See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When zinc (Zn) reacts with a vicinal dihalide (a compound having halogen atoms on adjacent carbons), it abstracts both halogen atoms simultaneously to form an alkene. This is called dehalogenation. Step 1: Identify the substrate. 2,3-Dibromobutane has bromine atoms on C2 and C3, which are adjacent (vicinal) carbons. Step 2: Identify the reagent. Zinc (Zn) is a metal that acts as a reducing agent and can insert between the two C-Br bonds in a concerted or stepwise fashion, removing both bromine atoms as ZnBr2. Step 3: Classify the elimination. In beta-elimination, the leaving group (Br) departs from one carbon and a hydrogen (or in this case another leaving group) is removed from the adjacent (beta) carbon. When Zn removes two halogens from adjacent carbons (C2 and C3), it is a 1,2-elimination — i.e., beta-elimination — forming the double bond between C2 and C3, giving 2-butene. Step 4: Rule out other options. - (a) Redox reaction: While Zn is oxidized from 0 to +2, this reaction is specifically classified in organic chemistry as a beta-elimination (dehalogenation), not simply a redox reaction. - (b) Alpha-elimination: Alpha-elimination involves removal of two groups from the SAME carbon to form a carbene or carbenoid intermediate. That is not happening here. - (d) Both alpha-elimination and redox: Alpha-elimination is not occurring here, so this is incorrect. Step 5: The removal of halogens from two adjacent carbons by Zn is the classic example of beta-elimination (dehalogenation), producing an alkene. Therefore, the correct answer is C.