The correct order of reactivity of the given chlorides with acetate in acetic acid is :

Solution

Chemistry ma’am · 3 steps

Acetate in acetic acid means SN1, so rank each chloride by the carbocation it leaves behind.

1. Spot the mechanism. Acetate is a weak nucleophile and acetic acid is a polar protic solvent, so these conditions favour SN1. The slow step is loss of Cl⁻, so the chloride that forms the most stable carbocation reacts fastest.
2. Check each cation. See whether the cation is allylic (spread over the C=C by resonance), and whether a resonance form puts the positive charge on a carbon that carries the methyl group.

   Chloride	Carbocation formed	Rank
   3-chloro-1-methylcyclohexene (Cl on the allylic carbon, CH₃ on the far alkene carbon)	Allylic, and one resonance form is tertiary (+ on the carbon with CH₃)	1, fastest
   3-chloro-4-methylcyclohexene (Cl allylic, CH₃ next to it)	Allylic, but both resonance forms are only secondary	2
   5-chloro-3-methylcyclohexene (Cl two carbons away from the C=C)	Plain secondary, no resonance with the C=C	3
   (chloromethyl)cyclohexene, -CH₂Cl	Primary, no resonance	4, slowest

3. So. The order is 3-chloro-1-methylcyclohexene > 3-chloro-4-methylcyclohexene > 5-chloro-3-methylcyclohexene > the -CH₂Cl compound, which is option (d).

The idea to keep. Weak nucleophile in a polar protic solvent means SN1, so rank by carbocation stability: allylic with a tertiary resonance form, then allylic secondary, then plain secondary, then primary.

Answer from the NTA final answer key. Solution by PrepWiser.