Question

Write the mechanism of the following reaction:
$nBuBr + KCN\xrightarrow{{EtOH - {H_2}O}}nBuCN$

Answer 1

Nucleophile is a chemical species that donates its electron pair to form a bond with other atoms and thus, the nucleophile is also called an electron-rich species. When nucleophile undergoes substitution, based on the nature of reactant, it comes under two categories such as Unimolecular Nucleophilic substitution (${S_N}^1$) and Bimolecular Nucleophilic substitution (${S_N}^2$) reaction.

Complete step by step answer:
When n-butyl bromide is reacted to potassium cyanide in presence of ethanol in water, then n-butyl cyanide will be formed. The reaction can be written as,
$nBuBr + KCN\xrightarrow{{EtOH - {H_2}O}}nBuCN$
The above reaction is an example of an unimolecular Nucleophilic substitution reaction ${S_N}^1$. ${S_N}^1$ is a two-step process and carbocation will be formed as an intermediate. The n-butyl group is bulky and therefore it won't undergo bimolecular Nucleophilic substitution reaction ${S_N}^2$. ${S_N}^2$ is a single step process and it is affected by the steric effect. The steric effect is the crowding of atoms and it leads to creating disturbance for upcoming atoms.
The mechanism for ${S_N}^1$ reaction of n-butyl cyanide formation can occur in two steps.
Step-1: Formation of carbocation
The n-butyl bromide will break into n-butyl carbocation and bromide ion. This step is a slow step process and thus, it is a rate-determining step. Here, bromide ion acts as the leaving group. The reaction can be written as,

Step-2: Formation of n-butyl cyanide
Cyanide ion acts as a nucleophile. when cyanide ion approaches n-butyl carbocation, then cyanide ion gets substituted and it forms n-butyl cyanide. This step is the fast step process. The reaction can be written as follows,

Thus, the above mechanism is the correct approach for the ${S_N}^1$ reaction.

Note: Carbocation is affected by electronic factors. Since the rate-determining step is carbocation formation, then the electronic factor is one of the important factors in ${S_N}^1$ reaction. Thus, tertiary alkyl favours ${S_N}^1$ reaction than secondary alkyl groups.
The order of reactivity for ${S_N}^1$ reaction can be written as,
Tertiary alkyl> Secondary alkyl> Primary alkyl