Question: Predict the products of the following acid-base reactions, or indicate if no significant reaction would take place.

$$\begin{gathered} {\mathbf{\text{(a)H-C≡C-H+NaNH}}}_{\mathbf{2}} \\ \mathbf{\text{(b)}}\mathbf{H}\mathbf{-}\mathbf{C}\mathbf{\equiv }\mathbf{C}\mathbf{-}\mathbf{H}\mathbf{+}{\mathbf{CH}}_{\mathbf{3}}\mathbf{Li} \\ \mathbf{\text{(c)}}\mathbf{H}\mathbf{-}\mathbf{C}\mathbf{\equiv }\mathbf{C}\mathbf{-}\mathbf{H}\mathbf{+}{\mathbf{NaOCH}}_{\mathbf{3}} \\ \mathbf{\text{(d)}}\mathbf{H}\mathbf{-}\mathbf{C}\mathbf{\equiv }\mathbf{C}\mathbf{-}\mathbf{H}\mathbf{+}\mathbf{NaOH} \\ \mathbf{\text{(e)}}\mathbf{H}\mathbf{-}\mathbf{C}\mathbf{\equiv }\mathbf{C}\mathbf{}{\mathbf{Na}}^{\mathbf{+}}\mathbf{+}\mathbf{NaOH} \\ \mathbf{\text{(f)}}\mathbf{H}\mathbf{-}\mathbf{C}\mathbf{\equiv }\mathbf{C}\mathbf{}{\mathbf{Na}}^{\mathbf{+}}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}} \\ \mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{H}\mathbf{-}\mathbf{C}\mathbf{\equiv }\mathbf{C}\mathbf{}{\mathbf{Na}}^{\mathbf{+}}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathit{C}\mathbf{=}\mathit{C}{\mathit{H}}_{\mathbf{2}} \\ \mathbf{\left(}\mathit{h}\mathbf{\right)}{\mathbf{\text{H}}}_{\mathbf{2}}\mathit{C}\mathbf{=}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{+}\mathit{N}\mathit{a}\mathit{N}{\mathit{H}}_{\mathbf{2}} \\ \mathbf{\left(}\mathit{i}\mathbf{\right)}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\mathbf{+}\mathit{N}\mathit{a}\mathit{N}{\mathit{H}}_{\mathbf{2}} \end{gathered}$$

A chemical reaction in which both acid and base take part to produce salt and water is an acid-base reaction. In this reaction, the exchange of hydrogen ions occurs between chemical species for neutralization, so it is also termed a neutralization reaction.

Due to the acidic character of terminal alkynes, they can form acetylide ions through the abstraction of an acetylenic proton. And the removal of an acetylenic proton from terminal alkynes can be possible only by using very strong bases like sodium amide. Bases that are not strong, like hydroxides and alkoxides, could not deprotonate terminal alkynes.

Hence, bases likeand can’t react with terminal alkynes.

So there will be no reaction in the case of options c and d.

is also not a strong base so there will be no reaction for option g.

In option h, there is an alkene, so there will be no reaction.

The reactions are given below.

$$\begin{gathered} {\text{(a)H-C≡C-H+NaNH}}_2\text{→}H-C\equiv C^{-}N{a}^{+}+N{H}_3 \\ \text{(b)}\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}+{\mathrm{CH}}_3\mathrm{Li}\to H-C\equiv C^{-}{\text{Li}}^{+}{\text{+CH}}_4 \\ \text{(c)}\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}+{\mathrm{NaOCH}}_3\text{→Noreaction} \\ \text{(d)}\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}+\mathrm{NaOH}\to Noreaction \\ \text{(e)}\mathrm{H}-\mathrm{C}\equiv \mathrm{C}{\mathrm{Na}}^{+}+\mathrm{NaOH}\to H-C\equiv C-H\text{+}{\mathrm{NaOCH}}_3 \\ \text{(f)}\mathrm{H}-\mathrm{C}\equiv \mathrm{C}{\mathrm{Na}}^{+}+{\text{H}}_2\text{O}\to H-C\equiv C-H+NaOH \\ \left(g\right)\mathrm{H}-\mathrm{C}\equiv \mathrm{C}{\mathrm{Na}}^{+}+{\text{H}}_{2}C=C{H}_2\to Noreaction \\ \left(h\right){\text{H}}_{2}C=C{H}_2+NaN{H}_2\to Noreaction \\ \left(i\right)C{H}_{3}OH+NaN{H}_2\to {\text{NaOCH}}_3+N{H}_3 \end{gathered}$$