In acetylene (ethyne, HC≡CH), the carbons are sp-hybridised, which make the attached protons more acidic than those in alkenes or alkanes. (See also: Relative acidities of alkanes, alkenes, and alkynes)

Why doesn't acetylene react with strong bases like NaOH or KOH?


1 Answer 1


The $\mathrm{p}K_\mathrm{a}$ of acetylene is $25$. It is very weakly acidic, and hence, $\ce{KOH}$ is not sufficiently alkaline to react with it. Even methanol, with a $\mathrm{p}K_\mathrm{a}$ of $15.5$ does not react much with hydroxide ion. On other hand, $\ce{NaH}$ or $\ce{NaNH2}$ have more basic anions than hydroxide, and weak acids such as acetylene or alcohols can react with them.

Predicting the relative basicities can be done qualitatively by invoking the Brønsted conjugate acid-base theory. The rule to remember is, a base gives its corresponding conjugate acid in solution, and stronger the conjugate acid, weaker its conjugate base (and vice versa). This can be understood from the following

$$\ce{HA<=>H^+ + A-}$$

Here, $\ce{A-}$ is the conjugate base of the acid $\ce{HA}$. Now, if $\ce{A-}$ is highly unstable, it will have a stronger tendency to react with a proton and get converted to its conjugate acid which is relatively more stable. Since the equilibrium lies more to the left in that case, we say that the conjugate base is strong (and consequently the conjugate acid is weak). Now consider the following anions for the sake of example.

$$\ce{HC#C-} > \ce{NH2-} > \ce{OH-} > \ce{F-}$$

These are listed in decreasing order of stability, and hence also their strength as bases. This is because as the electronegativity increases from carbon to fluorine, the negative charge becomes increasingly stable on the atom. More electronegative elements are increasingly stable with the extra electron due to higher effective nuclear force of attraction. Correspondingly, by the dictum "stronger base is weaker conjugate acid", the order of the acidity is:

$$\ce{HC#CH} < \ce{NH3} < \ce{H2O} < \ce{HF}$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.