Considering the reaction sequence given below | vedclass

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(A) The reaction sequence is as follows:$1$. $CH_3CH_2COOH \xrightarrow{Br_2, 	ext{red } P} CH_3CH(Br)COOH$ $(P)$.$2$. $CH_3CH(Br)COOH$ reacts with p

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$B, C, D$

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(A) The reaction sequence is as follows:$1$. $CH_3CH_2COOH \xrightarrow{Br_2, 	ext{red } P} CH_3CH(Br)COOH$ $(P)$.$2$. $CH_3CH(Br)COOH$ reacts with potassium phthalimide followed by hydrolysis to give $CH_3CH(NH_2)COOH$ $(Q)$ and phthalic acid.Analysis of statements:$A$. $P$ is $CH_3CH(Br)COOH$. $NaBH_4$ reduces carboxylic acids very slowly or not at all,and it does not reduce the $\alpha$-bromo group to an alcohol. This statement is incorrect.$B$. Treating $P$ $(CH_3CH(Br)COOH)$ with conc. $NH_4OH$ (ammonolysis) followed by acidification gives $CH_3CH(NH_2)COOH$,which is $Q$. This statement is correct.$C$. $Q$ is an $\alpha$-amino acid $(CH_3CH(NH_2)COOH)$. Treatment with $NaNO_2/HCl$ (diazotization) converts the $-NH_2$ group to $-N_2^+Cl^-$,which is unstable and decomposes to release $N_2$ gas,forming $\alpha$-hydroxy acid. This statement is correct.$D$. $P$ $(CH_3CH(Br)COOH)$ has an electron-withdrawing $-Br$ group at the $\alpha$-position,which stabilizes the conjugate base through the inductive effect,making it more acidic than $CH_3CH_2COOH$. This statement is correct.Thus,statements $B, C,$ and $D$ are correct.

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