The final product A, formed in the following | vedclass

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(B) $1$. Bromobenzene reacts with $Mg$ in ether to form phenylmagnesium bromide $(C_6H_5MgBr)$.$2$. Phenylmagnesium bromide reacts with $CO_2$ followe

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Aniline

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(B) $1$. Bromobenzene reacts with $Mg$ in ether to form phenylmagnesium bromide $(C_6H_5MgBr)$.$2$. Phenylmagnesium bromide reacts with $CO_2$ followed by acid hydrolysis $(H^+)$ to yield benzoic acid $(C_6H_5COOH)$.$3$. Benzoic acid reacts with $NH_3$ and heat $(\Delta)$ to form benzamide $(C_6H_5CONH_2)$.$4$. Benzamide undergoes the Hoffmann bromamide degradation reaction with $Br_2$ and $NaOH$ to form aniline $(C_6H_5NH_2)$.Therefore, the final product $A$ is aniline.

The final product $A$, formed in the following multistep reaction sequence is:

Similar Questions

What are $X$ and $Y$ respectively in the following set of reactions?

An organic compound $A$ reacts with $NH_3$ on heating to give $B$. Compound $B$ reacts with $Br_2$ in the presence of $KOH$ to give $CH_3CH_2NH_2$. What is $A$?

$CH_3CH_2CONH_2$ $\xrightarrow{Br_2 / KOH_{(aq)}} X$ $\xrightarrow{C_6H_5COCl} Y$. Which statement is true?

How will you convert:
$(i)$ Benzene into aniline
$(ii)$ Benzene into $N,N$-dimethylaniline
$(iii)$ $Cl-(CH_2)_4-Cl$ into hexane-$1,6$-diamine?

The product $(A)$ of the reaction is:

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