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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: CH3CH2CH=C=CH2

Solution

The correct product, CH3CH2CH=C=CH2, is formed through the elimination reactions facilitated by KOH in alcohol and NaNH2 in liquid ammonia, leading to the formation of a conjugated diene. This reaction pathway promotes the formation of a double bond between the carbon atoms adjacent to the bromine substituents, resulting in the desired product.

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