Radical Halogenation of Alkanes

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This page is a draft outline. It is visible to admins and still needs a full lesson.

Teach chlorination and bromination of alkanes, and why the two halogens do not give the same product mixture.

The chain mechanism for RH + X₂ with light or heat.
Selectivity from radical stability: 3° > 2° > 1° > methyl.
Chlorination is fast and unselective. Bromination is slower and highly selective for the weakest C–H bond.
Predicting the major monohalogenation product, and when a mixture is expected.
Practice: the same alkane with Cl₂/hv versus Br₂/hv.

Radical Halogenation of Alkanes

Alkanes are generally unreactive due to strong, nonpolar $\text{C}-\text{C}$ and $\text{C}-\text{H}$ bonds. However, in the presence of heat or light ($h\nu$), halogens ($\text{Cl}_2$ or $\text{Br}_2$) can react with alkanes to replace a hydrogen atom with a halogen, yielding an alkyl halide. While both chlorination and bromination share the same fundamental mechanism, their reaction profiles and product distributions differ drastically.

[ Insert Image: Overview of Alkane Halogenation ]

Figure 1. General radical substitution of an alkane to form an alkyl halide.

1. The Radical Chain Mechanism

Halogenation proceeds via a standard three-stage radical chain mechanism:

Initiation: Ultraviolet light or heat homolytically cleaves the weak halogen-halogen ($\text{X}-\text{X}$) bond, generating two reactive halogen radicals ($\text{X}^\bullet$).
Propagation (Two Repeating Steps):
- Step A (Hydrogen Abstraction): A halogen radical abstracts a hydrogen atom from the alkane ($\text{RH}$), forming a hydrogen halide ($\text{HX}$) and an alkyl radical ($\text{R}^\bullet$).
- Step B (Halogen Transfer): The alkyl radical reacts with a molecule of halogen ($\text{X}_2$), forming the alkyl halide product ($\text{RX}$) and regenerating a halogen radical ($\text{X}^\bullet$) to continue the chain.
Termination: Any combination of two radical species colliding removes them from the cycle and ends the chain.

[ Insert Mechanism: Halogenation Propagation Cycle ]

Figure 2. Propagation steps in radical halogenation.

2. Radical Stability and Selectivity

The rate-determining step in propagation is hydrogen abstraction. The ease with which a hydrogen atom is abstracted depends directly on the stability of the resulting carbon radical.

Radical Stability Order: Carbon radicals are stabilized by hyperconjugation and inductive electron donation from adjacent alkyl groups. Stability follows the order:
Tertiary ($\text{3}^\circ$) > Secondary ($\text{2}^\circ$) > Primary ($\text{1}^\circ$) > Methyl radical.
Consequently, $\text{3}^\circ \text{C}-\text{H}$ bonds are weaker and react faster than $\text{2}^\circ \text{C}-\text{H}$ bonds, which in turn react faster than $\text{1}^\circ \text{C}-\text{H}$ bonds.

[ Insert Image: Carbon Radical Stability Hierarchy ]

Figure 3. Stability trends of alkyl radicals.

3. Why Chlorination and Bromination Give Different Product Mixtures

Although both halogens prefer to substitute at more substituted carbons, chlorination and bromination differ vastly in their selectivity due to thermodynamics (explained by the Hammond postulate):

Chlorination ($\text{Cl}_2$)

Exothermic & Fast: The hydrogen abstraction step is strongly exothermic.
Early Transition State: Because the transition state occurs early (resembling the reactants), there is very little radical character developed yet.
Unselective: Energy differences between primary, secondary, and tertiary pathways are small. Chlorination reacts nearly indiscriminately, yielding complex statistical mixtures of isomers.

Bromination ($\text{Br}_2$)

Endothermic & Slow: The hydrogen abstraction step is endothermic.
Late Transition State: The transition state closely resembles the carbon radical intermediate.
Highly Selective: Radical stability dictates the activation energy. Bromine is extremely selective, almost exclusively attacking the weakest $\text{C}-\text{H}$ bond to form the most stable possible radical product.

[ Insert Graph: Hammond Postulate Comparison for Chlorination vs. Bromination ]

Figure 4. Energy profile and transition state differences driving chlorination versus bromination selectivity.

4. Predicting Products and Managing Isomeric Mixtures

When predicting monohalogenation products for an alkane:

Count all distinct types of hydrogen atoms ($\text{1}^\circ$, $\text{2}^\circ$, $\text{3}^\circ$) in the carbon skeleton.
With $\text{Cl}_2 / h\nu$: Expect a mixture of all possible constitutional isomers, weighted by both reactivity factors and statistical probability (number of hydrogens of each type).
With $\text{Br}_2 / h\nu$: Predict a single major product corresponding to substitution at the most substituted carbon (tertiary over secondary, secondary over primary) with high regiochemical purity.

5. Practice Comparison: 2-Methylbutane

Consider subjecting 2-methylbutane to radical halogenation:

Comparison of Reagents

With $\text{Cl}_2$ and light ($h\nu$): Yields a complex mixture of all four possible monochlorinated isomers (1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, and 1-chloro-3-methylbutane) because chlorine is too reactive to care greatly about stability differences.
With $\text{Br}_2$ and light ($h\nu$): Yields 2-bromo-2-methylbutane as the overwhelming major product (>90% yield) because bromine selectively abstracts the single tertiary hydrogen to form the stable tertiary radical.

[ Insert Reaction: Chlorination vs. Bromination of 2-Methylbutane ]

Figure 5. Divergent product distributions in the halogenation of 2-methylbutane.