See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When an allylic or vinylic bromide ionizes, it forms an allylic carbocation. Allylic carbocations are resonance-stabilized, meaning the positive charge is delocalized over two carbon atoms. Two different compounds can give the same allylic carbocation if their bromides are located at opposite ends of the same allylic system. Step 1: Identify the allylic carbocation from Compound 1. Compound 1 has a bromine on the carbon adjacent to a double bond (allylic position). When Br leaves, the carbocation is stabilized by resonance across the allylic system. The resulting allylic cation has the positive charge delocalized over two specific carbons with ethyl and methyl substituents. Step 2: Identify the allylic carbocation from Compound 2. Compound 2 has bromine directly on one of the sp2 carbons of an internal double bond (vinylic-type) or on the allylic carbon at one end of the double bond system. When ionization occurs, the resulting allylic carbocation has the charge delocalized over both ends of the double bond. Step 3: Compare the resonance structures. Compound 1 and Compound 2 are constitutional isomers where the Br is placed at opposite ends of the same allylic system (one has Br at one terminus, the other has Br at the other terminus of the same allylic unit). Because allylic carbocations are resonance hybrids, both compounds generate identical delocalized carbocations with the same connectivity and substitution pattern. Step 4: Why not other pairs? - Compounds 2 and 3: These differ in the position of Br relative to the double bond substituents, giving carbocations with different substitution patterns. - Compound 4: Has a different carbon skeleton or substitution pattern, giving a different allylic carbocation. - 1 and 3, 1 and 4: The carbocations produced differ in substitution. Conclusion: Compounds 1 and 2 both ionize to give the same allylic carbocation because they are related as the two resonance contributors of the same allylic system, with Br at opposite ends. Therefore, the correct answer is C.