Match List-I with List-II.
| List-I (Reaction) | List-II (Key step of the reaction) |
|---|---|
| (A) Kolbe’s reaction | (I) A substituted benzal chloride is hydrolysed in the presence of alkali. |
| (B) Reimer-Tiemann reaction | (II) A hydroperoxide is converted into the products by dilute acid. |
| (C) Preparation of phenol from cumene | (III) An alkoxide ion attacks a primary alkyl halide by SN2. |
| (D) Williamson synthesis | (IV) The phenoxide ion undergoes electrophilic substitution with carbon dioxide. |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-I, C-II, D-III
Kolbe’s and Reimer-Tiemann reactions both start from phenol in sodium hydroxide and put a group at the ortho position; they differ in the reagent and so in the step: carbon dioxide gives –COOH directly, while chloroform gives a benzal chloride that is then hydrolysed to –CHO.
(A) Kolbe’s reaction — This matches (IV) The phenoxide ion undergoes electrophilic substitution with carbon dioxide. Acidification then gives ortho-hydroxybenzoic acid.
\( \displaystyle \ce{C6H5OH} \xrightarrow{\ce{NaOH}} \ce{C6H5O^-Na+} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2}} \text{o-}\ce{HOC6H4COOH} \)
(B) Reimer-Tiemann reaction — This matches (I) A substituted benzal chloride is hydrolysed in the presence of alkali. Phenol with chloroform and sodium hydroxide first gives the substituted benzal chloride (–CHCl2 at the ortho position); its hydrolysis gives salicylaldehyde.
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CHCl3}\text{, aq. }\ce{NaOH}} \text{o-}\ce{HOC6H4CHO} \)
(C) Preparation of phenol from cumene — This matches (II) A hydroperoxide is converted into the products by dilute acid. Air oxidises cumene to cumene hydroperoxide, which dilute acid converts into phenol and acetone.
\( \displaystyle \ce{C6H5CH(CH3)2} \xrightarrow{\ce{O2}\text{ (air)}} \ce{C6H5C(CH3)2OOH} \xrightarrow{\ce{H+}/\ce{H2O}} \ce{C6H5OH + CH3COCH3} \)
(D) Williamson synthesis — This matches (III) An alkoxide ion attacks a primary alkyl halide by SN2. The alkoxide ion is the nucleophile and the halide ion leaves, giving an ether.
\( \displaystyle \ce{R-X + R'-O^-Na+ -> R-O-R' + NaX} \)
NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “Kolbe’s reaction Phenoxide ion generated by treating phenol with sodium hydroxide is even more reactive than phenol towards electrophilic aromatic substitution. Hence, it undergoes electrophilic substitution with carbon dioxide, a weak electrophile. Ortho hydroxybenzoic acid is formed as the main reaction product.”
“3. Reimer-Tiemann reaction On treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at ortho position of benzene ring. This reaction is known as Reimer - Tiemann reaction.” … “The intermediate substituted benzal chloride is hydrolysed in the presence of alkali to produce salicylaldehyde.”
7.4.2 Preparation of Phenols – “Cumene (isopropylbenzene) is oxidised in the presence of air to cumene hydroperoxide. It is converted to phenol and acetone by treating it with dilute acid.”
7.6.1 Preparation of Ethers – “2. Williamson synthesis It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.” … “The reaction involves SN2 attack of an alkoxide ion on primary alkyl halide.”