Arrange the following compounds in the decreasing order of reactivity towards $S_N1$ reaction

(A) $(Ph)_3C-I$
(B) $(Ph)_3C-Br$
(C) $(Ph)_3C-Cl$
(D) $(Ph)_3C-F$

Choose the correct answer from the options given below:

(A), (B), (C), (D)

The correct answer is Option (1) → (A), (B), (C), (D)

For an SN1 reaction, the rate-determining step is leaving group departure, so better leaving group ⇒ higher reactivity.

Leaving group ability:

$\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$

All compounds form the same triphenylmethyl (trityl) carbocation, so stability is the same; only the leaving group matters.

Decreasing order of SN1 reactivity:

$(Ph)_3C\!-\!I > (Ph)_3C\!-\!Br > (Ph)_3C\!-\!Cl > (Ph)_3C\!-\!F$