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:

When phenol is treated with bromine water (\(\text{Br}_2\) in \(\text{H}_2\text{O}\)), it undergoes electrophilic substitution to form:

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

The product is 2,4,6-Tribromophenol, a white precipitate.

The hydroxyl group (\(-\text{OH}\)) on phenol is a strong electron-donating group that activates the benzene ring, especially at the ortho and para positions, making it highly susceptible to electrophilic attack.

---

(ii) Reaction with Concentrated Nitric Acid:

When phenol is treated with concentrated nitric acid (\(\text{HNO}_3\)), it undergoes nitration to yield:

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

The product is 2,4,6-Trinitrophenol, commonly known as picric acid.

Again, the \(-\text{OH}\) group activates the ring toward electrophilic substitution, facilitating the introduction of nitro groups at the ortho and para positions.

Answer 2

Reaction Between an Alcohol and a Carbocation

The reaction between an alcohol and a carbocation, such as \({CH3C^+}\), is a typical example of a nucleophilic attack.

Mechanism:

1. The oxygen atom in the alcohol has a lone pair of electrons, making it a nucleophile.
2. The alcohol attacks the electrophilic carbocation \({CH3C^+}\) using its lone pair.
3. This results in the formation of a new bond between the oxygen and carbon atoms.
4. The final product is an ether of the form: \({R-O-CH3}\)

General Reaction:

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

This type of reaction is a key step in ether synthesis under acidic or carbocation-generating conditions.

Answer 3

Phenols do not undergo reactions involving the cleavage of the C – OH bond because the phenoxide ion (formed when phenol loses a proton) is stabilized by resonance. The negative charge on the oxygen atom can be delocalized into the aromatic ring, making the C – OH bond stronger and less likely to break under normal conditions. This stabilization by resonance prevents the cleavage of the C – OH bond in phenols.

Answer 4

The test used is the Lucas Test, which distinguishes alcohols based on their reactivity with Lucas reagent (conc. \({HCl}\) + anhydrous \({ZnCl_2}\)).

• 2-Methylpropan-2-ol is a tertiary alcohol. It reacts immediately at room temperature with Lucas reagent to form a cloudy solution due to the formation of an insoluble alkyl chloride: \[ {(CH3)3COH + HCl -> (CH3)3CCl + H2O} \]
• Butan-1-ol is a primary alcohol. It reacts very slowly or may not react at room temperature. No cloudiness is observed immediately.

Conclusion:
2-Methylpropan-2-ol gives an immediate turbidity with Lucas reagent, while Butan-1-ol does not, allowing easy distinction between the two.