Question

The 'X' in the following conversion is $CH_3CH_2COOH \xrightarrow{X} CH_3CHBrCOOH$

• A. (i) Br$_2$/P red., (ii) H$_2$O
• B. (i) Br$_2$/CCl$_4$, (ii) H$_2$O
• C. Br$_2$ / OH$^-$
• D. PBr$_3$

Hint:

The Hell-Volhard-Zelinsky (HVZ) reaction is the specific method for $\alpha$-halogenation of carboxylic acids. Remember the reagents: bromine (or chlorine), red phosphorus, followed by hydrolysis with water. This reaction selectively introduces the halogen atom at the carbon adjacent to the carboxyl group.

Answer 1

The Correct Option is A

The given conversion involves the bromination of propanoic acid $CH_3CH_2COOH$ to 2-bromopropanoic acid $CH_3CHBrCOOH$. The bromine atom is introduced at the $\alpha$-carbon (the carbon adjacent to the carboxyl group). This type of $\alpha$-halogenation of carboxylic acids is known as the Hell-Volhard-Zelinsky (HVZ) reaction.
The Hell-Volhard-Zelinsky reaction proceeds in two steps: 
Step 1: The carboxylic acid reacts with bromine (\(Br_2\)) in the presence of red phosphorus (P red.). The red phosphorus reacts with bromine to form phosphorus tribromide (\(PBr_3\)), which then converts the carboxylic acid to an acyl bromide. The acyl bromide undergoes tautomerization to form an enol, which is then brominated at the $\alpha$-position. 
Step 2: The $\alpha$-bromoacyl bromide is hydrolyzed with water (\(H_2O\)) to yield the $\alpha$-bromocarboxylic acid.
The overall reaction sequence is:
$$CH_3CH_2COOH \xrightarrow{Br_2/P} \, red.CH_3CHBrCOBr \xrightarrow{H_2O} CH_3CHBrCOOH$$ 
Therefore, the reagent 'X' required for this conversion is (i) Br$_2$/P red., followed by (ii) H$_2$O. 
Analyzing the other options:
Option (2) Br$_2$/CCl$_4$ is used for the bromination of alkenes or alkanes (radical substitution), not for $\alpha$-halogenation of carboxylic acids.
Option (3) Br$_2$ / OH$^-$ is used for the haloform reaction with methyl ketones or secondary alcohols that can be oxidized to methyl ketones.
Option (4) PBr$_3$ is used to convert carboxylic acids to acyl bromides or alcohols to alkyl bromides, but it does not directly introduce a bromine at the $\alpha$-position of a carboxylic acid. 
Thus, the correct reagent 'X' is $ \boxed{(i) Br_2/P \, red., (ii) H_2O} $.