Chemistry · Lesson

Carbocations, radicals and other reaction intermediates

Distinguish a reaction species from a transition state or resonance drawing. Follow explicit electron counts and finite intermediate accumulation before deciding whether a steady-state approximation is justified.

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By the end, you can

Compare cation, radical and anion electron counts without equating radical with neutral.
Separate resonance contributors, ion pairs and distinct intermediates.
Read a finite species balance for an intermediate.
Check the time and error scope of a steady-state approximation.

Charge and an unpaired electron are independent descriptors

Selected ordinary methyl Lewis representations CH₃⁺, CH₃· and CH₃⁻ each have three H nuclei. Their total valence electrons are six, seven and eight, respectively: carbon contributes four, H three, then subtract or add the net charge. The cation has no nonbonding electron, radical one and anion a pair in these explicit representations.

A radical is defined by unpaired-electron character rather than by compulsory neutrality. A selected formaldehyde ketyl radical-anion contributor, ·CH₂–O⁻, has net−1 and an unpaired electron in its stated Lewis bookkeeping. Actual spin/electron distributions can be delocalized, so a drawing does not establish a measured charge distribution, stability, lifetime or isolability.

Two contributor drawings need not be two species

The ordinary allylic cation and allylic radical each admit contributors that shift π bond order and the drawn charge or unpaired-electron location while preserving the same labeled nuclei, H attachments and total electrons. They describe one delocalized entity in each case. Treating those contributors as consecutive reaction intermediates would invent a species population and a step.

An ion pair instead describes oppositely charged ions associated under the relevant medium conditions; it is not automatically a new ordinary covalent bond. The closeness and persistence of ionic partners can influence capture and stereochemistry. A comparison of contributors, ion pairs and separate species therefore needs both the atomic/electronic description and the physical model being used.

A finite intermediate occupies a species state

A reaction intermediate is formed and consumed along a mechanism and can have a finite population. A transition state instead describes the relevant dividing/saddle region in a rate model; it is not counted as a stable reservoir species merely because a diagram labels it. A potential-energy minimum, a solution free-energy basin and an experimentally detected intermediate are related but distinct kinds of evidence.

An intermediate is not automatically negligibly populated or too transient to detect. Whether it accumulates depends on formation/removal rates and observation time. A relative-stability slogan alone does not supply a rate, and the existence of one detectable species does not uniquely establish every step around it.

Compare formation with removal explicitly

For the invented closed first-order network A→I→P, let k₁=0.100 s⁻¹, k₂=0.300 s⁻¹ and initial normalized amounts A=1, I=P=0. At fixed volume, A=e^(−k₁t), I=[k₁/(k₂−k₁)](e^(−k₁t)−e^(−k₂t)), and P=1−A−I. One constituent unit is conserved in each species, so their normalized amounts sum to one.

The intermediate peaks at t=ln(k₂/k₁)/(k₂−k₁)≈ 5.49306 s, reaching about 0.192450 of the total. At that instant formation and removal match, but that instantaneous zero derivative is not a statement that the whole reaction is at equilibrium. The graph is an exact solution of the stipulated model, not measured organic-reaction data.

A finite intermediate can accumulate to a measurable model fraction

Normalized amount fraction

Elapsed time / s

Starting species A
Intermediate I
Final species P
— Exact intermediate fraction

Underlying data

Elapsed time / s	Normalized amount fraction
0	1 · Starting species A
1	0.9048374180359595 · Starting species A
3	0.7408182206817178 · Starting species A
5.493061443340547	0.5773502691896258 · Starting species A
10	0.36787944117144233 · Starting species A
30	0.049787068367863944 · Starting species A
0	0 · Intermediate I
1	0.08200959867712082 · Intermediate I
3	0.1671242804705593 · Intermediate I
5.493061443340547	0.19245008972987526 · Intermediate I
10	0.1590461864017892 · Intermediate I
30	0.024831829281888634 · Intermediate I
0	0 · Final species P
1	0.013152983286919662 · Final species P
3	0.09205749884772294 · Final species P
5.493061443340547	0.2301996410804989 · Final species P
10	0.4730743724267685 · Final species P
30	0.9253811023502474 · Final species P

Original closed first-order A→I→P model with k₁=0.100 s⁻¹, k₂=0.300 s⁻¹, fixed volume and initial normalized amounts(1,0,0). Markers show all three conserved species fractions; the dashed curve draws only I. One constituent unit is retained in each species. This is not a transition-state population, a steady-state elimination or real experimental data.

A small accumulation derivative is an approximation condition

A steady-state elimination of I in this network sets dI/dt≈ 0 and estimates I≈(k₁/k₂)A. It needs a compatible regime, such as fast removal after an initial transient, and should be checked against the quantities of interest. It does not set all forward/reverse fluxes to equilibrium or say I is exactly constant at every time.

For a separate invented case k₁=0.00100 and k₂=0.100 s⁻¹ at 100 s, exact A≈ 0.904837 and I≈ 0.00913931; the approximation gives I≈ 0.00904837, about 0.995% low. At t=0 it would assign I=0.0100 while the specified exact I=0, so applying it from startup would be unjustified. Keep the approximation, its error and the model assumptions explicit.

Worked examples

Retain the same nuclei while changing charge and electron count

Count total valence electrons for CH₃⁺, CH₃· and CH₃⁻.

The neutral nuclear valence sum is 4+3=7.
For+1, subtract one:6.
For the neutral radical retain7; for−1 add one:8.

Six, seven and eight electrons in the selected ordinary Lewis representations, with identical C₁H₃ nuclei.

Find when formation and removal match

For the stated k₁=0.100, k₂=0.300 s⁻¹ network, find the peak time and I fraction.

t_peak=ln(3)/0.200≈ 5.493061 s.
At that time A≈ 0.577350.
I=0.500(e^(−0.100t)−e^(−0.300t))≈ 0.192450.
P=1−A−I≈ 0.230200.

The intermediate peaks at about 5.49 s with 19.2% of the conserved normalized amount. The whole network is not thereby at equilibrium.

Quantify the separate steady-state approximation error

For the separate fast-removal case at 100 s, compare I_exact=0.0091393133 with I_approx=0.0090483742.

Relative error=(I_approx−I_exact)/I_exact.
The result is about−0.00995032.
Multiplying by 100 gives−0.995032%.

The approximate I is about 0.995% low at this specified time; its startup behaviour still fails the exact initial condition.

Try it yourself

Must a radical have net charge zero?
Show the answer and reasoning
No.
Charged radicals have unpaired-electron character and a nonzero net charge.
Should the two ordinary allylic radical contributors each receive a separate concentration in this one-entity description?
Show the answer and reasoning
No.
The contributors describe electron redistribution within the same nuclear framework, not separate species.
What is A+I+P throughout the stipulated normalized closed network?
Show the answer and reasoning
One.
Each modeled species retains one conserved constituent unit and there is no external feed or loss.
May the stated I≈(k₁/k₂)A approximation be imposed as the exact startup value when I(0)=0?
Show the answer and reasoning
No.
It predicts nonzero I initially and needs a justified later-time regime.

What to carry forward

Count charge and unpaired electrons separately.
Do not turn resonance contributors into intermediate species.
A transition state is not a finite reservoir in this model.
Write formation, removal and conserved balances.
Check an approximation’s regime and error before using it.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.