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

The correct option/compound that gives the Hell-Volhard-Zelinsky (HVZ) reaction is: R-CH₂-COOH.

Step-by-Step Explanation:

1. Understanding the Hell-Volhard-Zelinsky (HVZ) Reaction:
The Hell-Volhard-Zelinsky reaction is a classic organic chemistry reaction used for the selective halogenation (typically bromination or chlorination) at the α-position of carboxylic acids. The reaction uses a halogen (like ${\mathrm{Br}}_2$ or ${\mathrm{Cl}}_2$) in the presence of a small amount of red phosphorus ($P$).

2. Requirement of α-Hydrogen Atoms:
For a compound to undergo the HVZ reaction, it must satisfy two main criteria:
- It must be a carboxylic acid containing the carboxyl functional group ($-\mathrm{COOH}$).
- The carbon atom adjacent to the carbonyl carbon (referred to as the α-carbon) must have at least one hydrogen atom directly attached to it (an α-hydrogen).

3. Analyzing the Given Compounds:

- Option 1: R-CH₂-COOH
This is a carboxylic acid. The α-carbon is the $-{\mathrm{CH}}_2-$ group, which contains two α-hydrogen atoms. Therefore, it satisfies all the conditions and undergoes the HVZ reaction to yield an α-halo carboxylic acid ($R-\mathrm{CH}\left(X\right)-\mathrm{COOH}$, where $X$ is a halogen).

- Option 2: R₃C-CHO
This compound is an aldehyde ($-\mathrm{CHO}$), not a carboxylic acid. Hence, it does not undergo the HVZ reaction.

- Option 3: R₂CO
This compound is a ketone ($-\mathrm{CO}-$), not a carboxylic acid. Therefore, it cannot undergo the HVZ reaction.

- Option 4: H-COOH
This is formic acid. Although it is a carboxylic acid, it does not have any α-carbon or α-hydrogen atom (the hydrogen is directly attached to the carbonyl carbon). Consequently, it does not show the HVZ reaction.

Thus, only R-CH₂-COOH undergoes the Hell-Volhard-Zelinsky reaction due to the presence of both the carboxylic acid group and α-hydrogen atoms.