What is the product of the following reaction?

$α$-Halocarboxylic acid

The correct answer is Option (1) → $α$-Halocarboxylic acid

The given reaction uses halogen (Cl₂ or Br₂) in the presence of red phosphorus followed by hydrolysis with water. This set of reagents corresponds to the Hell–Volhard–Zelinsky (HVZ) reaction.

The HVZ reaction is a well-known reaction in organic chemistry used for α-halogenation of carboxylic acids that contain at least one α-hydrogen.

General reaction:

R–CH₂–COOH → R–CHX–COOH

(X = Cl or Br)

Steps involved:

1. Formation of acyl halide

Red phosphorus reacts with halogen to form PX₃, which converts the carboxylic acid into an

acid halide (R–CH₂–COX).

2. α-Halogenation

The acid halide undergoes enol formation, and the halogen substitutes the α-hydrogen to produce R–CHX–COX.

3. Hydrolysis

When treated with water, the acid halide converts back into the carboxylic acid, giving α-halocarboxylic acid.

Final product:

R–CHX–COOH

Thus, halogen substitution occurs specifically at the α-carbon (carbon adjacent to the carboxyl group).

Option-wise Explanation

Option 1: α-Halocarboxylic acid

In the HVZ reaction, halogen substitution occurs at the α-position of the carboxylic acid due to enol formation and electrophilic halogenation. After hydrolysis, the product becomes R–CHX–COOH, which is an α-halocarboxylic acid. Therefore, this option is correct.

Option 2: β-Halocarboxylic acid

β-Halogenation would involve substitution at the second carbon away from the carboxyl group. The HVZ reaction specifically halogenates the α-carbon, not the β-carbon. Hence this option is incorrect.

Option 3: α, β-Dihalocarboxylic acid

The reaction introduces only one halogen atom at the α-position. It does not produce halogen substitution at both α and β positions. Therefore this option is incorrect.

Option 4: 1,1-Dihalocarboxylic acid

This structure would require two halogens attached to the same carbon atom. The HVZ reaction introduces only one halogen atom at the α-carbon, so this option is incorrect.