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Radical Halogenation
Selectivity & Mechanism

Master the three-step chain mechanism of free-radical substitution and learn why bromine is more selective than chlorine.

Organic Chemistry 1

What Are Free Radicals?

A free radical is an atom or molecule with an unpaired electron. Radicals are highly reactive because they seek another electron to complete their octet.

\(\text{Cl}\)
\(\text{Cl}\)
→
\(h\nu\) / heat
\(\text{Cl}\cdot\)
\(\text{Cl}\cdot\)
Bond Two radicals

Why Radical Halogenation Matters

Radical halogenation is one of the first substitution reactions you encounter in organic chemistry. It converts a C-H bond into a C-X bond (X = Cl or Br), creating a useful functional handle for further synthesis.

Functionalization
Converts unreactive alkanes into reactive alkyl halides
Selectivity
Different halogens prefer different hydrogen types
Industry
Used to manufacture solvents, polymers, and pharmaceuticals

Homolysis vs Heterolysis

Bond breaking can occur in two ways. Understanding this distinction is essential before studying radical mechanisms.

Homolytic Cleavage
Each atom keeps one electron. Produces two radicals. Uses fishhook arrows. Favored by heat or UV light.
Heterolytic Cleavage
One atom keeps both electrons. Produces ions (cation + anion). Uses full curved arrows. Favored by polar solvents.

The Chain Mechanism: Three Steps

1
Initiation: The halogen molecule undergoes homolysis (via UV or heat) to form two halogen radicals: \( \text{X}_2 \xrightarrow{h\nu} 2\,\text{X}\!\cdot \)
2
Propagation: Two repeating steps that carry the chain forward. A halogen radical abstracts an H from the alkane, forming HX and a carbon radical. The carbon radical then reacts with \( \text{X}_2 \) to form the product and regenerate \( \text{X}\!\cdot \).
3
Termination: Two radicals combine, destroying reactive intermediates. Examples: \( \text{X}\!\cdot + \text{X}\!\cdot \to \text{X}_2 \), or \( \text{R}\!\cdot + \text{X}\!\cdot \to \text{R-X} \).

Propagation Cycle

Step A
\( \text{X}\cdot + \text{R-H} \to \text{HX} + \text{R}\cdot \)
Step B
\( \text{R}\cdot + \text{X}_2 \to \text{R-X} + \text{X}\cdot \)
\(\text{R}\cdot\) carries chain
\(\text{X}\cdot\) regenerated
Input:
\(\text{R-H} + \text{X}_2\)
Output:
\(\text{R-X} + \text{HX}\)
Initiation: \( \text{X}_2 \to 2\,\text{X}\cdot \)
(\(h\nu\) or heat starts the chain)
Termination: radicals combine

Bond Dissociation Energy & \(\Delta H\)

Each propagation step has a characteristic \(\Delta H\) determined by bond dissociation energies (BDE). We compute the overall \(\Delta H\) by summing the two propagation steps:

Types of Hydrogen

The carbon a hydrogen is attached to determines its classification. Radical stability increases from primary to tertiary:

1° (Primary)
H on a carbon bonded to one other carbon. Least stable radical.
2° (Secondary)
H on a carbon bonded to two other carbons. Moderate stability.
3° (Tertiary)
H on a carbon bonded to three other carbons. Most stable radical.

Selectivity: Cl vs Br

Not all halogens are equally picky. Bromine strongly prefers the most substituted hydrogen, while chlorine is far less selective.

Chlorination Bromination
Relative reactivity: 3° 5.2 1640
Relative reactivity: 2° 3.9 82
Relative reactivity: 1° 1.0 1.0
Selectivity Low Very high
Rate Fast Slow

Why Is Bromine More Selective?

Energy Reaction Coordinate
\(\text{R-H} + \text{X}\cdot\)
\(\text{R}\cdot + \text{HCl}\)
\(\text{R}\cdot + \text{HBr}\)
\(\text{Cl}\cdot\) TS (early)
small \(E_a\)
\(\text{Br}\cdot\) TS (late)
large \(E_a\)

How to Predict Products

1
Identify all unique H types: Classify each set of equivalent hydrogens as 1°, 2°, or 3°.
2
Count the hydrogens in each category (statistical factor).
3
Multiply the number of H's by the relative reactivity for the halogen (Cl or Br).
4
Normalize to get the percentage of each product. Sum all weighted values, then divide each by the total.
5
Identify the major product. For Br, it is almost always the most substituted position.

Example: Chlorination of Propane

\(\text{CH}_3\)
\(\text{CH}_2\)
\(\text{CH}_3\)
6 × 1° H 2 × 2° H (same as left) →
\(\text{Cl}_2\), \(h\nu\)
1-chloropropane 2-chloropropane

Example: Bromination of Propane

Example: Bromination of Isobutane

\(\text{CH}_3\)
\(\text{CH}_3\)
\(\text{CH}\)
\(\text{CH}_3\)
9 × 1° H 1 × 3° H →
\(\text{Br}_2\), \(h\nu\)
1-bromo-2-methyl propane (1°) 2-bromo-2-methyl propane (3°)

Common Mistakes

✕ Ignoring Statistics
Only using reactivity ratios without multiplying by the number of equivalent H's. Both factors matter!
✕ Using Br Ratios for Cl
Mixing up selectivity values. Chlorine's 3° ratio is 5.2, not 1640. Always check which halogen is used.
✕ Confusing Steps
Writing ionic arrows in a radical mechanism. Use single-barbed fishhook arrows for one-electron movement.

Summary: Halogenation at a Glance

Property Chlorination Bromination
Propagation Step A \(\Delta H\) Slightly Endothermic Endothermic
Transition state resembles Reactant (early) Product (late)
Selectivity Low High
3° : 2° : 1° ratio 5.2 : 3.9 : 1 1640 : 82 : 1
Governs selectivity via Statistics (# of H) Radical stability
Best for synthesis? Product mixtures Single major product

Question 1 of 5

In radical halogenation, what does the initiation step produce?

Question 2 of 5

Why is bromine more selective than chlorine in radical halogenation?

Question 3 of 5

Propane has 6 primary H's and 2 secondary H's. Using chlorine's relative reactivities (1° = 1.0, 2° = 3.9), what is the approximate percentage of 2-chloropropane?

Question 4 of 5

In radical halogenation of isobutane with \(\text{Br}_2\), the major product is 2-bromo-2-methylpropane even though there is only 1 tertiary H vs 9 primary H's. Why?

Question 5 of 5

Which of the following is a termination step in radical halogenation?