CH3CH2COOHCl2,alcoholicKOH−−−−−−−−−−−−→redP End product of the above given reaction is:
A
CH3CH(OH)COOH
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B
CH2(OH)CH2COOH
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C
CH2=CHCOOH
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D
CH2(Cl)CH(OH)COOH
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Solution
The correct option is CCH2=CHCOOH Here first propionic acid undergoes chlorination to form 2-chloro propionic acid which on further reaction with alc. KOH undergoes β-elimination reaction to form prop-2-enoic acid.