CBSE Class 12 Chemistry sample paper 2023-24: question 20(b)

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Step-by-step solution

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1. (a) Gattermann–Koch reaction: ${\mathrm{C}}_6{\mathrm{H}}_6\stackrel{\mathrm{CO},\mathrm{HCl},anhyd.\mathrm{AlC}{\mathrm{l}}_3/CuCl}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$. Nitration: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}\stackrel{conc.\mathrm{HN}{\mathrm{O}}_3/conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4,273-283\mathrm{K}}{\to }\mathrm{m}\text{-}{\mathrm{O}}_2\mathrm{N}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}$. The –CHO group is meta-directing, so m-nitrobenzaldehyde forms.

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2. (b) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Br}\stackrel{\mathrm{Mg},\mathrm{dry}\mathrm{ether}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{MgBr}\stackrel{(\mathrm{i})\mathrm{C}{\mathrm{O}}_2(\mathrm{dry}\mathrm{ice})}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOMgBr}\stackrel{(\mathrm{ii}){\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$

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Answer: (a) Benzene → (CO/HCl, AlCl₃) benzaldehyde → (HNO₃/H₂SO₄) m-nitrobenzaldehyde; (b) Bromobenzene → (Mg, dry ether) PhMgBr → (CO₂, then H₃O⁺) benzoic acid

Revise the chapter

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