Question

Which reagent gives as major product from phenol

• A. \( \text{Br}_2 + \text{CS}_2, \text{ at } 273 \, \text{K} \)
• B. \( \text{Br}_2 + \text{heat} \)
• C. Bromine water
• D. \( \text{Br}_2 + \text{CCl}_4 \text{ at } 273 \, \text{K} \)
• E. \( \text{Br}_2 + \text{acetone at } 273 \, \text{K} \)

Hint:

When performing bromination on phenol, using a non-polar solvent like carbon disulfide (CS2) at lower temperatures leads to the substitution primarily at the para position, giving the major product.

Answer 1

The Correct Option is A

Step 1: Understanding the reaction.
When phenol reacts with bromine, substitution occurs at the ortho and para positions of the benzene ring due to the activating effect of the hydroxyl group. The reagent that gives the major product with phenol is typically bromine in a non-polar solvent, which prevents multiple substitutions and allows for substitution at the most reactive sites.

Step 2: Analysis of the options.
- (A) \( \text{Br}_2 + \text{CS}_2 \) at 273 K: Correct. In the presence of carbon disulfide (\( \text{CS}_2 \)) and bromine at 273 K, the reaction leads to a major substitution at the para position, giving the para-bromophenol as the major product. This reaction is common for phenol bromination.
- (B) \( \text{Br}_2 + \text{heat} \): Incorrect. Heating bromine with phenol leads to a more complicated reaction with multiple substitutions, resulting in the formation of a mixture of products rather than a major one.
- (C) Bromine water: Incorrect. Bromine water reacts with phenol, but it leads to a series of products, including polybrominated phenols, with no clear major product.
- (D) \( \text{Br}_2 + \text{CCl}_4 \) at 273 K: Incorrect. Although CCl4 is a non-polar solvent, it does not work as efficiently in this reaction as CS2, resulting in a mixture of products.
- (E) \( \text{Br}_2 + \text{acetone at 273 K} \): Incorrect. Acetone is polar, and the reaction does not efficiently lead to a major product.

Step 3: Conclusion.
The correct reagent to give the major product of bromophenol is \( \text{Br}_2 \) in \( \text{CS}_2 \) at 273 K.
\[ \boxed{\text{Br}_2 + \text{CS}_2 \text{ at } 273 \, \text{K}} \]
Final Answer: \( \text{Br}_2 + \text{CS}_2 \text{ at } 273 \, \text{K} \)