obtained by replacement of different $H$ atoms are identical.


2) - Alkane halogenation is a type of substitution reaction.
To have three isomeric monochlorides, the isomer of the alkane of the molecular formula $C_5{H}_{12}$ should contain three different types of $H-\text{atoms.}$

We can see that there are three different types of $H$ atoms labelled as a, b and c in $n-\text{pentane.}$ Hence the isomer is $n-\text{pentane.}$


The three possible structures are drawn below:


Here, $\left(a\right)\&\left(b\right)$ are identical


Here, $\left(a\right)\&\left(b\right)$ are identical

3) Alkane halogenation is a type of substitution reaction.

To have four isomeric monochlorides, the isomer of the alkane of the molecular formula $C_5{H}_{12}$ should contain four different types of $H$ atoms.

We can see that there are four types of $H-\text{atoms}$ labelled as a, b, c, and d in $2-\text{methylbutane.}$


Therefore, four isomeric monochlorides are given below.