Acidic cleavage of ether with a 2° alkyl group

I know that an ether with a 3° alkyl group and a 1° alkyl group is cleaved by $$\mathrm{S_N1}$$ mechanism to give a tertiary alkyl halide and a 1° alcohol.

And an ether with a two 1° alkyl groups is cleaved by $$\mathrm{S_N2}$$ mechanism to give a 1° alkyl halide and a 1° alcohol.

But what happens if a ether with a 2° alkyl group and a 1° alkyl group is cleaved with a halogen acid?

To this question I have drawn a scheme below. The given ether is $$\ce{2^0}$$ group to the left and methyl to right. Acid cleavage with a halogenic acid involves an initial protonation of ether. In an $$\ce{S_N^2}$$ reaction of this ether, nucleophile $$\ce{I^-}$$ attacks via path 1 and path2.
• In path 1, $$\ce{I^-}$$ approach is sterically hindered by phenyl and Methyl groups. This leads to a high energy transition state.
• In path 2, $$\ce{I^-}$$ approach is from right of protonated ether where steric hindrance is comparatively less( only $$\ce{H}$$). A more stable transition state favors $$\ce{S_N^2}$$ reaction .