10.2 Preparing Alkyl Halides from Alkanes: Radical Halogenation

Radical Halogenation of Alkanes

Radical halogenation is one of the few ways to directly functionalize an unreactive alkane, replacing a C–H bond with a C–X bond (where X = Cl or Br). Because alkanes lack functional groups, this radical chain reaction is a key entry point for converting them into more useful alkyl halides. The challenge is that alkanes have many C–H bonds, so the reaction often produces a mixture of products. Predicting and controlling that mixture requires understanding radical stability, halogen selectivity, and the mechanism itself.

The Mechanism: Initiation, Propagation, and Termination

Radical halogenation follows a three-stage chain mechanism.

1. Initiation

Heat ($\Delta$) or UV light ($h\nu$) causes homolytic cleavage of the diatomic halogen molecule ($X_2$), splitting the bond evenly to generate two halogen radicals:

$X_2 \xrightarrow{h\nu \text{ or } \Delta} 2 \; X\cdot$

Each atom takes one electron from the shared pair, producing two reactive species with unpaired electrons.

2. Propagation (two repeating steps)

These two steps form a self-sustaining cycle that produces the alkyl halide product:

1. Hydrogen abstraction: A halogen radical ($X\cdot$) pulls a hydrogen atom off the alkane, breaking a C–H bond and forming HX plus an alkyl radical ($R\cdot$).
2. Halogen abstraction: The alkyl radical ($R\cdot$) collides with another $X_2$ molecule, breaking the X–X bond to form the alkyl halide product ($R{-}X$) and regenerating a halogen radical ($X\cdot$).

The regenerated $X\cdot$ feeds back into step 1, so a single initiation event can cycle through propagation thousands of times before the chain breaks.

3. Termination

Any step that destroys radicals without generating new ones ends the chain:

• Combination: Two radicals collide and form a covalent bond (e.g., $R\cdot + X\cdot \rightarrow R{-}X$, or $R\cdot + R\cdot \rightarrow R{-}R$).
• Disproportionation: One radical abstracts a hydrogen from the beta-carbon of another radical, producing an alkane and an alkene (e.g., two propyl radicals yield propane and propene).

Termination products are usually minor because radical concentrations are low at any given moment.

Why Radical Halogenation Gives Product Mixtures

Several factors lead to mixtures rather than a single clean product:

• Multiple hydrogen types. Most alkanes contain 1°, 2°, and 3° C–H bonds. Each type can be abstracted, generating a different alkyl radical and ultimately a different alkyl halide isomer. For example, butane can yield both 1-chlorobutane (from a 1° radical) and 2-chlorobutane (from a 2° radical).
• Multiple propagation cycles. Because the chain keeps running, polyhalogenation can occur: a monohalogenated product re-enters the cycle and picks up a second halogen.
• Termination side products. Combination of two alkyl radicals produces higher-molecular-weight alkanes (e.g., two propyl radicals combine to form hexane). Disproportionation gives alkane/alkene pairs.

Reactivity of Hydrogens: The Selectivity Factor

Not all C–H bonds are equally easy to break. The reactivity order for hydrogen abstraction is:

3° > 2° > 1° > methyl

This trend exists because the resulting radical's stability follows the same order. A tertiary radical is stabilized by hyperconjugation with three adjacent alkyl groups, lowering the activation energy for its formation. A primary radical has only one alkyl group providing hyperconjugation, making it less stable and harder to form.

The practical result: in a molecule like 2-methylbutane, the lone 3° hydrogen is abstracted preferentially over the nine 1° hydrogens, even though 1° hydrogens vastly outnumber it.

Chlorination vs. Bromination: Selectivity Matters

This is one of the most important distinctions in this topic. Chlorine and bromine both undergo radical halogenation, but they differ dramatically in selectivity.

Why the difference? It comes down to the Hammond postulate and the energy of the transition state:

• Chlorination has an early, reactant-like transition state. The radical's stability doesn't matter much yet, so chlorine is relatively indiscriminate.
• Bromination has a late, product-like transition state. The stability of the forming radical heavily influences the activation energy, so bromine is far more selective for the most stable (most substituted) radical.

Predicting Products: A Step-by-Step Approach

When asked to predict the major product of a radical halogenation:

1. Identify all distinct types of hydrogen in the alkane (1°, 2°, 3°). Count how many of each type exist.
2. Determine the halogen. If it's $Br_2$, selectivity dominates and the most substituted position wins. If it's $Cl_2$, both selectivity and statistical probability (number of each H type) matter.
3. For chlorination, calculate relative amounts. Multiply the number of each hydrogen type by its relative reactivity factor (1 for 1°, 4 for 2°, 5 for 3°). The ratio of these products gives you the approximate product distribution.
4. For bromination, the most substituted hydrogen almost always wins, regardless of how few of that type are present. The selectivity factor is so large that statistical corrections barely matter.
5. Draw the product by replacing the abstracted H with X, and check for stereochemistry if relevant (radical halogenation at a stereocenter gives a racemic mixture, since the planar radical can be attacked from either face).

Key Factors Affecting Radical Halogenation (Summary)

• Bond dissociation energy (BDE): Weaker C–H bonds (3° < 2° < 1°) are easier to break, favoring abstraction at more substituted carbons.
• Radical stability: More substituted radicals are more stable (hyperconjugation), lowering the activation energy for their formation.
• Halogen identity: Bromine is far more selective than chlorine. Fluorine is too reactive and unselective (often dangerous). Iodine is too unreactive (the reaction is endothermic overall and doesn't proceed).
• Statistical factor: The number of equivalent hydrogens of each type matters, especially for chlorination where selectivity differences are small.