Match List-I with List-II
|
List-I Name of the Reaction |
List-II Compound synthesized |
|
(A) Kolbe's Reaction |
(I) Salicylaldehyde |
|
(B) Reimer Tiemann Reaction |
(II) Salicylic acid |
|
(C) Etard Reaction |
(III) Chlorobenzene |
|
(D) Sandmeyer Reaction |
(IV) Benzaldehyde |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
These named reactions introduce a group into a benzene ring or change one group into another:
On phenol (as phenoxide, in NaOH): Kolbe’s reaction with \( \ce{CO2} \) puts a –COOH group at the ortho position; Reimer–Tiemann reaction with \( \ce{CHCl3} \) puts a –CHO group there.
On a methyl side chain: Etard reaction with chromyl chloride stops the oxidation of toluene at the aldehyde.
On a diazonium salt: Sandmeyer reaction with Cu(I) halides replaces –\( \ce{N2^{+}} \) by Cl or Br (Gattermann reaction does the same with Cu powder and HX).
(A) Kolbe’s Reaction — This matches (II) Salicylic acid. The phenoxide ion, more reactive than phenol, undergoes electrophilic attack by carbon dioxide at the ortho position; acidification gives 2-hydroxybenzoic acid.
\( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2},\ \text{400 K, 4-7 atm}} o\text{-}\ce{HOC6H4COOH} \)
(B) Reimer Tiemann Reaction — This matches (I) Salicylaldehyde. Chloroform and aqueous NaOH introduce a –\( \ce{CHCl2} \) group at the ortho position, which is hydrolysed by the alkali to –CHO; acidification gives 2-hydroxybenzaldehyde.
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CHCl3},\ \text{aq. }\ce{NaOH}} o\text{-}\ce{HOC6H4CHO} \)
(C) Etard Reaction — This matches (IV) Benzaldehyde. Chromyl chloride oxidises the methyl group of toluene to a chromium complex, which is hydrolysed to benzaldehyde.
\( \displaystyle \ce{C6H5CH3} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{CrO2Cl2},\ \ce{CS2}} \ce{C6H5CHO} \)
(D) Sandmeyer Reaction — This matches (III) Chlorobenzene. Copper(I) chloride in HCl replaces the diazonium group by chlorine, with loss of nitrogen gas.
\( \displaystyle \ce{C6H5N2^+Cl^-} \xrightarrow{\ce{Cu2Cl2}/\ce{HCl}} \ce{C6H5Cl + N2 ^} \)
Shortcut: (A)-(II) appears only in Option (3).