CBSE Class 12 Chemistry 2026: question 27

Hints

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Step-by-step solution

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1. (a) Acetophenone to benzoic acid, by oxidation with alkaline $\mathrm{KMn}{\mathrm{O}}_4$ (or haloform reaction) followed by acidification:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3\stackrel{(\mathrm{i})\mathrm{KMn}{\mathrm{O}}_4/KOH,\mathrm{\Delta };(\mathrm{ii}){\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$

   Alternatively: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3+3NaOI\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+\mathrm{CH}{\mathrm{I}}_3+2NaOH$, then ${\mathrm{H}}_3{\mathrm{O}}^{+}$ gives ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$.

   1 mark

2. (b) Acetonitrile to acetone, by Grignard reagent followed by hydrolysis:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{N}+\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}\stackrel{\mathrm{dry}\mathrm{ether}}{\to }(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{NMgBr}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3+\mathrm{N}{\mathrm{H}}_3+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$

   1 mark

3. (c) Benzoic acid to benzene, by forming the sodium salt and then decarboxylating with soda-lime:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{NaOH}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+\mathrm{NaOH}\stackrel{\mathrm{CaO},\mathrm{\Delta }}{\to }{\mathrm{C}}_6{\mathrm{H}}_6+\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$

   1 mark

Answer: (a) Alk. KMnO₄ (or NaOI), then H₃O⁺; (b) CH₃MgBr, then H₃O⁺; (c) Na salt heated with soda-lime (NaOH/CaO)

Revise the chapter

• Chapter: Aldehydes, Ketones and Carboxylic Acids (notes, formulas, where marks are lost and every past question)