Question

$$Propanoicacid\underset{H_{2}O}{\overset{C{l}_2/RedP}{\to }}\mathrm{X}$$.
What is $$\mathrm{X}$$?
A) Propanal
B) Propanol
C) propane
D) $$\alpha -$$ chloro propanoic acid

Answer 1

Hint: We know that the reaction of carboxylic acid which must have $$\alpha -$$ hydrogen atom with bromine or chlorine results in the formation of $$2-$$ bromo carboxylic acid or $$2-$$ chloro carboxylic acid in the presence of red phosphorus. The halogenation takes place during the reaction at the $$\alpha -$$ 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 $$\alpha -$$ hydrogen atom. The reaction of propanoic acid with chlorine in the presence of red phosphorus leads to the formation of $$2-$$chloropropanoic acid which is also named as $$\alpha -$$chloro propanoic acid because the chlorination will take place at $$\alpha -$$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.