Organic Chemistry · Organohalides
Stability of the Allyl Radical: Resonance Revisited
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In 30 seconds
The Allyl radical CH₂=CH–CH₂•, unpaired electron delocalized over both terminal carbons Full entry →, CH₂=CH–CH₂•, is the simplest radical whose unpaired electron is delocalized over more than one carbon. You met resonance for anions and cations in Chapter 2; this topic applies the same logic to radicals, showing that resonance stabilization is general — it works for any electron-poor or electron-rich center next to a π bond. The allylic C–H bond is weaker than an ordinary C–H bond, the allylic radical is more stable than a secondary alkyl radical, and both facts trace back to one picture: the unpaired electron is spread over two equivalent carbons instead of stuck on one.
The two resonance forms of the allyl radical are:
CH2=CH-CH2• ↔ •CH2-CH=CH2
The real molecule is a Resonance hybrid The real structure, intermediate between resonance forms Full entry → of these forms — a single structure in which the unpaired electron (Spin density The probability of finding the unpaired electron on a given atom Full entry →) is shared between the two terminal carbons and the two C–C bonds are identical, intermediate between single and double bonds. This Delocalization Spreading an electron (or charge) over multiple atoms Full entry → lowers the radical's energy and explains allylic bromination (Topic 3) and the whole family of allylic reactions.
Why this matters
Radical stability is the master key to radical reactivity: the more stable the radical, the easier it is to form, and the more selective its formation. The allylic radical sits at the top of the alkyl-radical stability ladder (alongside benzylic radicals), which is exactly why allylic bromination with NBS works so cleanly and why allylic C–H bonds are the preferred site of attack in so many reactions. Beyond this chapter, resonance-stabilized radicals appear in polymerization, antioxidant chemistry (vitamin E), the biosynthesis of terpenes and steroids, and the ozone cycle. Learning to draw and reason with radical resonance forms now — fishhook-arrow rules included — is a skill you will reuse for the rest of the book.
The college version
Core Concepts
Radical stability order and its evidence
Alkyl substituents stabilize radicals through Hyperconjugation σ(C–H) donation into the electron-poor radical center Full entry →, giving the familiar order:
methyl < primary < secondary < tertiary ≈ allylic ≈ benzylic
The evidence is in the C–H bond dissociation energies: the weaker the C–H bond, the more stable the radical formed. Typical values (kJ/mol):
| Bond broken | BDE (kJ/mol) | Radical |
|---|---|---|
| CH₃–H (methane) | 438 | methyl |
| 1° C–H (ethane-type) | 423 | primary |
| 2° C–H (propane middle) | 413 | secondary |
| 3° C–H (isobutane center) | 404 | tertiary |
| Allylic C–H (propene CH₃) | 368 | allyl |
| Benzylic C–H (toluene CH₃) | 377 | benzyl |
| Vinylic C–H (ethylene) | 444 | vinyl |
The allylic C–H bond is ~45 kJ/mol weaker than a secondary C–H bond and ~76 kJ/mol weaker than a vinylic C–H bond — a huge difference. Notice the striking contrast: allylic radicals are among the most stable alkyl radicals, while vinylic radicals are among the least stable. Both sit next to a double bond; the difference is that the allylic radical is conjugated with the π bond, while the Vinylic radical Radical centered on an sp² carbon of a double bond Full entry → sits on the sp² carbon, gains no resonance benefit, and must break a stronger sp²–H bond.
Drawing resonance forms of radicals
Resonance forms of radicals follow the same rules as for ions, with one extra rule: the number of unpaired electrons must be conserved in every form (always one unpaired electron for a radical). To interconvert allylic resonance forms, move the unpaired electron and the π bond as a pair:
- Form 1: CH₂=CH–CH₂• — the radical is on the terminal carbon, the double bond is between C1–C2.
- Move the π electron pair down: the double bond shifts to C2–C3, and the unpaired electron shifts to C1.
- Form 2: •CH₂–CH=CH₂ — same atoms, radical relocated, π bond relocated.
Every atom keeps its valence; each carbon still has four bonds-or-lone-pairs. The two forms are equivalent in energy for the allyl radical, so the hybrid is perfectly symmetric: equal spin density on both terminal carbons and two identical C–C bonds of bond order 1.5.
For a substituted allyl radical, the forms are not equivalent; the more substituted form (radical on the more substituted carbon) is lower in energy and contributes more to the hybrid — which is why unsymmetrical alkenes give both 1-bromo and 3-bromo products in allylic bromination, with relative amounts reflecting the stability of the two radical ends.
Why resonance stabilizes: the orbital picture
In the allyl radical, the p orbital on the central carbon overlaps with p orbitals on both terminal carbons, so the unpaired electron occupies a molecular orbital spread over all three carbons rather than an atomic orbital on one — and spreading an electron over more atoms lowers its energy. The same logic applies to allylic cations (empty p orbital over three carbons) and allylic anions (lone pair over three carbons): resonance stabilizes radicals, cations, and anions alike. That is why this topic is "revisited": what you learned for cations in Chapter 2 transfers directly to radicals here.
Consequences for reactivity
Stability shows up in reactivity three ways:
- Selective abstraction: allylic H is removed selectively because the allylic radical forms fastest (Topic 3's NBS bromination).
- Product mixtures from delocalization: the radical reacts at either end of the delocalized system, giving both allylic isomers.
- Chain behavior: longer-lived stabilized radicals readily combine with other radicals (Termination Two radicals combining Full entry →) — exploited in polymerization and in antioxidants, where stable radicals stop destructive chains.
Common Confusions
| Do not confuse | With | Difference |
|---|---|---|
| Allylic radical | Vinylic radical | Allylic: radical on carbon next to C=C, resonance-stabilized (stable); vinylic: radical on the sp² carbon, no stabilization (unstable, BDE ~444) |
| Resonance forms of a radical | Resonance forms of a cation | Same delocalization idea, but radicals keep one unpaired electron in every form; use fishhook arrows for the electron |
| Resonance hybrid | One resonance form | The hybrid is the real, averaged structure; forms are only drawings. Don't picture the radical "flipping" between forms |
| Allylic radical stability | Alkyl radical stability via hyperconjugation | Both real, but allylic/benzylic stabilization (resonance, ~45 kJ/mol) is larger than tertiary-vs-secondary hyperconjugation differences (~9 kJ/mol) |
| "Weaker bond" | "More reactive bond" | Weaker C–H bond = easier homolysis = more stable radical; but stability of the radical and its reactivity are inverse |
| Allylic position | Allylic radical | Position = the carbon next to C=C; radical = the species formed when its H is removed. Same location, different concepts |
| Spin density on both ends | Two different molecules | The two ends are the same molecule's hybrid; products form at either end, giving isomers — not separate intermediates |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A radical's unpaired electron is like a hot marble you have to hold. With one hand (one carbon), your hand gets very hot — that radical is unstable and eager to react. An allylic radical is like holding the marble with three hands in a row: the middle carbon passes the heat to both ends, so no single hand gets too hot. Spreading the "heat" over more atoms is called resonance, and it makes the radical calmer, more stable, and easier to create.
Worked example
Example 1: Drawing the resonance forms of the allyl radical and locating the spin density
Start: propene's radical, CH₂=CH–CH₂•, drawn as a chain C1=C2–C3 with the dot on C3.
Step 1 — identify the movable pieces: the π bond (C1=C2) and the unpaired electron on C3.
Step 2 — move them together: push the π electron pair from C1=C2 onto C2–C3 (a fishhook arrow shows the single electron of the radical moving from C3 to C1 simultaneously). Result: the double bond is now C2=C3 and the radical sits on C1: •CH₂–CH=CH₂.
Step 3 — verify valence: C1 now has a double bond's partner (one bond to C2) plus one H plus the radical — valence 3, correct. C3 now has the double bond plus CH₃ hydrogens — valence 4, correct.
Step 4 — interpret: the two forms are equivalent, so the hybrid places half the spin density on C1 and half on C3, with two equal C–C bonds. Any reagent that traps the radical (Br₂ in allylic bromination) can attack either C1 or C3 — hence two products for unsymmetrical alkenes, but only one (allyl bromide) here because the ends are identical.
Example 2: Comparing BDEs to rank radical stability
Using the table above, rank these radicals from most to least stable and justify with bond strengths: allyl, 2° (isopropyl-type), vinyl, tertiary, methyl.
Read BDEs from the table: allyl 368, tertiary 404, secondary 413, primary 423, methyl 438, vinyl 444. The largest BDE belongs to the vinylic C–H bond, which means the vinylic radical is the least stable. Order of stability (most to least): allyl > tertiary > secondary > primary > methyl > vinyl.
Note the two surprises at the ends: allylic outranks even tertiary, and vinylic drops below methyl. Both follow from the same rule — weaker C–H bond, more stable radical — but the reason differs: allylic gains resonance stabilization; vinylic gains none and pays the price of an sp² carbon with a stronger bond. This ranking is worth memorizing because exam questions routinely ask "which radical is more stable, allylic or tertiary?" (answer: allylic) and "allylic or vinylic?" (allylic, by a wide margin).
Example 3: Predicting allylic bromination products of 2-methyl-2-butene
Substrate: (CH₃)₂C=CH–CH₃ (double bond between C2 and C3; C2 bears two methyls, C3 bears one). Removing an allylic H from a methyl on C2 gives the radical (CH₃)(CH₂•)C=CH–CH₃, whose unpaired electron delocalizes onto C3: (CH₃)₂C=CH–CH₂•. Bromine can be delivered to either end:
Product A (Br at the methylene): CH₂(Br)–C(CH₃)=CH–CH₃ = 1-bromo-2-methyl-2-butene. Product B (Br at C4): (CH₃)₂C=CH–CH₂Br = 1-bromo-3-methyl-2-butene.
Both are allylic bromides, differing only in which end of the delocalized radical trapped the bromine. The lesson, repeated from Topic 3: for allylic bromination of an unsymmetrical alkene, draw both resonance forms and report both products.
Key takeaways
- Allyl radical: CH₂=CH–CH₂• ↔ •CH₂–CH=CH₂ — two equivalent resonance forms; the hybrid has equal spin density on both terminal carbons.
- Radical stability order: methyl < 1° < 2° < 3° ≈ allylic ≈ benzylic; vinylic radicals are less stable than primary.
- BDE evidence: allylic C–H ≈ 368 kJ/mol vs 2° C–H ≈ 413 kJ/mol, 3° ≈ 404 kJ/mol, vinylic ≈ 444 kJ/mol.
- Resonance rules for radicals: conserve the number of unpaired electrons (always one); move the radical and π bond together; every atom keeps its valence.
- Resonance stabilizes radicals, cations, and anions alike — same delocalization principle as Chapter 2.
- Delocalization means reaction at either end → two allylic products from unsymmetrical alkenes (Topic 3).
- Hyperconjugation explains the 3° > 2° > 1° > methyl trend; resonance explains why allylic/benzylic outrank tertiary.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Draw the two resonance forms of the allyl radical and state what the hybrid looks like.
Show answer
CH₂=CH–CH₂• ↔ •CH₂–CH=CH₂. The hybrid has the unpaired electron (spin density) shared equally between the two terminal carbons and two identical C–C bonds of intermediate order (1.5).
Which is more stable: an allylic radical or a tertiary alkyl radical? What evidence supports your answer?
Show answer
Allylic. The allylic C–H BDE (~368 kJ/mol) is far lower than the tertiary C–H BDE (~404 kJ/mol), so the allylic radical forms more easily.
Why is the vinylic radical so unstable even though it is adjacent to a double bond?
Show answer
Because the radical sits on the sp² carbon of the double bond — no conjugation, no resonance benefit — and the sp² C–H bond is stronger, so its BDE (~444 kJ/mol) is the highest of the series.
What is the stability order of alkyl radicals, and what interaction explains it?
Show answer
Methyl < primary < secondary < tertiary. The trend comes from hyperconjugation: alkyl substituents donate electron density from adjacent C–H σ bonds into the electron-poor radical center, stabilizing it.
How many allylic bromination products does 2-methyl-2-butene give, and why?
Show answer
Two: 1-bromo-2-methyl-2-butene and 1-bromo-3-methyl-2-butene. The allylic radical delocalizes the unpaired electron over two carbons, and bromine can be delivered to either end.
What arrow type draws electron movement in radical resonance forms, and what must be conserved?
Show answer
Single-headed (fishhook) arrows; the number of unpaired electrons (one, for a radical) must be conserved in every form.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Allyl radical
- CH₂=CH–CH₂•, unpaired electron delocalized over both terminal carbons
- Resonance hybrid
- The real structure, intermediate between resonance forms
- Spin density
- The probability of finding the unpaired electron on a given atom
- Bond dissociation energy (BDE)
- Energy to break a bond homolytically
- Hyperconjugation
- σ(C–H) donation into the electron-poor radical center
- Benzylic radical
- C₆H₅–CH₂•, radical on the carbon next to a benzene ring
- Vinylic radical
- Radical centered on an sp² carbon of a double bond
- Fishhook (single-headed) arrow
- Shows movement of a single electron
- Delocalization
- Spreading an electron (or charge) over multiple atoms
- Termination
- Two radicals combining
Sources & references
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