Chemistry - Why can't alcohols form hydrogen-bonded dimers like carboxylic acids?

Solution 1:

The strength of a hydrogen bond somewhat depends on the $\ce{X-H\bond{...}X}$ angle that the hydrogen-bonding hydrogen forms with the two electronegative elements $\ce{X}$. In our case, carboxylic acids or alcohols, $\ce{X} = \ce{O}$ so the angle is $\ce{O-H\bond{...}O}$. The ideal angle for this fragment is $180^\circ$.

As you have drawn for carboxylic acids, it is very easy to allow this linear arrangements. If you wish, you could describe the entire $\ce{(R-COOH)2}$ feature as a benzene-like ring streched in a single direction. Most importantly, the acceptors and donors line up nicely, the carboxyl function features an angle of $120^\circ$ and the $\ce{C-O-H}$ angle (and the $\ce{C=O\bond{...}H}$ one!) is also close to $120^\circ$. These arae the theoretically predicted, unstrained angles.

For two alcohol (e.g. ethanol) molecules to attempt a similar arrangement, we would end up with only four atoms that have to form a rectangle with $\ce{O, H, O, H}$ at the four corners. Thus, the $\ce{O-H\bond{...}O}$ hydrogen bond angle would be much closer to $90^\circ$ — a very bad arrangement. Furthermore, this would put the two electronegative oxygens (which thus feature a negative partial charge) closer together with nothing in-between which would cause destabilisation due to like charges approaching. The hydrogen-bonding hydrogens cannot alleviate this unfavourable interaction since they are at the corners of the square.

The situation for alcohols is much better if they create a network of a number of molecules that allow for much more favourable angles. With four molecules, you could already create a cyclic structure of approximately $100^\circ$ $\ce{H\bond{...}O-H}$ angles and $170^\circ$ $\ce{O-H\bond{...}O}$ angles — much more favourable. Of course, in the actual solution this number will fluctuate strongly as hydrogen bonds are broken and reformed constantly and different ring sizes happen all the time.

Solution 2:

The carboxylic acid dimer is more likely than the alcohol dimer because the carboxylic acid dimer has two attachment points, whereas the alcohol dimer only has one. While the hydrogen bonds may be of comparable strength in the two cases, the carboxylic acid case is also favored entropically.

Consider, by analogy, the chelate effect. Chelating (multidentate - more than one binding site) ligands bind more strongly to metals than monodentate ligands.

Here are some data for the comparable systems of copper/ethylenediamine $\ce{Cu(en)_2^2+}\ (\ce{en}=\ce{H2NCH2CH2NH2})$ and copper/methylamine $\ce{Cu(CH3NH2)_4^2+}$. Here, I am presenting the overall formation constants $\beta$ (whcih are the products of the equilibrium/formation/binding constants of each individual association step). The formation constant for $\ce{Cu(en)_2^2+}$ is higher than for $\ce{Cu(CH3NH2)_4^2+}$ by a factor of $\sim 10^4$, which corresponds to the chelated complex having a more negative free energy of binding by 23 kJ/mol. The major component in this lower energy state is entropy. Data from here.

$$\begin{array}{|c|c|c|c|c|}\hline \mathrm{Equilibrium} & \beta & \Delta G^\circ& \Delta H^\circ& -T \Delta S^\circ\\ \hline \ce{Cu^2+ + 2en <=> Cu(en)_2^2+} & 4.17 \times 10^{10} &-60.67 & -56.48 & -4.19 \\ \hline \ce{Cu^2+ + 4CH3NH2 <=> Cu(CH3NH2)_4^2+} & 3.55 \times 10^6 & -37.4 & -57.3 & 19.9 \\ \hline \end{array}$$ (data of last three columns in $\mathrm{kJ\ mol^{-1}}$).