Question

What happens when phenol is treated with the following?
(i) Br2 water
(ii) Conc. HNO3

Answer 1

Electrophilic Substitution Reactions of Phenol

(i) Reaction with Bromine Water:

Phenol reacts with bromine water (\(\text{Br}_2\) in \(\text{H}_2\text{O}\)) via electrophilic substitution to produce 2,4,6-Tribromophenol, a white precipitate.

\[ \text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr} \]

The hydroxyl group (\(-\text{OH}\)) activates the benzene ring at the ortho and para positions, making it highly reactive to electrophilic attack.

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(ii) Reaction with Concentrated Nitric Acid:

Phenol undergoes nitration when treated with concentrated nitric acid (\(\text{HNO}_3\)) to form 2,4,6-Trinitrophenol, also known as picric acid.

\[ \text{C}_6\text{H}_5\text{OH} + 3\text{HNO}_3 \rightarrow \text{C}_6\text{H}_2(\text{NO}_2)_3\text{OH} + 3\text{H}_2\text{O} \]

The \(-\text{OH}\) group's activating effect facilitates the introduction of nitro groups at the ortho and para positions.

Answer 2

Alcohol-Carbocation Interaction

The reaction of an alcohol with a carbocation, exemplified by \({CH3C^+}\), demonstrates a classic instance of a nucleophilic assault.

Process:

1. The alcohol's oxygen atom, possessing a lone pair of electrons, functions as a nucleophile.
2. The alcohol targets the electrophilic carbocation \({CH3C^+}\) via its lone pair.
3. A new bond forms between the oxygen and carbon atoms.
4. The resultant product is an ether with the structure \({R-O-CH3}\).

Overall Transformation:

\[ {R-OH + CH3C^+ -> R-O-CH3} \]

This reaction is fundamental in forming ethers, particularly in acidic or carbocation-generating environments.

Answer 3

Phenols are resistant to C–OH bond cleavage due to the resonance stabilization of the phenoxide ion. When phenol loses a proton, the resulting negative charge on the oxygen atom delocalizes into the aromatic ring. This delocalization strengthens the C–OH bond, making it less prone to breaking under typical conditions and thus preventing C–OH bond cleavage in phenols.

Answer 4

The Lucas Test, employing Lucas reagent (conc. \({HCl}\) + anhydrous \({ZnCl_2}\)), differentiates alcohols by their reactivity.

• 2-Methylpropan-2-ol, a tertiary alcohol, reacts instantly at room temperature with Lucas reagent. This forms an insoluble alkyl chloride, resulting in a cloudy solution: \[ {(CH3)3COH + HCl -> (CH3)3CCl + H2O} \]
• Butan-1-ol, a primary alcohol, exhibits minimal or no reaction at room temperature, with no immediate cloudiness observed.

Conclusion:
The immediate turbidity produced by 2-Methylpropan-2-ol with Lucas reagent, contrasted with the absence of such a reaction in Butan-1-ol, facilitates their clear distinction.