Question

What is ‘B’ in the given reaction sequence?
 $ C{H_3}C{H_2}COOH\xrightarrow[{{\text{Red P}}}]{{C{l_2}}}A\xrightarrow{{alc.KOH}}B $
A. $ C{H_3}C{H_2}COCl $
B. $ C{H_3}C{H_2}CHO $
C. $ C{H_2} = CHCOOH $
D. $ C{H_3}C{H_2}C{H_2}COOH $

Answer 1

Hint : We know that the hydrogen bonded to a carbon atom adjacent to the carbonyl centre is referred to as alpha hydrogen. Unlike aldehydes and ketones, the carboxylic acids don’t generally form stable enols, so alpha addition is comparatively more difficult to achieve in case of carboxylic acids than aldehydes and ketones. One of the most common reactions which demonstrate the alpha addition of halogen with carboxylic acid is known as the Hell Volhard Zelinsky (HVZ) reaction.

Complete Step By Step Answer:
In the given reaction sequence, ethanoic acid reacts with chlorine in the presence of red phosphorus to show the Hell Volhard Zelinsky reaction. In this reaction, the aliphatic carboxylic acids react smoothly with chlorine or bromine in the presence of red phosphorus to yield a compound in which the chlorine atom replaces the alpha hydrogen of the given carboxylic acid. The reaction takes place as follows:
 $ C{H_3}C{H_2}COOH\xrightarrow[{{\text{Red P}}}]{{C{l_2}}}C{H_3}CH(Cl)COOH $
So, the product ‘A’ formed is 2-chloropropanoic acid. Further, product ‘A’ reacts with alcoholic potassium hydroxide which is a dehydrohalogenation reagent which basically removes a hydrogen and a halogen group from the compound to undergo elimination reaction and form respective alkenes. The reaction takes place as follows:
 $ C{H_3}CH(Cl)COOH\xrightarrow[{ - HCl}]{{alc.KOH}}C{H_2} = CHCOOH $
Hence, the product ‘B’ formed in the given reaction sequence is acrylic acid.
Thus, option (C) is the correct answer.

Note :
It is important to note that alcoholic $ KOH $ dissociates in water to form $ R{O^ - } $ ions which acts as a strong base which extract hydrogen and results in elimination reaction but aqueous $ KOH $ dissociates in water to give $ O{H^ - } $ ions which act as a good nucleophile and prefers substitution reaction. Thus, in case of aqueous $ KOH $ formation of alcohol takes place instead of alkenes.