See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: In an SN1/E1-type ionization step, the transition state resembles the carbocation being formed (Hammond's postulate for endothermic steps). Therefore, the stability of the transition state mirrors the stability of the carbocation product. Step 1 – Identify the carbocations formed: - Reaction 1: CH3+ (methyl carbocation, primary, 0 alkyl substituents) - Reaction 2: (CH3)3C+ (tertiary carbocation, 3 alkyl substituents) - Reaction 3: (CH3)2CH+ (secondary carbocation, 2 alkyl substituents) Step 2 – Rank carbocation stability: Carbocation stability increases with more alkyl groups due to hyperconjugation and inductive effects: methyl (1) < secondary (3) < tertiary (2) So stability order: CH3+ < (CH3)2CH+ < (CH3)3C+ Step 3 – Apply Hammond's postulate: Because the ionization steps are endothermic (breaking a C–O bond to form a high-energy carbocation), the transition state is product-like. Therefore, transition state stability follows the same order as carbocation stability: TS of reaction 1 < TS of reaction 3 < TS of reaction 2 In numerical terms: 1 < 3 < 2 Step 4 – Eliminate wrong options: (a) 1 < 2 < 3 incorrectly places tertiary TS less stable than secondary. (b) 2 < 3 < 1 incorrectly places tertiary TS as least stable. (d) 2 < 1 < 3 incorrectly places tertiary TS as least stable. Only (c) 1 < 3 < 2 correctly reflects methyl < secondary < tertiary TS stability. Therefore, the correct answer is C.