Organic Chemistry · Part 3 of 9

Reactive Intermediates II — Radicals, Carbenes, Nitrenes & Arynes

Reaction Mechanisms, Part 3 · 10 sections · about 19,093 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

The intermediates in this part behave nothing like ions. Radicals ignore most of what you learned about nucleophiles and electrophiles and follow their own selectivity rules. Carbenes come in two electronic flavours that give different stereochemistry from the same reaction. Nitrenes drive four different named rearrangements. Arynes make substitution happen where no leaving group is. Each is treated the same way: structure first, then stability, then how it is made, then what it does. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 10 sections in Part 3

1Radical structure, geometry and stability Free below
2Chain reactions — initiation, propagation, termination
3Selectivity — halogenation, NBS, and radical addition
4Carbene electronic structure — singlet versus triplet
5Making carbenes — α-elimination, diazo compounds, carbenoids
6Carbene reactions — cyclopropanation, insertion, Wolff
7Nitrenes — generation and structure
8Hofmann, Curtius, Lossen and Schmidt — one mechanistic family
9Arynes, ylides and radical ions
10Where this leads — a pointer to Part 4

Radical structure, geometry and stability

Free extract

Section C.1 of Part 3, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

What a radical looks like, why the answer to ‘planar or pyramidal?’ is ‘almost planar, and it hardly matters’, and how bond-dissociation energies measure stability directly.

Geometry: the pyramidal / planar question

A carbon radical has three σ bonds and one electron left over. There are two ways to arrange that. Either the carbon is sp², the three bonds are trigonal planar, and the odd electron sits in the leftover pure 2p orbital perpendicular to that plane. Or the carbon is sp³, the three bonds form a shallow pyramid, and the odd electron sits in the fourth sp³-like hybrid pointing away from them. The first is called a π radical, the second a σ radical.

The two limiting geometries. In the planar case the singly occupied molecular orbital (SOMO) is a pure 2p orbital, exactly like the empty orbital of a carbocation — and it is perpendicular to the three σ bonds, which is why the plane of those bonds is drawn here almost edge-on (the dashed ellipse) with one lobe above it and one below; in the pyramidal case the SOMO acquires s character and leans to one side. Simple alkyl radicals sit at the planar end but the potential is extremely flat — the methyl radical inverts through planarity with essentially no barrier. Electronegative substituents (F, OR) pull the radical towards pyramidal, and a bridgehead carbon is forced pyramidal by the ring system, which a radical tolerates and a carbocation does not. Schematic drawing; lobe sizes are illustrative, not computed.

The experimental answer for simple alkyl radicals is: planar or very nearly so, and — more importantly — the potential energy surface for bending is exceptionally flat. The methyl radical •CH₃ is planar. The ethyl, isopropyl and tert-butyl radicals are slightly pyramidal but invert so fast that for every practical purpose they behave as planar. This has one enormous stereochemical consequence:

Radicals racemise. A radical generated at a stereogenic carbon loses its stereochemical information, because the planar (or rapidly inverting) radical is attacked with equal probability from both faces. A radical reaction at a stereocentre therefore gives a racemic product, exactly as an S_N1 reaction does — and for the same geometric reason.

What pushes a radical towards pyramidal

Two things.

Electronegative substituents. •CF₃ is decidedly pyramidal, with F–C–F angles around 111°. The reason is the same one that governs Bent’s rule: electronegative substituents prefer to be bonded through orbitals of high p character, which pushes s character into the remaining orbital — the one holding the odd electron. Putting the unpaired electron into an orbital with s character lowers its energy, because s orbitals penetrate closer to the nucleus. •C(OR)₃ and •CCl₃ behave the same way.
Ring constraints. A bridgehead carbon in a bicyclic cage physically cannot become planar. This is fatal for a carbocation and merely inconvenient for a radical.

Note: The bridgehead contrast is worth memorising as a one-line exam answer. 1-Bromobicyclo[2.2.1]heptane (1-bromonorbornane) is essentially inert to S_N1 solvolysis, because the bridgehead cation cannot achieve the planar geometry it needs. But bridgehead radicals form readily and bridgehead C–H bonds undergo radical halogenation without difficulty. A radical does not need planarity; it merely prefers it slightly, and gives that preference up cheaply.

σ radicals that cannot choose

Vinyl, aryl and acyl radicals are a different case again. In •C₆H₅ (phenyl radical) the odd electron is in an sp² orbital lying in the plane of the ring, pointing outwards where the C–H bond used to be. It is orthogonal to the aromatic π system and therefore gets no delocalisation from it. That is why the phenyl radical is so reactive: it is a σ radical with nowhere to put its electron. Hold on to this geometry — it comes back, unchanged, when we derive aryne regiochemistry in C.9.

Simple alkyl radicals. RDKit draws the odd electron as a dot. All four are effectively planar at carbon; the small pyramidalisation that exists in the last three is washed out by inversion faster than any chemistry can happen.

Stability, and how we know: bond-dissociation energies

You cannot weigh a radical. What you can measure is how much energy it costs to make one by pulling a bond apart homolytically, and that is exactly what a bond-dissociation energy reports.

Bond-dissociation energy (BDE), D(A–B): the enthalpy change for the gas-phase homolysis A–B → A• + B• at 298 K. It is always positive. A lower BDE for a C–H bond means the resulting carbon radical is more stable, because the hydrogen atom produced is the same in every case.

D(R–H) = ΔH_f(R•) + ΔH_f(H•) − ΔH_f(R–H)

That equation is why BDE is a clean measure of radical stability. Comparing D(R–H) across different R changes only one thing — the radical — because H• is a constant. So the differences between C–H BDEs are the differences in radical stability, offset by whatever differences exist in the stabilities of the parent hydrocarbons (usually small for this comparison).

C–H bond broken	Radical formed	BDE / kJ mol⁻¹	BDE / kcal mol⁻¹	Stabilisation vs •CH₃
HC≡C–H	ethynyl (σ)	558	133.3	−119 (much less stable)
C₆H₅–H (benzene)	phenyl (σ)	473	112.9	−34
CH₂=CH–H	vinyl (σ)	465	111.2	−26
CH₃–H	methyl	439	105.0	0 (reference)
CH₃CH₂–H	ethyl (1°)	421	100.5	+18
(CH₃)₂CH–H	isopropyl (2°)	412	98.6	+27
(CH₃)₃C–H	tert-butyl (3°)	404	96.5	+35
N≡C–CH₂–H	cyanomethyl	397	94.8	+42
C₆H₅CH₂–H (toluene)	benzyl	375	89.7	+64
CH₃C(O)–H	acetyl	374	89.4	+65
CH₂=CHCH₂–H (propene)	allyl	371	88.8	+68
(C₆H₅)₃C–H	trityl	≈339	≈81	≈+100

Homolytic C–H bond-dissociation enthalpies at 298 K, gas phase. The final column is the difference from methane and is the practical measure of how much a substituent stabilises the radical. Read the table as a single continuous scale: sp carbon at the top is the hardest place to make a radical, a delocalised benzylic or allylic position the easiest.

The stability order, read off the table

Radical stability: allyl ≈ benzyl > 3° > 2° > 1° > CH₃• > vinyl > aryl > alkynyl

Two mechanisms are doing the work, and they are the same two you already know from carbocations, with one addition.

Hyperconjugation. A C–H σ bond on an adjacent carbon overlaps with the half-filled SOMO. Because the SOMO holds only one electron, it can accept density from the σ bond; the two-orbital interaction is stabilising. More alkyl groups, more σ bonds available, more stabilisation — hence 3° > 2° > 1°. The increments are small: roughly 18 kJ mol⁻¹ for the first methyl, then 9, then 8. Compare that with carbocations, where each alkyl group is worth many tens of kJ mol⁻¹. Radical stabilisation by alkyl groups is real but modest.
Delocalisation. Allyl and benzyl radicals put the odd electron into an extended π system. This is worth 60–70 kJ mol⁻¹ — three to four times what a full set of alkyl groups buys — and it dominates every selectivity question where an allylic or benzylic position is available.
Adjacent π acceptors also help. This is the addition. Notice N≡C–CH₂–H at 397 kJ mol⁻¹: a cyano group stabilises a radical by about 42 kJ mol⁻¹, which is more than a full complement of methyl groups. A cyano group destabilises a carbocation catastrophically. Radicals are stabilised by donors and by acceptors, because a half-filled orbital can both give and take.

The four radicals whose relative stability is examined most often. The first two are stabilised by delocalisation into a π system (60–70 kJ mol⁻¹); the third only by hyperconjugation (about 35 kJ mol⁻¹ relative to methyl); the fourth is a plain primary radical, with the bulky tert-butyl group too far away to help electronically at all.

The allyl radical is not “a radical next to a double bond”; it is a three-centre π system holding three electrons. Two go into the bonding ψ₁ and one into the non-bonding ψ₂. Because ψ₂ has a node at the central carbon, the odd electron is shared equally between C1 and C3 and there is no net spin density in the middle. That is why allylic radicals give products at both ends, why ESR shows four equivalent terminal protons and one distinct central proton rather than five equivalent ones, and why the allylic C–H bond of propene is about 68 kJ mol⁻¹ weaker than the C–H bond of methane. Schematic MO drawing.

Advanced / reference layer

The spectroscopic evidence for radical geometry, the thermodynamic-versus-kinetic distinction, and the cases where radical stability and carbocation stability come apart.

How the geometry is actually known: ESR hyperfine coupling

Electron spin resonance measures the interaction between the unpaired electron and magnetic nuclei. The isotropic hyperfine coupling constant to a nucleus is proportional to the s-orbital spin density at that nucleus, because only an s orbital has non-zero amplitude at the nuclear position. This gives a direct experimental handle on hybridisation.

For •¹³CH₃ the ¹³C hyperfine coupling is small. For •¹³CF₃ it is several times larger. The interpretation is immediate: in the methyl radical the SOMO is essentially a pure p orbital, which has a node at carbon and therefore contributes almost no s density; in the trifluoromethyl radical the SOMO has substantial s character, so the carbon nucleus sits inside the orbital. •CF₃ is pyramidal; •CH₃ is not.

Thermodynamic stability is not kinetic persistence

This distinction is examined and is routinely got wrong.

Stabilised radical: low in energy relative to its precursor — a thermodynamic statement, measured by BDE.
Persistent radical: long-lived in solution — a kinetic statement, usually the result of steric shielding of the radical centre so that dimerisation is slow.

Where radical and cation stabilities part company

The examinable trap is the assumption that anything stabilising a carbocation stabilises a radical, in the same order and by a comparable amount. The order is often the same. The amounts are not, and in several important cases the direction reverses.

Structural feature	Effect on carbocation	Effect on radical	Do they agree?
Alkyl substitution (3° vs 1°)	Very large — tens of kJ mol⁻¹ per group	Small — about 35 kJ mol⁻¹ in total from CH₃• to t-Bu•	Same order, wildly different magnitude
α-OR, α-NR₂ (π donor)	Enormous — a full octet is restored (oxocarbenium, iminium)	Modest — roughly 20–35 kJ mol⁻¹	Same direction, not comparable
α-CN, α-C(O)R (π acceptor)	Strongly destabilising	Stabilising (+42 kJ mol⁻¹ for CN)	Opposite
α-F	Stabilising (lone-pair donation)	Essentially neutral to slightly destabilising	Effectively opposite
Bridgehead carbon	Prohibitive — planarity impossible	Tolerated — radicals pyramidalise cheaply	Opposite
Cyclopropylcarbinyl	Spectacular stabilisation (bisected conformer, σ donation)	Small stabilisation, and the radical ring-opens in nanoseconds	Opposite in practice
Adjacent π system (allyl, benzyl)	Large	Large	Yes — the one clean parallel

The trap is not that the two series are unrelated — for plain alkyl substitution they run parallel. The trap is transferring a carbocation intuition to a substituent that has a lone pair or a π acceptor, where the physics is genuinely different because a SOMO is half-filled and an empty p orbital is not.

Easy

Rank these C–H bonds in order of increasing BDE, and give the reasoning in one sentence each: (i) the C–H of ethyne, (ii) the benzylic C–H of toluene, (iii) the tertiary C–H of 2-methylpropane, (iv) the vinylic C–H of ethene.

Show solution

(ii) benzylic 375 < (iii) tertiary 404 < (iv) vinylic 465 < (i) alkynyl 558 kJ mol⁻¹.
(ii) Benzylic: the resulting radical is delocalised over the ring, so the bond is weakest.
(iii) Tertiary: hyperconjugation from nine C–H bonds, but no π delocalisation, so a smaller effect.
(iv) Vinylic: the radical is a σ radical in an sp² orbital, orthogonal to the π bond, so no delocalisation at all; and the higher s character makes the original bond strong.
(i) Alkynyl: the same σ-radical problem with an sp orbital — 50% s character, the strongest C–H bond in ordinary organic chemistry.
The general lesson: s character raises BDE, delocalisation lowers it, and delocalisation only helps if the SOMO can actually reach the π system.

Hard

A radical is generated at C-2 of optically pure (R)-2-bromobutane by treatment with Bu₃SnH/AIBN in the presence of a small amount of a chiral solvent. The product 2-substituted butane is racemic. Explain, and say what would have to be true for it not to be.

Show solution

The 2-butyl radical is planar or rapidly inverting at the radical carbon, so its two faces are enantiotopic and equally exposed. Whatever traps it — here a hydrogen atom from Bu₃SnH — approaches from either face with equal probability, giving a 50:50 racemate. A chiral solvent does not change this to any measurable extent because solvation is weak and non-covalent, and radicals are famously insensitive to the medium.
For a non-racemic product you would need one of: (a) a covalently attached chiral auxiliary close enough to block one face sterically; (b) a chiral Lewis acid bound to a coordinating group on the substrate, which is how modern enantioselective radical chemistry actually works; or (c) a radical whose inversion is slower than its trapping — achievable at a bridgehead or with strongly electronegative substituents, where the pyramidal form is a genuine minimum with a real barrier.

Read the rest of Part 3

The remaining 9 sections of this part — Chain reactions — initiation, propagation, termination, Selectivity — halogenation, NBS, and radical addition, Carbene electronic structure — singlet versus triplet, Making carbenes —… — and all nine parts of Reaction Mechanisms are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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