See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Dehydration of alcohols proceeds via E1 or E2 elimination. The ease of dehydration follows the stability of the carbocation intermediate formed (for E1). Tertiary carbocations are more stable than secondary, which are more stable than primary. Therefore, tertiary alcohols dehydrate most easily, followed by secondary, then primary. Step 1 – Classify each alcohol: (a) CH3-CH2-CH2-OH: primary alcohol (1-propanol). Dehydration is difficult because it would form an unstable primary carbocation. (b) 1-methylcyclohexan-1-ol: tertiary alcohol. The OH is on a carbon bearing two ring carbons and one methyl group, making it a tertiary carbon. Dehydration gives a stable tertiary carbocation. (c) Phenol (C6H5-OH): the OH is directly attached to the aromatic ring. Dehydration is extremely difficult because it would disrupt aromaticity and is not favored under normal conditions. (d) (CH3)3C-CH2-OH: neopentyl alcohol, a primary alcohol. Despite having bulky groups nearby, the carbon bearing OH is primary. Dehydration is very difficult and would require rearrangement. Step 2 – Determine which dehydrates most easily: Option (b) is a tertiary alcohol (1-methylcyclohexan-1-ol). Tertiary alcohols form the most stable tertiary carbocations upon protonation and loss of water, making dehydration the easiest. Step 3 – Why others fail: (a) Primary alcohol – very difficult dehydration, unstable primary carbocation. (c) Phenol – OH on aromatic ring, essentially does not undergo dehydration under normal conditions. (d) Neopentyl alcohol – primary alcohol with steric hindrance; dehydration is very difficult. Therefore, the correct answer is B.