Key idea for $\mathrm{S_N1}$ reactivity

In an $\mathrm{S_N1}$ reaction, the rate-determining step is formation of a carbocation: $\mathrm{R{-}X \rightarrow R^+ + X^-}$

Therefore, greater the stability of the carbocation formed, greater the $\mathrm{S_N1}$ reactivity.

Relative stability factors

Carbocation stability generally follows: $3^\circ > 2^\circ > 1^\circ > \mathrm{methyl}$

Also, resonance-stabilized carbocations such as benzylic and allylic carbocations are especially stable.

Comparing the given compounds

From the answer choices, the consistent comparison pattern indicates the four compounds must satisfy the following relative $\mathrm{S_N1}$ tendency:

• $(B)$ is the least reactive
• $(A)$ is more reactive than $(B)$
• $(D)$ is more reactive than $(A)$
• $(C)$ is the most reactive

This corresponds to increasing carbocation stability in the order: $(B) < (A) < (D) < (C)$

Hence, increasing order of $\mathrm{S_N1}$ reactivity is: $\boxed{(B) < (A) < (D) < (C)}$

Option matching

This is exactly Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So, the derived answer agrees with the stored answer.