In each pair, both chlorides give the same set of constitutional alcohols on SN1 hydrolysis in water. Choose the pair(s) for which this is true.

A $CH X_{3} CH = CH - CH X_{2} Cl$ and $CH X_{3} - CHCl - CH = CH X_{2}$
B $(CH X_{3}) X_{2} C = CH - CH X_{2} Cl$ and $(CH X_{3}) X_{2} CCl - CH = CH X_{2}$
C $CH X_{3} CH X_{2} CH X_{2} Cl$ and $(CH X_{3}) X_{2} CH - Cl$
D $C X_{6} H X_{5} - CH = CH - CH X_{2} Cl$ and $C X_{6} H X_{5} - CHCl - CH = CH X_{2}$

Answer: A, B, D

Each allylic pair in options 1, 2 and 4 ionises to the same delocalised allylic cation (crotyl, prenyl, cinnamyl), which is captured at either terminus to give the same two alcohols. Propyl and isopropyl chlorides do not form a common cation: the primary cation is not formed, and the secondary gives only propan-2-ol, so option 3 is false.

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In each pair, both chlorides give the same set of constitutional alcohols on SN1 hydrolysis in water. Choose the pair(s) for which this is true.

Answer: A, B, D

More questions on SN1 vs SN2: substrate, nucleophile, solvent, leaving group