PhNH_2 + Ph_3COH xrightarrow{H^{+}} P (major) | vedclass

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(B) $1$. In the presence of an acid $(H^{+})$,triphenylmethanol $(Ph_3COH)$ undergoes protonation followed by the loss of water to form a stable triph

Vedclass · Accepted Answer

p-tritylaniline

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(B) $1$. In the presence of an acid $(H^{+})$,triphenylmethanol $(Ph_3COH)$ undergoes protonation followed by the loss of water to form a stable triphenylmethyl carbocation $(Ph_3C^{+})$.$2$. Aniline $(PhNH_2)$ acts as a nucleophile. The amino group $(-NH_2)$ is a strongly activating group and is ortho/para-directing.$3$. Due to the significant steric hindrance provided by the bulky triphenylmethyl group $(Ph_3C^{+})$,the electrophilic aromatic substitution occurs primarily at the para-position.$4$. Therefore,the major product is $p$-tritylaniline.

$PhNH_2 + Ph_3COH \xrightarrow{H^{+}} P$ (major) (not an $N$-derivative). The product $(P)$ is:

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