See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting material is 2,2-dimethylhexan-1-ol, a primary neopentyl-type alcohol where the carbon bearing -OH (C1) is primary, and the adjacent carbon (C2) bears two methyl groups and a butyl chain. Step 2 - Mechanism under acidic conditions (H+, heat): The -OH is protonated to give a good leaving group (water). Direct ionization of a primary carbocation is highly unfavorable. Instead, the system undergoes a 1,2-hydride or 1,2-alkyl/hydride shift simultaneously with or after ionization. The primary C1 loses water to transiently form a primary carbocation at C1, which immediately undergoes a 1,2-hydride shift from C2 to C1, generating a tertiary carbocation at C2 (which bears two methyl groups and the n-butyl chain). This tertiary carbocation at C2 is much more stable. Step 3 - Elimination (E1) from the tertiary carbocation at C2: The carbocation at C2 can lose a proton from an adjacent carbon. The adjacent carbons are C1 (now CH2 after hydride shift, i.e., =CH2 formation) or C3 (the CH2 of the butyl chain). Loss of a proton from C1 gives CH3-CH2-CH2-CH2-CH=C(CH3)2, which is a trisubstituted (more substituted) alkene. Loss of a proton from C3 gives CH3-CH2-CH2-CH=C(CH3)-... a different product. By Zaitsev's rule, the more substituted alkene is the major product. Step 4 - Identify the major product: Loss of H+ from C1 (the -CH2- originally bearing OH, now =CH2) gives CH3CH2CH2CH2-CH=C(CH3)2, which is option (b). This is a trisubstituted alkene and represents the Zaitsev (more substituted, more stable) product. Step 5 - Why other options fail: - Option (a) would require a different carbon skeleton rearrangement not consistent with a simple 1,2-hydride shift followed by E1. - Option (c) CH3-CH2-CH2-CH2-C(CH3)=CH2 is a less substituted (terminal) alkene - this would be the minor product (anti-Zaitsev). - Option (d) involves a skeletal rearrangement of the butyl chain (methyl shift) leading to a branched product, which is less favorable than the straightforward 1,2-hydride shift giving the tertiary carbocation followed by Zaitsev elimination. Therefore, the correct answer is B.