Question

Elimination reaction of 2-bromopentane to form pent-2-ene is:
(A) $\beta$- elimination reaction
(B) Follows zaitsev's rule
(C) Dehydrohalogenation reaction
(D) Dehydration reaction

• A. (A), (B), (C)
• B. (A), (C), (D)
• C. (B), (C), (D)
• D. (A), (B), (D)

Answer 1

The correct option is A (A), (B), (C)

$\stackrel{5}{C}{H}_3-\stackrel{4}{C}{H}_2-\stackrel{3}{C}H-\underset{Br}{\underset{|}{\stackrel{2}{C}}}H-\stackrel{1}{C}{H}_3\underset{-HBr}{\overset{\beta \text{elimination}}{\to }}\stackrel{5}{C}{H}_3-\stackrel{4}{C}{H}_2-\stackrel{3}{C}H=\stackrel{2}{C}H-\stackrel{1}{C}{H}_3$

(A) $\beta -$hydrogen is removed so it is a $\beta -$ elimination reaction.
(B) Major product is most substituted alkene which is formed by Zaitsev's rule.
(C) In this elimination rection hydrogen halide $\left(HBr\right)$ is getting removed, hence it is a dehydrohalogenation reaction.