See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The stability of a transition state in an SN1/E1-type ionization (loss of water from a protonated alcohol) mirrors the stability of the carbocation being formed. This is because the transition state resembles the carbocation product (Hammond's postulate for endothermic/late transition state steps). A more stable carbocation means a lower-energy (more stable) transition state. Step 1 - Identify the carbocations formed: - Reaction 1: CH3(+) → primary (methyl) carbocation — least stable - Reaction 3: (CH3)2CH(+) → secondary carbocation — intermediate stability - Reaction 2: (CH3)3C(+) → tertiary carbocation — most stable Step 2 - Apply Hammond's Postulate: For each ionization step, the transition state has significant carbocation character. The more stable the carbocation product, the more stable the transition state leading to it. Step 3 - Rank transition states by increasing stability (least to most stable): - TS1 (leads to methyl cation) < TS3 (leads to secondary cation) < TS2 (leads to tertiary cation) - This gives: 1 < 3 < 2 Step 4 - Match to options: Option (c) states 1 < 3 < 2, which matches our ranking. Why other options fail: - (a) 1 < 2 < 3: Wrong order; places tertiary before secondary. - (b) 2 < 3 < 1: Completely reversed; tertiary carbocation TS is most stable, not least. - (d) 2 < 1 < 3: Wrong; places tertiary TS as least stable. Therefore, the correct answer is C.