Regarding the conversion, ${CH}_{3} {CHBrCH}_{2} {CH}_{3} \overset{alc . KOH}{\to} {CH}_{3} CH = {CHCH}_{3}$ correct statements are
A) It is dehyrohalogenation reaction
B) It is $β - elimination$
C) Produced Zaitsev product
D) Follows $E_{1}$ mechanism

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A, B, C, D

A, B, D only

A, C only

A,B,C only

answer is A.

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Detailed Solution

(A) loss of HX … Dehydrohalogenation
(B) H–atom is removed from $β - carbon$
(C) more substituted alkene is the major product, Zaitsev’s rule
(D) $E_{2}$ mechanism is followed

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