Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
How does one explain the trend in bond lengths and acidity of the following hydrocarbons?
$$\begin{array}{ccc} \hline \text{Species} & \ce{C-H}\text{ bond length / Å} & \mathrm{p}K_\mathrm{a} \\ \hline \text{Ethane, }\ce{C2H6} & 1.102 & 50 \\ \text{Ethene, }\ce{C2H4} & 1.085 & 44 \\ \text{Ethyne, }\ce{C2H2} & 1.061 & 25 \\ \hline \end{array}$$
Shouldn't the shorter bonds between $\ce{C-H}$ be easier to break than longer ones, making ethane the most acidic?
You're right in that bond length, and therefore bond strength does affect acidity (see: $\ce{H2S}$, $\mathrm{p}K_\mathrm{a} = 7$ and $\ce{H2O}$, $\mathrm{p}K_\mathrm{a} = 15.7$). If we defined acidity with the following equation
$$\ce{HX -> H + X}$$
then the bond strength would indeed be the only deciding factor in the acidity of $\ce{HX}$, since the enthalpy change of that process is literally the bond dissociation energy.
However, acidity actually corresponds to the heterolytic dissociation of the $\ce{H-X}$ bond, with both electrons in the bond going to $\ce{X}$:
$$\ce{HX -> H+ + X-}$$
so bond strength isn't the only factor. In the case of ethane, ethene, and ethyne, the most important factor is the type of orbital that the electrons from the $\ce{C-H}$ bond end up in after deprotonation. For ethane, the two electrons end up in a carbon $\mathrm{sp^3}$ orbital; for ethene, an $\mathrm{sp^2}$ orbital; and for ethyne, an $\mathrm{sp}$ orbital.
Now, of these three, the $\mathrm{sp}$ lone pair is the most stable, since the $\mathrm{sp}$ hybrid orbital contains the most s character ($50\%$) and therefore has the lowest energy. So, of the three carbanions, the lone pair in the acetylide ion $\ce{HC2-}$ is most stabilised, making ethyne the most acidic molecule.
The acidity of terminal alkynes is in fact pretty useful from a synthetic viewpoint, since you can deprotonate them with a strong base like $\ce{NaNH2}$, forming a carbon-based nucleophile which can then be used to make new $\ce{C-C}$ bonds.
As for why the electronic stabilisation outweighs the bond strength, it's worth looking at some quantitative data. The $\ce{H-O}$ and $\ce{H-S}$ bonds have mean bond enthalpies of $463$ and $338~\mathrm{kJ~mol^{-1}}$ respectively (source: Physical Chemistry 9th ed., Atkins & de Paula, p 932), which is a difference of $\mathbf{125~kJ~mol^{-1}}$. This can, to some extent, explain the difference in acidities of $\ce{H2O}$ and $\ce{H2S}$.
On the other hand, the dissociation energies of the $\ce{H-C\mathrm{(sp)}}$ and $\ce{H-C\mathrm{(sp^2)}}$ bonds are given in J. Phys. Chem. 1987, 91, 17-19 as $132.6$ and $116~\mathrm{kcal~mol^{-1}}$ respectively. The difference is roughly $17~\mathrm{kcal~mol^{-1}}$, which is equivalent to $\mathbf{69.5~kJ~mol^{-1}}$ - a somewhat smaller value than before. You can see why bond length variations might therefore play a smaller role in the hydrocarbons.
Another equivalent way of looking at it is that an $\mathrm{sp}$-hybridised carbon is more electronegative than an $\mathrm{sp^2}$-hybridised carbon, which is in turn more electronegative than an $\mathrm{sp^3}$-hybridised carbon.
As mentioned earlier, an $\mathrm{sp}$ orbital has the greatest s-character, and consequently electrons in a carbon $\mathrm{sp}$ orbital experience a greater effective nuclear charge and are more tightly bound to the nucleus than electrons in $\mathrm{sp^2}$ or $\mathrm{sp^3}$ orbitals.
Of course, the $\ce{C-H}$ bond also involves a contribution from the hydrogen 1s orbital. However, the above means that the electrons in $\ce{C(sp)-H}$ bonds are more strongly attracted to the carbon nucleus than electrons in $\ce{C(sp^2)-H}$ or $\ce{C(sp^3)-H}$ bonds. This means that the $\ce{C(sp)-H}$ bond is more polarised towards carbon, leaving less electron density on hydrogen, and a greater acidity.
The amount of electron density on the hydrogen can in fact be probed using $\ce{^1H}$ NMR spectroscopy. A lower electron density translates into a higher chemical shift. Even though there is a second, unrelated, factor - anisotropic shielding - that serves to decrease the chemical shift of acetylenic hydrogens, they still tend to show up at higher chemical shifts than hydrogens on saturated hydrocarbons (~2 ppm, compared to ~1 ppm). This is a clear indication that the local electron density on a proton attached to $\ce{C(sp)}$ is less than that of a proton attached to $\ce{(sp^3)}$.
