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The major product of the following reaction is - CH3CH2CHBrCH2Br (i) KOH alc. (ii) NaNH2 in liq. NH3

1. CH3CH2C≡CH
2. CH3CH2CH=C=CH2
3. CH3CH2CH(NH2)CH2NH2
4. CH3CH=CHCH2NH2

Correct answer: CH3CH2C≡CH

Solution

1,2-dibromobutane on double dehydrohalogenation with alc. KOH gives a bromoalkene/alkyne mixture; NaNH2 in liq. NH3 then drives it fully to the terminal alkyne but-1-yne, CH3CH2C(triple)CH. So the product is CH3CH2C≡CH.

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