Which of the following is not expected to be an intermediate during t…

The correct answer is Option (2) → the cyclohexylmethyl cation, a primary carbocation (positive charge on the \( \ce{CH2} \) outside the ring)

In water, a polar protic solvent, 1-chloro-2-methylcyclohexane reacts by the S_N1 route: the C–Cl bond breaks first to give a carbocation, and water then attacks it. A carbocation can rearrange to a more stable one (3° > 2° > 1°) before water attacks, which is why the –OH ends up on the carbon that carries the methyl group.

• Ionisation: the C–Cl bond breaks slowly and gives the 2-methylcyclohexyl cation, a secondary carbocation (Option 4), and \( \ce{Cl^-} \).

• Hydride shift: the H atom on the next ring carbon (the one carrying \( \ce{CH3} \)) moves over with its electron pair. The positive charge passes to that carbon and gives the 1-methylcyclohexyl cation, a more stable tertiary carbocation (Option 1).

• Attack of water: a water molecule adds to the tertiary carbocation and gives the protonated alcohol (an oxonium ion) (Option 3).

• Deprotonation: the oxonium ion loses \( \ce{H^+} \) to the solvent and gives 1-methylcyclohexan-1-ol, the product shown.

How each option fits the reaction

Option (1) the 1-methylcyclohexyl cation (tertiary) — An intermediate in this reaction. It forms by the hydride shift and is the carbocation that water attacks.

Option (2) the cyclohexylmethyl cation (primary) — Not an intermediate in this reaction. No C–Cl bond sits on the \( \ce{CH2} \), and the shift goes from the secondary to the more stable tertiary carbocation, never to a less stable primary one. Had it formed, the product would be cyclohexylmethanol, not the alcohol shown.

Option (3) protonated 1-methylcyclohexan-1-ol — An intermediate in this reaction. It forms when water attacks the tertiary carbocation and gives the product on losing \( \ce{H^+} \).

Option (4) the 2-methylcyclohexyl cation (secondary) — An intermediate in this reaction. It is the first carbocation, formed when the C–Cl bond breaks.

NCERT (2026-27 edition):
(b) Substitution nucleophilic unimolecular (S_N1) – “S_N1 reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid, etc.). … It occurs in two steps. In step I, the polarised C—Br bond undergoes slow cleavage to produce a carbocation and a bromide ion. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction.”

Exercises, 7.33 (3-methylbutan-2-ol treated with HBr) – “Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.”