Answer

In the presence of sulphuric acid $\left({H_2} {SO_4}\right)$, ${KI}$ produces ${HI}$

$ 2 {KI}+{H_2} {SO_4} \longrightarrow 2 {KHSO_4}+2 {HI} $

Since ${H_2} {SO_4}$ is an oxidizing agent, it oxidizes ${HI}$ (produced in the reaction to ${I_2}$.

$ 2 {HI}+{H_2} {SO_4} \longrightarrow {I_2}+{SO_2}+{H_2} {O} $

As a result, the reaction between alcohol and ${HI}$ to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as ${H_3} {PO_4}$ is used.

Answer

There are four different dihalogen derivatives of propane. The structures of these derivatives are shown below.

Answer

(i) To have 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 ${H}$-atom leads to the formation of the same product. The isomer is neopentane.

${CH_3 - \stackrel{\substack{{CH_3} \\ |}}{\underset{\substack{ | \\ {CH_3}}}{C}} -CH_3}$

Neopentane

Therefore, the isomer is $n$-pentane. It can be observed that there are three types of ${H}$ atoms labelled as $a, b$ and $c$ in $n$-pentane.

(iii) 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. Therefore, the isomer is 2-methylbutane. It can be observed that there are four types of ${H}$ atoms labelled as $a, b, c$, and $d$ in 2-methylbutane.

Answer

(i)

(ii)

(iii)

(iv)

(v)

(vi)