Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
In alcohols, the oxygen atom of the group is hybridized. The bond angle is slightly less than the tetrahedral angle of (e.g., in methanol) due to the repulsion between the lone pairs of electrons on oxygen.
In phenol, the group is attached to an hybridized carbon of an aromatic ring. The bond length in phenol is , which is shorter than the in methanol. This is due to (i) partial double bond character resulting from the conjugation of the unshared electron pair of oxygen with the aromatic ring and (ii) the hybridized state of carbon to which oxygen is attached.
In ethers, the oxygen atom is hybridized. The bond angle is slightly greater than the tetrahedral angle (e.g., in methoxymethane) because of the repulsive interactions between the two bulky alkyl groups.
The bond length in ethers ( in methoxymethane) is almost the same as in alcohols ( in methanol).
📐Formulae
💡Examples
Problem 1:
Explain why the bond angle in alcohols like is slightly less than the tetrahedral angle, whereas in ethers like , it is slightly greater.
Solution:
In , . In , .
Explanation:
In alcohols, the repulsion between the two lone pairs on the oxygen atom pushes the and bonds closer together, decreasing the angle. In ethers, although lone pair repulsion is present, the steric hindrance (repulsion) between the two bulky alkyl groups () outweighs the lone pair effect, forcing the bond angle to expand beyond .
Problem 2:
Compare the bond length in and .
Solution:
Explanation:
The bond in phenol () is shorter than in methanol () because in phenol, the lone pair on oxygen undergoes resonance with the benzene ring, giving the bond partial double bond character. Additionally, the carbon in phenol is hybridized (more -character, more electronegative), making the bond shorter and stronger than the carbon-oxygen bond in methanol.