In the reaction, (CH3)₃C − O − CH₃ + HI → Products, CH₃OH and (CH₃)₃CCl are the products and not CH₃I and (CH₃)₃COH. It is because

The correct option is: In step 2 of the reaction, the departure of leaving group (HO–CH₃) creates more stable carbocation.

Let us understand the step-by-step mechanism of this reaction to explain why this option is correct.

Step 1: Protonation of the ether
In the first step, the oxygen atom of the ether methyl tert-butyl ether, $({\mathrm{CH}}_3{)}_{3}C-O-{\mathrm{CH}}_3$, gets protonated by the strong acid hydrogen iodide (HI) to form a protonated ether intermediate (oxonium ion):
$({\mathrm{CH}}_3{)}_{3}C-O^{+}(H)-{\mathrm{CH}}_3$

Step 2: Departure of the leaving group and formation of the carbocation
Because one of the alkyl groups attached to the oxygen is a bulky tert-butyl group, the reaction proceeds via an S_N1 mechanism rather than S_N2.
In this step, the C - O bond between the tert-butyl carbon and the protonated oxygen breaks. The leaving group in this cleavage is neutral methanol (${\mathrm{CH}}_3\mathrm{OH}$).
The departure of this leaving group (${\mathrm{CH}}_3{\mathrm{OH}}^{+}-$ group) creates a highly stable tert-butyl carbocation, $({\mathrm{CH}}_3{)}_3{C}^{+}$, which is a tertiary (3°) carbocation stabilized by nine hyperconjugative hydrogens and inductive effects.
If the bond had cleaved on the methyl side instead, it would have generated a highly unstable methyl carbocation ($\mathrm{CH}_3{}^{+}$). Therefore, the reaction selectively forms the more stable carbocation.

Step 3: Nucleophilic attack
Finally, the halide nucleophile (iodide ion, $I^{-}$) attacks the stable tert-butyl carbocation to yield tert-butyl iodide:
$({\mathrm{CH}}_3{)}_3{C}^{+}+I^{-}\to ({\mathrm{CH}}_3{)}_{3}C-I$
(Note: The question text mentions the formation of $({\mathrm{CH}}_3{)}_3\mathrm{CCl}$ as a typographical representation of the alkyl halide product, which corresponds to the tertiary butyl halide formed from this path.)

Thus, the reaction yields ${\mathrm{CH}}_3\mathrm{OH}$ and the tertiary alkyl halide because the departure of the leaving group in step 2 generates the more stable tertiary carbocation.