Radical Stability

Introduction

When a bond breaks homolytically, each fragment takes one electron and a radical is formed. Radicals drive combustion, polymerisation, atmospheric chemistry and the halogenation of alkanes. Like carbocations, carbon radicals are electron-deficient, with only seven valence electrons, so they respond to similar stabilising effects. Understanding radical stability explains why bromine picks out tertiary hydrogens, why allylic and benzylic positions are so reactive, and why some radicals persist while others vanish in microseconds.

Core explanation

Structure. A simple alkyl radical such as CH₃• is planar or very nearly planar, with the unpaired electron in a p orbital. Because it has seven valence electrons, it is short of an octet and is stabilised by electron donation, but far less dramatically than a carbocation, which is short by two electrons.

Measuring stability. Radical stability is judged from C–H bond dissociation energies (BDEs). If less energy is needed to break a C–H bond, the radical produced is more stable. Approximate values are:

Bond broken Radical formed BDE / kJ mol⁻¹ --- --- --- CH₃–H methyl 439 CH₃CH₂–H primary 420 (CH₃)₂CH–H secondary 413 (CH₃)₃C–H tertiary 404 CH₂=CHCH₂–H allyl 369 C₆H₅CH₂–H benzyl 375 CH₂=CH–H vinyl 464

Hyperconjugation. Adjacent C–H and C–C σ bonds overlap slightly with the half-filled p orbital, donating electron density. More alkyl groups mean more hyperconjugation, giving the order tertiary > secondary > primary > methyl. The differences are real but modest, around 10–15 kJ mol⁻¹ per step.

Resonance. When the unpaired electron is next to a π bond, it is delocalised. The allyl radical has two equivalent resonance forms that place the electron on either end carbon, and the benzyl radical spreads it into the ring. This lowers the BDE by roughly 50 kJ mol⁻¹, a much larger effect than hyperconjugation, so allylic and benzylic radicals are more stable than even tertiary alkyl radicals.

Vinyl and aryl radicals. A radical on an sp² carbon of a double bond or ring is not delocalised, because its orbital lies perpendicular to the π system. These radicals are very unstable, and vinyl and aryl C–H bonds are hard to break.

Radical substitution. In the halogenation of alkanes, the key step is a halogen atom removing a hydrogen atom to give a carbon radical. The more stable the radical, the faster this step, so the product distribution reflects radical stability. Bromine atoms are much more selective than chlorine atoms: bromination of 2-methylpropane gives almost entirely 2-bromo-2-methylpropane, while chlorination gives a significant mixture despite the tertiary preference.

Step-by-step reasoning

To predict the major monohalogenation product of an alkane:

1. Identify every different type of hydrogen. 2. For each, decide which radical forms when it is removed. 3. Rank those radicals: allylic/benzylic > tertiary > secondary > primary > methyl. 4. With bromine, the most stable radical dominates the product. 5. With chlorine, weigh both radical stability and the number of each type of hydrogen.

Visual explanation

Draw the allyl radical as two resonance structures joined by a double-headed arrow, using single-headed fishhook arrows to move one electron at a time. The single electron sits on carbon 1 in one structure and carbon 3 in the other. A spin-density picture shows equal lobes on the two end carbons and almost none in the middle.

Real-world analogy

A radical is like a person with one glove looking for the other. Standing alone in a crowd (a primary radical), they are restless. Surrounded by helpful friends offering support (alkyl groups), they are calmer. If they can share the problem around a group (resonance), the urgency drops even further.

Real-world example

Food and fuel oils go rancid through radical chain oxidation that begins at allylic hydrogens in unsaturated fats, since these C–H bonds are relatively weak. Antioxidants such as vitamin E and BHT protect the oils by donating a hydrogen atom to form a highly resonance-stabilised phenoxyl radical, which is too stable to continue the chain.

Why?

Why are brominations more selective than chlorinations? Hydrogen abstraction by Br• is endothermic, so its transition state is late and resembles the radical product (Hammond postulate). Differences in radical stability therefore show up strongly in the activation energies. Abstraction by Cl• is exothermic, with an early, reactant-like transition state that barely senses radical stability.

Common misconception

"Radicals are stabilised in exactly the same way and to the same extent as carbocations." The trend is the same, but the effects are far smaller for radicals, because a radical is only one electron short of an octet. Radical rearrangements by alkyl shift are consequently rare.

Worked example

Question: Which C–H bond in propene, CH₂=CH–CH₃, is weakest, and why?

Reasoning: Removing H from CH₃ gives the allyl radical, stabilised by resonance over two carbons (BDE about 369 kJ mol⁻¹). Removing a vinyl hydrogen gives a vinyl radical with no delocalisation (about 464 kJ mol⁻¹).

Answer: The allylic C–H bond on the CH₃ group is weakest, because its radical is resonance-stabilised.

Quick check

1. Which is more stable, a tertiary alkyl radical or a benzyl radical, and what effect explains this? Answer: The benzyl radical, because resonance delocalisation into the ring outweighs hyperconjugation.

Exam focus

Be ready to rank radicals, justify the ranking with hyperconjugation or resonance, and use fishhook arrows correctly. In halogenation questions, name bromine as the more selective reagent and explain the result using radical stability and the Hammond postulate.

Advanced insight

Some radicals are so stabilised by resonance and steric shielding that they can be isolated. The triphenylmethyl radical, first reported by Gomberg in 1900, exists in solution in equilibrium with its dimer. Nitroxide radicals such as TEMPO are stable enough to be bottled and are used as oxidation catalysts and spin labels.

Summary

Carbon radicals are planar, seven-electron species stabilised by electron donation. Bond dissociation energies show the order allylic ≈ benzylic > tertiary > secondary > primary > methyl > vinyl. Hyperconjugation gives modest stabilisation; resonance gives much more. Radical stability controls selectivity in radical halogenation, with bromine being especially selective.

Practice questions

1. Use bond dissociation energies to explain why a tertiary radical is more stable than a primary radical. Answer: The tertiary C–H bond (about 404 kJ mol⁻¹) needs less energy to break than a primary C–H bond (about 420 kJ mol⁻¹), so the tertiary radical formed is lower in energy. 2. Predict the major product of monobromination of methylbenzene in light and explain. Answer: (Bromomethyl)benzene, because removing a methyl hydrogen gives the resonance-stabilised benzyl radical. 3. Why is the vinyl radical so unstable? Answer: Its unpaired electron is in an orbital perpendicular to the π bond, so it cannot be delocalised, and the C–H bond broken is very strong. 4. Explain why chlorination of propane gives substantial amounts of 1-chloropropane. Answer: Chlorine atoms are not very selective, and there are six primary hydrogens but only two secondary ones, so primary substitution is statistically favoured.