LearnChem · Part V — Why that product and not another?
When a reaction can give two products, something has to decide which one you get.
Why does a cook get caramel at one temperature and something bitter and black at another, from the same sugar?
Same sugar, same pan, two very different results. So what is the extra heat actually changing?
Module 25 drew the curly arrow. An arrow is electrons caught in the act of moving. You followed a pair of electrons out of one bond and into a space that was empty. That gave you a mechanism, step by step, with a barrier on each step. A mechanism tells you how a product gets made. It does not yet tell you which product you end up holding. Two arrows can leave the same starting material and land in different places. This page is about choosing between them.
Picture a fork in the road. Two roads leave the same point. One of them is quicker to walk. The other ends somewhere you would rather be. Which road you take depends on more than speed. Chemistry has forks like this one. A cook meets one every time sugar goes into a hot pan. Low and slow, the sugar turns golden and sweet. Hotter, it turns dark, bitter and smoky. Here is one thing to go on. The golden stage and the black stage smell of quite different things. Make a guess about what the extra heat is really doing.
Caramel first, then bitter black, in the same pan. One reaction carried further, or two different reactions?
Answer from your gut. The reveal follows.
Caramelisation is a whole family of reactions. Sugar molecules break apart and join up again in new ways. That family needs a certain amount of push to get going. The charring family needs a great deal more. Turn the heat up and the second family gets going too. The char does come out of the pan the caramel was in. But it is not caramel pushed further along one road. It is a second road, opened by the extra heat. Neither road runs backwards. Burnt sugar never turns back into syrup in the pan. So the cook is not choosing between a fast product and a stable one. The cook is choosing which family of reactions gets to run at all.
Now hold a picture in your head for the rest of the page. You are walking, and the road forks. Whichever branch you take, there is a gate at the end of it. The only question that matters is whether your gate swings. A gate that swings both ways lets you come back out and try again. A gate that opens one way keeps you where it put you. Some roads are fitted with one kind and some with the other. Which kind you walked through decides whether anything can still change.
You get the product that forms fastest, unless it can go back — then waiting gives the most stable product.
That sentence is the whole module. The rest of this page tests it against real reactions, starting with the commonest case of all.
— Two different questions: what forms first, and what survives on standing.
Two answers need two names. Chemists call them kinetic control and thermodynamic control. IUPAC is the body that writes chemistry's definitions. Its definition of kinetic control turns on the relative rates of the paths running forward. The same wording names reaction times among the conditions. Its definition of thermodynamic control turns on an equilibrium instead, with the products able to turn into each other. Now read both definitions again and hunt for the word temperature. It is in neither of them. One is about speed. The other is about a reaction that can run both ways and settle. Hold that, and start with the simplest addition in organic chemistry. Propene and hydrogen bromide give 2-bromopropane.
Heat the propene and hydrogen bromide mixture hard. Does the major product switch to something else?
Predict, then read the arithmetic below.
Hydrogen bromide adds to propene in two steps. The first step makes a cation. A cation is a piece of a molecule that carries a positive charge. That charge can land on the middle carbon or on an end carbon. The middle one is far more stable, so its path has the lower barrier. That path is the faster one, and it gives 2-bromopropane. This is Markovnikov's rule, reached from the mechanism rather than learned as a slogan. No source found gives a measured ratio for this reaction, so this page gives none. What it can say is which product forms, and that very little of the other one turns up.
The panel above does the subtraction in the open, in two lines. The first line takes the ethyl cation's formation enthalpy and subtracts the isopropyl cation's value from it. Both figures are gas-phase measurements in kcal per mole. The ethyl cation stands in for the primary cation, which that table never measured. The difference comes out positive, so the isopropyl cation is the more stable of the two. The panel then converts that difference from kcal per mole into kJ per mole. The second line subtracts the two bromopropanes' formation enthalpies in the same direction. That second gap is printed as a range, because two sets of measurements disagree about its size. It is far smaller than the cation gap, in the same units. So the faster path and the more stable product are the same product here.
Could 2-bromopropane come apart again and give the other product a chance? The panel's last line is the barrier for that reverse step. It was measured on a close relative, 2-chloropropane losing hydrogen chloride. Now compare that barrier with the energy an ordinary collision carries at room temperature. The barrier is bigger by a factor of many tens. The team who measured it worked hundreds of kelvin above a hot oven. Below that the reaction barely moved. So this product stays where it is, and the exception clause in our sentence never fires. The fast product is the answer at every temperature you can reach.
One reaction can look a lot like another one running backwards. The check below asks you to tell those two apart.
A strong base turns 2-bromopropane into propene at room temperature. Is that the addition running backwards?
— One product, at every temperature.
Now a system where two products really are on offer. 1,3-butadiene carries two double bonds, one bond apart. Add hydrogen bromide and the first step makes a cation again. Here the positive charge is spread over three carbons at once. Chemists call that an allylic cation. The bromide then lands on one of two carbons. Landing next door gives the 1,2-adduct. Landing at the far end gives the 1,4-adduct. For this section only, we forbid either product to turn back. Section 4 takes that rule away again. Two things decide a race like this one. One is the barrier each path has to climb. The other is how often the molecules meet facing the right way. Chemists call that second one the A-factor.
Heat that race hard, with neither product allowed back. Where does the ratio end up?
Predict, then drive the bench.
The bench below runs that race for you. Three controls drive it. The first sets the barrier gap between the two paths, in kJ per mole. The second sets the A-factor ratio, which compares how often each path's molecules meet facing the right way. Nobody has measured that ratio for this system, so it is yours to explore. The third sets the temperature, in kelvin. At 193 kelvin the measured split is 81 to 19 in favour of the 1,2-adduct. Call that the cold anchor. It is the one setting here that matches a real experiment. Watch the two bars, then read the two check lines underneath.
The barrier control is a range and not a single setting, on purpose. Nobody has measured that gap directly. It is worked backwards from the measured splits, and only by assuming both paths meet just as often. What comes out is about 2 to 6 kJ per mole, or roughly 0.5 to 1.5 kcal per mole. Here is the arithmetic the bench runs on it. The rate ratio is e raised to the gap divided by R times the temperature. R is the gas constant, 8.314 joules per kelvin per mole, and the gap goes in as joules per mole. Multiply that by the A-factor ratio and you have the answer the bars show. At 193 kelvin with equal A-factors it returns the cold anchor.
The two lines under the bench check that arithmetic, in public. The first asks the bench to reproduce the cold anchor from the barrier gap alone. It would go wrong if the sign or the units were backwards. The second drives the temperature to an absurd height, far past anything the slider offers. It sets the A-factor ratio away from one and watches where the ratio lands. It would go wrong if the code ignored that control. Now look back at the hot pan. The charring family has the higher barrier, so it only gets going when the pan is hot. Nothing in that pan goes back either. The fast product is still what you get. The heat changed which path was fastest, and it never handed over a stable one.
The bench stops at 500 kelvin, and the arithmetic does not. The check below asks you about the gap between those two things.
Set the A-factor ratio to 2 and push the temperature slider to its top. Is the split 2:1 yet?
— With no way back, heat only erodes the lead. It never hands over the win.
Section 3 forbade either product to turn back. Now take the forbidding away. Butadiene and hydrogen bromide is the one reaction on this page that really can go back. The 1,2-adduct can lose its bromide again and return to the shared cation. Section 3 left that out on purpose, so that one thing at a time changed. Chemists have known this case for decades. Mixed cold, one adduct dominates. Left alone warm, the books report the other one. That usually gets filed under temperature and then forgotten. A run has more in it than the number on a thermometer. Make your prediction before you touch the bench.
Two flasks sit at 40 degrees Celsius. Read one straight after mixing and one much later. Same answer?
Predict, then set the controls below and look.
Here is the measured passage, in plain numbers. Mix butadiene with hydrogen bromide at 0 degrees Celsius and you get a 71:29 mixture. The first figure is the 1,2-adduct and the second is the 1,4-adduct. Now take that cold mixture, hold it at 40 degrees Celsius, and wait. The hydrogen bromide has to still be there, and that turns out to be the clue. The textbook says the ratio slowly changes to 15:85. Stop on that sentence. Product that had already formed turned into the other product. The only way that happens is if the 1,2-adduct comes apart again. The bromide leaves, the cation gets a second chance, and this time it is caught at the far end. Nobody had to assume a way back. The changing ratio forces it. Now the word slowly. Read the flask early at that same 40 degrees Celsius and it has barely moved. That early reading follows from the word slowly, and is not a measurement anyone reported. No source found gives a time or a half-life for that change. So the elapsed control below is an ordering, not a clock.
Now clear the go-back box and set the temperature to 40 degrees Celsius. The bench falls back to bench 1's arithmetic, with no way back at all. The split barely moves off the cold one, and it never approaches 15:85. Think about what a way back actually does. Every time the 1,2-adduct comes apart, that cation is dealt again. The 1,4-adduct comes apart far less often, because it is the steadier of the two. Its double bond carries more carbon groups, and that arrangement sits lower in energy. So the mixture drains, slowly, into the product that is hardest to undo. That is what most stable means here, and it is why waiting is the thing that buys it. Now the sentence the rest of this page needs. Heat is usually how you buy a way back, and time is how you spend it. Buy it without spending it and you still hold the fast product. That is why the hot-gives-stable story works so often, and why it is not the rule. One last thing about the gate picture. A real gate is open or shut, and a way back is not. The same reaction at the same temperature can look shut over five minutes and open over five hours.
That rule is not about this flask alone. The check below takes it back to a flask from earlier on the page.
You leave section 2's flask of 2-bromopropane alone for a week. What do you find in it?
— One flask, one temperature, two readings.
Back to propene and hydrogen bromide, from section 2. For years chemists got inconsistent results from what looked like one experiment. Some flasks behaved one way and some another, in the same hands. The culprit turned out to be traces of peroxide. A peroxide is a compound with a weak oxygen-to-oxygen bond in it. Old solvents pick peroxides up from the air on their own. Add some on purpose and the run switches, every time. This section is about what that small addition really does. Decide first what kind of change you expect.
A peroxide changes how propene and hydrogen bromide react. Is that a faster route, or a different one?
Predict, then read the chain steps below.
Here is what the peroxide does. Its oxygen-to-oxygen bond is the weakest bond in the flask, so it breaks first. That gives two fragments, each carrying one unpaired electron. A fragment like that is called a radical. One radical pulls the hydrogen off hydrogen bromide and leaves a bromine atom behind. That whole opening is called initiation, and it happens once per chain. Then two steps repeat over and over. The bromine atom adds to the double bond first. The carbon radical left behind takes a hydrogen from another hydrogen bromide molecule. That second step makes the product and hands back a bromine atom. Without peroxide, a proton attacks the double bond first. With peroxide, a bromine atom does.
The panel above prints the two repeating steps for hydrogen bromide. Both of them release energy, and that is what keeps a chain alive. A chain dies the moment one of its two steps has to be pushed uphill. The other hydrogen halides are not alike here. The bond from hydrogen to chlorine is much stronger than the bond from hydrogen to bromine. The bond from hydrogen to iodine is weaker than both. The carbon to iodine bond it would form is weak too. So bond strengths decide which halide can run a chain. Temperature decides nothing at all here.
Why does the bromine atom go to the end carbon? Look at the last two figures in the panel, in kJ per mole. They come from bond strengths, and they say a secondary radical is the more stable one. Adding at the end carbon leaves the unpaired electron on the middle carbon, which is secondary. Adding at the middle carbon would leave it on an end carbon, which is primary. The first route has the lower barrier, so it is the faster one. Chemists call what it gives the anti-Markovnikov product. Can that product go back? It is an alkyl bromide like section 2's, and undoing that kind of addition needs a barrier far above any bench. So the fast product is the only product here too. The rule did not change. The species that attacks first did.
This trick does not carry over to every hydrogen halide. The check below asks you where the trouble sits.
Run the same peroxide trick with hydrogen chloride and the chain dies. Which repeating step kills it?
— Fastest wins, whichever step starts the chain.
One more family, and the module's hardest test. Benzene is a ring of six carbons that share their electrons all round. Put a group on that ring and it still reacts. Nitration puts a nitro group on. Sulfonation puts a sulfonic acid group on instead. The new group does not land at random. It can sit beside the group already there, across from it, or in between. Chemists call those three positions ortho, para and meta. Usually one or two of them take nearly everything. The question here is whether that needs a rule of its own. Or whether the sentence you have been testing already covers it.
Heating toluene nitration never moves the product to meta. Can heat move naphthalene sulfonation's product?
Predict, then query the table below.
Start with why any position wins at all. The first step is the slow one, and it leaves the ring carrying a positive charge. Whatever steadies that charge best gives the lowest barrier and the fastest path. A group that pushes electron density in can steady the charge from the ortho and para positions. From meta it cannot reach it. So the ortho and para paths form fastest. A group that pulls electron density out leaves meta as the least bad option. Now the second half of our sentence. Nitration and halogenation cannot undo themselves. So the fastest product is the product you get, at every temperature anyone has tried. The bench below holds seven measured examples. Stock and Brown collected them in 1963, and they are quoted from Rablen's 2025 paper. Nobody states the acid, the temperature or the solvent behind them. Read them as a pattern rather than a recipe.
| substrate | ortho, percent | meta, percent | para, percent |
|---|
The bench prints a check line under each row. The source gave a second column as well, counting each result position by position. A ring has two ortho positions, two meta positions and one para position. So the ortho percentage is two hundred times the ortho figure, divided by the whole count. Meta is worked the same way, and para is one hundred times its own figure. It is a way of checking one column of the paper against the other. They agree to about one percentage point in every row. One more thing about this table. Published percentages often do not add up to exactly a hundred, and these do not. Four of the seven rows miss it, and anisole misses by the most. They print here exactly as published.
Now the one ring reaction that breaks the pattern. Naphthalene is two benzene rings fused along one edge. Sulfonating it can put the group in two places. The 1-position is the faster one, because the charged ring it goes through is the steadier of the two. The 2-position gives the more stable product, because the 1-position sits crowded up against the second ring. Here the fast product and the stable product are different molecules. That is the fork this module opened with, on a ring. Now the proof, and it is not a temperature. Take the pure 1-isomer, put it back in sulfuric acid and heat it, and it turns into the 2-isomer. Product that had already formed came apart and formed again somewhere else. Heat the product with dilute watery acid instead and the group comes off altogether, leaving naphthalene. Sulfonation can be undone, and nitration cannot. That one difference is why this is the only ring reaction here that switches. The readout above gives the temperatures the textbooks report, and two of those sources disagree about the low one. Read those temperatures as the price of the way back, not as the rule.
The nitrobenzene row deserves one more line. Stock and Brown's figures are the ones the bench prints. Hoggett and colleagues published a different set in 1971, as ranges rather than single values. Their ortho figure runs from 5 to 8 percent, their meta from 91 to 93, and their para from 0 to 2. The two sets overlap, and they are not the same. When two careful groups disagree, you are better off seeing both.
One warning before the check. How fast a ring reacts and where the next group lands are two different questions. A group can slow the whole ring down and still decide the position. So do not read the position off the speed, or the speed off the position. The table in front of you holds both kinds of behaviour.
One row of that table is worth predicting before you look at it. The check below names the substrate and asks you for the shape of its answer.
Before you look, predict this one:
Bromobenzene's published row and its check line agree, and both sit on the ortho and para side.
— Ortho and para at every temperature, unless the group can come off again.
Two paths race each other with equal A-factors, at 250 kelvin. The barrier gap between them is {t} kJ per mole. Give the rate ratio of the faster path to the slower one, to two decimal places. Then say how this race differs from the one-electron steps that rust a metal in module 21. This box will not mark that sentence.
1.
2. Cyclopentadiene reacts with itself to make a dimer, and there are two possible dimers. Here are three facts, and nothing else. The transition state leading to the endo dimer sits about 3.3 kcal per mole lower than the one leading to the exo dimer. The exo dimer is about 0.7 kcal per mole more stable once it exists. Both of those came in kcal per mole, while this page has worked in kJ per mole. The reverse reaction, which splits a dimer back into two cyclopentadienes, only gets going above about 150 degrees Celsius. Name the main product of a short run at room temperature. Then name the main product of a long run held well above 150 degrees Celsius. Last, before you open the worked answer, say to yourself what a short run above 150 degrees Celsius would give. Nothing marks that one.
Use this module's rule and nothing else. In a short run at room temperature the reverse step is not available at all. So the only thing that matters is which transition state is lower. That is the endo one, so the endo dimer is the main product. In a long run above 150 degrees Celsius the reverse step opens. Dimers come apart and form again, and time is on offer. The more stable dimer wins that exchange, and that is the exo one. Now the third run. Above 150 degrees Celsius the way back is open, but a short run never spends it. So the rule predicts mostly endo again. That one is a prediction from the rule, not a reported measurement. Heat bought the way back. Only time spends it.
You get the product that forms fastest, unless it can go back — then waiting gives the most stable product.