See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is 2-methylbut-3-en-2-ol, i.e., (CH3)2C(OH)-CH=CH2. It has a tertiary alcohol adjacent to a vinyl group. Step 2 - Mechanism with HBr (allylic rearrangement via carbocation): When HBr reacts with an allylic alcohol, the OH is protonated first to give water as a leaving group. Loss of water generates a carbocation. The initial carbocation would form at C2 (tertiary position): (CH3)2C(+)-CH=CH2. However, this carbocation is allylic and can be represented as a resonance hybrid with the positive charge delocalized: (CH3)2C(+)-CH=CH2 <-> (CH3)2C=CH-CH2(+). Step 3 - Resonance structures of the allylic carbocation: The two resonance contributors are: - Tertiary allylic carbocation: (CH3)2C(+)-CH=CH2 - Primary allylic carbocation: (CH3)2C=CH-CH2(+) Step 4 - Attack by Br-: Bromide can attack at either end of the allylic system. Attack at the tertiary carbon gives (CH3)2CBr-CH=CH2 (option a, 3-bromo-3-methylbut-1-ene). Attack at the primary (terminal) carbon gives (CH3)2C=CH-CH2Br (option c, 1-bromo-3-methylbut-2-ene). Step 5 - Determine the major product: Although the tertiary carbocation end is more stable, Br- (a good nucleophile under thermodynamic or SN1 conditions) preferentially attacks the less hindered primary carbon in an allylic system under typical conditions. More importantly, product (c) is the more stable product because it has a more substituted (trisubstituted) double bond (the double bond is between C2 and C3 in (CH3)2C=CHCH2Br, which is disubstituted/trisubstituted), making it thermodynamically favored. The major product is thus the allylic rearrangement product: (CH3)2C=CH-CH2Br, which is option (c). Step 6 - Why other options fail: (a) Would be the non-rearranged direct addition product at the tertiary carbon; it is the minor product due to steric hindrance at the tertiary center. (b) Represents an elimination product with no bromine; HBr is an addition reagent not an elimination reagent here. (d) Retains the OH group; HBr replaces OH, so retention of OH is incorrect. Therefore, the correct answer is C.