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During the Reimer-Tiemann reaction, when phenol is treated with chloroform in the presence of aqueous sodium hydroxide, an electrophile is generated in situ which attacks the aromatic ring. The identity of this electrophile is:

Options

  1. A^+CHCl₂
  2. B^+CCl₃
  3. C:CCl₂
  4. DCCl₃^-

Correct answer

C. :CCl₂

Step-by-step solution

In the Reimer-Tiemann reaction, the base ( OH^- ) abstracts a proton from chloroform ( CHCl₃ ) to form the trichloromethyl carbanion ( CCl₃^- ). CHCl₃ + OH^- CCl₃^- + H₂O This carbanion readily undergoes alpha-elimination by losing a chloride ion to generate dichlorocarbene ( :CCl₂ ), which is a neutral, electron-deficient species. CCl₃^- :CCl₂ + Cl^- The dichlorocarbene acts as the electrophile and attacks the electron-rich phenoxide ring to eventually yield 2-hydroxybenzaldehyde after hydrolysis. Answer: :CCl₂

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