Question

Why is the C - O - H bond angle in alcohols slightly less than the tetrahedral angle whereas the C - O - C bond angle in ether is slightly greater?

Answer 1

The bond angle in $C-\stackrel{..}{\underset{..}{O}}-H$ in alcohols is slightly less than tetrahedral angle $\left({108.9}^{\circ }\right)$. It is due to the repulsion between the unshared electron pairs of oxygen. In alcohols, two lone pair of electrons are present. Therefore, there is comparatively more repulsion and less bond angle.
The C - O - C bond angle $\left({111.7}^{\circ }\right)$in ether is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky (- R) groups.