NEET Solved Paper 2016 Question 2
Question: In the reaction $ H-C\equiv CH\frac{(1)NaNH _2/liq.NH _3}{(2)CH _3CH _2Br}X\frac{(1)NaNH _2/liq.NH _3}{(2)CH _3CH _2Br}Y $ X and Y are:
Options:
A) X = 1-Butyne; Y = 3-Hexyne
B) X = 2-Butyne; Y = 3-Hexyne
C) X = 2-Butyne; Y = 2-Hexyne
D) X = 1-Butyne; Y = 2-Hexyne
Show Answer
Answer:
Correct Answer: A
Solution:
- $NaNH_2$ is a strong base which on reacting with terminal alkyne removes terminal acidic hydrogen leaving alkynyl carbanion, which on reaction with alkyl halide produces longer alkyne by nucleophilic addition. Thus, X is 1-Butyne and Y is 3-Hexyne.
