JEE Main 2024 · 30 January · evening · Single correct · +4 / −1
Salicylaldehyde is synthesized from phenol, when reacted with
- (a) HCCl₃, NaOH
- (b) CO₂, NaOH
- (c) H-C(∥(O))-Cl , NaOH
- (d) CCl₄, NaOH
Answer and solution
Answer (a) HCCl₃, NaOH
Solution
Chemistry ma’am · 4 steps
An aldehyde group on phenol's ortho position with chloroform and alkali is the Reimer–Tiemann reaction.
- Make the electrophile. In aqueous NaOH, chloroform loses a proton and then a chloride ion to give dichlorocarbene, :CCl₂. This carbon has only six electrons, so it is a strong electrophile.
- Attack at ortho. NaOH also turns phenol into the phenoxide ion, which is very electron-rich at the ortho and para positions. It attacks :CCl₂, mainly at the ortho position, and puts a -CHCl₂ group on the ring.
- Hydrolysis. The alkali hydrolyses the two C–Cl bonds of -CHCl₂, and the carbon becomes -CHO. Acidifying at the end gives salicylaldehyde, o-hydroxybenzaldehyde.
Reagent with phenol Group placed at ortho Product CHCl₃, NaOH -CHCl₂, hydrolysed to -CHO salicylaldehyde CO₂, NaOH (Kolbe) -COO⁻ salicylic acid CCl₄, NaOH -CCl₃, hydrolysed to -COOH salicylic acid HCOCl, NaOH none on the ring unstable reagent; an acid chloride acylates the O, not the ring - So. Phenol gives salicylaldehyde with CHCl₃ and NaOH, which is option (a).
The idea to keep. Count the chlorines on the carbon that goes onto the ring. CHCl₃ has two C–Cl bonds left after the carbon attaches, so you get -CHO. CCl₄ has three left, so you get -COOH.
Answer from the NTA final answer key. Solution by PrepWiser.