Question

Given a reaction ${C_6}{H_5}OH\xrightarrow[{N{H_3}}]{{ZnC{l_2}}}X$
Identify X.

Answer 1

Hint: The starting reactant present here is alcohol with formula ${C_6}{H_5}OH$. It is called phenol. We know that phenol is acidic in nature as the carbon on the benzene ring attached to the $ - OH$ group is more electron-withdrawing. Phenol can give away its ${H^ + }$ ion and as a result, it is acidic in nature. We know that $ZnC{l_2}$ is a dehydrating agent that removes water from a compound.

Complete step by step answer:
If water is present in a reaction mixture, anhydrous $ZnC{l_2}$ absorbs the water. This is the reason why it is called a dehydrating agent.
In this reaction also, water is removed.
One hydrogen from the $N{H_3}$ and the $ - OH$from phenol forms a water molecule. When one Hydrogen is broken from the $N{H_3}$, the $N{H_2}$ formed gets attached to the benzene ring.
Thus, Aniline is formed.
The complete reaction can be written as ${C_6}{H_5}OH\xrightarrow[{N{H_3}}]{{ZnC{l_2}}}{C_6}{H_5}N{H_2} + {H_2}O$
Therefore, X is Aniline with a molecular formula - ${C_6}{H_5}N{H_2}$. It is a method of formation of aniline.

Note: There are other methods of formation of aniline from phenol too. Instead of anhydrous $ZnC{l_2}$, if Zn dust is used, phenol gets reduced to benzene. The benzene can then be treated with a nitrating mixture which will form nitrobenzene. Nitrobenzene is reduced to Aniline using Tin and HCl.
${C_6}{H_5}OH\xrightarrow{{Zn\;dust}}{C_6}{H_6}\xrightarrow{{HN{O_3} + {H_2}S{O_4}}}{C_6}{H_5}N{O_2}\xrightarrow{{Sn/HCl}}{C_6}{H_5}N{H_2}$.
$ZnC{l_2}$ is a dehydrating agent whereas Zn dust is a reducing agent.
By reductive amination of phenol, phenol can be converted to aniline. Aniline can be converted to phenol by the process of diazotization.
Ethanol is also used as a dehydrating agent. Hot aluminium oxide, concentrated phosphoric acid are other examples of dehydrating agents.