Q.Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields ____________.

Concept: Free-radical benzylic halogenation, followed by nucleophilic substitution.

Reasoning:

1. Sunlight promotes free-radical substitution at the benzylic position (side-chain), not electrophilic substitution on the ring. Toluene gives benzyl chloride (C6H5CH2Cl).
2. Hydrolysis with aqueous NaOH replaces the chlorine with an -OH group: C6H5CH2Cl + NaOH -> C6H5CH2OH + NaCl.
3. The product is benzyl alcohol (phenylmethanol).

The key here is recognising that sunlight + chlorine triggers a free-radical mechanism, not electrophilic aromatic substitution. Toluene's methyl group has benzylic C-H bonds that are unusually weak, because the resulting radical is resonance-stabilised by the aromatic ring, so chlorine radicals preferentially abstract a benzylic hydrogen rather than substituting on the ring.

1. Initiation: Cl2, with sunlight (hv), splits homolytically into 2 Cl radicals.
2. Propagation (H-abstraction): C6H5CH3 + Cl(radical) -> C6H5CH2(radical) + HCl -- the benzylic radical is resonance-stabilised by the ring.
3. Propagation (chain continuation): C6H5CH2(radical) + Cl2 -> C6H5CH2Cl + Cl(radical) -- giving benzyl chloride as the major product.

4. Hydrolysis: Benzyl chloride, a benzylic halide, reacts with aqueous NaOH by nucleophilic substitution: C6H5CH2Cl + NaOH (aq) -> C6H5CH2OH + NaCl -- giving benzyl alcohol.

Checking the options: (i) o-cresol and (ii) m-cresol are ring-hydroxylated products, which would require electrophilic substitution on the ring, not the free-radical path; (iii) 2,4-dihydroxytoluene is even further from this mechanism; (iv) benzyl alcohol is exactly what forms.

Concept: Free Radical Halogenation & Nucleophilic Substitution

This problem tests two sequential reactions:

1. Free radical chlorination (in sunlight) — occurs at the benzylic position due to high stability of the benzylic radical.
2. Nucleophilic substitution (hydrolysis with aq. NaOH) — replaces Cl with OH.

Method: Reaction Sequence Analysis

Step 1: Identify the reactive site for chlorination

• Toluene has a methyl group attached to benzene.
• In sunlight, chlorination follows a free radical mechanism.
• The benzylic C–H bond is weakest because the resulting radical is resonance-stabilized by the benzene ring.
• Result: Chlorine substitutes at the benzylic carbon, not on the ring.

Step 2: Hydrolysis with aq. NaOH

• The product from Step 1 is benzyl chloride ($\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$).
• Aqueous NaOH causes nucleophilic substitution (SN1 or SN2, depending on conditions).
• The Cl is replaced by an –OH group.

Step 3: Identify the final product

• The product is benzyl alcohol ($\text{C}_6\text{H}_5\text{CH}_2\text{OH}$).
• It is not a cresol (which would have –OH on the ring).

Final Answer

Key takeaway: Sunlight directs chlorination to the benzylic position, not the aromatic ring. Hydrolysis then gives the corresponding alcohol.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing the Reaction Conditions (Sunlight vs. Catalyst)

The mistake: Students see "toluene + chlorine" and immediately think of electrophilic aromatic substitution (using $FeCl_3$ or $AlCl_3$ catalyst), which would give ortho/para chlorotoluene. They then hydrolyse that to get cresols (options A, B, C).

Why it's wrong: The condition is sunlight — this triggers a free radical substitution at the benzylic position (side chain), not on the ring.

How to avoid: Always check the reaction conditions first:

• Sunlight / UV / heat → free radical substitution (side chain)
• Lewis acid catalyst ($FeCl_3$, $AlCl_3$) → electrophilic substitution (ring)

Mistake 2: Forgetting the Benzylic Radical Stability

The mistake: Students think chlorine could attack any C–H bond randomly.

Why it's wrong: Free radical chlorination is highly selective for the benzylic position because the benzylic radical is resonance-stabilised by the aromatic ring.

How to avoid: Remember the stability order of radicals:

So in toluene, the methyl group is the only benzylic site — that's where Cl attacks.

Mistake 3: Misinterpreting "Hydrolysis with aq. NaOH"

The mistake: Students think hydrolysis of a chlorinated ring compound gives a phenol (cresol).

Why it's wrong: The product after chlorination is benzyl chloride ($C_6H_5CH_2Cl$), not a ring-chlorinated product. Hydrolysis of benzyl chloride with aqueous NaOH gives benzyl alcohol via $S_N2$ substitution.

Reaction sequence:

How to avoid: Track the carbon where the chlorine is attached:

• If Cl is on the ring → hydrolysis gives phenol/cresol
• If Cl is on the side chain → hydrolysis gives alcohol

Mistake 4: Not Recognising the Final Product's Functional Group

The mistake: Students pick o-cresol or m-cresol without checking if the product is actually an alcohol or a phenol.

Why it's wrong: Benzyl alcohol ($C_6H_5CH_2OH$) is a primary alcohol, not a phenol. The –OH is on the side chain, not directly on the ring.

How to avoid: Identify the functional group:

• Phenol: –OH directly attached to benzene ring
• Alcohol: –OH attached to an alkyl (side chain) carbon

Here, the –OH is on the –CH$_2$– group → benzyl alcohol.

Final Answer

The correct product is benzyl alcohol → option (D).

Key takeaway: Sunlight + Cl$_2$ on toluene = side chain chlorination → hydrolysis gives benzyl alcohol, not cresols.