1.

CH_(3)OC_(2)H_(5) and (CH_(3))_(3)COCH_(3) are treated with hydriodic acid. The fragments after reaction obtained are:

Answer»

`CH_(3)I+HOC_(2)H_(5), (CH_(3))_(3)C-I+HOCH_(3)`
`CH_(3)OH+C_(2)H_(5)I,(CH_(3))_(3)C-I+HOCH_(3)`
`CH_(3)OH+C_(2)H_(5)I,(CH_(3))_(3)C-OH+CH_(3)I`
`CH_(3)I+HOC_(2)H_(5),CH_(3)I+(CH_(3))_(3)C-OH`

Solution :`CH_(3)OC_(2)H_(5)` undergoes CLEAVAGE by `S_(N)2` mechanism to give `CH_(3)I and C_(2)H_(5)OH` while `(CH_(3))_(3)COCH_(3)` undergoes cleavage by `S_(N)1` mechanism to give `(CH_(3))_(3)Cl and CH_(3)OH`


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