IIT-JEE 2007 · Paper 2 · Chemistry Q41

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ChemistrySingle correct+4 / -1Question 41 of 66

Passage

Paragraph for Question Nos. 39 to 41

Reimer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.

(Scheme: 4-methylphenol (p-cresol) → intermediate I → II → III, where III is 2-hydroxy-5-methylbenzaldehyde; the last step is acidification.)

(The paragraph is not printed in the official archive copy (which reproduces only the questions with their answers); its text is restored from the JEE 2007 question paper.)

The structure of the intermediate I is

Sodium 2-(chloromethyl)-4-methylphenoxide: O⁻Na⁺ on the ring, CH₂Cl ortho to it and CH₃ para to it.

Sodium 2-(dichloromethyl)-4-methylphenoxide: O⁻Na⁺ on the ring, CHCl₂ ortho to it and CH₃ para to it.

Sodium 4-methyl-2-(trichloromethyl)phenoxide: O⁻Na⁺ on the ring, CCl₃ ortho to it and CH₃ para to it.

Sodium 2-(hydroxymethyl)-4-methylphenoxide: O⁻Na⁺ on the ring, CH₂OH ortho to it and CH₃ para to it.

Answer and worked solution

Official answer: B

Attack of the carbene. The phenoxide of p-cresol attacks $:\ce{CCl2}$ at the free ortho position; after proton transfer this gives a dichloromethyl group ortho to O⁻:
$\textbf{I}=\text{sodium 2-(dichloromethyl)-4-methylphenoxide}.$
Hydrolysis of $\ce{CHCl2}$ then gives the aldehyde (via II), and acidification gives 2-hydroxy-5-methylbenzaldehyde (III).

(B)

Topics:Phenols Reactive intermediates

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IIT-JEE 2007 · Paper 2, Chemistry question 41