Animated Solution for Chemistry - Organic Chemistry: The hydrocarbon which can react with sodium in liquid ammonia is

When we encounter sodium metal dissolved in liquid ammonia (${\text{Na/liq. NH}}_3$) in organic chemistry, we must immediately pause and ask ourselves: what role is it playing here?

This reagent is famous for its role in the Birch reduction, where it acts as a powerful source of solvated electrons to reduce internal alkynes into trans-alkenes.

However, this reagent has a second, equally important personality. It is a strong base. When it encounters a molecule with a sufficiently acidic proton, the acid-base reaction will outpace any reduction process.

To understand why some hydrocarbons have acidic protons, we need to dive into the concept of hybridization and s-character.

Let's look at the carbon atoms in different types of bonds. In an alkane, the carbons are $s{p}^3$ hybridized (25% s-character). In an alkene, they are $s{p}^2$ hybridized (33% s-character). But in an alkyne, the triple-bonded carbons are $sp$ hybridized, meaning they possess a whopping 50% s-character.

Why does this matter? The s-orbital is spherical and sits closer to the nucleus than the p-orbitals. Therefore, a higher s-character means the electrons in that hybrid orbital are held more tightly by the positively charged nucleus.

This makes the $sp$ hybridized carbon highly electronegative. When this carbon is attached to a hydrogen atom (as in a terminal alkyne), it pulls the electron density of the $\text{C-H}$ bond strongly towards itself. This polarization leaves the hydrogen atom electron-deficient and prone to leaving as an $H^{+}$ ion. Thus, the terminal hydrogen becomes distinctly acidic.

Armed with this knowledge, let's evaluate the options provided in the question. We are looking for a hydrocarbon that can react with our strong base, which means we are hunting for an acidic hydrogen.

Option (a): ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{C}\equiv {\text{CCH}}_2{\text{CH}}_2{\text{CH}}_3$ This is an internal alkyne. The triple bond is buried deep within the carbon chain. There are no hydrogens attached to the $sp$ carbons. Therefore, it has no acidic protons.

Option (c): ${\text{CH}}_3\text{CH}={\text{CHCH}}_3$ This is an alkene. The carbons are $s{p}^2$ hybridized, which is not electronegative enough to make the attached hydrogens acidic.

Option (d): ${\text{CH}}_3{\text{CH}}_2\text{C}\equiv {\text{CCH}}_2{\text{CH}}_3$ Similar to option (a), this is another internal alkyne. It lacks the crucial terminal hydrogen.

Option (b): ${\text{CH}}_3{\text{CH}}_2\text{C}\equiv \text{CH}$ This is 1-butyne, a terminal alkyne. The triple bond is at the end of the chain, meaning one of the $sp$ carbons is directly bonded to a hydrogen atom. This is our culprit! It possesses the acidic hydrogen we are looking for.

When 1-butyne is introduced to sodium in liquid ammonia, a rapid acid-base reaction occurs. The active sodium metal reacts with the acidic terminal hydrogen.

The sodium displaces the hydrogen, forming a stable salt known as sodium butynide, and in the process, liberates hydrogen gas.

The chemical equation for this elegant transformation is: ${\text{CH}}_3{\text{CH}}_2\text{C}\equiv \text{CH}+\text{Na}\stackrel{{\text{liq. NH}}_3}{\to }{\text{CH}}_3{\text{CH}}_2\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}+\frac{1}{2}{\text{H}}_2\uparrow$

This evolution of hydrogen gas is a classic, definitive laboratory test to distinguish terminal alkynes from internal alkynes and other hydrocarbons.

Therefore, the only hydrocarbon among the choices that will react in this manner is the terminal alkyne, making Option (b) the correct answer.