See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: 1,4-elimination (also called dehalogenation across positions 1 and 4) using zinc dust removes two halogen atoms from positions 1 and 4 of a carbon chain, generating a conjugated diene. Step 1: Identify the starting material. The compound is 1,4-dibromo-2-butene: Br-CH2-CH=CH-CH2-Br. The bromine atoms are on C1 and C4, and there is already a double bond between C2 and C3. Step 2: Apply the 1,4-elimination mechanism. Zinc dust acts as a reducing agent and abstracts both bromine atoms simultaneously (or stepwise) from C1 and C4. This removes the two C-Br bonds and generates two new pi electrons. Step 3: Determine the product. Removing Br from C1 and Br from C4 of BrCH2-CH=CH-CH2Br gives H2C=CH-CH=CH2 (1,3-butadiene). The existing double bond between C2 and C3 remains, and new double bonds form at C1-C2 and C3-C4, yielding the conjugated diene 1,3-butadiene. Step 4: Evaluate other options. (a) CH3-CH=C=CH2 is an allene (cumulated diene) - this would require a rearrangement, not a simple 1,4-elimination. (b) CH3-C≡C-CH3 is 2-butyne - this would require elimination of two H atoms along with Br atoms, not consistent with the reagent and substrate. (c) CH3-CH2-C≡CH is 1-butyne - similarly requires loss of H atoms, inconsistent with the substrate structure. (d) H2C=CH-CH=CH2 is 1,3-butadiene - directly obtained by removing both Br atoms from C1 and C4, consistent with 1,4-elimination. Therefore, the correct answer is D.