Question

In the reaction of p-chlorotoluene with $KN{H_2}$ in liquid $N{H_3}$, the major product is.
A) O-toluidine.
B) m-toluidine.
C) p-toluidine.
D) P-chloroaniline.

Answer 1

Hint: We know that the reaction of p-chlorotoluene with $KN{H_2}$ in liquid $N{H_3}$,proceeds via benzyne mechanism. The primary stage in the mechanism is a base promoted Dehydrohalogenation of chlorobenzene and the intermediate formed in the step will contain a triple bond in an aromatic ring called benzyne.

Complete step by step answer:
We know that para-Chlorotoluene initially forms benzyne with methyl as substituent. When -\[N{H_2}\] attacks at carbon which is bearing triple bond it gives a mixture of Meta and para-toluidine. The major product formed is m-toluidine because of the formation of carbanion at meta position by the attack of nucleophile and it is more stable compared to para position. Hence option B is correct.
The mechanism at which the reaction takes place,
First step: Removal of proton by \[KN{H_2}\] from the Meta position of p-chlorotoluene which provides benzyne as an intermediate.
Second step: Since benzyne is susceptible to nucleophilic attack and hence the nucleophile present within the solution is \[N{H_2}\]which may attack on the each side of the Benzene bond
Meta product is major as just in case of attack on 1 the negative charge formed is going to be farthest from the methyl group which facilitates less $ + 1$ and hence more stable negative ions.
We can write chemical equation for this mechanism as,

So, the correct answer is Option B.

Note:
Two mechanisms were proposed for the nucleophilic aromatic substitution, one among the mechanisms involves a benzyne because the intermediate formed is a benzyne and so, it is named as benzyne mechanism. Arynes or benzyne are extremely reactive groups resulting from an aromatic ring by elimination of two substituents.