1.

Find the product for CH_(3)CH_(2)-O-CH_(2)-CH_(2)-O-CH_(2)-C_(6)H_(5)+HI(exess)

Answer»

`HO-CH_(2)CH_(2)OH, C_(6)H_(5)CH_(2)-I, CH_(3)CH_(2)-I`
`C_(6)H_(5)CH_(2)-OH, CH_(3)CH_(2)-I,I-CH_(2)CH_(2)-OH`
`I-CH_(2)-CH_(2)-I, C_(6)H_(5)CH_(2)-I, CH_(3)CH_(2)-OH`
`HO-CH_(2)CH_(2)-OH, C_(6)H_(5)CH_(2)-I, CH_(3)CH_(2)-OH`

Solution :Presence of EXCESS of HI FAVOURS `SN^(1)` mechanism.
So formation of produce is controlled by the stablity of the carbocation resulting in the cleavage of C-O bond in protonated ether. Thus the PRODUCT for GIVEN equation are
`C_(6)H_(5)CH_(2)I, CH_(3)CH_(2)I, HOCH_(2)-CH_(2)OH`


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