Q.Give equations of the following reactions:

The key idea here is electrophilic substitution in phenol — the -OH group activates the ring strongly, directing incoming groups to the ortho and para positions. For oxidation, the alcohol converts to a carboxylic acid.

(i) Propan-1-ol is a primary alcohol. Alkaline $KMnO_4$ oxidises it to propanoic acid.

(ii) Bromine in $CS_2$ (a non-polar solvent) gives mono-bromination at the para position (major product).

(iii) Dilute $HNO_3$ nitrates phenol to a mixture of ortho and para nitrophenols.

$C_6H_5OH + HNO_3 (\text{dil.}) \rightarrow \text{o-Nitrophenol} + \text{p-Nitrophenol} + H_2O$ …

Let’s go through each reaction one by one, focusing on why the product forms the way it does.

1. Oxidation of propan-1-ol with alkaline KMnO₄

This is not a phenol reaction — it’s a primary alcohol oxidation. Alkaline KMnO₄ is a strong oxidising agent. For a primary alcohol, the first oxidation gives an aldehyde, but under these conditions the aldehyde is further oxidised to a carboxylic acid.

Propan-1-ol: $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$

The reaction:

Equation:

2. Bromine in CS₂ with phenol

Phenol undergoes electrophilic substitution. The –OH group is strongly activating and ortho/para-directing. In a non-polar solvent like CS₂, the reaction is controlled — you get monobromination at the para position (because the para position is less sterically hindered than ortho).

The product is 4-bromophenol (p-bromophenol).

3. Dilute HNO₃ with phenol

Again, phenol’s high reactivity means even dilute nitric acid (at room temperature or slightly warm) gives nitration. But dilute HNO₃ is not as strongly nitrating as the concentrated acid mixture (HNO₃ + H₂SO₄). With dilute HNO₃, you get a mixture of ortho- and para-nitrophenol.

The ortho product is steam-volatile (intramolecular H-bonding), while the para product is not — this is used to separate them.

$\ce{C6H5OH + HNO3 (dil) ->[\text{room temp}] o-NO2-C6H4OH + p-NO2-C6H4OH + H2O}$ …

Concept: Acidity of Phenol

Method Name: Resonance Stabilisation & Electron-Withdrawing Effect Analysis

Steps:

1. Draw the conjugate base — Remove the phenolic H⁺ to form the phenoxide ion ($C_6H_5O^-$).
2. Analyse resonance — Show that the negative charge on oxygen is delocalised into the benzene ring (ortho and para positions). This stabilises the phenoxide ion.
3. Compare with alcohol — In alcohols (e.g., ethanol), the alkoxide ion ($RO^-$) has no such resonance — charge is localised on oxygen, making it less stable.
4. Conclusion — Greater stability of phenoxide ion means phenol loses H⁺ more easily → phenol is more acidic than alcohols.

Key result: Phenol ($pK_a \approx 10$) is $10^6$ times more acidic than ethanol ($pK_a \approx 16$).

Reactions of Phenol

(i) Oxidation of propan-1-ol with alkaline $KMnO_4$

Reaction type: Oxidation of primary alcohol to carboxylic acid

Product: Propanoic acid

(ii) Bromine in $CS_2$ with phenol

Reaction type: Electrophilic substitution (monobromination at low temperature)

Product: Mixture of ortho- and para-bromophenol

Note: In aqueous medium, phenol gives 2,4,6-tribromophenol (white precipitate).

(iii) Dilute $HNO_3$ with phenol …

Here are the common mistakes students make with these specific reactions, along with the correct equations and strategies to avoid errors.

General Mistake: Confusing Reagent Strength & Conditions

Students often treat all oxidizing agents the same or forget that alkaline $KMnO_4$ is a strong oxidant (cleaves the chain), while acidic $K_2Cr_2O_7$ is milder.

How to avoid: Memorize the "Oxidation Ladder":

• Primary alcohol $\xrightarrow{\text{alk. KMnO}_4}$ Carboxylic acid (not aldehyde).
• Primary alcohol $\xrightarrow{\text{Cu/573K}}$ Aldehyde.

(i) Oxidation of propan-1-ol with alkaline $KMnO_4$

Common Mistake: Writing the product as propanal ($CH_3CH_2CHO$) or propanoic acid with the wrong carbon count.

Why it's wrong: Alkaline $KMnO_4$ is a strong oxidizing agent. It does not stop at the aldehyde stage. It cleaves the C–C bond next to the –OH group, giving a carboxylic acid with one less carbon (plus $CO_2$).

Correct Equation:

(Propan-1-ol → Propanoic acid + Carbon dioxide)

How to avoid: Remember: Strong oxidant + Primary alcohol = Acid with one less carbon (due to decarboxylation of intermediate).

(ii) Bromine in $CS_2$ with phenol

Common Mistake: Writing the product as 2,4,6-tribromophenol (the usual aqueous bromine product).

Why it's wrong: $CS_2$ is a non-polar, non-aqueous solvent. In this medium, bromination is mono-substitution (not tri-substitution). The –OH group directs to the ortho and para positions, but the major product is para-bromophenol (due to steric hindrance in $CS_2$).

Correct Equation:

How to avoid: Note the solvent. Aqueous $\text{Br}_2$ → tribromo. Non-aqueous ($CS_2$, $CCl_4$) → mono-bromo (para major).

(iii) Dilute $HNO_3$ with phenol

Common Mistake: Writing the product as 2,4,6-trinitrophenol (picric acid).

Why it's wrong: Dilute $HNO_3$ gives mono-nitration (mainly ortho and para). Picric acid requires concentrated $HNO_3$ + $H_2SO_4$ (nitrating mixture).

Correct Equation:

How to avoid: Remember: Dilute → mono-nitro. Conc. + $H_2SO_4$ → tri-nitro (picric acid).

(iv) Treating phenol with chloroform in presence of aqueous NaOH

Common Mistake: Writing the product as salicylaldehyde (correct) but forgetting the Reimer-Tiemann mechanism or writing the wrong byproduct. …