The earlier sections of this chapter were about structure: which atom is electron-poor, which group can leave, which lone pair does the attacking. This section adds energy. One picture, the reaction-coordinate diagram, answers the two questions you can ask about any reaction: how fast does it go, and how far does it go? The picture answers them in two different places, and neither answer predicts the other.
What the picture plots
Start with a reaction from Leaving groups. Hydroxide ion attacks bromomethane, and bromide leaves: HO− + CH₃Br → CH₃OH + Br−. It happens in a single step, with the new O–C bond forming while the old C–Br bond breaks. The figure plots the energy of the system as that step runs.
The vertical axis is free energy (G), the kind of energy that decides which side of a reaction is favored. Higher on the page means less stable.
The horizontal axis is the reaction coordinate. It tracks how far the bonds have changed, from the bonds the reactants have to the bonds the products have. It is not time, and it is not concentration. A molecule does not move along it at a steady speed; the axis simply puts the stages of the reaction in order.
Now read the curve from left to right. The flat stretch on the left is the reactants, and the flat stretch on the right is the products. The highest point between them is the transition state: the arrangement of atoms with the most energy along the path, caught with the old bond half broken and the new one half made.
Two heights on the picture have names. Both are measured from the reactants, but they end in different places:
- ΔG° is the height of the products relative to the reactants, the difference between the two plateaus. When it is negative (products lower), the reaction is exergonic. When it is positive (products higher), the reaction is endergonic.
- ΔG‡, read “delta G double dagger,” is the activation energy: the height of the peak above the reactants. The double dagger marks a quantity that belongs to a transition state rather than to something you could put in a bottle.
The hydroxide reaction runs downhill. Other reactions run uphill, and the next figure draws one of each.
How fast and how far are different questions
Diamond is less stable than graphite, the black carbon in pencil lead, at room temperature and pressure. So ΔG° for diamond turning into graphite is negative, and the change is downhill. Yet no one has ever seen a diamond turn into graphite on a shelf. The barrier is enormous, so the downhill change never gets started.
That one example carries the whole rule. ΔG‡ controls the rate. ΔG° controls the equilibrium. Neither one predicts the other. A reaction can be steeply downhill and take a century, or barely downhill and be over in a second.
One equation links ΔG° to the equilibrium:
Here R is the gas constant, T is the temperature in kelvin, and K is the equilibrium constant. The equation says that stability and the position of equilibrium are the same fact written two ways. At 25 °C it boils down to a handy rule: every 1.4 kcal/mol of ΔG° is a factor of ten in K. A reaction favored by 3 kcal/mol has a K of about 150, so more than 99% of the material ends up as product. One favored by 6 kcal/mol has a K of about 25,000.
Barriers follow the same arithmetic. Lowering ΔG‡ by 1.4 kcal/mol makes a reaction about ten times faster. A few kcal/mol may sound like a small difference, but it changes both rates and yields by large factors.
Transition state or intermediate?
Compare two species from Leaving groups. When hydroxide attacks bromomethane, there is a moment when the O–C bond is half formed and the C–Br bond is half broken. When (CH₃)₃C–Br ionizes instead, bromide departs first and leaves a carbocation behind: (CH₃)₃C⁺, a carbon with only three bonds and a positive charge. (Carbocations, the next section, covers them in detail.) The first species is a transition state. The second is an intermediate.
The energy diagram tells the two apart without any chemistry at all: a transition state is a maximum, and an intermediate is a minimum.
A transition state sits at the top of a hill. Its bonds are partly broken and partly formed, so they are drawn dotted, and any charge appears as partial charges (δ+ or δ−) on the atoms that share it. The whole structure goes in square brackets with a double dagger outside the top right corner. A transition state is not a compound. You cannot put it in a bottle or record its spectrum, and it lasts only about as long as one bond vibration.
An intermediate sits at the bottom of a valley between two hills. It is a real species with every bond fully made. Carbocations, radicals (species with an unpaired electron) and protonated alcohols are all intermediates. An intermediate may react within nanoseconds, but it exists, and it can sometimes be detected or trapped. Because it sits in a well, it has a barrier of its own to climb before it can go anywhere.
| Transition state | Intermediate | |
|---|---|---|
| On the diagram | a peak | a well between two peaks |
| Bonds | partial, dotted | all fully made |
| Drawn as | in brackets with ‡ | an ordinary structure |
| Lifetime | none to speak of | short but real |
| Isolable? | never | sometimes |
Multi-step reactions: one hump per step
The ionization of (CH₃)₃C–Br is the first step of a two-step reaction: bromide leaves, and then a nucleophile attacks the carbocation. Its diagram has two humps with a valley between them. Each hump is one step with its own transition state, labeled TS1 and TS2, and the valley is the carbocation.
So count the humps to count the steps, and count the valleys between them to count the intermediates.
The rate-determining step is the step that holds the whole reaction back. It is not always the step with the biggest climb of its own, so finding it takes some care.
All energies are in kcal/mol, measured from the reactants at 0. The figure below draws both cases.
Case 1. TS1 is at 18, the intermediate at 14, TS2 at 22 and the products at −10. Step 2’s own climb, from the intermediate up to TS2, is only 22 − 14 = 8, less than step 1’s climb of 18. Now look at what the intermediate does once it forms. Going back over TS1 costs it 18 − 14 = 4, while going on over TS2 costs 8. It falls back to the reactants faster than it moves forward. So the intermediate crosses back and forth over TS1 many times, and step 1 becomes a pre-equilibrium: a fast step that runs both ways. To become product, a molecule has to climb all the way from 0 to 22. Step 2 is rate-determining.
Case 2. This time the intermediate sits in a deep well. TS1 is at 15, the intermediate at −10, TS2 at 8 and the products at −20. Molecules cross TS1 faster than they can leave the well, so they pile up there. From there, the climb to TS2 is 8 − (−10) = 18, more than the 15 it took to reach TS1. Step 2 again sets how fast product forms, even though TS1 is now the higher peak.
Both cases follow one rule. Find the largest climb from the reactants, or from any valley, up to a peak that comes after it. That climb sets the rate, and the step whose transition state sits at its top is the rate-determining step.
Most courses teach a shorter version: the rate-determining step is the one with the highest transition state on the diagram. The shortcut gives the same answer whenever every intermediate lies above the reactants, which is the usual case and is Case 1 above. It can go wrong only when an intermediate sits below the starting level, as in Case 2. Then you have to measure from the bottom of the well.
Once you have found the rate-determining step, the steps after it do not affect the rate. Suppose TS1 sits at 25, the intermediate at 10 and TS2 at 15. Molecules wait at the first hump. Once over it, they cross the second (a climb of only 5, against 15 back over TS1) almost the moment they arrive. Making the second step faster changes nothing, because nothing was waiting there.
Catalysts and temperature: two ways to go faster
A bottle of hydrogen peroxide solution keeps for months. Add a pinch of manganese dioxide, or pour the peroxide on a cut, and it fizzes at once as it breaks down into water and oxygen. The manganese dioxide, like the enzyme catalase in the cut, is a catalyst: a substance that speeds up a reaction and comes out unchanged at the end.
A catalyst gives the reaction a different route with a lower peak, often by binding or protonating the reactant, which stabilizes the transition state. It leaves the two plateaus alone. The reactants and products are the same compounds at the same energies with or without the catalyst, so ΔG° does not change, and neither does K. A catalyst changes how quickly equilibrium is reached, never where it lies.
A catalyst also lowers the barrier by the same amount in both directions, because the reverse reaction climbs the same peak from the product side. That is why a catalyst for a reaction is automatically a catalyst for its reverse.
Real catalysts often replace one hill with two smaller ones, because the reactant first binds to the catalyst. The two end plateaus stay fixed all the same.
Temperature speeds a reaction up in a different way. It moves nothing on the diagram; it changes the molecules instead. The molecules in a sample carry a spread of energies, and only those in the high-energy tail of that spread have enough to cross the barrier. Warming the sample stretches the tail, so many more molecules can cross.
This is the Arrhenius picture, and only its qualitative form is needed here. The fraction of molecules that can cross depends exponentially on both the barrier height and the temperature, so a modest warming buys a large increase in rate. A common rule of thumb says that near room temperature, a rise of about 10 °C roughly doubles the rate of a typical organic reaction. Treat it as a rough guide rather than a law, since the true factor depends on how high the barrier is.
The Hammond postulate
The numbers in this part of the section are enthalpy changes, ΔH, rather than free-energy changes, because they come from bond strengths, and bond strengths are enthalpies. A diagram with enthalpy on the vertical axis has the same shape; only the words change. A downhill step is exothermic (ΔH negative, heat given out), and an uphill step is endothermic (ΔH positive, heat taken in). If a question gives ΔH instead of ΔG, read the diagram the same way.
Take a step that Radical halogenation depends on. A bromine atom pulls a hydrogen off an alkane: Br• + H–CR₃ → H–Br + •CR₃. The dot marks an unpaired electron, so Br• and •CR₃ are both radicals, and R stands for any carbon group. The step breaks a C–H bond and makes a weaker H–Br bond, so it is uphill, by about 10.5 kcal/mol for a secondary hydrogen.
The products already sit 10.5 kcal/mol up, and the peak only has to be a little higher, so the peak lies close to the products in energy. It lies close to them in structure too. At the top of the hill, the C–H bond is nearly broken and the H–Br bond is nearly made, so the transition state already looks much like the products.
The Hammond postulate turns that reasoning into a rule: a transition state resembles, in structure and in energy, whichever side of its step it lies closer to in energy. It works because two species on the same path that are close in energy are usually close in structure as well. This matters because no one can observe a transition state, yet its energy sets the rate. Hammond lets you reason about its structure anyway.
- An exothermic (downhill) step has its peak early, near the reactants and only a little above them. The transition state is reactant-like: its bonds have barely begun to change.
- An endothermic (uphill) step has its peak late, near the products and just above them. The transition state is product-like: the new species is almost fully formed at the top of the hill.
Chlorine shows the early case. Cl• + H–CR₃ is downhill by about 5 kcal/mol for a secondary hydrogen, because the new H–Cl bond is stronger than the C–H bond it replaces. Its peak comes early, and at the top the C–H bond has barely stretched.
The same reasoning applies to any uphill step that makes a reactive intermediate. Forming a carbocation is uphill, so the transition state of that step looks like the carbocation, and whatever stabilizes the carbocation lowers the barrier as well. Carbocations puts this to use.
The payoff of the postulate is a prediction about selectivity, which is how strongly a reaction prefers one product over another. Suppose a step can give two products, one more stable than the other. In an endothermic step, the transition state already looks like the product. Whatever stabilizes the product stabilizes the transition state almost as much, so the barrier drops by nearly the full amount and the two rates end up far apart. The step is selective.
In an exothermic step, the transition state looks like the reactants, and the reactants are the same for both routes. The product’s extra stability arrives too late to lower the barrier, so the two routes run at nearly the same rate. The step is unselective.
A halogen atom X• pulls a hydrogen off an alkane: X• + R–H → H–X + R•. The more carbons attached to the radical carbon, the more stable the radical left behind. Bond dissociation energies, the energies needed to break each bond into two radicals, put numbers on this: a primary C–H bond takes about 101 kcal/mol, a secondary about 98, and a tertiary about 96. A weaker C–H bond leaves behind a more stable radical, so the 3° radical sits about 5 kcal/mol below the 1° one. Both halogens face the same 5 kcal/mol gap, yet they use it very differently.
Step 1 — get ΔH for each abstraction. The step breaks a C–H bond and makes an H–X bond, so ΔH = (bond broken) − (bond made). For a secondary hydrogen with chlorine, 98 − 103 = −5, exothermic. With bromine, 98 − 87.5 = +10.5, endothermic. The difference comes entirely from H–Cl being a much stronger bond than H–Br.
Step 2 — place each peak. Chlorine’s step runs downhill, so by Hammond its peak is early, and the structure at the top still looks like the intact alkane. Bromine’s step runs uphill, so its peak is late, and the structure at the top already looks like the carbon radical.
Step 3 — ask what the 5 kcal/mol can act on. At chlorine’s early transition state the radical has barely begun to exist, so almost none of its stability is present yet, and the 3° and 1° barriers come out nearly equal. At bromine’s late transition state the radical is almost fully formed, so nearly the whole 5 kcal/mol already shows up in the barrier.
Step 4 — turn energy into rate. Every 1.4 kcal/mol is a factor of ten at room temperature, and a little less when hot, so a barrier difference approaching 5 kcal/mol is a rate ratio in the hundreds to thousands. Measured per hydrogen, bromination (usually run hot, near 125 °C) prefers a 3° hydrogen over a 1° one by roughly 1600 to 1, and chlorination by only about 5 to 1. The alkane is the same and so is the 5 kcal/mol of radical stability; the outcomes differ only because one step runs uphill and the other downhill.
Read the shortcut off it. If a question tells you a step is strongly exothermic, expect poor selectivity, with the product mix set mostly by how many hydrogens of each kind there are. If it tells you a step is endothermic, expect the stability order to win almost completely. Radical halogenation covers the full chemistry.
A preview: kinetic and thermodynamic products
This is a preview, not the full treatment. When one starting material can give two products, the one that forms faster, over the lower peak, is the kinetic product. The one that is more stable, in the deeper well, is the thermodynamic product. They need not be the same compound.
Which product you actually isolate depends on whether the reaction can run backward. If it cannot, you get mostly whichever product forms faster. If it can, the mixture keeps re-sorting itself until it settles in the deeper well. Low temperature tends to give the first situation and heat the second. Conjugation & Pericyclic Reactions develops this properly with the addition of HBr to a conjugated diene; for now, the two names simply attach to the two heights you have learned to read.
Four mistakes this diagram invites
What carries forward
Almost every mechanism from here on comes with one of these diagrams, and three habits will keep paying off. First, when a question asks about rate, find the climb that sets the rate and ask what stabilizes the transition state at its top; when it asks about yield or equilibrium, compare the plateaus and ignore the peaks. Second, whenever one product forms in preference to another, ask whether the preference comes from a barrier or from a stability, because Hammond says the answer depends on which way the step runs. Third, expect the diagram to be the question itself. Labeling the intermediate, marking the rate-determining step and choosing the right transition-state structure are all standard exam items, and you can read all three straight off a picture you can now draw.