Question

Give the stereochemical products
$HC\equiv CH+2NaN{H}_2+2EtBr⟶\left(A\right)\stackrel{Cs,N{H}_3}{⟶}\left(B\right)\stackrel{B{r}_2}{⟶}\left(C\right)$
Identify A?

• A. $EtC\equiv CEt$
• B. $MeC\equiv CMe$
• C. $MeC{H}_{2}C\equiv CC{H}_{2}Me$
• D. $EtC{H}_{2}C\equiv CC{H}_{2}Et$

Answer 1

Answer: (A)

$EtC\equiv CEt$

Terminal alkyne can itself be converted into an alkynide anion with $NaN{H}_2$, and can be alkylated a second time to yield an internal alkyne by reacting with alkyl halide.

$CH\equiv CH\stackrel{NaN{H}_2}{\to }N{a}^{+}{C}^{-}\equiv C^{-}N{a}^{+}$

$N{a}^{+}{C}^{-}\equiv C^{-}N{a}^{+}\stackrel{C_2{H}_{5}Br}{\to }{C}_2{H}_{5}C\equiv C{C}_2{H}_5$

Thus the product is :

$EtC\equiv CEt$

Thus optiion A is correct.