In the reaction [phenol + CHCl ₃ + NaOH → sodium salicylaldehyde (o-O ^- Na ^+ C ₆ H ₄ CHO), shown as a structure] the electrophile involved is 1. dichloromethyl cation ( oversetoplusC HCl ₂ ) 2. dichloromethyl anion ( oversetominusC HCl ₂ ) 3. formyl cation ( oversetoplusC HO) 4. dichlorocarbene (:CCl ₂ )

Answer: 4. dichlorocarbene (:CCl ₂ ) This is the Reimer-Tiemann reaction. The electrophile is dichlorocarbene. The base removes a proton from CHCl₃ to give mathrm^(-C)Cl₃ , which loses Cl^- to form the neutral, electron-deficient :CCl₂ . It attacks the electron-rich phenoxide ring at the ortho position, and the product hydrolyses to the -CHO group.

Topic: Phenols: acidity, electrophilic substitution, Reimer-Tiemann· Organic Compounds Containing Oxygen

In the reaction[phenol + CHCl_3 + NaOH → sodium salicylaldehyde (o-O^-Na^+ C_6H_4 CHO), shown as a structure]the electrophile involved is

In the reaction

[phenol + CHCl\(_3\) + NaOH \(\rightarrow\) sodium salicylaldehyde (o-O\(^-\)Na\(^+\) C\(_6\)H\(_4\) CHO), shown as a structure]

the electrophile involved is

reaction of phenol with CHCl3 and NaOH giving sodium salicylaldehyde

1. dichloromethyl cation (\(\overset{\oplus}{C}\)HCl\(_2\))

2. dichloromethyl anion (\(\overset{\ominus}{C}\)HCl\(_2\))

3. formyl cation (\(\overset{\oplus}{C}\)HO)

4. dichlorocarbene (:CCl\(_2\))

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Answer: 4. dichlorocarbene (:CCl\(_2\))

This is the Reimer-Tiemann reaction. The electrophile is dichlorocarbene.

The base removes a proton from \(\mathrm{CHCl_3}\) to give \(\mathrm{{}^-CCl_3}\), which loses \(\mathrm{Cl^-}\) to form the neutral, electron-deficient \(\mathrm{:CCl_2}\). It attacks the electron-rich phenoxide ring at the ortho position, and the product hydrolyses to the -CHO group.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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In the reaction[phenol + CHCl_3 + NaOH → sodium salicylaldehyde (o-O^-Na^+ C_6H_4 CHO), shown as a structure]the electrophile involved is

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