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Alcohols, Phenols and Ethers - Structures of Functional Groups

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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In alcohols, the oxygen atom of the −OH-OH group is sp3sp^3 hybridized. The C−O−HC-O-H bond angle is slightly less than the tetrahedral angle of 109.5∘109.5^\circ (e.g., 108.9∘108.9^\circ in methanol) due to the repulsion between the lone pairs of electrons on oxygen.

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In phenol, the −OH-OH group is attached to an sp2sp^2 hybridized carbon of an aromatic ring. The C−OC-O bond length in phenol is 136 pm136 \text{ pm}, which is shorter than the 142 pm142 \text{ pm} 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 sp2sp^2 hybridized state of carbon to which oxygen is attached.

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In ethers, the oxygen atom is sp3sp^3 hybridized. The C−O−CC-O-C bond angle is slightly greater than the tetrahedral angle (e.g., 111.7∘111.7^\circ in methoxymethane) because of the repulsive interactions between the two bulky alkyl groups.

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The C−OC-O bond length in ethers (141 pm141 \text{ pm} in methoxymethane) is almost the same as in alcohols (142 pm142 \text{ pm} in methanol).

📐Formulae

Bond Angle in Methanol (CH3OH)≈108.9∘\text{Bond Angle in Methanol (CH}_3\text{OH)} \approx 108.9^\circ

Bond Angle in Methoxymethane (CH3OCH3)≈111.7∘\text{Bond Angle in Methoxymethane (CH}_3\text{OCH}_3\text{)} \approx 111.7^\circ

Bond Length C−O (Phenol)=136 pm\text{Bond Length } C-O \text{ (Phenol)} = 136 \text{ pm}

Bond Length C−O (Methanol)=142 pm\text{Bond Length } C-O \text{ (Methanol)} = 142 \text{ pm}

💡Examples

Problem 1:

Explain why the bond angle in alcohols like CH3OH\text{CH}_3\text{OH} is slightly less than the tetrahedral angle, whereas in ethers like CH3OCH3\text{CH}_3\text{OCH}_3, it is slightly greater.

Solution:

In CH3OH\text{CH}_3\text{OH}, ∠C−O−H=108.9∘<109.5∘\angle C-O-H = 108.9^\circ < 109.5^\circ. In CH3OCH3\text{CH}_3\text{OCH}_3, ∠C−O−C=111.7∘>109.5∘\angle C-O-C = 111.7^\circ > 109.5^\circ.

Explanation:

In alcohols, the repulsion between the two lone pairs on the oxygen atom pushes the O−HO-H and C−OC-O bonds closer together, decreasing the angle. In ethers, although lone pair repulsion is present, the steric hindrance (repulsion) between the two bulky alkyl groups (−CH3-\text{CH}_3) outweighs the lone pair effect, forcing the bond angle to expand beyond 109.5∘109.5^\circ.

Problem 2:

Compare the C−OC-O bond length in CH3OH\text{CH}_3\text{OH} and C6H5OH\text{C}_6\text{H}_5\text{OH}.

Solution:

d(C−O)phenol<d(C−O)methanold(C-O)_{\text{phenol}} < d(C-O)_{\text{methanol}}

Explanation:

The C−OC-O bond in phenol (136 pm136 \text{ pm}) is shorter than in methanol (142 pm142 \text{ pm}) because in phenol, the lone pair on oxygen undergoes resonance with the benzene ring, giving the C−OC-O bond partial double bond character. Additionally, the carbon in phenol is sp2sp^2 hybridized (more ss-character, more electronegative), making the bond shorter and stronger than the sp3sp^3 carbon-oxygen bond in methanol.

Structures of Functional Groups Class 12 Notes & Examples