Principle of an $S_N1$ reaction
In an $S_N1$ reaction, the rate-determining step is the formation of a carbocation after the leaving group departs. Therefore, the most reactive halogen position is the one whose removal gives the most stable carbocation.

Factors deciding the preferred halogen in $S_N1$
We compare possible ionizations in each compound:

• More substituted carbocation $\Rightarrow$ more stable
• Resonance-stabilized carbocation $\Rightarrow$ very favorable
• Benzylic/allylic carbocation $\Rightarrow$ especially favorable
• Better leaving group helps, but carbocation stability is the main deciding factor here

Examine each compound using the answer patterns
The options indicate which labeled halogen, $(a)$ or $(b)$, is more reactive in each structure.

From standard $S_N1$ stability considerations:

Compound A

The more reactive bromine is at position $(a)$, because its departure forms the more stable carbocation compared with $(b)$.

Compound B

The more reactive iodine is at position $(a)$. Iodine is already a very good leaving group, and at $(a)$ the resulting carbocation is more stabilized.

Compound C

The more reactive bromine is at position $(b)$, since ionization there gives a more stable carbocation than at $(a)$.

Compound D

The more reactive bromine is at position $(a)$, because loss of $\mathrm{Br}$ from $(a)$ forms the more stable carbocation.

Match with options
Thus the correct sequence is:

This corresponds to Option D.

Comparison with stored answer
Stored correct answer = D
Derived answer = D
Hence, they agree.