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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

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Benzanilide

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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 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

Consider the following reaction:
$C_6H_5NO_2$ $\xrightarrow{Sn / HCl} X$ $\xrightarrow{C_6H_5COCl} Y + HCl$
What is $Y$?

Which of the following compounds is not soluble in $NaOH$?

Which of the following products is a Schiff base?
$C_2H_5NH_2 \xrightarrow{HNO_2} a$
$C_2H_5NH_2 \xrightarrow{C_6H_5CHO} b$
$C_2H_5NH_2 \xrightarrow{NOCl} c$
$C_2H_5NH_2 \xrightarrow{C_6H_5SO_2Cl} d$

When a primary amine is heated with $NaNO_2/HCl$,a gas is evolved with brisk effervescence. Identify the gas.
$R-NH_2 \xrightarrow{NaNO_2/HCl} ..................$

The compound with the lowest boiling point is:

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