Question

The treatment of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH}}\,$ with chlorine in presence of phosphorus gives:
A.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COC}}{{\text{H}}_{\text{3}}}$
B.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Cl}}$
C.${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(Cl)COOH}}$
D.${\text{ClC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH}}$

Answer 1

Hint: The product can be identified with the help of Hell Volhard Zelinsky’s reaction. The reaction is used for alpha halogenation of carboxylic acids.

Complete answer:
Hell Volhard Zelinsky reaction is a halogenation reaction that is used for the alpha halogenation of carboxylic acids mainly.
The reaction is a nucleophilic acyl substitution type reaction.
In this reaction, phosphorus is added in the form of phosphorus trihalide that reacts with acid.
The phosphorus trihalide replaces the $ - {\text{OH}}$ group of carboxylic acid forming a halide derivative of carboxylic acid. This keto form of halide derivative of carboxylic acid tatumerize to enol form.
This enol form reacts with the halogen to give halogenation at the alpha position.
So, the product of the Hell Volhard Zelinsky reaction is alpha halogenated acid.
The reaction is given as follow:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH}}\,\,\,{\text{ + }}\,\,\,{\text{C}}{{\text{l}}_2}\mathop \to \limits^{{\text{Phosphorous}}} \,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(Cl)COOH}}$
The phosphorous trichloride replaces the $ - {\text{OH}}$ group of carboxylic acid forming a chloride derivative of carboxylic acid. This keto form of chloride derivative of carboxylic acid tatumerize to enol form.
This enol form reacts with the chlorine to give chlorination at alpha position.
So, the product of the Hell Volhard Zelinsky reaction is alpha chlorinated ethanoic acid. So, option (C) is correct.

Therefore, option (C)${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH(Cl)COOH}}$, is correct.

Note: Alpha position is the carbon atom at which the carboxylic group is attached. The formic acid does not give the Hell Volhard Zelinsky reaction due to the absence of an alkyl group.