Identify the following name reaction.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Haloform reaction
Several named reactions of carbonyl compounds and carboxylic acids can be recognised from the group that reacts and the reagent:
Haloform reaction: a methyl ketone (\( \ce{CH3CO} \)– group) with sodium hypohalite (NaOX) gives the sodium salt of an acid with one carbon fewer and a haloform (\( \ce{CHX3} \)).
Hell–Volhard–Zelinsky reaction: a carboxylic acid with an α-H, with \( \ce{X2} \) and red phosphorus, gives an α-halo acid.
Gattermann–Koch reaction: benzene with CO and HCl (anhydrous \( \ce{AlCl3} \)/CuCl) gives benzaldehyde.
Esterification: a carboxylic acid with an alcohol and a little acid gives an ester.
In the given scheme, a methyl ketone (\( \ce{R-CO-CH3} \)) reacts with NaOCl, and the products are the sodium carboxylate R–COONa and chloroform (\( \ce{CHCl3} \)). The \( \ce{CH3} \) group is converted into the haloform:
\( \displaystyle \ce{R-CO-CH3 + 3NaOCl -> R-COONa + CHCl3 + 2NaOH} \)
So it is the haloform reaction.
Why the other options are incorrect
Option (1) Hell-Volhard- Zelinsky reaction — Incorrect. It halogenates the α-carbon of an acid; no haloform is formed.
\( \displaystyle \ce{R-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}/\text{red P}} \ce{R-CH(X)-COOH} \)
Option (2) Gattermann- Koch reaction — Incorrect. It puts a –CHO group on benzene.
\( \displaystyle \ce{C6H6 + CO + HCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}/\ce{CuCl}} \ce{C6H5CHO} \)
Option (4) Esterification — Incorrect. An acid and an alcohol give an ester and water; a ketone is not involved.
\( \displaystyle \ce{RCOOH + R'OH <=> RCOOR' + H2O} \)
NCERT (2026-27 edition):
Oxidation of Methyl Ketones by Haloform Reaction – “Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform. …”