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In the equation, $C{{H}_{3}}COOH+C{{l}_{2}}\xrightarrow[-HCl]{\text{Red P}}A$, the compound A is:

Last updated date: 20th Jun 2024
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Hint: The chemical reaction of halogenation of carboxylic acid in the presence of Red P takes place at $\alpha$- carbon atom to form a mono form of acetic acid attached to a halogen. Now, we can find the product.

Complete step by step answer:
-Firstly, let us discuss the chemical reaction. It is a HVZ reaction i.e. Hell-Volhard-Zelinsky reaction.
-As we know, in the HVZ reaction, the carboxylic acid reacts with halogen, and a catalytic amount of phosphorus is added, leading to the -halogenation of carboxylic acids.
- The halogen that gives this reaction is generally Chlorine (Cl) and Bromine (Br).
-Now, we are given with the carboxylic acid i.e. acetic acid and halogen used is chlorine.
-Thus, when the acetic acid reacts with chlorine in the presence of Red P, $\alpha$-H atom is replaced with Cl atom.
-It leads to the formation of monochloroacetic acid, its chemical formula is ClCH$_2$COOH.
-Therefore, the chemical reaction can be written as:

\[C{{H}_{3}}COOH+C{{l}_{2}}\xrightarrow[-HCl]{\text{Red P}}ClC{{H}_{2}}COOH\]

The compound A is ClCH$_2$COOH.

So, the correct option is B.

Additional Information:
-If this reaction is carried out at extremely high temperature, then different products would be produced. It would result in the elimination of hydrogen halide, thereupon generating the beta unsaturated carboxylic acids.
-This reaction has been developed by three chemists-Carl Magnus Von Hell, Jacob Volhard and Nikolay Zelinsky. That is it has been named as Hell-Volhard-Zelinsky reaction.

The reaction temperature is maintained at above 373K in Hell-Volhard-Zelinsky reaction. Also, chlorination and bromination is only accomplished through this reaction at the $\alpha $-carbon. It should be noted that fluorination and iodination is not executed through these reaction conditions.