Question

The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle whereas the C-O-C bond angle in ether is slightly greater because:
(A) of repulsion between the two bulky -R groups
(B) O atom in both alcohol and ethers is \[s{p^3}\] hybridized
(C) lone pair-lone pair repulsion is greater than bond pair-bond pair repulsion
(D) none of these

Answer 1

Hint: Lone pair- lone pair repulsion lesser than the repulsion between two alkyl groups bonded with oxygen atoms in ethers. A Hydrogen atom is not a bulky group but it is the smallest group or atom possible.

Complete answer:
We know that in alcohols and ethers, the oxygen atom is \[s{p^3}\] hybridized. So, according to hybridization, the geometry around oxygen atoms should be tetrahedral and angle should be equal to \[{109^ \circ }28'\]. But in reality it is not seen. The reason is the two lone pairs on the oxygen atom. They cause repulsion to each other and resultive bond angle is slightly less in those compounds.
Hence, the C-O-H bond angle in alcohols is slightly less than the tetrahedral angle which is equal to \[{109^ \circ }28'\].
But in the case of ethers, there are two alkyl groups bonded with oxygen atoms. Now alkyl groups are bulky enough to compensate for lone pair-lone pair repulsion and in fact they repulse each other more than that of two lone pairs of oxygen. So, as a result there is an increase in C-O-C bond angle in ethers.

Additional Information:
Remember that when \[s{p^3}\] hybridization is present in the molecule then, the molecule has always tetrahedral geometry and the bond angles between atoms is generally \[{109^ \circ }28'\].
Now, certain repulsions like lone pair- lone pair repulsions, lone pair- bond pair repulsions and repulsion between two bulky groups decide the slight changes in the bond angles.

Note: Statement given in option (C) is absolutely correct but do not treat it as the right answer because this statement has nothing to do with the increase in bond angles of C-O-C in ethers.