Which of the following compounds will give Hell-Volhard-Zelinsky reaction?

\(R-CH_2-COOH\)

The correct answer is option 1.\(R-CH_2-COOH\)

The Hell-Volhard-Zelinsky (HVZ) reaction is a specific reaction that allows for the halogenation of the alpha carbon of carboxylic acids. For the HVZ reaction to occur, the compound must contain an alpha hydrogen adjacent to a carboxylic acid group.

Let us analyze each compound:

1. \(R-CH_2-COOH\) (Primary carboxylic acid):

This compound has a carboxylic acid group (\(-COOH\)) and at least one hydrogen atom on the alpha carbon (the carbon adjacent to the carboxyl group).
Eligible for HVZ reaction: Yes, because it has alpha hydrogens adjacent to the carboxylic acid group.

2. \(R_3C-CHO\) (Tertiary aldehyde):

This compound is an aldehyde with a tertiary carbon group. It does not contain a carboxylic acid group and thus is not a carboxylic acid.

Eligible for HVZ reaction: No, because it is not a carboxylic acid.

3. \(R_2CO\) (Ketone):

This compound is a ketone and does not contain a carboxylic acid group.

Eligible for HVZ reaction: No, because it is not a carboxylic acid.

4. \(H-COOH\) (Formic acid):

Formic acid is a carboxylic acid but does not have an alpha carbon with hydrogen atoms (the carbon directly attached to the carboxyl group is already part of the carboxyl group itself).

Eligible for HVZ reaction: No, because it lacks alpha hydrogens.

Given this analysis, the only compound that will undergo the Hell-Volhard-Zelinsky reaction is: 1. \(R-CH_2-COOH\)