See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type. The starting material is 1-iodo-2-methylcyclohexane reacting with H2O (a polar protic solvent/nucleophile). This is an SN1 solvolysis reaction because the substrate can form a stabilized carbocation. Step 2 - Determine the first intermediate. When iodide leaves, a secondary carbocation forms at C1 (the carbon that bore the iodine), adjacent to C2 which bears the methyl group. This gives a secondary carbocation — option (d), which is the 2-methylcyclohex-1-yl cation (secondary). This IS an expected intermediate. Step 3 - Determine if a 1,2-hydride or 1,2-methyl shift occurs. The secondary carbocation at C1 (adjacent to the methyl-bearing C2) can undergo a 1,2-methyl shift, migrating the methyl from C2 to C1, generating a tertiary carbocation at C1 with the methyl now on C1. This gives option (b) — wait, let us re-examine the structures. Option (a): positive charge on the exocyclic carbon bearing the methyl group — this appears to be a primary/methyl carbocation (the + is on the methyl carbon itself, i.e., a primary carbocation on a CH2+ group attached to cyclohexane). This would NOT be a expected intermediate because it is a primary carbocation, which is highly unstable and would never form preferentially. Option (b): tertiary carbocation at C1 of the ring with methyl also at C1 — this is the rearranged tertiary carbocation formed after a 1,2-methyl shift. This IS expected. Option (c): protonated alcohol (oxocarbenium / protonated water) at C1 with methyl at C1 — this is the intermediate after water attacks the tertiary carbocation, before deprotonation. This IS expected. Option (d): secondary carbocation at C1 with methyl at C2 — the initial carbocation after iodide departure. This IS expected. Step 4 - Identify the non-intermediate. Option (a) shows a primary carbocation (positive charge on the exocyclic methyl carbon, i.e., a –CH2+ group attached to cyclohexane ring carbon). Formation of a primary carbocation from a secondary one is not expected; the rearrangement always goes toward greater stability (secondary → tertiary via methyl shift), never toward a primary carbocation. Therefore option (a) is NOT expected to be an intermediate. Step 5 - Why other options are valid intermediates: (d) forms first upon ionization; (b) forms after 1,2-methyl shift giving the tertiary cation that explains the product; (c) forms when water attacks the tertiary cation and is deprotonated to give the product. None of these involve a primary carbocation. Therefore, the correct answer is A.