Synthesis and Properties of Intermediate Compounds in Reaction Sequence

To determine the correct statement(s), let us analyze the given reaction sequence step-by-step:

Step 1: Identification of Compound $\mathbf{P}$

Propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$) is treated with $\text{Br}_2$ in the presence of red phosphorus followed by hydrolysis ($\text{H}_2\text{O}$). This is the Hell-Volhard-Zelinsky (HVZ) reaction, which selectively substitutes an $\alpha$-hydrogen with a bromine atom:

$\text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{1.\text{ Br}_2/\text{red P}, \; 2.\text{ H}_2\text{O}} \text{CH}_3\text{CH(Br)COOH}$

Thus, compound $\mathbf{P}$ is $2$-bromopropanoic acid ($\text{CH}_3\text{CH(Br)COOH}$).

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Step 2: Identification of Compound $\mathbf{Q}$

Compound $\mathbf{P}$ undergoes a modified Gabriel Phthalimide Synthesis:

1. Reaction of $2$-bromopropanoic acid with potassium phthalimide results in nucleophilic substitution ($\text{S}_\text{N}2$) of the bromide group by the phthalimide ion, yielding $N$-phthaloyl alanine.
2. Subsequent alkaline hydrolysis ($\text{NaOH}$) followed by acidification ($\text{H}_3\text{O}^+$) cleaves the phthalimide adduct to form phthalic acid as a byproduct and alanine ($\alpha$-aminopropanoic acid) as compound $\mathbf{Q}$.

$\mathbf{P} \; (\text{CH}_3\text{CH(Br)COOH}) \xrightarrow{1.\text{ Potassium phthalimide}, \; 2.\text{ NaOH}, \; 3.\text{ H}_3\text{O}^+} \mathbf{Q} \; (\text{CH}_3\text{CH(NH}_2\text{)COOH}) + \text{Phthalic acid}$

Thus, compound $\mathbf{Q}$ is alanine ($2$-aminopropanoic acid).

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Evaluation of Options:

• (A) $\mathbf{P}$ can be reduced to a primary alcohol using $\text{NaBH}_4$: $\text{NaBH}_4$ (sodium borohydride) is a mild reducing agent that reduces aldehydes, ketones, and acyl chlorides, but does not reduce carboxylic acids ($-\text{COOH}$) to primary alcohols (which require stronger reducing agents like $\text{LiAlH}_4$ or $\text{B}_2\text{H}_6$). Thus, Statement (A) is INCORRECT.

• (B) Treating $\mathbf{P}$ with conc. $\text{NH}_4\text{OH}$ solution followed by acidification gives $\mathbf{Q}$: Treatment of an $\alpha$-halo acid like $2$-bromopropanoic acid ($\mathbf{P}$) with excess concentrated ammonium hydroxide ($\text{NH}_4\text{OH}/\text{NH}_3$) leads to nucleophilic substitution of the $-\text{Br}$ group by an amino group ($-\text{NH}_2$), producing alanine ($\mathbf{Q}$). This is a standard laboratory method for synthesizing $\alpha$-amino acids. Thus, Statement (B) is CORRECT.

• (C) Treating $\mathbf{Q}$ with a solution of $\text{NaNO}_2$ in aq. $\text{HCl}$ liberates $\text{N}_2$: Compound $\mathbf{Q}$ contains a primary aliphatic amino group ($-\text{NH}_2$). Primary aliphatic amines react with nitrous acid ($\text{HNO}_2$, generated in situ from $\text{NaNO}_2 + \text{HCl}$) to form an unstable aliphatic diazonium salt, which spontaneously decomposes to liberate nitrogen gas ($\text{N}_2\uparrow$) quantitatively (Van Slyke reaction):

  $\text{CH}_3\text{CH(NH}_2\text{)COOH} + \text{HNO}_2 \rightarrow \text{CH}_3\text{CH(OH)COOH} + \text{N}_2\uparrow + \text{H}_2\text{O}$

  Thus, Statement (C) is CORRECT.

• (D) $\mathbf{P}$ is more acidic than $\text{CH}_3\text{CH}_2\text{COOH}$: $2$-bromopropanoic acid ($\mathbf{P}$) contains a highly electronegative bromine atom at the $\alpha$-position relative to the carboxyl group. The electron-withdrawing inductive effect ($-I$ effect) of bromine stabilizes the conjugate base ($\text{CH}_3\text{CH(Br)COO}^-$) to a greater extent than the unsubstituted propanoate ion. Hence, $\mathbf{P}$ is significantly more acidic than propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$). Thus, Statement (D) is CORRECT.

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Conclusion:

The correct statements are B, C, and D.