In the reaction H-C CH [(ii) CH3CH2Br](i) NaNH2/liq. NH3 X [(ii) CH3CH2Br](i) NaNH2/liq. NH3 Y, X and Y are
- X = 2-Butyne, Y = 2-Hexyne
- X = 1-Butyne, Y = 2-Hexyne
- X = 1-Butyne, Y = 3-Hexyne
- X = 2-Butyne, Y = 3-Hexyne
Answer
(3) X = 1-Butyne, Y = 3-Hexyne
Solution
Step 1: NaNH2 removes the acidic H of ethyne, HC CH NaNH2/liq. NH3 HC C-Na+, and the acetylide is then alkylated:
HC C-Na+ + CH3CH2Br HC C-CH2CH3 + NaBr
So X = CH3CH2C CH (1-Butyne).
Step 2: X still has a terminal H, so the same two steps repeat:
CH3CH2C CH NaNH2 CH3CH2C C-Na+
CH3CH2C C-Na+ + CH3CH2Br CH3CH2C CCH2CH3 + NaBr
So Y = CH3CH2C CCH2CH3 (3-Hexyne).
Answer: (3)