EPR of Organic Radicals

Introduction

Organic radicals appear as reaction intermediates, in polymers, in biological electron transfer and in materials such as organic semiconductors. Their g values lie very close to the free-electron value, so g alone rarely identifies them. Instead, their hyperfine patterns act as fingerprints. Because the unpaired electron is often delocalised over a conjugated framework, the size of each proton coupling maps how much spin sits on each carbon, giving a direct experimental picture of a singly occupied molecular orbital.

Core explanation

g values of organic radicals. Carbon-centred radicals have g ≈ 2.002–2.003. Spin density on nitrogen raises g slightly (nitroxides about 2.006), and spin on oxygen raises it further (phenoxyl and semiquinone radicals about 2.004–2.005). Sulfur-centred radicals can reach 2.01–2.03 or more. These small differences need careful calibration but are useful supporting evidence.

α-protons and the McConnell relation. In a planar π radical, a proton bonded to a spin-bearing carbon lies in the nodal plane of the carbon p orbital, so it cannot directly receive spin density. Instead, the unpaired π electron slightly polarises the C–H σ bond, placing a small excess of opposite spin at the proton. The result is the McConnell relation:

a H = Q ρ C

where ρ C is the π spin density on the carbon and Q is about −2.3 to −2.7 mT. The negative sign means the proton carries negative spin density, although ordinary EPR reports only the magnitude of a.

Aromatic radical ions. In the benzene radical anion, the extra electron is shared equally over six carbons, ρ = 1/6. Six equivalent protons give seven lines with intensities 1:6:15:20:15:6:1 and a = 0.375 mT, consistent with Q ≈ 2.25 mT in magnitude. In the naphthalene radical anion, the four α positions (1, 4, 5, 8) carry more spin than the four β positions, so couplings of about 0.49 mT and 0.18 mT produce 5 × 5 = 25 lines. The larger coupling marks the positions where the singly occupied orbital has larger coefficients, matching Hückel predictions.

β-protons and hyperconjugation. Protons one carbon further away couple through hyperconjugation with the singly occupied p orbital. Their coupling depends on geometry: a β ≈ B ρ C cos²θ, where θ is the dihedral angle between the C–H bond and the p-orbital axis and B is roughly 5 mT. Freely rotating methyl groups average the angle and give a β of about 2.5–2.7 mT, often larger than α-proton couplings.

Persistent and transient radicals. Stable radicals such as TEMPO or triarylmethyl derivatives survive in solution; most carbon radicals do not. Spin trapping solves this: a nitrone or nitroso compound captures the transient radical to form a longer-lived nitroxide adduct. The adduct's a N and a H(β) values depend on the trapped fragment, so comparing them with reference data helps identify the original radical.

Step-by-step reasoning

To assign an organic radical spectrum:

1. Measure g and confirm it lies near 2.002–2.006. 2. Count lines and measure the total spectral width. 3. Test combinations of equivalent proton sets against the line count and intensity pattern. 4. Use the McConnell relation to convert each coupling into carbon spin densities. 5. Compare spin densities with a molecular-orbital prediction or computed values.

Visual explanation

Draw the naphthalene skeleton and shade each carbon in proportion to its spin density, darker at positions 1, 4, 5 and 8 and lighter at 2, 3, 6 and 7. The shading pattern is essentially the square of the orbital coefficients, and each shade corresponds to one measured proton coupling.

Real-world analogy

A rumour spreading through a village is heard most often in some houses and rarely in others. By asking each household how often they heard it, you map where it circulated most. Proton couplings are those household reports: they reveal where the unpaired electron spends its time.

Real-world example

In photosynthesis research, EPR of the oxidised chlorophyll pair and of tyrosyl radicals in photosystem II uses proton hyperfine couplings to map spin distribution. In polymer science and food chemistry, spin trapping helps show which radicals form during degradation, informing the choice of stabilisers and antioxidants.

Why?

Why can a proton in the nodal plane show hyperfine coupling at all? Electron exchange favours parallel spins in the same region, so the σ electron near carbon tends to share the spin of the π electron, leaving the σ electron near hydrogen with the opposite spin. This spin polarisation places a small negative spin density at the proton.

Common misconception

"Spin density must always be positive and add up to one on each atom." Spin polarisation can create negative spin density on some atoms, as in the allyl radical's central carbon. The total spin density sums to one, but individual values can be negative.

Worked example

Question: A radical anion of an aromatic hydrocarbon shows seven lines with ratios 1:6:15:20:15:6:1 and a = 0.375 mT. Identify it and find the carbon spin density, taking Q = 2.25 mT.

Reasoning: Seven binomial lines mean six equivalent protons. ρ C = a/ Q = 0.375/2.25 = 0.167, or 1/6, so spin is shared equally over six carbons.

Answer: The benzene radical anion, with a spin density of 1/6 on each carbon.

Quick check

1. Why does a freely rotating methyl group attached to a radical centre often show a large proton coupling? Answer: Its β-protons couple by hyperconjugation, and rotation averages cos²θ to one half, giving a sizeable coupling of about 2.5 mT.

Exam focus

Apply a H = Qρ C with Q around 2.3–2.7 mT, predict line counts for sets of equivalent protons, and explain spin polarisation and hyperconjugation. Know the g range for carbon-, nitrogen-, oxygen- and sulfur-centred radicals and the purpose of spin trapping.

Advanced insight

The sign of hyperfine couplings, invisible in ordinary EPR, can be determined by ENDOR or by NMR of paramagnetic species. Modern density functional calculations reproduce spin densities well, enabling assignment of complex radicals. Line-width alternation in some spectra reveals dynamic processes such as conformational interconversion, analogous to exchange effects in NMR.

Summary

Organic radicals have g close to 2.002–2.006, so hyperfine patterns identify them. α-Proton couplings follow a H = Qρ C through spin polarisation, and β-proton couplings follow a cos²θ hyperconjugation law. Aromatic radical ions reveal orbital coefficients, and spin trapping converts transient radicals into persistent nitroxides with diagnostic couplings.

Practice questions

1. A carbon in a π radical has spin density 0.40. Estimate the coupling of its attached proton using Q = 2.5 mT. Answer: a = 2.5 × 0.40 = 1.0 mT. 2. Predict the number of lines for the naphthalene radical anion and explain. Answer: 25 lines: two sets of four equivalent protons each give five lines, and 5 × 5 = 25. 3. Why is a spin trap useful for detecting hydroxyl or alkyl radicals in solution? Answer: These radicals are too short-lived to accumulate; the trap converts them into a persistent nitroxide whose couplings identify the original radical. 4. What does a g value of about 2.005 suggest compared with 2.0025? Answer: Significant spin density on oxygen or nitrogen, as in a semiquinone or nitroxide, rather than a purely carbon-centred radical.