When `1-`alkyne is treated with `Na + Liq. NH_(3)` and product is reacted with methyl chloride, the end product of the reaction will be

Options

A. Lower alkyne having two carbon less than `1-` alkyne
B. Lower alkyne having one carbon less than `1-`alkyne
C. Higher alkyne having one carbon more than `1-` alkyne
D. Higher alkyne having two carbons more than `1-`alkyne

Answer

c

Explanation

`R-C-=C.Hoverset(Na //liq. NH_(3))rarrR-C-= barC.Na^(+)overset(ClCH_(3))rarrR-C-=C-CH_(3)`
Higher alkyne having one carbon more than `1-`alkyne