Question

$Propanoic\;acid \xrightarrow[H_2O]{Cl_2/Red\;P} {\rm{X}}$.
What is ${\rm{X}}$?
A) Propanal
B) Propanol
C) propane
D) $\alpha {\rm{ - }}$ chloro propanoic acid

Answer 1

We know that the reaction of carboxylic acid which must have $\alpha {\rm{ - }}$ hydrogen atom with bromine or chlorine results in the formation of ${\rm{2 - }}$ bromo carboxylic acid or ${\rm{2 - }}$ chloro carboxylic acid in the presence of red phosphorus. The halogenation takes place during the reaction at the $\alpha {\rm{ - }}$ carbon atom.

Complete answer:
This reaction is carried out on the basis of Hell - Volhard - Zelinsky halogenation reaction halogenated carboxylic acids at the $\alpha$ carbon atom.
We know that propanoic acid is the carboxylic acid with $\alpha {\rm{ - }}$ hydrogen atom. The reaction of propanoic acid with chlorine in the presence of red phosphorus leads to the formation of ${\rm{2 - }}$chloropropanoic acid which is also named as $\alpha {\rm{ - }}$chloro propanoic acid because the chlorination will take place at $\alpha {\rm{ - }}$carbon atom.
We can write the chemical equation for the reaction carried as follows.

Hence, we can conclude that the correct option is D.

Note: The reaction which is carried out in the above question is commonly known as Hell-Volhard-Zelinsky reaction. The reaction takes place by the formation of phosphorous trichloride as the in-situ which catalysis the reaction. Therefore, we can say that the reaction is acid catalyzed enolization which is followed by the chlorination at alpha position.