CBSE Class 12 Chemistry 2026: question 19

Hints

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Step-by-step solution

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1. (a) Coupling takes place at the para position of phenol in alkaline medium:
   ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}+{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\stackrel{\mathrm{O}{\mathrm{H}}^{-}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\text{–}\mathrm{N}=\mathrm{N}\text{–}{\mathrm{C}}_6{\mathrm{H}}_4\text{–}\mathrm{OH}(\mathrm{p})+\mathrm{C}{\mathrm{l}}^{-}+{\mathrm{H}}_2\mathrm{O}$
   The product is p-hydroxyazobenzene, an orange dye.

   1 mark

2. (b) First reduce nitromethane to methanamine, then carry out the carbylamine reaction:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{O}}_2\stackrel{Sn/HCl(\mathrm{or}{\mathrm{H}}_2/Pd)}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$
   $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2+\mathrm{CHC}{\mathrm{l}}_3+3KOH(alc.)\stackrel{\mathrm{\Delta }}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{NC}+3KCl+3{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (a) p-Hydroxyazobenzene (+ Cl⁻ + H₂O); (b) Reduction to CH₃NH₂, then CHCl₃/alc. KOH gives CH₃NC

Revise the chapter

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