The correct order of the rate of reaction of the following reactants with nucleophile by S_N 1 mechanism is :
(Given : Structures I and II are rigid)

Solution

Chemistry ma’am · 3 steps

All four are tertiary, so judge each carbocation by one thing: can it go flat and spread its charge?

1. What decides SN1 rate. The slow step is C-Br breaking to give a carbocation. A carbocation is stable only when it can become planar (sp²), so its empty p orbital can be stabilised by neighbouring groups.
2. Bridgehead problem. In I and II the Br sits at a bridgehead of a rigid cage. The cage stops that carbon from flattening, so the cation is very unstable. The tighter cage of II (bicyclo[2.2.1]heptane) distorts more than the roomier cage of I (bicyclo[2.2.2]octane).

   Substrate	Carbocation formed	Can it be planar?	Stability
   II	bridgehead of bicyclo[2.2.1]	no, badly strained	lowest
   I	bridgehead of bicyclo[2.2.2]	no, but less strained	low
   III	(CH₃)₃C⁺	yes	good: hyperconjugation and +I effect
   IV	Ph₃C⁺	yes	highest: resonance over three rings

3. So. The SN1 rate follows carbocation stability: II < I < III < IV, which is option (b).

The idea to keep. For SN1, rank the carbocations, not the halides. A bridgehead cation in a rigid small cage cannot go planar, so it is very slow, and the smaller the cage, the slower it is.

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