Alcohols, Phenols and Ethers
Alcohols and phenols are formed when a hydrogen atom in a hydrocarbon, aliphatic and aromatic respectively, is replaced by –OH group. These classes of compounds find wide applications in industry as well as in day-to-day life. For instance, have you ever noticed that ordinary spirit used for polishing wooden furniture is chiefly a compound containing hydroxyl group, ethanol. The sugar we eat, the cotton used for fabrics, the paper we use for writing, are all made up of compounds containing –OH groups. The common name of an alcohol is derived from the common name of the alkyl group and adding the word alcohol to it. For example, CH3OH is methyl alcohol. The simplest hydroxy derivative of benzene is phenol. It is its common name and also an accepted IUPAC name. As structure of phenol involves a benzene ring, in its substituted compounds the terms ortho (1,2-disubstituted), meta (1,3-disubstituted) and para (1,4-disubstituted) are often used in the common names.
In ethers, the four electron pairs, i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement. The bond angle is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky (–R) groups.
Consider the following chemical reaction and identify the intermediate involved in the reaction.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → sodium 2-(dichloromethyl)phenoxide
Note: The given answer is as per the official NTA Answer Key. However, there is more than one correct option for this question, as explained below.
In the Reimer–Tiemann reaction phenol is treated with chloroform and aqueous NaOH, and acidification (H+) then gives salicylaldehyde: a –CHO group is introduced at the ortho position. The reaction is carried out in strongly alkaline solution, so the phenol is present as the phenoxide ion, whose ring is highly activated towards electrophiles. The electrophile is dichlorocarbene (:\( \ce{CCl2} \)), formed from chloroform and alkali:
\( \displaystyle \ce{CHCl3 + OH^- -> CCl2 + Cl^- + H2O} \)
The steps are:
\( \displaystyle \ce{C6H5OH + NaOH -> C6H5O^-Na^+ + H2O} \)
\( \displaystyle \ce{C6H5O^-Na^+} \xrightarrow{\ce{CHCl3/NaOH}} \text{sodium 2-(dichloromethyl)phenoxide} \)
\( \displaystyle \text{sodium 2-(dichloromethyl)phenoxide} \xrightarrow{\text{aq. NaOH}} \text{sodium salt of salicylaldehyde} \)
\( \displaystyle \text{sodium salt of salicylaldehyde} \xrightarrow{\ce{H^+}} \text{salicylaldehyde (2-hydroxybenzaldehyde)} \)
Option (1) sodium salt of salicylaldehyde — Is also correct. It is formed when the –\( \ce{CHCl2} \) group is hydrolysed, and it is present before acidification, so it is also an intermediate of the reaction.
Option (2) sodium 2-(dichloromethyl)phenoxide — Correct. It is formed first, when dichlorocarbene attacks the ortho position of the phenoxide ion. It is the compound shown as the intermediate in the textbook scheme.
Option (3) 2-(dichloromethyl)phenol — Incorrect. In the strongly alkaline medium the –OH group stays as –\( \ce{O^{-}Na^{+}} \). The free phenol appears only after acidification, and by then –\( \ce{CHCl2} \) has already become –CHO.
Option (4) sodium 2-(chloromethyl)phenoxide — Incorrect. The carbene carries two chlorine atoms, so the group introduced is –\( \ce{CHCl2} \), not –\( \ce{CH2Cl} \). A –\( \ce{CH2Cl} \) group would be hydrolysed to –\( \ce{CH2OH} \), not to –CHO.
NCERT (2026-27 edition):
Reimer-Tiemann Reaction
