Ph-NC xrightarrow{LiAlH_4} A xrightarrow{NaNO | vedclass

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(B) $1$. The reduction of phenyl isocyanide $(Ph-NC)$ with $LiAlH_4$ yields $N$-methylaniline $(Ph-NH-CH_3)$ as product $A$.$2$. $N$-methylaniline is

Vedclass · Accepted Answer

$N$-nitroso-$N$-methylaniline

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(B) $1$. The reduction of phenyl isocyanide $(Ph-NC)$ with $LiAlH_4$ yields $N$-methylaniline $(Ph-NH-CH_3)$ as product $A$.$2$. $N$-methylaniline is a secondary aromatic amine.$3$. Secondary amines react with nitrous acid $(NaNO_2 + HCl)$ at low temperatures $(0-5 \ ^\circ C)$ to form $N$-nitroso compounds.$4$. Therefore,the reaction of $N$-methylaniline with $NaNO_2 + HCl$ gives $N$-nitroso-$N$-methylaniline as product $B$.

$Ph-NC$ $\xrightarrow{LiAlH_4} A$ $\xrightarrow{NaNO_2 + HCl} B$. Identify product $B$.

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