Consider the following reaction : [figure] The product Y formed is :
- 2-methylhex-3-yne
- 5-methylhex-2-yne
- Isopropylbut-1-yne
- 2-methylhex-2-yne
Answer
(1) 2-methylhex-3-yne
Solution
The reactant is 1,1-dibromobutane (a geminal dihalide). With 2 mol NaNH2 it loses two molecules of HBr:
CH3CH2CH2CHBr2 + 2NaNH2 CH3CH2C CH + 2NaBr + 2NH3
X = but-1-yne (a terminal alkyne)
(i) NaNH2 removes the acidic ≡C–H hydrogen:
CH3CH2C CH + NaNH2 CH3CH2C C-Na+ + NH3
(ii) The acetylide ion is alkylated by 2-bromopropane:
CH3CH2C C-Na+ + (CH3)2CHBr CH3CH2C C-CH(CH3)2 + NaBr
Y = CH3-CH(CH3)-C C-CH2-CH3
Longest chain = 6 C; the triple bond gets locant 3 from either end, so the methyl group gets the lower locant 2.
Y = 2-methylhex-3-yne
Answer: (1)