CBSE Class 12 Chemistry 2023: question 23(b)

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. (i) Step 1 – Protonation of the alcohol:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\ddot{\mathrm{O}}\mathrm{H}+{\mathrm{H}}^{+}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}{\mathrm{H}}_2$ (fast)

   ½ mark

2. Step 2 – Formation of carbocation (slow, rate-determining):
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{O}}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_3\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$
   Step 3 – Loss of proton:
   $\mathrm{H}-\mathrm{C}{\mathrm{H}}_2-\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}^{+}$

   ½ mark

3. (ii) Reimer–Tiemann reaction: Phenol is heated with chloroform and aqueous NaOH, forming a dichloromethyl (–CHCl₂) intermediate at the ortho position. Hydrolysis then gives sodium salicylaldehyde, and acidification gives salicylaldehyde:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\underset{\text{340 K}}{\overset{\mathrm{CHC}{\mathrm{l}}_3,aq.NaOH}{\to }}\mathrm{o}\text{-}{\mathrm{C}}_6{\mathrm{H}}_4({\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+})(\mathrm{CHO})\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{o}\text{-}{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{OH})\mathrm{CHO}$ (2-hydroxybenzaldehyde)

   1 mark

Answer: (i) Protonation → carbocation (slow) → loss of H⁺ giving ethene; (ii) phenol + CHCl₃/aq NaOH, then H⁺ → salicylaldehyde

Revise the chapter

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