Question

Which one of the following compounds undergoes HVZ reaction?

• A. \( \text{C}_6\text{H}_5\text{COOH} \)
• B. \( \text{CH}_3\text{CH}_2\text{COOH} \)
• C. \( \text{CCl}_3\text{COOH} \)
• D. \( \text{CH}_3\text{CH}_2\text{CHO} \)
• E. \( \text{HCOOH} \)

Hint:

The HVZ reaction is characteristic of carboxylic acids, particularly for halogenation followed by decarboxylation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The Hell-Volhard-Zelinsky (HVZ) reaction is a chemical reaction that involves 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. The alpha-hydrogen is a hydrogen atom on the carbon atom adjacent to the carboxyl group.
Step 2: Key Formula or Approach:
The general reaction is:
\[ \text{RCH}_2\text{COOH} \xrightarrow[\text{2. H}_2\text{O}]{\text{1. X}_2\text{, Red P}} \text{RCH(X)COOH} \] where X = Cl or Br.
We need to examine the structure of each compound to see if it has an alpha-hydrogen.
Step 3: Detailed Explanation:
Let's analyze each compound:
(A) C\(_6\)H\(_5\)COOH (Benzoic acid): The carboxyl group is attached to a carbon atom of the benzene ring. This alpha-carbon is sp\(^2\)-hybridized and has no hydrogen atoms attached to it. Therefore, benzoic acid does not have an alpha-hydrogen and cannot undergo the HVZ reaction.
(B) CH\(_3\)CH\(_2\)COOH (Propanoic acid): The carbon atom adjacent to the -COOH group (the alpha-carbon) is -CH\(_2\)-. It has two alpha-hydrogen atoms. Therefore, propanoic acid will undergo the HVZ reaction.
(C) Cl\(_2\)C-COOH (Dichloroacetic acid): The alpha-carbon is bonded to two chlorine atoms and the carboxyl group. It has no alpha-hydrogen atoms. Thus, it cannot undergo the HVZ reaction.
(D) CH\(_3\)CH\(_2\)CHO (Propanal): This is an aldehyde, not a carboxylic acid. The HVZ reaction is specific to carboxylic acids. Aldehydes undergo different alpha-halogenation reactions (e.g., in acid or base catalysis), but not the HVZ reaction itself.
(E) HCOOH (Formic acid): The carboxyl group is bonded directly to a hydrogen atom. There is no alpha-carbon atom, and therefore no alpha-hydrogen. Formic acid cannot undergo the HVZ reaction.
Step 4: Final Answer:
Only propanoic acid (CH\(_3\)CH\(_2\)COOH) has alpha-hydrogens and is a carboxylic acid, making it the only compound on the list that can undergo the Hell-Volhard-Zelinsky reaction. This corresponds to option (B).