10.3 Preparing Alkyl Halides from Alkenes: Allylic Bromination

Allylic Bromination

Process of Allylic Bromination

Allylic bromination selectively places a bromine atom at the allylic position, the carbon directly adjacent to a double bond. The alkene itself stays intact, which makes this reaction distinct from addition reactions that consume the double bond.

The reagent combo is N-bromosuccinimide (NBS) plus a radical initiator (benzoyl peroxide or UV light). A typical setup looks like this:

• Alkene substrate + NBS (1 equiv.) dissolved in $\text{CCl}_4$
• Catalytic benzoyl peroxide, or irradiation with light
• Heat to reflux (~77 °C for $\text{CCl}_4$)

NBS serves a specific purpose here: it maintains a very low, steady concentration of $\text{Br}_2$ in solution. That low concentration favors the substitution pathway (allylic bromination) over electrophilic addition across the double bond.

Mechanism (radical chain):

1. Initiation: Heat or light causes homolytic cleavage, generating a bromine radical ($\text{Br}\cdot$).
2. Propagation, step 1 (H-abstraction): $\text{Br}\cdot$ selectively abstracts an allylic hydrogen from the substrate, forming a resonance-stabilized allylic radical and HBr.
3. Propagation, step 2 (Br transfer): The allylic radical reacts with $\text{Br}_2$ (produced in small amounts when HBr reacts with NBS) to give the allylic bromide product and regenerate $\text{Br}\cdot$, continuing the chain.
4. Termination: Two radicals combine, ending the chain.

The role of NBS in step 3 is worth emphasizing: NBS reacts with the HBr byproduct to regenerate a small amount of $\text{Br}_2$, keeping its concentration low throughout the reaction.

Selectivity in Allylic Bromination

Bromination occurs at the allylic position because the allylic radical is far more stable than alternative radical intermediates. This stability traces back to two factors: bond dissociation energies and resonance.

Bond dissociation energies (BDEs) of C–H bonds:

A lower BDE means less energy is needed to break that bond. The allylic C–H bond is significantly weaker, so $\text{Br}\cdot$ preferentially abstracts the allylic hydrogen.

Why is the allylic radical so stable? The unpaired electron is delocalized across the $\pi$ system through resonance. You can draw two resonance structures for the allylic radical, spreading the radical character over two carbons. Radicals at alkyl or vinylic positions can't delocalize this way, so they're higher in energy and don't form competitively.

One consequence of this resonance: if the two resonance contributors place the radical on non-equivalent carbons, you can get a mixture of constitutional isomers (bromine ends up on either end of the allylic system). Always check whether the two ends of the allylic radical are the same or different.

Allylic Bromination vs. Other Alkyl Halide Preparations

These reactions all make alkyl halides from alkenes, but they work through completely different mechanisms and give different products.

• Electrophilic addition of HX ($\text{X} = \text{Cl, Br, I}$): Adds H and X across the double bond, consuming it. Follows Markovnikov's rule (halogen goes to the more substituted carbon). The product is a simple alkyl halide with no remaining double bond.
• Bromination with $\text{Br}_2$: Adds two bromines across the double bond via a bromonium ion intermediate, giving a 1,2-dibromide with anti stereochemistry (bromines on opposite faces). Again, the double bond is consumed.
• Allylic bromination (NBS): Replaces an allylic C–H bond with a C–Br bond. The double bond is preserved. This is a radical substitution, not an ionic addition.

Mechanistic Considerations and Product Formation

Because allylic bromination is a free radical process, its behavior differs from ionic reactions in several ways:

• Regioselectivity is governed by C–H bond strengths and radical stability, not by carbocation stability or Markovnikov's rule.
• The allylic radical is planar ($sp^2$-hybridized at the radical carbon). Bromine can attack from either face, so if a stereocenter forms at the allylic position, expect a racemic mixture.
• If the allylic system is unsymmetrical (the two resonance forms put the radical on different carbons), you'll get a mixture of regioisomeric products. The ratio depends on the substitution and stability at each end of the allylic system.