Question

The order of activity of the various o and p-director is :

• A. \( -O^->-OH>-OCOCH_3>-COCH_3 \)
• B. \( -OH>-O^->-OCOCH_3>-COCH_3 \)
• C. \( -OH>-O^->-COCH_3>-OCOCH_3 \)
• D. \( -O^->-COCH_3>-OCOCH_3>-OH \)

Hint:

Stronger +R effect $\Rightarrow$ stronger activating o,p-director.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Activating groups (ortho/para directors) increase the electron density of the benzene ring, making it more reactive toward electrophilic aromatic substitution. The degree of activation depends on how easily the group can donate electrons via resonance (\( +M \)) or induction (\( +I \)).
Step 2: Detailed Explanation:
1. \( \text{O}^{-} \): The phenoxide ion has a negative charge on Oxygen. It is a very strong electron donor via resonance because the negative charge is directly pushed into the ring. This is the strongest activator among the options.
2. \( -\text{OH} \): The hydroxyl group has lone pairs on Oxygen. It donates electrons via resonance (\( +M \)), but not as strongly as the negatively charged \( \text{O}^{-} \) because the Oxygen atom also exerts a significant \( -I \) effect.
3. \( -\text{OCOCH}_{3} \): The lone pair on Oxygen is cross-conjugated with the carbonyl group (\( \text{C=O} \)). This means the lone pair is less available to be donated to the benzene ring, making it a weaker activator than \( -\text{OH} \).
4. \( -\text{COCH}_{3} \): The acetyl group is actually a deactivating group and a meta-director because the carbonyl carbon pulls electron density out of the ring (\( -M \) and \( -I \)). However, in the provided sequence of "relative activity," it sits at the bottom as the least activating (most deactivating).
The correct order of decreasing activating power is \( \text{O}^{-}>-\text{OH}>-\text{OCOCH}_{3}>-\text{COCH}_{3} \).
Step 3: Final Answer:
The order of activity is \( \text{O}^{-}>-\text{OH}>-\text{OCOCH}_{3}>-\text{COCH}_{3} \).