Question:
Identify the major product formed when 4-methyl phenol is treated with chloroform in presence of NaOH.
Identify the major product formed when 4-methyl phenol is treated with chloroform in presence of NaOH.
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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
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The Correct Option is D
Solution and Explanation
Step 1: Recognizing the reaction type.
Phenol (here 4-methyl phenol) reacts with chloroform in presence of NaOH via Reimer–Tiemann reaction, which introduces a formyl group (\(-CHO\)) at ortho and para positions.
Step 2: Activation of phenoxide ion.
NaOH converts phenol into phenoxide ion, which strongly activates the benzene ring toward electrophilic substitution, especially at ortho and para positions.
Step 3: Role of chloroform.
CHCl\(_3\) generates dichlorocarbene (:CCl\(_2\)) in basic medium, which acts as the electrophile in the reaction.
Step 4: Orientation effect of substituents.
In 4-methyl phenol, both OH and CH\(_3\) are ortho/para directing groups. However, steric hindrance favors substitution at the para position relative to OH group.
Step 5: Product formation.
Thus, formyl group enters mainly at the para position forming p-hydroxybenzaldehyde derivative.
Step 6: Final conclusion.
Hence, the major product is para-substituted aldehyde derivative.
Final Answer: \[ \boxed{\text{p-hydroxybenzaldehyde derivative}} \]
Phenol (here 4-methyl phenol) reacts with chloroform in presence of NaOH via Reimer–Tiemann reaction, which introduces a formyl group (\(-CHO\)) at ortho and para positions.
Step 2: Activation of phenoxide ion.
NaOH converts phenol into phenoxide ion, which strongly activates the benzene ring toward electrophilic substitution, especially at ortho and para positions.
Step 3: Role of chloroform.
CHCl\(_3\) generates dichlorocarbene (:CCl\(_2\)) in basic medium, which acts as the electrophile in the reaction.
Step 4: Orientation effect of substituents.
In 4-methyl phenol, both OH and CH\(_3\) are ortho/para directing groups. However, steric hindrance favors substitution at the para position relative to OH group.
Step 5: Product formation.
Thus, formyl group enters mainly at the para position forming p-hydroxybenzaldehyde derivative.
Step 6: Final conclusion.
Hence, the major product is para-substituted aldehyde derivative.
Final Answer: \[ \boxed{\text{p-hydroxybenzaldehyde derivative}} \]
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