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

Hint: 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.