Question

What are the intermediates formed during the reactions A and B ?
A. Reimer-Tiemann reaction.
B. Dehydration of alcohols in the slow rate determining step.

• A. Option A
• B. Option B
• C. Option C
• D. Option D

Hint:

Reimer-Tiemann reaction proceeds via dichlorocarbene intermediate, while dehydration of alcohols proceeds via carbocation intermediate in the slow step.

Answer 1

The Correct Option is B

Step 1: Identify intermediate in Reimer-Tiemann reaction.
Reimer-Tiemann reaction involves treatment of phenol with \( CHCl_3 \) and \( NaOH \).
In this reaction, dichlorocarbene \( (:CCl_2) \) is generated as the reactive intermediate.
The phenoxide ion reacts with this intermediate to give ortho-hydroxybenzaldehyde.

Step 2: Structure of intermediate A.
The intermediate involves phenoxide ion \( (C_6H_5O^-) \) and dichlorocarbene attached to the ring.
Thus, A corresponds to the structure containing \( O^-Na^+ \) and \( CHCl_2 \) group.

Step 3: Identify intermediate in dehydration of alcohols.
Dehydration of alcohols occurs via elimination reaction.
The slow step involves formation of carbocation after loss of water.

Step 4: Structure of intermediate B.
Thus, intermediate B is a carbocation.
For example, in ethanol dehydration, ethyl carbocation \( CH_3CH_2^+ \) is formed.

Step 5: Match with given options.
Option (B) shows:
- A as phenoxide ion with \( CHCl_2 \) substituent
- B as a carbocation
This matches the required intermediates.

Step 6: Conclusion.
Thus, the correct intermediates are given in option (B).
\[ \boxed{\text{Option B}} \]