CBSE Class 12 Chemistry 2026: question 27

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Step-by-step solution

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1. (a) Reimer–Tiemann reaction: phenol treated with chloroform in aqueous NaOH introduces –CHO at the ortho position. The intermediate (substituted benzal chloride, then sodium salicylaldehyde) is acidified to give salicylaldehyde:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\underset{\text{aq. NaOH}}{\overset{\mathrm{CHC}{\mathrm{l}}_3}{\to }}o\text{-}{\mathrm{C}}_6{\mathrm{H}}_4({\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+})\mathrm{CHO}\stackrel{{\mathrm{H}}^{+}}{\to }o\text{-}\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}$ (salicylaldehyde)

   1 mark

2. (b) Kolbe's reaction: sodium phenoxide reacts with $\mathrm{C}{\mathrm{O}}_2$ (about 400 K, 4–7 atm) to give sodium salicylate, which is acidified to give salicylic acid:
   ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}\underset{400K,4\text{–}7atm}{\overset{\mathrm{C}{\mathrm{O}}_2}{\to }}o\text{-}\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CO}{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}\stackrel{{\mathrm{H}}^{+}}{\to }o\text{-}\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COOH}$ (2-hydroxybenzoic acid)

   1 mark

3. (c) Friedel–Crafts acylation of anisole:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{COCl}\stackrel{\text{anhyd. }\mathrm{AlC}{\mathrm{l}}_3}{\to }p\text{-}\mathrm{C}{\mathrm{H}}_3\mathrm{O}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COC}{\mathrm{H}}_3(\text{major})+o\text{-}\mathrm{C}{\mathrm{H}}_3\mathrm{O}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COC}{\mathrm{H}}_3(\text{minor})+\mathrm{HCl}$

   The products are 4-methoxyacetophenone (major) and 2-methoxyacetophenone (minor).

   1 mark

Answer: (a) Phenol → salicylaldehyde; (b) sodium phenoxide → salicylic acid; (c) anisole → 4-methoxyacetophenone (major) + 2-methoxyacetophenone (minor)

Revise the chapter

• Chapter: Alcohols, Phenols and Ethers (notes, formulas, where marks are lost and every past question)