In the reaction sequence: C_6H_5NH_2 xrightar | vedclass

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Question

(A) The reaction sequence is as follows:$1$. Aniline $(C_6H_5NH_2)$ reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ under

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$C_6H_5NHCH_3$

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(A) The reaction sequence is as follows:$1$. Aniline $(C_6H_5NH_2)$ reacts with chloroform $(CHCl_3)$ and alcoholic potassium hydroxide $(KOH)$ under heating to undergo the carbylamine reaction,forming phenyl isocyanide $(C_6H_5NC)$ as product $(X)$.$C_6H_5NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O$$2$. Phenyl isocyanide $(C_6H_5NC)$ is then reduced using lithium aluminium hydride $(LiAlH_4)$ followed by hydrolysis to yield $N$-methylaniline $(C_6H_5NHCH_3)$ as product $(Y)$.$C_6H_5NC \xrightarrow{LiAlH_4/H_2O} C_6H_5NHCH_3$Therefore,the correct option is $(A)$.

In the reaction sequence: $C_6H_5NH_2$ $\xrightarrow{CHCl_3/KOH, \Delta} (X)$ $\xrightarrow{LiAlH_4/H_2O} (Y)$,the product $(Y)$ is:

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