Question

Which of the following can undergo Hell-Volhard-Zelinsky reaction?

• A. Benzoic acid
• B. Phenylacetic acid
• C. Phenylacetaldehyde
• D. Methyl phenyl ketone

Hint:

The essential condition for the Hell-Volhard-Zelinsky (HVZ) reaction is the presence of an $\alpha$-hydrogen in a carboxylic acid. Carboxylic acids where the carboxyl group is attached to a carbon with no hydrogens (like benzoic acid or trimethylacetic acid) cannot undergo this reaction.

Answer 1

The Correct Option is B

The Hell-Volhard-Zelinsky (HVZ) reaction is a chemical reaction for the alpha-halogenation of a carboxylic acid.
The key requirement for a carboxylic acid to undergo the HVZ reaction is the presence of at least one alpha-hydrogen atom. An alpha-hydrogen is a hydrogen atom bonded to the carbon atom adjacent to the carboxyl group.
The reaction typically involves treating the carboxylic acid with a halogen (Br$_2$ or Cl$_2$) in the presence of a catalytic amount of phosphorus or a phosphorus halide (like PBr$_3$).
Let's analyze the given options:
(A) Benzoic acid (C$_6$H$_5$-COOH): The carboxyl group (-COOH) is directly attached to the benzene ring. The alpha-carbon is part of the benzene ring and does not have any hydrogen atoms attached to it (it is bonded to another carbon in the ring and to the carboxyl group). Therefore, benzoic acid has no alpha-hydrogens and cannot undergo the HVZ reaction.
(B) Phenylacetic acid (C$_6$H$_5$-CH$_2$-COOH): The carboxyl group (-COOH) is attached to a -CH$_2$- group. This -CH$_2$- group is the alpha-carbon. It has two alpha-hydrogen atoms. Therefore, phenylacetic acid can undergo the HVZ reaction.
(C) Phenylacetaldehyde (C$_6$H$_5$-CH$_2$-CHO): This is an aldehyde, not a carboxylic acid. The HVZ reaction is specific to carboxylic acids.
(D) Methyl phenyl ketone (C$_6$H$_5$-CO-CH$_3$): This is a ketone, not a carboxylic acid. The HVZ reaction is specific to carboxylic acids.
Therefore, only phenylacetic acid can undergo the Hell-Volhard-Zelinsky reaction.