The following sequence of reactions on A give | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The reaction sequence is as follows:$1$. The starting material is methyl $2$-(carbamoylmethyl)benzoate.$2$. Treatment with $Br_2/NaOH$ (Hofmann br

Vedclass · Accepted Answer

$A$ cyclic amide (lactam)

Vedclass · Answer

(C) The reaction sequence is as follows:$1$. The starting material is methyl $2$-(carbamoylmethyl)benzoate.$2$. Treatment with $Br_2/NaOH$ (Hofmann bromamide degradation) converts the amide group $(-CONH_2)$ into a primary amine group $(-NH_2)$.$3$. The resulting intermediate is methyl $2$-(aminomethyl)benzoate.$4$. Upon heating,this intermediate undergoes intramolecular cyclization (nucleophilic acyl substitution) where the amino group attacks the ester carbonyl,eliminating methanol $(CH_3OH)$ to form a cyclic amide,specifically a lactam (isoindolin$-1-$one).

The following sequence of reactions on $A$ gives:

Similar Questions

The correct order of basic strength for the given compounds is:

What product is formed when benzamide is reacted with $POCl_3$?

Consider the following statements:
Statement-$I$: $CH_3NH_2$ is more basic than $NH_3$, but $C_6H_5NH_2$ is less basic than $NH_3$.
Statement-$II$: The order of basic strength of ethyl-substituted amines in the aqueous phase follows the order $(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2$.

Starting from propanoic acid,the following reactions were carried out: $CH_3-CH_2-COOH$ $\xrightarrow{SOCl_2} X$ $\xrightarrow{NH_3} Y$ $\xrightarrow{Br_2 + KOH} Z$. What is the compound $Z$?

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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module