Chemistry Class 12 NCERT Solutions: Chapter 11 Alcohols Phenols and Ethers Part 8 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q: 22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.


Ethanol undergoes intermolecular H-bonding due to the presence of − OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the boiling point of ethanol is higher than that of methoxymethane.

Q_22_The Boiling Point of Ethanol is Higher Than That of Met …

Q: 23. Give IUPAC names of the following ethers:






Q_23_v_IUPAC Names


Q_23_vi_IUPAC Names


(i) 1-Ethoxy-2-methylpropane

(ii) 2-Chloro-1-methoxyethane

(iii) 4-Nitroanisole

(iv) 1-Methoxypropane

(v) 4-Ethoxy-1,1-dimethylcyclohexane

(vi) Ethoxybenzene

Q: 24. Write the names of reagents and equations for the preparation of the following ethers b Williamson՚s synthesis:

(i) 1-Propoxypropane

(ii) Ethoxybenzene

(iii) 2-Methoxy-2-methylpropane

(iv) 1-Methoxyethane








Q: 25. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.


The reaction of Williamson synthesis involves attack of an alkoxide ion on a primary alkyl halide.

Q_25_1_The Limitations of Williamson Synthesis

But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, the elimination would compete over substitution. As a result, alkenes would be produced This is because alkoxides are nucleophiles as well as strong bases. Hence, they reac with alkyl halides, which results in an elimination reaction.

Q_25_2_The Limitations of Williamson Synthesis

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