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

Carboxylic acids such as acetic acid are capable of forming dimers:

I'm wondering why alcohols like ethanol don't generally form dimers. In the diagram below, the oxygen atom on the left ethanol molecule should be capable of accepting a hydrogen bond from the partially positive $\ce{H}$ atom on the right ethanol molecule. Does it have something to do with the $\ce{-OH}$ bond angle?

• Because the ring would be too strained. – DHMO Oct 10 '16 at 12:01
• That I don't believe in, hydroxyl groups are just not acidic/basic enough to engage in that type of hydrogen bonding, no matter the ring size. If you would add in water molecules the rings would get extended and still you wouldn't get dimers with incorporated water. – logical x 2 Oct 18 '16 at 11:58

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.

• "As you have drawn for carboxylic acids, it is very easy to allow this linear arrangements. If you wish, you could describe the entire (R−COOH)2 feature as a benzene-like ring streched in a single direction." Well except that in benzene, no bond is linear as they are all tilted by 60° (so that the bond angle is 120°)." Also the two alcohols dont necessarily have to form a rectangle, as the h bonds might just as well be longer than the covalent O-H bonds. Im still not convinced... – logical x 2 Nov 20 '16 at 23:45
• @ketbra Re angle: The first sentence is about a general $\ce{X-H\bond{...}X}$ angle. The second sentence tells you which atoms we are dealing with. And since $\ce{X}$ is $\ce{O}$, it also specifies that the angle is $\ce{O-H\bond{...}O}$ in this case. An angle will always be made up by three points in space; in the context of molecules/chemistry, typically three atoms. – Jan Nov 20 '16 at 23:52
• Re streched benzene ring: Yes, benzene consists of six atoms while a carboxylic acid dimer consists of eight. Hoever, imagine two bonds of benzene which are transformed onto one another by a mirror image parallel to both extended in a way such that an additional atom fits between the two carbons. These eight atoms are now a ‘benzene-like ring structure’; six $120^\circ$ angles and two $180^\circ$. – Jan Nov 20 '16 at 23:55
• Re rectangle: A rectangle is a shape with four corners, two sets of parallel sides and four $90^\circ$ angles. I don’t see how that should be unable to accomodate H bonds ‘longer than the covalent $\ce{O-H}$ bonds’? I explicitly said rectangle, not square. – Jan Nov 20 '16 at 23:57
• Four 90 ° angles would be the definition of a square though... But then I dont get the next statement that the "hydrogen bond angle would be much closer to 90° — a very bad arrangement". How do you come up with that? I would say that there is considerable stabilization between hydroxyl groups (i.e. alcohols) by hydrogen bonding e.g. in water and there also exists the water dimer (or methanol) so that more or less refutes this statement. – logical x 2 Nov 21 '16 at 7:36

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 $\mathrm{kJ\ mol^{-1}}$).

It's all a matter of relative energies. Carboxylic acids form two aligned hydrogen bonds, so the $\Delta E$ associated with this process is fairly negative. There's also some extra electron density on the carbonyl O that likely contributes to a better hydrogen bond. Now, alcohols can form one hydrogen bond. You almost certainly make the hydrogen bond in solution, but any collision from a solvent molecule could break the hydrogen bond.

Edit: Apparently, my answer was not clear. No, the bond angle is not the problem.

The main reason for the higher extend of H bonding in carboxylic acids is due the higher acidity of the carboxyl group (e.g. ethanol: $~\mathrm pK_\mathrm a = 16$, acetic acid: $~\mathrm pK_\mathrm a=4.5$) making it a good hydrogen bond donor, and the ability of the carbonyl oxygen to act as a good hydrogen bond acceptor. Thus the extend of H bonding can be explained solely based on acidity/basicity arguments. Compared to that a hydroxide is just too basic to be formed out of the hydroxyl group, so hydrogen bonded dimers of alcohols make no sense. What is also important to point out, is that the resulting dimer will be completely symmetric (due to resonance) making the situation energetically even more favourable.

• I have a feeling you were rather confused here... – Mithoron Nov 19 '16 at 22:14
• @Mithoron Thanks for getting personal : ) Maybe try finding arguments for your position next time, if you have any. But I really don't believe in the points mentioned in the other answers. It just has to do with the strength of the O-H bond, i.e. its lower aciditiy, compared to carboxylic acids and not with the alignment. – logical x 2 Nov 20 '16 at 8:52
• Well we could talk about it in chat (chat.stackexchange.com/rooms/3229/the-periodic-table) or maybe separate room. Or you could just have faith in Jan's answer :D – Mithoron Nov 20 '16 at 21:07