CBSE Class 12 Chemistry 2025: question 26

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Step-by-step solution

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1. (a) ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OC}{\mathrm{H}}_3+\mathrm{HI}\to {\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}+\mathrm{C}{\mathrm{H}}_3\mathrm{I}$

   The products are phenol and methyl iodide. The $\mathrm{O}-\mathrm{C}{\mathrm{H}}_3$ bond breaks because the $\mathrm{O}-{\mathrm{C}}_6{\mathrm{H}}_5$ bond has partial double bond character.

   1 mark

2. (b) Phenol + conc. $\mathrm{HN}{\mathrm{O}}_3$ gives 2,4,6-trinitrophenol (picric acid). Its structure is a benzene ring with –OH at C-1 and $-N{O}_2$ groups at C-2, C-4 and C-6.

   1 mark

3. (c) ${\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{MgBr}+\mathrm{HCHO}\to {\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{C}{\mathrm{H}}_2\mathrm{OMgBr}\stackrel{{\mathrm{H}}^{+}/{\mathrm{H}}_2\mathrm{O}}{\to }{\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$

   The product is cyclohexylmethanol, a primary alcohol: a cyclohexane ring bearing a $-C{H}_{2}OH$ group.

   1 mark

Answer: (a) Phenol + CH₃I; (b) 2,4,6-trinitrophenol (picric acid); (c) Cyclohexylmethanol, C₆H₁₁CH₂OH

Revise the chapter

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