The following sequence of reactions on A give | vedclass

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

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$A$ cyclic amide (lactam)

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(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 final product of the given sequence is :

Identify the final product $Z$ in the following reaction sequence:

Assertion : Aniline does not undergo Friedel-Crafts reaction.
Reason : $-NH_2$ group of aniline reacts with $AlCl_3$ (Lewis acid) to give acid-base reaction.

Aniline in a set of reactions yielded a product $D$. The structure of product $D$ would be:
Aniline $\xrightarrow{NaNO_2/HCl} A$ $\xrightarrow{CuCN} B$ $\xrightarrow{H_2/Ni} C$ $\xrightarrow{HNO_2} D$

Complete the following acid-base reactions and name the products:
$(i)$ $CH_{3}CH_{2}CH_{2}NH_{2} + HCl \rightarrow$
$(ii)$ $(C_{2}H_{5})_{3}N + HCl \rightarrow$

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