# What would be the intermolecular forces between the molecules of methanetetrol?

Methanetetrol has the following chemical formula: $$\ce{CO4H4}$$. It is also referred to as orthocarbonic acid.

Since the molecule has a molecular geometry of a tetrahedron, similar to that of a molecule of methane, I would assume that the dipoles at the bonds cancel each other out. Hence, making the molecule non-polar with no dipole or charge separation overall. Thereby, the intermolecular force between the molecules of methanetetrol should be London dispersion forces.

However, I've been told that the intermolecular force between the methanetetrol molecules is hydrogen bonding since the four $$\ce{OH}$$; hydroxyl groups in the molecule are capable of making 4 hydrogen bonds with 4 subsequent methanetetrol molecules.

Hence, my question is how can methanetetrol have hydrogen bondings as the intermolecular force between its molecules if the molecules do not have a dipole; or so I believe, i.e. what do you think is the intermolecular force between methanetetrol molecules?

• You do not need a molecular dipole. The polarization of an individual bond could suffice. According to Wikipedia the molecule remains hypothetical! so somebody told you a hypothetical fact. – Alchimista Jan 9 at 9:38
• This molecule is very polar, despite having zero dipole moment. Besides, it does not exist. – Ivan Neretin Jan 9 at 9:46

Let's try to guess a reasonable structure for $$\ce{C(OH)_4}$$. The oxygen atoms and the central carbon would probably form a tetrahedron, therefore, those bond dipoles would basically cancel (as in e.g. $$\ce{CCl_4}$$). However, the carbon-oxygen-hydrogen angle would probably not be 180°. Whether the hydroxyl group would rotate freely about the carbon-oxygen bond or there would be some intramolecular hydrogen bonding preventing that, I would assume that intermolecular hydrogen bonding could still occur. (Which would be the first step towards eliminating water and forming $$\ce{CO_2}$$.)