Question:medium

Identify the major product formed when 4-methyl phenol is treated with chloroform in presence of NaOH.

Show Hint

Reimer–Tiemann reaction introduces –CHO group mainly at ortho/para positions of phenol, para is favored when steric hindrance is high.
Updated On: Jun 19, 2026
  • 1
  • 2
  • 3
  • 4
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Identifying the transformation.
4-Methylphenol treated with CHCl₃ and NaOH undergoes the Reimer–Tiemann reaction, introducing a formyl (–CHO) group onto the aromatic ring.

Step 2: Phenoxide activation.

NaOH deprotonates phenol to phenoxide, powerfully activating the ring for electrophilic attack at ortho and para positions.

Step 3: Electrophile generation.

Chloroform in basic medium produces dichlorocarbene (:CCl₂), which serves as the electrophilic species.

Step 4: Directing influences.

Both –OH and –CH₃ are ortho/para directors; however, steric factors favor reaction at the less hindered para position relative to –OH.

Step 5: Major product.

The formyl group predominantly enters para to the hydroxyl, affording a p-hydroxybenzaldehyde derivative.

Step 6: Conclusion.

The major product is a para-substituted hydroxybenzaldehyde.
Was this answer helpful?
0