Bond angle about oxygen in alcohol

Question

What should be the C-O-H bond angle in an alcohol be, would it be greater or smaller than 109°28’? According to me, the angle should be greater since the steric factor of R will overcome the lone pair-lone pair and lone pair-bond pair repulsion.

Answer 1

The $\ce{C-O-H}$ angle in ethanol is reported as 105.4°, smaller than the tetrahedral angle of 109.5°. As I mentioned in my comment, Bent's Rule provides a reasonable explanation for this slight angular compression.

Simply stated, Bent's Rule tells us that orbitals with higher electron density will have more s character than orbitals with lower electron density. This is because s orbitals are lower in energy than p orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more s character mixed in.

The electrons in the $\ce{C-O}$ and $\ce{O-H}$ bonds are shared between oxygen and another atom; the two lone pairs on oxygen are not shared, they are concentrated on the oxygen atom. According to Bent's Rule, we would expect the lone pair orbitals, with their higher electron density, to have more s character mixed in and the $\ce{C-O}$ and $\ce{O-H}$ bonds to contain less s character (consequently more p character).

Placing more p character in two bonds will reduce the angle between them (consider the change in bond angle as we move from sp to sp² to sp³ hybridization; the ultimate case is pure p hybridization where the bond angle is reduced to 90°). Therefore, as we mix more p character into the $\ce{C-O}$ and $\ce{O-H}$ bonds in ethanol we would expect the $\ce{C-O-H}$ angle to be smaller than the tetrahedral angle as experimentally observed.

Answer 2

The $\ce{COH}$ bond angle is $104$°. This can be explained by stating that the two lone-pairs in Oxygen atom are more "cumbersome" than the two covalent bonds $\ce{C-O}$ and $\ce{O-H}$.