Everything in this chapter so far has been a story about structure: which atom is electron-poor, which group is willing to leave, which lone pair goes where. This section supplies the other half of the account, which is about energy. One picture holds it — the reaction-coordinate diagram — and it answers the only two questions anybody ever asks about a reaction: how fast does it go, and how far does it go. The single most useful thing you can learn here is that those are separate questions with separate answers, and that the diagram keeps them in separate places.
What the picture actually plots
The vertical axis is free energy. Higher means less stable. The horizontal axis is the reaction coordinate, and it is not time and not concentration — it is "how far along the structural change has got," from the bonds the starting material had to the bonds the product has. A molecule does not travel along it at a steady speed; the axis exists only to put the species of the reaction in order.
On that canvas a one-step reaction is a single hill. The flat stretch on the left is the reactants, the flat stretch on the right is the products, and the maximum between them is the transition state. Two heights are worth naming, and they are measured from different places:
- ΔG° is the height of the products relative to the reactants — the difference between the two plateaus. Negative (products lower) is exergonic; positive (products higher) is endergonic.
- ΔG‡, read "delta G double dagger," is the activation energy: the height of the peak above the reactants. The double dagger is the mark that a quantity belongs to a transition state rather than to something you could put in a bottle.
Introductory courses often draw the same diagram with enthalpy on the axis instead, in which case the words become exothermic and endothermic and the label is ΔH. For most of what follows the distinction does not change the reasoning, and the radical numbers later in this section are quoted as ΔH because bond dissociation energies are enthalpies. Use whichever letter the question uses, and keep the shape of the picture.
How fast and how far are different questions
Here is the whole point of the section, stated as plainly as it can be. ΔG‡ controls the rate. ΔG° controls the equilibrium. Neither one predicts the other. A reaction can be enormously downhill and take a century, and it can be barely downhill and be over before you have capped the flask.
The equilibrium side has one equation worth carrying, and this section will not ask for another:
All it says is that stability and equilibrium position are the same fact written two ways. At 25 °C the useful version is that about 1.4 kcal/mol of ΔG° is a factor of ten in K, so a reaction favored by 3 kcal/mol sits at roughly 99:1 and one favored by 6 kcal/mol has essentially nothing left on the reactant side. The same arithmetic applies to barriers: lowering ΔG‡ by 1.4 kcal/mol multiplies the rate by about ten. Small energy differences are never small.
Transition state or intermediate?
These are the two things on the diagram that students mix up, and the diagram itself tells them apart without any chemistry at all: a transition state is a maximum and an intermediate is a minimum.
A transition state is the top of a hill. Bonds there are partly broken and partly formed, which is drawn with dotted lines and partial charges, and the whole structure is enclosed in square brackets with a double dagger outside the corner. It is not a compound. There is no bottle of it, no spectrum of it, no lifetime to quote — it is the single highest arrangement the system passes through, and it lasts about as long as one vibration of a bond.
An intermediate is the bottom of a valley between two hills. It is a real species with all its bonds fully made: a carbocation, a radical, a protonated alcohol. It may be wildly reactive and last only nanoseconds, but it exists, it can in favorable cases be detected or trapped, and — this is the useful part — it sits in a well, so it has its own barrier 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
Count the humps and you have counted the steps; count the wells between them and you have counted the intermediates. A two-step reaction is two hills with one valley in the middle, and each hill has its own transition state, conventionally labeled TS1 and TS2.
The rate-determining step is the slow one, and the slow one is the one with the biggest hill — but "biggest" has to be measured from the right place. The barrier for a step is the height of its peak above the valley immediately before it, not above the floor of the whole diagram. Step one is measured from the reactants; step two is measured from the intermediate. Getting this wrong is the classic multi-step diagram error, and it usually happens because the eye goes to whichever peak is drawn highest on the page rather than to whichever climb is longest.
Once the rate-determining step is identified, everything after it is invisible to the rate. If the first step has a 25 kcal/mol barrier and the second has a 5 kcal/mol barrier, molecules queue at the first hill and then fall over the second one the instant they arrive; making the second step faster still changes nothing, because nothing was waiting there. A chain is as slow as its slowest link, and speeding up the fast links does not help.
Catalysts and temperature: two ways to go faster
A catalyst gives the reaction a different route with a lower peak — usually by binding, protonating or otherwise stabilizing the transition state — and it is regenerated at the end. What matters for reading diagrams is what a catalyst does not touch: the two plateaus. Reactants and products are the same compounds with the same energies whether the catalyst is present or not, so ΔG° is unchanged, and therefore K is unchanged. A catalyst changes how quickly equilibrium is reached and never where it lies. It also lowers the barrier in both directions by the same amount, which is why a good catalyst for a reaction is automatically a good catalyst for its reverse.
Temperature works the other way: it moves nothing on the diagram and instead changes the molecules. Energy is distributed over a population, and only the molecules in the high tail of that distribution carry enough to clear the barrier. Warming the flask fattens the tail, so more collisions succeed. This is the Arrhenius idea, and the qualitative form is all that is needed here: the fraction of molecules that can clear a barrier depends exponentially on the barrier height and on temperature, so a modest warming buys a large factor in rate. The familiar rule of thumb is that near room temperature a rise of about 10 °C roughly doubles the rate of a typical organic reaction; treat it as an order-of-magnitude guide rather than a law, since the true factor depends on how big the barrier is.
The Hammond postulate
A transition state is the one species in the whole story that cannot be observed, and yet its energy is the thing that sets the rate. The Hammond postulate is the bridge: a transition state resembles, in both structure and energy, whichever species it is closer to in energy. Because two species close in energy are usually also close in structure, the peak of a hill looks like whichever end of that hill it sits nearer to.
- An exothermic (downhill) step has its peak early — near the reactants and only a little above them. The transition state is reactant-like: 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 nearly fully formed at the top of the hill.
The consequence is the reason the postulate is worth having, and it is a statement about selectivity. Suppose a step can give two different products and one of them is more stable than the other. For an endothermic step, the transition state already looks like the product, so whatever stabilizes the product stabilizes the transition state by almost as much — the barrier falls nearly the full amount, and the rates diverge sharply. The reaction is selective. For an exothermic step the transition state looks like the reactants, which are identical in the two cases; the product's extra stability arrives too late to show up in the barrier, so both routes have nearly the same rate and the reaction is unselective.
A halogen radical — an atom or group with one unpaired electron, written with a dot and taken up properly in the next section — pulls a hydrogen off an alkane: X· + R–H → H–X + R·. The carbon radical left behind is more stable the more substituted it is, and bond dissociation energies put a number on that: a primary C–H is about 101 kcal/mol, a secondary about 98, a tertiary about 96. A weaker C–H means a more stable radical, so the 3° radical sits about 5 kcal/mol below the 1° one. Both halogens face that same 5 kcal/mol gap. They exploit it completely differently.
Step 1 — get ΔH for each abstraction. The step breaks a C–H and makes an H–X, so ΔH = (bond broken) − (bond made). Taking 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 — put each on a hill. 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 C–H has barely stretched and the radical has barely begun to exist, so almost none of its stability is present yet: the 3° and 1° barriers end up nearly equal. At bromine's late transition state the radical is essentially fully formed, so almost the whole 5 kcal/mol is already in the barrier.
Step 4 — turn energy into rate. Using the 1.4 kcal/mol-per-factor-of-ten conversion from earlier, a barrier difference of nearly 5 kcal/mol is a rate ratio in the thousands. Measured per hydrogen, bromination prefers 3° over 1° by roughly 1600 to 1, chlorination by only about 5 to 1. Same alkane, same 5 kcal/mol of radical stability, opposite outcomes — and the only thing that changed was whether the step was uphill or downhill.
Read the shortcut off it. If a question tells you a step is strongly exothermic, expect poor selectivity and expect the product distribution to be governed by how many of each kind of hydrogen there are. If it tells you a step is endothermic, expect the stability order to dominate almost completely. The full radical chemistry is in the next section.
A preview: kinetic and thermodynamic products
This is a preview, not the treatment. When one starting material can give two products, the one that forms faster (the lower barrier) is the kinetic product, and the one that is more stable (the deeper well) is the thermodynamic product. There is no rule making them the same compound. Which one you actually isolate depends on whether the reaction can run backward: if it cannot, the product is whatever formed first, and if it can, the system keeps re-sorting itself until it settles in the deeper well. Low temperature tends to enforce the first situation and heat the second. Conjugation & Pericyclic Reactions develops this properly, with the diene case that makes it concrete; all that is needed now is that the two words name the two heights you have just learned to read.
Four mistakes this diagram invites
What carries forward
Almost every mechanism after this point comes with one of these diagrams attached, and three habits will keep paying. First, when you are asked about rate, find the rate-determining barrier and ask what stabilizes that one transition state; when you are asked about yield or equilibrium, compare the plateaus and ignore the peaks. Second, whenever you meet a claim that one product forms preferentially, ask whether the selectivity 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 transition state structure are all standard exam items, and all three are read straight off a picture you can now draw yourself.