The major product of the following reaction is CH3CH2C(Br)H-C(Br)H2 (i) KOH(alc.) → (ii) NaNH2 in liq NH3

The given reaction involves a vicinal dibromide. When vicinal dibromides are treated with alcoholic KOH, they undergo dehydrohalogenation to form alkynes. In this case, the product of the first step is CH3CH=CHCH2Br. The subsequent reaction with NaNH2 in liquid ammonia leads to dehydrohalogenation, resulting in the formation of an alkyne. Therefore, the major product is CH3CH2C≡CH. The other options are not formed as major products due to the reaction mechanism. The reaction proceeds as follows:

Step 1: Dehydrohalogenation with alcoholic KOH

CH3CH2C(Br)H-C(Br)H2 + 2KOH(alc.) → CH3CH=CHCH2Br + 2KBr + 2H2O

Step 2: Dehydrohalogenation with NaNH2 in liquid ammonia

CH3CH=CHCH2Br + NaNH2 → CH3CH2C≡CH + NaBr + NH3