(A) xrightarrow{KOBr} (B) xrightarrow[KOH]{CH | vedclass

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(B) $1$. The reaction sequence starts with $(A)$,which is $p$-methylbenzamide $(CH_3-C_6H_4-CONH_2)$.$2$. Hofmann bromamide degradation: $(A) \xrighta

Vedclass · Accepted Answer

p-methylphenyl isocyanide

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(B) $1$. The reaction sequence starts with $(A)$,which is $p$-methylbenzamide $(CH_3-C_6H_4-CONH_2)$.$2$. Hofmann bromamide degradation: $(A) \xrightarrow{KOBr} (B)$. This converts the amide group $(-CONH_2)$ to a primary amine group $(-NH_2)$. Thus,$(B)$ is $p$-toluidine $(CH_3-C_6H_4-NH_2)$.$3$. Carbylamine reaction: $(B) \xrightarrow{CHCl_3/KOH} (C)$. Primary amines react with chloroform and alcoholic $KOH$ to form isocyanides (isonitriles). Thus,$(C)$ is $p$-methylphenyl isocyanide $(CH_3-C_6H_4-NC)$.$4$. Reduction: $(C) \xrightarrow{LiAlH_4} 	ext{Product}$. Isocyanides are reduced to secondary amines $(R-NH-CH_3)$. Therefore,the final product is $N$-methyl-$p$-toluidine $(CH_3-C_6H_4-NHCH_3)$.$5$. The question asks for the structure of $(C)$,which is $p$-methylphenyl isocyanide.

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

$(A)$ $\xrightarrow{KOBr} (B)$ $\xrightarrow[KOH]{CHCl_3} (C)$ $\xrightarrow{LiAlH_4} \text{Structure of } (C) \text{ is}$

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