Metallic sodium does not react normally with:

Metallic sodium is a strong reducing agent and reacts with acidic hydrogen atoms in compounds. The key is to identify which compound among the given options does not have a sufficiently acidic hydrogen for sodium to react with under normal conditions.

Gaseous ammonia ($$\text{NH}_3$$) has N-H bonds that react with sodium: $$2\text{Na} + 2\text{NH}_3 \rightarrow 2\text{NaNH}_2 + \text{H}_2$$, though this reaction requires heating.

Ethyne ($$\text{HC \equiv CH}$$) has a terminal C-H bond that is acidic ($$pK_a \approx 25$$) because the carbon is $$sp$$ hybridised: $$2\text{Na} + 2\text{HC \equiv CH} \rightarrow 2\text{NaC \equiv CH} + \text{H}_2$$.

tert-Butyl alcohol has an O-H bond, which readily reacts with sodium: $$2\text{Na} + 2(CH_3)_3\text{COH} \rightarrow 2(CH_3)_3\text{CONa} + \text{H}_2$$.

But-2-yne ($$\text{CH}_3\text{C \equiv CCH}_3$$) is an internal alkyne. It has no terminal C-H bond on the triple-bond carbon. Both carbons of the triple bond bear methyl groups, so there is no acidic terminal alkynyl hydrogen. Sodium reacts with terminal alkynes by abstracting the acidic acetylenic proton, but since but-2-yne has no such proton, it does not react with metallic sodium under normal conditions.

Therefore, the correct answer is option (2): But-2-yne.