See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: Under acidic conditions with heat, alcohols undergo dehydration (elimination of water) to form carbocations as intermediates. A resonance-stabilized carbocation (allylic or benzylic) indicates a resonance-stabilized product pathway. Step 1 – Analyze option (a): The substrate is a cyclohexenyl methanol (CH2-OH attached to a carbon adjacent to a double bond). Under H+/Δ, the primary alcohol loses water to generate a primary carbocation at the CH2 position. However, this carbon is allylic to the ring double bond, so the resulting carbocation is resonance-stabilized as an allylic cation. Thus, option (a) produces a resonance-stabilized (allylic) carbocation intermediate/product. Step 2 – Analyze option (b): The substrate is an allylic alcohol on a cyclohexene ring (OH directly on the allylic carbon next to the double bond). Under H+/Δ, protonation of OH and loss of water directly gives an allylic carbocation that is resonance-stabilized between two carbons of the ring. Thus, option (b) produces a resonance-stabilized allylic carbocation. Step 3 – Analyze option (c): The substrate is a tertiary benzylic alcohol (the carbon bearing OH and CH3 is at the benzylic position of a tetrahydronaphthalene system). Under H+/Δ, loss of water gives a tertiary benzylic carbocation, which is extensively resonance-stabilized by both the tertiary nature and conjugation with the aromatic ring. Thus, option (c) produces a resonance-stabilized benzylic carbocation. Step 4 – Since all three reactions (a), (b), and (c) proceed through or produce resonance-stabilized carbocation intermediates/products (allylic in a and b, benzylic in c), the answer is that all of these reactions form resonance-stabilized products. Why other options fail: Options (a), (b), and (c) individually are each correct, so none of the single-option choices is complete; the all-inclusive option (d) is the only fully correct answer. Therefore, the correct answer is D.