1.

Dehydration of 1 -propanol by the use of H_(2)SO_(4) and subsequent treatment with HI gives :

Answer»

`CH_(3)CH_(2)CH_(2)I`
`CH_(3)CH(I)CH_(3)`
`CH_(2)=CHCH_(2)I`
`ICH=CHCH_(3)`

Solution :`CH_(3)CH_(2)CH_(2)OHoverset(H_(2)SO_(4))rarrCH_(3)CH=CH_(2)OVERSET(HI)rarrCH_(3)underset(I)underset("| ")(CH)-CH_(3)`


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