Salicylaldehyde is formed as a main product from phenol by :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Reimer Tiemann reaction
The –OH group makes the ring of phenol very reactive towards electrophilic substitution at the ortho and para positions. Phenol and phenoxide give several named reactions:
Reimer-Tiemann reaction: phenol with chloroform and aqueous NaOH, then acid, puts a –CHO group at the ortho position and gives salicylaldehyde (2-hydroxybenzaldehyde).
Kolbe’s reaction: sodium phenoxide with CO2, then acid, puts a –COOH group at the ortho position and gives salicylic acid.
\( \ce{C6H5OH} \xrightarrow[\text{(ii) H}^+]{\text{(i) CHCl}_3\text{, aq. NaOH}} \ce{C6H4(OH)CHO} \) (salicylaldehyde)
Why the other options are incorrect
Option (1) Kolbe’s reaction — Incorrect. The phenoxide ion reacts with carbon dioxide, a weak electrophile, and the product is salicylic acid, not the aldehyde.
\( \ce{C6H5OH} \xrightarrow{\text{(i) NaOH, (ii) CO}_2\text{, (iii) H}^+} \ce{C6H4(OH)COOH} \)
Option (3) Kolbe’s electrolytic reaction — Incorrect. Electrolysis of an aqueous solution of the sodium salt of a carboxylic acid gives an alkane (R–R); phenol is not involved.
\( \ce{2RCOONa + 2H2O} \xrightarrow{\text{electrolysis}} \ce{R-R + 2CO2 + H2 + 2NaOH} \)
Option (4) Friedel Craft’s reaction — Incorrect. It puts an alkyl or acyl group on an aromatic ring with anhydrous AlCl3; it does not introduce a –CHO group. Acylation of benzene, for example, gives a ketone:
\( \ce{C6H6 + CH3COCl} \xrightarrow{\text{anhyd. AlCl}_3} \ce{C6H5COCH3 + HCl} \)
NCERT (2026-27 edition):
Reimer-Tiemann Reaction
