Which of the following compounds do not react with $NaNH_2$? Why?
$CH_3-C \equiv C-CH_3, CH_3-CH_2-C \equiv CH, C_2H_5-C \equiv C-C_2H_5, HC \equiv CH, CH_3-CH=CH_2, CH_2=CH-C \equiv CH$
$CH_3-C \equiv C-CH_3, CH_3-CH_2-C \equiv CH, C_2H_5-C \equiv C-C_2H_5, HC \equiv CH, CH_3-CH=CH_2, CH_2=CH-C \equiv CH$
(A, C, E) $NaNH_2$ is a strong base that reacts with terminal alkynes (alkynes containing an acidic hydrogen atom,i.e.,$R-C \equiv CH$) to form acetylides.
$1$. $CH_3-C \equiv C-CH_3$ (Internal alkyne,no acidic hydrogen)
$2$. $C_2H_5-C \equiv C-C_2H_5$ (Internal alkyne,no acidic hydrogen)
$3$. $CH_3-CH=CH_2$ (Alkene,no acidic hydrogen)
These compounds do not react with $NaNH_2$ because they lack the terminal $-C \equiv CH$ group,which possesses the necessary acidic proton.
$1$. $CH_3-C \equiv C-CH_3$ (Internal alkyne,no acidic hydrogen)
$2$. $C_2H_5-C \equiv C-C_2H_5$ (Internal alkyne,no acidic hydrogen)
$3$. $CH_3-CH=CH_2$ (Alkene,no acidic hydrogen)
These compounds do not react with $NaNH_2$ because they lack the terminal $-C \equiv CH$ group,which possesses the necessary acidic proton.
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