1.

The reagents (s) for the following conversion .

Answer»

alcoholic `KOH`
alcoholic `KOH` followed by `NaNH_(2)`
aqueous `KOH` followed by `NaNH_(2)` .
`Zn//CH_(3)OH`

Solution :`BrCH_(2)-CH_(2)-CH_(2)Br overset(Alc.KOH)RARR CH_(2)=CHBR overset(NaNH_(2))rarr CH-=CH`
Elimination of `HBr` from `CH_(2) -CHBr` requires a stronger base because here Br acquires partial double bond CHARACTER due to resonance .


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