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Question

(A) $1$. Aniline $(Ph-NH_2)$ reacts with Caro's acid $(H_2SO_5)$ to form nitrosobenzene $(Ph-N=O)$ as product $(A)$.$2$. Nitrosobenzene then undergoes

Vedclass · Accepted Answer

$Ph-N=C(CN)-Ph$

Vedclass · Answer

(A) $1$. Aniline $(Ph-NH_2)$ reacts with Caro's acid $(H_2SO_5)$ to form nitrosobenzene $(Ph-N=O)$ as product $(A)$.$2$. Nitrosobenzene then undergoes a condensation reaction with benzyl cyanide $(Ph-CH_2CN)$ in the presence of a base $(EtO^-)$ and heat.$3$. The base abstracts a proton from the $\alpha$-carbon of benzyl cyanide to form a carbanion $(Ph-CH^-(CN))$.$4$. This carbanion attacks the electrophilic nitrogen of the nitroso group in $Ph-N=O$.$5$. Subsequent steps involving proton transfer and elimination of water lead to the formation of the final product,$Ph-N=C(CN)-Ph$.

List-$I$ (Mixture of Compounds)	List-$II$ (Reagent used to distinguish)
$A$. Diethyl amine + Ethyl amine	$I$. Bromine water
$B$. Acetaldehyde + Acetone	$II$. $CHCl_3 + KOH, \Delta$
$C$. Ethanol + Phenol	$III$. Neutral $FeCl_3$
$D$. Benzoic acid + Cinnamic acid	$IV$. Ammonical silver nitrate

The product $(B)$ is:

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