Given the products of the following ethers with:
a. Aqueous HI b. Anlydrous HI + ether c. Excess of aq. HI
(I)

Explanation

(I) With reagent (a) and (b), products are same, since no `3^(@)C^(o+)` is formed.

(II)
(III) In regent (a), the high polarity of solvent `(H_(2)O)` favours `SN^(1)` reaction, whereas in reagent (b), the low polarity of solvent (ether) favours `SN^(-2)` reaction. Since benzyl `C^(o+)` is more stable (also stabilised by + I and hyperconjugative (H.C.) effect of (Me) group at p-position) than `3^(@)C^(o+)`, with benzyl `C^(o+)` both `SN^(-1)` and `SN^(2)` reactions are favoured. So the products form reagents (a) and (b) would be same.

(IV)