Question

Which of the following will not undergo H V Z reaction?
a. Propanoic acid
b. Ethanoic acid
c. 2-methyl propanoic acid
d. 2,2-dimethyl propanoic acid

Answer 1

Hint: Hell- Volhard- Zelinsky reaction is a type of odd substitution reaction in which carboxylic acid which contains $\alpha $hydrogen gets converted to $\alpha $ halo carboxylic acid.

Complete step by step answer:
First of all, let us understand what the Hell- Volhard- Zelinsky [H V Z] reaction?
Hell- Volhard- Zelinsky [H V Z] reaction is also known as H V Z reaction. It is a type of substitution reaction in which carboxylic acids are converted to $\alpha $halo carboxylic acid. The reaction is initiated by the addition of phosphorus tribromide [Catalytic amount] and the further addition of one molar equivalent of di-atomic bromine.
Since, it is clear that $\alpha $hydrogen present in carboxylic acid is substituted by a halo group so the presence of $\alpha $hydrogen in the corresponding carboxylic acid is very important.
Let us draw the structure of four carboxylic acids given to us in the question.
(a) Propanoic acid

$(2 \propto \, - H)$
(b) Ethanoic acid

$(3 \propto \, - H)$
(c) 2-methyl propanoic acid

$(1 \propto \,H)$
(d) 2,2-dimethyl propanoic acid

$(no \propto \, - H)$

Thus, 2,2-dimethyl propanoic acid does not contain any $\alpha $hydrogen. Hence it will not participate in Hell- Volhard- Zelinsky [H V Z] reaction.

Option (d) is the correct answer.

Note:
Students should note that the H V Z reaction generally accomplishes bromination but fails in case of fluorination and iodination of carboxylic acid. The following reaction is not conducted at extremely high temperature, as there may be an elimination of hydrogen halide from the product thereby resulting in the formation of beta unsaturated carboxylic acid.