Free radical halogenation is a substitution reaction of alkanes in which a hydrogen atom is replaced by a halogen (Cl or Br) in the presence of UV light (hv) or heat.

(X = Cl or Br)

Mechanism:

Step 1

Step 2: Propagation (Chain continues)

Step 3:Termination (Radicals combine)

Order of Reactivity of Halogens

$$F_2 > Cl_2 > Br_2 > I_2$$

But in practice:

• Fluorination → Explosive
• Chlorination → Fast, less selective
• Bromination → Slow, more selective
• Iodination → Not feasible (endothermic)

Selectivity Rule (Very Important for JEE)

Relative Reactivity of Hydrogens

$$3^\circ > 2^\circ > 1^\circ$$

Because:
✔ Stability of radical
✔ Hyperconjugation
✔ Inductive effect

Energy Profile Concept

✔ Chlorination → Low activation difference
✔ Bromination → Large activation difference
👉 Bromination gives major substituted product (Hammond postulate)

JEE / NEET Special Points

✔ Reaction requires UV light
✔ Radical inhibitor (like O₂) slows reaction
✔ Tertiary hydrogen is most reactive
✔ Allylic and benzylic positions are highly reactive
✔ Reaction gives mixture of products

Allylic Halogenation (Very Important Exception)

When alkene is treated with halogen under radical conditions → substitution occurs at allylic position, not addition.

Example:
Propene + Br₂ (hv) → Allylic bromide

Best reagent: NBS (N-Bromosuccinimide)
Reason:
✔ Allylic radical is resonance stabilized
✔ More stable than normal 3° radical sometimes

Benzylic Halogenation

Example:
Toluene + Br₂ (hv) → Benzyl bromide

Reason:
✔ Benzyl radical is highly resonance stabilized
✔ Very reactive position

Reactivity order (special case):$$Benzylic > Allylic > 3^\circ > 2^\circ > 1^\circ$$

Energy Profile Concept (JEE Advanced Angle)

• Chlorination → Exothermic
• Bromination → Slightly exothermic
• Iodination → Endothermic

Iodination in Presence of HIO₃ (Why It Works?)

Iodination is carried out in presence of HIO₃ because it oxidizes HI to I₂, removing HI and shifting equilibrium forward, making the reaction feasible.

The Problem (Normal Iodination

$$RH + I_2 \xrightarrow{hv} RI + HI$$

✔ Propagation step is endothermic
✔ Reaction is reversible
✔ HI formed shifts equilibrium backward
👉 So iodination is normally very slow / does not proceed.

---

⚡ Role of HIO₃ (Iodic Acid)

When HIO₃ is added:$$5HI + HIO_3 \rightarrow 3I_2 + 3H_2O$$

What happens?

✔ HI (product) is removed
✔ Equilibrium shifts forward (Le Chatelier principle)
✔ Reaction becomes feasible

Stability of Free Radicals (Including Vinylic Radical) – JEE / NEET

Chlorination – Relative Reactivity of Different Hydrogens (JEE / NEET)

Meaning of the Ratio

• 3° hydrogen reacts about 5 times faster than 1° hydrogen
• 2° hydrogen reacts about 3.8 times faster than 1° hydrogen
• 1° hydrogen ≈ methyl hydrogen (taken as 1)

Why This Order?

Hydrogen abstraction forms a radical:$$R-H + Cl^\bullet \rightarrow R^\bullet + HCl$$

Since radical stability order is:$$3^\circ > 2^\circ > 1^\circ > CH_3^\bullet$$

More stable radical → lower activation energy → faster reaction.

But remember
Chlorination is less selective, so differences are moderate.

Comparison with Bromination (Very Important for JEE)

For bromination:$$3^\circ : 2^\circ : 1^\circ = 1600 : 80 : 1$$

👉 Bromination is highly selective.
👉 Chlorination is fast but less selective.

Numerical 1

Calculate percentage of products formed when propane undergoes chlorination.

Step 1: Identify hydrogens

Propane (CH₃–CH₂–CH₃)

• 6 primary H
• 2 secondary H

Step 2: Multiply by reactivity ratio

Primary:$$6 × 1 = 6$$

Secondary:$$2 × 3.8 = 7.6$$

---

Step 3: Total

$$6 + 7.6 = 13.6$$

Final Answer

✔ 1-chloropropane ≈ 44%
✔ 2-chloropropane ≈ 56%

Order of C–H Bond Dissociation Energy (BDE)

Bond strength decreases as radical stability increases

Question:

Ans: IV > III> 1 > II >V