Animated Solution for Chemistry - Organic Chemistry: Acetylene does not react with

When we look at acetylene ($HC\equiv CH$), the simplest of all alkynes, we are looking at a molecule with a very special property. Unlike alkanes or alkenes, terminal alkynes possess hydrogens that are noticeably acidic. But why is this the case? It all boils down to the concept of hybridization.

In acetylene, the carbon atoms involved in the triple bond are $sp$ hybridized. This means their hybrid orbitals are composed of 50% s-character and 50% p-character. Because s-orbitals are closer to the nucleus than p-orbitals, a higher s-character means the electrons in that orbital are held more tightly by the carbon nucleus.

Consequently, an $sp$ hybridized carbon is highly electronegative—almost as electronegative as a nitrogen atom! It pulls the electron density of the $C-H$ bond strongly towards itself, leaving the terminal hydrogen atom with a partial positive charge. This makes the hydrogen susceptible to removal by a sufficiently strong base or a reactive metal.

A Redox Perspective

Let's test this acidity by introducing sodium metal ($Na$). Sodium is an alkali metal, highly electropositive, and a powerful reducing agent. When it encounters the slightly acidic terminal hydrogen of acetylene, a redox reaction occurs.

Sodium readily donates an electron to reduce the hydrogen ion into hydrogen gas ($H_2$), while forming sodium acetylide:

This vigorous evolution of hydrogen gas is a classic confirmation of the acidic nature of terminal alkynes.

A Visual Confirmation

Next, we consider ammoniacal silver nitrate, widely known as Tollens' reagent ($AgN{O}_3$ in $N{H}_3$). This reagent is famous for distinguishing aldehydes from ketones, but it also serves as a definitive test for terminal alkynes.

The silver ions ($A{g}^{+}$) in the solution replace the acidic protons of acetylene, resulting in the formation of silver acetylide. This compound is highly insoluble and immediately crashes out of the solution as a distinct white precipitate:

What happens when we introduce hydrochloric acid ($HCl$)? Here, the reaction takes a different path. Instead of an acid-base proton transfer, the electron-rich triple bond acts as a nucleophile and undergoes electrophilic addition.

The pi electrons attack the electrophilic hydrogen of $HCl$, and the chloride ion attaches to the adjacent carbon. This process typically requires a heavy metal catalyst like $H{g}^{2+}$ to stabilize the highly reactive intermediate, ultimately forming vinyl chloride:

A Battle of pKa Values

Finally, we arrive at sodium hydroxide ($NaOH$). We know $NaOH$ is a strong base, and we've established that acetylene is acidic. So, why doesn't a reaction occur? The answer lies in the delicate balance of chemical equilibrium and $p{K}_a$ values.

For an acid-base reaction to proceed in the forward direction, the reacting base must be stronger than the conjugate base produced, or conversely, the reacting acid must be stronger than the conjugate acid formed.

The $p{K}_a$ of acetylene is approximately 25. If the hydroxide ion ($O{H}^{-}$) were to deprotonate acetylene, it would form water ($H_{2}O$) as the conjugate acid. However, the $p{K}_a$ of water is about 15.7. Because a lower $p{K}_a$ indicates a stronger acid, water is a much stronger acid than acetylene.

Therefore, the equilibrium heavily favors the reactants. The hydroxide ion is simply too weak to pull the proton off the alkyne:

Additionally, the Hard Soft Acid Base (HSAB) theory provides another layer of insight. The acetylide ion ($HC\equiv C^{-}$) is considered a 'soft' base due to its polarizable electron cloud, while the sodium ion ($N{a}^{+}$) is a 'hard' acid. Hard-soft interactions are generally less stable and less favored compared to hard-hard or soft-soft interactions, further explaining the lack of reactivity with $NaOH$.

Sodamide

If $NaOH$ fails, what can we use to deprotonate acetylene? We need a base whose conjugate acid has a $p{K}_a$ significantly higher than 25. Enter sodamide ($NaN{H}_2$).

The amide ion ($N{H}_2^{-}$) is an exceptionally strong base. Its conjugate acid is ammonia ($N{H}_3$), which boasts a $p{K}_a$ of around 38. Because ammonia is a much weaker acid than acetylene, the amide ion easily and irreversibly deprotonates the terminal alkyne:

Understanding these subtle differences in acidity and basicity is crucial for mastering organic synthesis and predicting reaction outcomes with confidence.