Subjects and lessons to explore.

Law

Organic Chemistry 1 · Radical Chemistry

Allylic Bromination

8 min read
8 key termsReport an error
Want it in plain words first? Jump to Eli explains — the same idea, no jargon.
On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

Allylic bromination replaces a hydrogen on the carbon next to a C=C double bond (the ) with bromine while the double bond is retained. It uses (N-bromosuccinimide) under radical conditions (heat, light, or a peroxide initiator) to keep a tiny, steady Br₂ concentration, favoring radical substitution over ionic addition across the double bond. Because the intermediate is a resonance-stabilized with two equivalent ends, an unsymmetrical alkene can give a mixture of two allylic bromides ().

Why this matters

Allylic bromides are versatile building blocks: the allylic C–Br bond is highly reactive toward nucleophilic substitution, so allylic bromination is a common early step in synthesizing pharmaceuticals, fragrances, and natural products (many contain allylic groups). Safety boundaries: NBS is a strong oxidizer and irritant, and radical reactions need controlled conditions; no operational, quantity, PPE, or disposal instructions are provided here — all such work must follow approved institutional safety documentation under trained supervision.

The college version

1. The Allylic Position and the Allylic Radical

The allylic position is the carbon directly attached to a C=C double bond (the neighbor carbon, one bond away). An allylic radical forms when a hydrogen is removed from that carbon (CH₂=CH–CH₂•). It is unusually stable because the unpaired electron is delocalized over two carbons by resonance, with two equivalent :

CH2=CH–CH2•  ⟷  •CH2–CH=CH2

The contributors are mirror-related (the radical and double bond swap positions), so they contribute equally — the classic "equivalent contributors, most stable hybrid" case (topic 3). This lowers the allylic C–H bond-dissociation energy (≈ 89 kcal/mol vs ≈ 98 for a typical alkyl C–H), which is why the allylic hydrogen is abstracted.

2. NBS and Avoiding Alkene Addition

Direct bromination of an alkene with Br₂ would give ionic addition across the double bond (a vicinal dibromide, topic 29) — not what we want. NBS supplies bromine in a controlled way: it reacts with the HBr byproduct to regenerate a low, steady Br₂ concentration:

NBS + HBr ⟶ succinimide + Br2

Because [Br₂] stays very low, fast ionic addition (which needs Br₂ to encounter the alkene) is suppressed, while the radical chain — needing only traces of Br• — proceeds at the allylic position. Net result: substitution at the allylic carbon with the double bond retained (an allylic bromide).

3. Radical Conditions and Product Prediction

: NBS in a nonpolar solvent (classically CCl₄) with heat, light (hν), or a radical initiator such as AIBN. The initiator/light produces a trace of Br•, which abstracts the allylic (weakest) hydrogen.

Product prediction hinges on the radical's two ends. After Br₂ delivers bromine to either end of the delocalized radical: equivalent ends (symmetric radical) give one product; different ends (unsymmetric radical) give a mixture of two allylic bromides — "allylic rearrangement," where the double bond appears to shift. For example, 1-butene (CH₂=CH–CH₂CH₃) gives 3-bromo-1-butene and 1-bromo-2-butene, because the allylic radical CH₂=CH–•CH–CH₃ has the radical/bond on two different carbons.

How it works

  1. NBS maintains a very low, steady Br₂ concentration (consuming HBr as it forms).
  2. Light, heat, or an initiator produces a trace of Br•, which abstracts the weakest allylic hydrogen to form a resonance-stabilized allylic radical.
  3. The allylic radical reacts with Br₂ to install bromine and regenerate Br•; the scarce Br₂ suppresses ionic addition to the double bond.
  4. Bromine attaches at either end of the delocalized radical — one product if symmetric, a mixture if not.

Common confusions

Do not confuseWithDifference
Allylic position (next to C=C)Vinylic position (on the C=C)Allylic H gives a stable radical; vinylic H does not
Allylic bromination (substitution)Bromination of the alkene (addition)Substitution keeps the double bond; addition destroys it
NBS (controlled Br₂ source)Br₂ (direct bromination)NBS avoids ionic addition by limiting [Br₂]
Two resonance contributorsTwo different moleculesThey are the same hybrid drawn two ways
Allylic rearrangement (mixture)A single productUnsymmetrical radicals give two bromides

Memory aids

"NBS = Next-to-the-Bond Substitution." The allylic position is "A-lying next to the C=C." For products: "symmetric = one, unsymmetric = two." The radical's two ends are like two doors on one room — if the room is symmetric, both doors lead to the same hallway (one product); if not, two hallways (two products).

Quick review

Topic Recap

Allylic bromination selectively substitutes a hydrogen on the carbon next to a double bond, retaining the double bond. It uses NBS under radical conditions to keep Br₂ concentration low, preventing ionic addition. The key intermediate is a resonance-stabilized allylic radical with two contributors; symmetrical radicals give one product, unsymmetrical ones a mixture through allylic rearrangement. The allylic C–H is the weakest, most easily functionalized C–H in an alkene.

Knowledge Check

  1. Where is the allylic position on CH₂=CH–CH₂CH₃, and why is its C–H especially reactive?
  2. What role does NBS play in allylic bromination, and why does it matter?
  3. Draw the two resonance contributors of the allylic radical from propene.
  4. Why does allylic bromination of 1-butene give a mixture while propene gives one product?
  5. How does allylic bromination avoid adding Br₂ across the double bond?

Answers and Rationales

  1. The allylic position is the CH₂ attached to the C=C (the –CH₂CH₃ carbon in CH₂=CH–CH₂CH₃). Its C–H is reactive because the resulting allylic radical is resonance-stabilized, lowering the bond's dissociation energy. Rationale: weaker bond + stable radical = easiest abstraction.
  2. NBS supplies a low, steady Br₂ concentration (consuming HBr and regenerating Br₂). This keeps enough Br• for the chain but too little Br₂ for fast ionic addition. Rationale: NBS channels the reaction into substitution instead of addition.
  3. CH₂=CH–CH₂• and •CH₂–CH=CH₂ (radical and double bond swap positions). Rationale: moving a π bond relocates the unpaired electron to the other terminal carbon.
  4. Propene's allylic radical has two equivalent ends (both CH₂), so brominating either gives the same product (CH₂=CH–CH₂Br). 1-Butene's radical (CH₂=CH–•CH–CH₃) has two different ends, giving 3-bromo-1-butene and 1-bromo-2-butene. Rationale: radical symmetry decides one vs two products.
  5. By keeping [Br₂] very low, ionic addition (which needs Br₂ to encounter the alkene) is suppressed, while the radical chain needs only trace Br• to proceed at the allylic position. Rationale: concentration control favors substitution over addition.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a house with a bright neon sign (the C=C double bond) on the front door. To change the wall next to the door (the allylic position) without touching the sign, you need a careful painter — not someone who rushes up and repaints the sign (ionic addition across the double bond). NBS is that careful painter: it removes a hydrogen from the wall beside the door and installs a bromine there, leaving the sign glowing.

The wall is special because the "unpaired electron" left behind can shift between two spots, like a squeezed water balloon — that spreading-out (resonance) makes the allylic radical extra stable.

Where this stops being exact: the balloon suggests the electron bounces back and forth in time, but resonance is a single averaged hybrid — both ends are equivalent simultaneously. And the "careful painter" makes it sound like the product is always clean; when the two ends differ, bromine can attach at either end, giving a mixture — the reaction is not always perfectly selective.

Simple Example

Brominating propene (CH₃–CH=CH₂) at the allylic position (the CH₃) gives 3-bromopropene (CH₂=CH–CH₂Br, allyl bromide). The double bond stays put. Because the allylic radical is symmetric (both ends are CH₂), only one product forms.

Worked example

Allylic bromination of propene, moving electrons before naming the product.

  1. Initiation — form Br•. Light/heat or an initiator generates a trace of bromine radicals (conceptually from the small Br₂ pool NBS maintains): Br–Br ⇀ 2 Br•.
  2. Propagation step 1 — abstract the allylic hydrogen (electrons first). A Br• uses its unpaired electron (fishhook ⇀) to pull one electron from the allylic C–H bond; the second C–H electron stays on carbon, giving HBr and an allylic radical CH₂=CH–CH₂•.
  3. Resonance of the intermediate. Move a π bond (double-headed arrow) to relocate the unpaired electron: CH₂=CH–CH₂• ↔ •CH₂–CH=CH₂. Both contributors are equivalent.
  4. Propagation step 2 — brominate the radical. The allylic radical grabs a bromine atom from Br₂ (fishhook on Br–Br), forming a C–Br bond and a new Br•. Because the radical is symmetric, bromination at either end gives the same molecule.
  5. State the product. 3-Bromopropene, CH₂=CH–CH₂Br — the double bond intact, bromine on the allylic carbon.
  6. Termination (minor): radical combination, e.g., Br• + Br• → Br₂ or allylic-radical coupling.

Key takeaways

  • High yield: Allylic bromination substitutes at the carbon next to the double bond and keeps the double bond.
  • High yield: The allylic radical is resonance-stabilized (two equivalent contributors), which is why the allylic C–H is attacked.
  • High yield: NBS provides a low, steady [Br₂], suppressing ionic addition to the double bond.
  • High yield: An unsymmetrical alkene gives a mixture of two allylic bromides (allylic rearrangement).
  • High yield: A symmetrical allylic radical (e.g., propene) gives one product.
  • The allylic C–H bond is weaker (≈ 89 kcal/mol) than an ordinary alkyl C–H, making it the easy target.
  • NBS regenerates Br₂ from the HBr byproduct — keeping bromine concentration low but nonzero.

Keep learning

Ready to build on this? Continue to the next lesson.

Practice Organic Chemistry 1

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • Identify the allylic position of an alkene and explain why an allylic radical is stabilized by resonance contributors.
  • Describe how NBS enables radical allylic bromination while avoiding alkene addition (ionic bromination of the double bond).
  • Draw the resonance contributors of an allylic radical and use them to predict products (including allylic rearrangement).
  • Recognize the radical conditions (conceptual) and the safety boundaries of working with NBS.

Key vocabulary

Allylic position
Carbon directly next to a C=C double bond
Allylic radical
Radical on the allylic carbon
NBS
N-bromosuccinimide, a controlled Br₂ source
Radical allylic bromination
Substituting an allylic H with Br via radicals
Resonance contributors
Structures differing only in π/electron placement
Allylic rearrangement
Double bond shifts as Br adds to either end
Avoiding alkene addition
Keeping [Br₂] low so the double bond is not brominated
Radical conditions (conceptual)
NBS + heat/light/initiator in nonpolar solvent

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.