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(B) $1$. Benzoic acid $(C_6H_5COOH)$ reacts with $NH_3$ followed by heating to form benzamide $(C_6H_5CONH_2)$,which is product $P$.$2$. Benzamide und

Vedclass · Accepted Answer

Benzanilide

Vedclass · Answer

(B) $1$. Benzoic acid $(C_6H_5COOH)$ reacts with $NH_3$ followed by heating to form benzamide $(C_6H_5CONH_2)$,which is product $P$.$2$. Benzamide undergoes the Hofmann bromamide degradation reaction with $NaOBr$ to form aniline $(C_6H_5NH_2)$,which is product $Q$.$3$. Aniline reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base to form $N$-phenylbenzamide,commonly known as benzanilide $(C_6H_5NHCOC_6H_5)$,which is the final product $R$.

The final product $R$ of the reaction is:

Similar Questions

The reaction is shown below. What is $X$ (major)?
$C_6H_5-CONH-C_6H_5 \xrightarrow{Br_2/Fe} X \text{ (major)}$

The reduction of benzenediazonium chloride to phenylhydrazine can be accomplished by:

$R-CONH_2 \xrightarrow[\Delta]{KOBr} R-NH_2$. The correct statement about the product is:

The reaction of $RCONH_2$ with bromine and $KOH$ gives $RNH_2$ as the end product. Which one of the following is the intermediate product formed in this reaction?

Which of the following reaction names is not associated with a scientist?

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