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 alcohols and ethers is $s{p}^3$ - hybridised.
• C. lone pair - lone pair repulsion is greater than bond pair-bond pair repulsion.
• D. none of these

Answer 1

Answer: (A)

of repulsion between the two bulky R groups.

The oxygen atom in ether and alcohol is $s{p}^3$ hybridised, therefore has a tetrahedral geometry where two positions are occupied by lone pairs of electrons and the other two by the alkyl groups in ether and one alkyl and one $H$ in alcohol. The bond angle in regular tetrahedral is ${109}^{\circ }{28}^{\prime }$.

In general, the lone pairs on $O$ tend to repel the bond pairs and results in lower bond angles than ${109}^{\circ }$ in case of alcohols

But in ethers, the $C-O-C$ bond angle is slightly more than this i.e nearly ${110}^{\circ }$. This is because of the presence of two bulky alkyl group on $O$ atom. They repel each other and counterbalance the repulsion due to lone pairs on $O$ atom on bond pairs. This results in the bond angle slightly greater than the tetrahedral angle.

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 of repulsion between the two bulky R groups.