Source: Baldwin, CHEM 2510, Ohio State University
Tags: bond dissociation energy, BDE, homolytic cleavage, heterolytic cleavage, radical stability, hyperconjugation, carbon radicals, sp2 hybridisation, alkane bond strength, organic chemistry
Difficulty: Introductory–Intermediate Prerequisites: General chemistry bond concepts, Lewis structures, orbital hybridisation (sp3, sp2), electronegativity basics.
This material sits right at the start of the reactivity portion of organic chemistry. Up to now you have been drawing structures and thinking about stability. Here the question shifts: what does it take to break a bond, and what happens to the pieces once it breaks? Bond dissociation energy gives you a quantitative handle on bond strength, and radical stability explains why some bonds break more easily than others. If you are behind, make sure you are comfortable with orbital hybridisation and electron-pair geometry before diving in.
There are two ways to break a covalent bond: heterolytic (both electrons go to one atom, giving ions) and homolytic (one electron to each atom, giving radicals). Bond dissociation energy (BDE) measures the energy for homolytic cleavage. The more substituted a carbon radical is, the more stable it is, which means the bond that produced it was weaker.
Heterolytic cleavage
The breaking of a covalent bond so that both bonding electrons travel to the same atom. This produces a cation and an anion. Drawn with a standard two-headed curved arrow. Think of it as: one atom takes the whole electron pair and leaves the other empty-handed.
Homolytic cleavage
The breaking of a covalent bond so that each atom receives one electron. This produces two radicals. Drawn with two single-headed (fish-hook) curved arrows. In simple terms, the bond splits evenly, one electron each.
Bond dissociation energy (BDE)
The enthalpy (ΔH°) required to break a covalent bond by homolytic cleavage in the gas phase. Reported in kcal/mol. A higher BDE means a stronger bond. Think of it as: the price tag for ripping a bond apart, measured per mole.
Radical
An atom or molecule with an unpaired electron. Carbon radicals produced by homolytic cleavage of C–H bonds are central to this chapter. In simple terms, a fragment with a lonely, unpaired electron that makes it highly reactive.
Hyperconjugation
A stabilising interaction in which electron density from filled, aligned C–H (sp3) sigma bonds donates into an adjacent empty or half-filled p orbital. It is a type of induction and is the primary reason more substituted radicals are more stable. Think of it as: neighbouring bonds quietly sharing their electron density with the radical centre to calm it down.
Heterolytic cleavage produces a cation and an anion from a single bond.
Both electrons travel together to one atom.
Represented by a standard two-headed curved arrow.
Homolytic cleavage produces two radicals from a single bond.
Each atom gets one electron.
Represented by two fish-hook (single-headed) curved arrows.
This is the type of cleavage measured by BDE.
H–H: ΔH° = 104 kcal/mol
H–CH₃: ΔH° = 104 kcal/mol
H₃C–CH₃: ΔH° = 88 kcal/mol
H–CH₂CH₃ (1° C–H): ΔH° = 98 kcal/mol
H–CH(CH₃)₂ (2° C–H): ΔH° = 95 kcal/mol
H–C(CH₃)₃ (3° C–H): ΔH° = 91 kcal/mol
Notice BDE drops as the resulting radical becomes more substituted. That is not a coincidence.
Stronger bonds form between overlapping orbitals of the same size (good overlap = strong bond).
Stronger bonds create less stable radicals upon homolytic cleavage.
Stronger bonds require more energy to break.
The practical takeaway: it is easier to break a bond when the resulting radical (or other intermediate) is stable.
A carbon radical is sp2-hybridised, with the unpaired electron in a p orbital perpendicular to the plane of the three substituents.
The radical centre is planar, which matters enormously for stereochemistry later.
Methyl < 1° < 2° < 3°
More substituted radicals are more stable. The dominant reasons:
Hyperconjugation: Aligned sp3 C–H bonds on adjacent carbons donate electron density into the half-filled p orbital of the radical. A 3° radical has more such interactions than a methyl radical.
Induction: Alkyl groups are weakly electron-donating, which helps disperse the electron deficiency.
Resonance can also stabilise a radical when the unpaired electron sits in a p orbital that can overlap with an adjacent pi system. The radical must be in a p orbital for resonance to operate. More stable radicals come from weaker bonds, which means those bonds are easier to break.
BDE relationship:
ΔH° (bond breaking) = BDE of bond broken
Radical stability order (carbon radicals):
methyl (·CH₃) < primary (1°) < secondary (2°) < tertiary (3°)
BDE vs. substitution table:
Bond cleaved | BDE (kcal/mol) | Radical class |
|---|---|---|
H–CH₃ | 104 | Methyl |
H–CH₂CH₃ | 98 | 1° |
H–CH(CH₃)₂ | 95 | 2° |
H–C(CH₃)₃ | 91 | 3° |
BDE values are not just exam fodder. Engineers and materials scientists use them to predict which bonds in a polymer chain will break first under heat or UV light, which directly informs decisions about plastic degradation, flame retardants, and coating durability. In combustion chemistry, knowing which C–H bonds break most easily helps model fuel ignition.
Students often assume a higher BDE means the bond is "more reactive." It is the opposite: higher BDE means harder to break, so less reactive toward homolytic cleavage.
Students frequently confuse heterolytic and homolytic cleavage arrows. Two-headed curved arrows are for heterolytic; fish-hook arrows are for homolytic. Using the wrong arrow type changes the meaning entirely.
Some students think radical stability and carbocation stability follow the same trend for the same reason. The trend is similar (3° > 2° > 1° > methyl), but the underlying physics differ: carbocations have an empty p orbital, radicals have a half-filled one. Hyperconjugation operates in both cases, but do not conflate them.
A common slip is thinking that BDE measures heterolytic cleavage. It does not. BDE is defined exclusively for homolytic cleavage.
⚠️ You will almost certainly be asked to rank radicals by stability or predict which C–H bond in a molecule is easiest to break. Know the BDE table cold.
⚠️ Be able to draw the fish-hook arrows for homolytic cleavage and explain why a 3° radical is more stable than a 1° radical (hyperconjugation).
⚠️ Expect a question linking BDE to reactivity: "Which bond breaks first?" The answer is the one with the lowest BDE, because it produces the most stable radical.
⚠️ Know that a carbon radical is sp2-hybridised and planar. This will come back in the stereochemistry of radical reactions.
True or false: Bond dissociation energy measures the energy to break a bond via heterolytic cleavage.
Fill in the blank: A 3° carbon radical is more stable than a 1° carbon radical primarily because of __________.
True or false: Homolytic cleavage of a bond produces one cation and one anion.
Fill in the blank: The BDE of the H–C(CH₃)₃ bond is ______ kcal/mol.
True or false: A carbon radical centre is sp3-hybridised and tetrahedral.
Answers: 1. False (homolytic). 2. Hyperconjugation. 3. False (it produces two radicals). 4. 91. 5. False (sp2 and planar).
Q: Rank the following radicals in order of increasing stability: ·C(CH₃)₃, ·CH₃, ·CH₂CH₃, ·CH(CH₃)₂.
A: ·CH₃ < ·CH₂CH₃ < ·CH(CH₃)₂ < ·C(CH₃)₃. Stability increases with substitution due to hyperconjugation.
Q: The BDE of the C–H bond in methane is 104 kcal/mol and the BDE of the 3° C–H bond in 2-methylpropane is 91 kcal/mol. Explain the difference.
A: Breaking the 3° C–H bond produces a tertiary radical, which is stabilised by hyperconjugation from the three adjacent methyl groups. Because the product radical is more stable, less energy is needed to form it, so the BDE is lower.
Q: Draw the fish-hook arrows for the homolytic cleavage of the C–H bond in ethane. What two species are produced?
A: Two single-headed curved arrows, each pointing from the bond to one of the two atoms. The products are a hydrogen radical (H·) and an ethyl radical (·CH₂CH₃).
Q: A student claims that because BDE measures bond strength, a bond with a high BDE will react fastest. Is this correct?
A: No. A high BDE means the bond is strong and requires more energy to break, making it less reactive toward homolytic cleavage, not more.
This material connects directly to radical halogenation (Ch. 3, Sections 3.4–3.6), where the relative ease of breaking different C–H bonds determines product distribution. It also links to carbocation stability (Ch. 6 and beyond): the same substitution trend (3° > 2° > 1°) appears, though the electronic reasons differ slightly. Understanding BDE is also useful when you reach thermodynamic analysis of reactions, since ΔH° for a reaction can be estimated as the sum of bonds broken minus bonds formed.
bond dissociation energy, BDE, homolysis, homolytic bond cleavage, heterolysis, heterolytic bond cleavage, radical, free radical, carbon radical, fish-hook arrow, single-headed curved arrow, hyperconjugation, sigma donation, radical stability order, methyl radical, primary radical, secondary radical, tertiary radical, sp2 radical, planar radical, bond strength, CHEM 2510, organic chemistry chapter 3