Key idea: Metallic sodium reacts readily with compounds that can provide an acidic hydrogen.

The general reaction is: $2\,\text{R-H} + 2\,\text{Na} \rightarrow 2\,\text{RNa} + \text{H}_2$ This happens when the hydrogen is sufficiently acidic.

Check each option:

A. Gaseous ammonia, $\mathrm{NH_3}$

Ammonia has hydrogen attached to nitrogen, and sodium reacts with liquid/gaseous ammonia under suitable conditions to form sodium amide eventually: $2\,\mathrm{Na} + 2\,\mathrm{NH_3} \rightarrow 2\,\mathrm{NaNH_2} + \mathrm{H_2}$ So sodium does react with ammonia.

B. But-2-yne, $\mathrm{CH_3-C\equiv C-CH_3}$

This is an internal alkyne. There is no terminal acidic hydrogen on the triple bond.

Only terminal alkynes, having the group $\mathrm{-C\equiv C-H}$, show acidity and react with active metals like sodium.

But-2-yne has structure: $\mathrm{CH_3-C\equiv C-CH_3}$ No acidic acetylenic hydrogen is present, so metallic sodium does not react normally.

C. Ethyne, $\mathrm{HC\equiv CH}$

Ethyne is a terminal alkyne and has acidic hydrogens. Hence it reacts with sodium: $\mathrm{HC\equiv CH + 2Na \rightarrow NaC\equiv CNa + H_2}$ So sodium does react.

D. tert-butyl alcohol, $(\mathrm{CH_3})_3\mathrm{COH}$

Alcohols react with sodium because the $\mathrm{O-H}$ hydrogen is acidic enough: $2(\mathrm{CH_3})_3\mathrm{COH} + 2\mathrm{Na} \rightarrow 2(\mathrm{CH_3})_3\mathrm{CONa} + \mathrm{H_2}$ So sodium does react.

Conclusion: The only compound that does not react normally with metallic sodium is but-2-yne.

Comparison with stored answer: Derived answer = B

Stored correct answer = B

Hence, they agree.