See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: This question involves allylic bromination of propene (with a 14C isotopic label at the terminal carbon C1) under conditions of low concentration of Br2 or high temperature, which favor free-radical substitution at the allylic position rather than electrophilic addition across the double bond. Step 1 - Identify the reaction conditions: Low concentration of Br2 or high temperature promotes free-radical (homolytic) allylic bromination rather than ionic addition. Under these conditions, a bromine radical abstracts an allylic hydrogen from the CH3 group (C3), generating an allylic radical. Step 2 - Allylic radical formation: The allylic C-H bond (at CH3, i.e., C3) is broken homolytically to give an allylic radical. The allylic radical in propene is delocalized over C2 and C3: 14CH2=CH-CH2(radical) <--> 14CH2(radical)-CH=CH2 Step 3 - Resonance structures of the allylic radical: Because the radical is delocalized, bromine can attack at either end of the allylic system: - Attack at C3 gives: 14CH2=CH-CH2Br (option a) - Attack at C1 (the 14C end) gives: BrCH2-CH=14CH2, which can be written as H2C=CH-14CH2-Br (option b) Step 4 - Both products are formed: Since the allylic radical is symmetric with respect to radical position (delocalized over both terminal carbons), both products (a) and (b) are obtained. The 14C label distinguishes them structurally but both are valid products of allylic bromination. Step 5 - Why option (c) fails: Option (c) shows a vicinal dibromide (addition product across the double bond), which would result from electrophilic addition of Br2, not from free-radical allylic substitution under low Br2 concentration or high temperature. Therefore, the correct answer is D.