Reactivity towards SN1 depends upon the stability of the carbocation.
Let's figure out which of these compounds will react the fastest via an $ \text{S}_\text{N}1 $ mechanism. Remember, the rate-determining step in an $ \text{S}_\text{N}1 $ reaction is the formation of a carbocation. The more stable the carbocation intermediate, the faster the reaction will proceed.
Analyzing the Carbocations Formed:
(1) Cyclohexylmethyl carbocation:
This is a primary carbocation ($ \text{1}^\circ $). Primary carbocations are generally unstable due to the lack of hyperconjugation and inductive stabilization from more alkyl groups.
(2) Cyclohexyl carbocation:
This is a secondary carbocation ($ \text{2}^\circ $). Secondary carbocations are more stable than primary carbocations due to more hyperconjugative and inductive effects.
(3) Bromobenzene:
If the bromine were to leave, it would form a phenyl carbocation. Phenyl carbocations are highly unstable because the positive charge resides in an $ sp^2 $ hybridized orbital, which is held more tightly to the nucleus, and the carbocation cannot be stabilized by resonance in the usual sense.
(4) 1-Bromo-1-phenylethane:
Loss of bromine would result in the formation of a benzylic secondary carbocation ($ \text{2}^\circ $ benzylic). Benzylic carbocations are particularly stable due to resonance with the benzene ring, which delocalizes the positive charge over several carbon atoms. Furthermore, it's also a secondary carbocation, benefiting from hyperconjugation and inductive effects from the methyl group.
Comparing the Stabilities of the Potential Carbocations:
The order of stability is as follows:
Benzylic $ \text{2}^\circ $ > $ \text{2}^\circ $ alkyl > $ \text{1}^\circ $ alkyl >> phenyl
Conclusion:
The carbocation formed from compound (4) is the most stable. This means that compound (4) will undergo $ \text{S}_\text{N}1 $ reaction at the fastest rate.
Final Answer:
The compound that reacts the fastest via an $ \text{S}_\text{N}1 $ mechanism is: $ \boxed{(4)} $