LearnChem · Part V — Why that product and not another?
Open almost any organic chemistry page and you will find curly arrows. A curly arrow is the full-headed arrow drawn across a reacting bond. It is drawn constantly and defined almost never. This module asks what that arrow is actually tracking. A proper answer needs four things. It needs a notation, a geometry, an energy and a slow step to name.
An equation shows what goes in and what comes out. What happens in between — or is there no in between?
Most equations hide the middle. Reactants on the left, products on the right, and nothing between them. Yet something did happen in between. Something moved, and it moved in a direction. This module asks what that something is. A notation for it already exists, drawn on millions of pages.
Module 24 left six electrons sitting still. Benzene shares them all round its ring. Staying put is exactly why that ring refuses to add bromine. That was the case where nothing moves. This module asks about the rest of the time. When electrons do move, something has to keep track of them.
You have met this arrow already. It curls from one place on a structure to another. Most pages never say what it means. Here is the rule. The tail sits on the electrons that are leaving. The head sits on the place they arrive. The arrow tracks electrons, never atoms. Where the atoms end up is worked out from that. One more thing is worth knowing first. Not every reaction moves its electrons two at a time. Predict which mark the reaction below needs.
Under UV light, Cl2 splits into two separate chlorine atoms. Curly arrow, or single-barbed fishhook?
Answer from your gut. The reveal follows.
Two marks, two meanings. A curly arrow has a full head, with two barbs. It always carries a pair. A fishhook has a single barb. It carries one electron only. Robinson drew the first such arrow for a mechanism in 1924. Chlorine's split under UV light is the second kind. The bond's two electrons separate, one going to each atom. Two fishhooks are needed, and no curly arrow will do. Chemists call that homolysis. Homolysis is a bond breaking with one electron going each way. This module is about the opposite case. It is about bonds being made, and about one pair arriving where there was nothing.
So where does the sentence you keep stand here? It says a pair moves when a new bond forms. Homolysis moves no pair at all. One electron goes each way. The sentence therefore says nothing about it. Module 21 carried a second case like this. A sodium atom hands one electron to another molecule. No bond forms in that step. Chemists call that single-electron transfer. Again only one electron moves. Again the sentence stays silent. Silent is not the same as wrong. The sentence was scoped on purpose, and these two cases sit outside that scope.
From here on, every bench on this page moves a pair. So the curly arrow is the notation for everything below. Read each arrow as two electrons travelling together.
The notation is settled. The harder question is where a pair can go. An arrow's head has to land somewhere. Somewhere is a place with a shape. The next question is therefore about geometry.
Two chlorine atoms meet and form Cl2. Which marks does that bond-making step need?
— The mark you draw is a claim about how many electrons travel together.
Start with the simplest reaction chemists have. A nucleophile attacks a carbon and a group leaves. A nucleophile is a species that brings a pair of electrons. The carbon receiving that pair is the electrophile. An electrophile is an atom that accepts a pair into a space that was empty. A leaving group is the group that departs, taking its bonding pair. Bromomethane, CH3-Br, is the standard example. The incoming pair has to arrive somewhere. A backside attack is an attack made from directly behind the leaving group. Whether that is the right picture is the bench's business, not this paragraph's. Now make the prediction below.
A nucleophile attacks from directly behind the bromine. Is (CH3)3C-Br faster or slower than CH3-Br?
Predict, then drive the bench.
Bench 1 is below. Set the substrate and the leaving group. Read the relative rate beside the verdict. Then look at the two sliders underneath. In 1935 Hughes and his co-workers ran the test those sliders reproduce. They mixed 2-iodooctane with radioactive iodide in acetone. R and S name the two mirror-image forms that molecule can take. The sample started as pure R, so it rotated polarised light one way. Two rates could then be followed at once. One was the rate at which labelled iodide exchanged into the molecule. The other was the rate at which the rotation died away. The bench calls the second divided by the first the rate ratio.
Three attack geometries are possible, and each predicts a different rate ratio. Start with a sample in which every molecule is R. Suppose the pair always arrives on the same side the iodide leaves. Then no molecule ever becomes S, the rotation never falls, and the rate ratio is 0. Suppose instead the pair arrives from either side at random. Each event then gives S half the time and R the other half. The sample turns racemic only once every molecule has exchanged, so the ratio is 1. Now suppose every event inverts, which is what arriving from directly behind would force. Then the sample is racemic once half of it has exchanged, not all of it. Rotation therefore falls twice as fast as label goes in, and the ratio is 2. Hughes measured 2. Only the third geometry predicts 2, so the pair arrives from directly behind. Dial the sliders and watch the rate ratio for yourself. A rate ratio above 2 would need more inversions than there were reactions. So 2 is the largest value this test can give, and Hughes found it.
The rule is not about how many groups sit on the reacting carbon. It is about whether the whole cone behind the leaving group is empty. Empty here means two things at once. There must be no electrons in that direction, and no atoms blocking the way. The reference geometry printed on the canvas shows why the direction is so tight. Both carbon-halogen bonds measure 2.147 angstrom, and the angle between them is 179 degrees. That is a symmetric halide exchange, not bromomethane with a nucleophile. It stands here as a reference for how straight the line has to be. Neopentyl bromide makes the point. Its own reacting carbon is a plain CH2, like any primary substrate. One bond further out sits a quaternary carbon, which is a carbon carrying four other carbons. That is far enough out to block the line anyway. Its rate comes from a separate experiment, on a separate scale, and the readout says so.
Picture a closed umbrella caught by a gust. The three retained C-H bonds start bowed away from the incoming pair, like ribs. As the pair arrives behind and the bromide pulls away in front, the ribs flatten and snap over. The analogy breaks in one place, and it matters. A flipped umbrella is damaged and does not reform. The molecule that comes out is whole, just mirror-imaged, and as sound as before.
Leaving groups have their own order, and it is not the bond-strength order. On Hoffmann's scale, chloride sits at 0.0024, bromide at 1, iodide at 1.9 and tosylate at 3.6. Tosylate has no carbon-halogen bond at all, and it still beats iodide. So the explanation cannot be how strong the broken bond was. It is how stable the anion is once it has left. A stable anion is a weak base. That is what the conjugate-acid pKa in the readout beside the rate tracks. Fluoride is the odd one out. Its bond to carbon is the shortest of the four, at 1.383 angstrom. HF's own pKa is 3.17, while HCl, HBr and HI are all strongly negative. Fluoride is therefore a genuinely strong base, and a strong base makes a poor leaving group. No measured rate for a primary alkyl fluoride was found, so no figure is offered here.
The geometry question is settled for the carbon being attacked. The pair needs an empty line behind the leaving group. But several different species could supply that pair. The next question is which of them gets there first.
Bench 1's readout gives neopentyl bromide's own rate. Does counting the groups on its reacting carbon predict it?
— The pair moves into the space directly behind the leaving group, and nowhere else.
A second question now opens. Several species can supply the pair, so which of them is quickest? Fluoride and iodide are the interesting pair to compare. Both carry a single negative charge. Fluoride is much the smaller of the two. Its charge is therefore packed into a far smaller volume. One line of reasoning follows from that. More charge in less space should mean a stronger pull on the carbon. The scale below measures the real thing, toward CH3-I in water. Predict the winner first.
Toward CH3-I in water, which is the better nucleophile, iodide or fluoride?
Predict, then read the scale below.
The panel below has one control and one table. Pick a nucleophile and the live line gives its relative rate toward CH3-I. The table lists all six on Pearson's protic scale. A protic solvent is one whose own hydrogens can hydrogen-bond to an anion. Water and methanol are both protic. Iodide wins, and not narrowly. Fluoride sits at the bottom of the same table.
| nucleophile | relative rate, protic scale |
|---|
Take the two halide entries from the table and divide the larger by the smaller. Iodide reads 25,100,000 and fluoride reads 500, so the quotient is 50,200. The hydration figures printed beside the table carry the reason for the first one. A hydration enthalpy is the energy released when water wraps itself round an ion. Fluoride's hydration enthalpy is -503 kJ/mol and iodide's is -298 kJ/mol. Subtract -298 from -503 and the difference is -205 kJ/mol. Water holds fluoride 205 kJ/mol more tightly, and that is the cost of freeing its pair. A second question sits in the same table, about basicity rather than size. Hydrosulfide reads 1,000,000,000 and hydroxide reads 3,163,000, which divide out to about 316. Hydrosulfide is the better nucleophile of the two, by roughly that factor. Keep the two comparisons apart, because they answer two different questions.
Water is the whole story here. Every dissolved anion is wrapped in water molecules. The smaller and more concentrated the charge, the tighter that wrapping. Fluoride is held hardest of the four. To attack a carbon, its pair must leave that shell first. Iodide's shell is loose, so iodide's pair is more available. In water, charge density matters the opposite way round from the guess.
Change the solvent and the order changes with it. An aprotic solvent has no hydrogens of its own to hold the anions. The printed order reverses, and the readout gives that order only. No reliable numbers across all four halides in one aprotic solvent were found. So none are shown.
The basicity contrast belongs here too. Hydroxide is a far stronger base than hydrosulfide. Hydrosulfide is still the better nucleophile. Those are two separate scales. One is measured by rate and the other at equilibrium.
Two scales have been in play in this section. Neither of them was the reaction's energy. The next section takes an energy diagram apart. The first thing it asks is what the numbers on it are measured from.
You need a fast substitution and you are choosing between two anions. Which scale should you read?
— A pair wrapped in water has to shed that water before it can move.
Module 16 gave every reaction a barrier to climb. It never said what that height is measured from. A height is always measured from something. Bench 2 below lets you change that something. Bench 2 is worked out, not measured. Its MODE control picks which reaction it draws. Set MODE to sn2_gas for chloride attacking chloromethane in the gas phase. Before you look, predict the shape of the path itself. Does it rise straight to its peak, or dip on the way?
In the gas phase, does the chloride plus chloromethane path rise straight to its peak, or dip first?
Predict, then set MODE to sn2_gas below and look.
Set MODE to sn2_gas and read the canvas. The path does not rise straight to its peak. It drops first. Then it climbs to the central barrier. Then it drops again on the far side. Those two dips are ion-dipole complexes. An ion-dipole complex is a loose pairing of an ion with a polar molecule. No bond holds it together. The ion's charge and the molecule's own uneven charge attract, and that is enough. They are real minima on the computed surface, not drawing flourishes. So the gas-phase path has one barrier with a genuine well each side.
Start with the thing this section is really about. Change ZERO_REFERENCE and the transition state does not move, but the printed number does. Measured from the free reactants it reads 11.5 kJ/mol. Measured from the first well it reads 55.5 kJ/mol. One peak, two numbers, and both of them correct. Subtract 11.5 from 55.5 and you get 44.0, which is how deep that well is. That is a derivation, and it is sound arithmetic. It is not evidence about the paper, because the same subtraction works for any two numbers. The paper also prints a second figure for the same well in its running text. Take 41.1 kJ/mol as the depth instead and the identity gives 14.4, not 11.5. It misses by 2.9 kJ/mol, and the readout says so.
Now switch MODE to sn2_solution. These figures are in kcal/mol, not kJ/mol, so they are not on the gas-phase scale. The page prints a computed span rather than one value. The cheap end of that span is 26.8 kcal/mol, and the experimental figure beside it is 26.6 kcal/mol. The more expensive calculation, which is the one that finds a dip at all, sits far above both.
A barrier is not one number until the reference point is named. Whenever you read one, look for the state the author called zero. The solution case adds a second caution. Whether any real dip survives in water depends on how the calculation was done. One method finds a shallow one. A cheaper method in the same paper finds none. That is not settled by a single calculation. SN2 in solution is usually drawn as one smooth hill. In the gas phase that drawing is wrong.
Both pictures so far show one mechanism. In it, the pair arrives as the leaving group departs. Both happen in one step. The next section breaks that order.
You read a barrier of 55.5 kJ/mol for this reaction somewhere else. What should you check first?
— A number printed in a real paper need not agree with that paper's own other numbers.
So far the pair has arrived as the group left. Suppose the group leaves first, with nothing waiting behind. The carbon it leaves is then one pair short. A carbocation is a carbon atom with a positive charge and three bonds. It is not a transition state. A transition state is a peak on the path, with no lifetime. An intermediate is a real species that lasts a measurable time. A carbocation is an intermediate. Tert-butyl chloride in water is the standard case. First predict which of its two steps is the slow one.
Tert-butyl chloride reacts with water. Which step is slower, losing the chloride or being caught by water?
Predict from what a peak and a real species each last.
Bench 2 has changed mode for you, to sn1. That mode is panel C, the third of the bench's three pictures. Read it from left to right. The reactants sit at zero. TS1 is the first peak, at its one sourced height of 85.7 kJ/mol. After it come two hollow points, joined by a dashed line. Hollow means the height was never measured. The first of them is the carbocation, and the second is the peak for capture by water. So ionisation is the slow step, and the rate gap below says by how much. Read those two points for their position in the order, not for a height. Note the solvent before reading too much into the one real number. TS1's 85.7 kJ/mol is an enthalpy in aqueous 2-propanol. Panel B, which is sn2_solution, gave free energies in pure water. They tell one story together. They are not one shared measurement.
| carbocation | formation enthalpy, kcal/mol | uncertainty, kcal/mol | alpha-hydrogen count |
|---|
Now the rate gap, which is what settles the slow step. Ionisation runs at 5.287 × 10−4 per second under those conditions. Capture of a simple tertiary cation by water runs at about 3 × 1012 per second. That second figure is a different cation in a different solvent. Divide the second by the first and take the base-ten logarithm of the quotient. The logarithm comes out at 15.75, and the inputs do not justify two decimals. Call it about sixteen orders of magnitude and leave it there. The well between the two peaks carries no depth number, and that is deliberate. No source gives that depth on the same scale as the barriers. One paper explains why, and the reason is physical rather than a missing measurement. The energy barely changes as the two ions drift apart. So there is no single bottom for a depth to be measured from.
Bench 3's table ranks the four cations by formation enthalpy. Lower formation enthalpy, steadier cation, so tert-butyl is the steadiest of the four. Methyl, ethyl, isopropyl and tert-butyl are the methyl, primary, secondary and tertiary cases in order. The last column of that table counts alpha-hydrogens. An alpha-hydrogen is a hydrogen on a carbon next to the charged one. The count rises down the table as the enthalpy falls, so it orders the ladder correctly. What it does not do is scale with the size of those gaps. Methyl minus tert-butyl is 261.8 minus 162.9, or 98.9 kcal/mol. Ethyl's 3 alpha-hydrogens are a third of tert-butyl's 9, so a third of 98.9 would be about 33. Ethyl's own gap from methyl is 261.8 minus 219.2, which is 42.6 kcal/mol. Set ION_A to ethyl and ION_B to tert-butyl and the readout says the two ratios are not equal. Counting alpha-hydrogens is an ordering, not a number.
Look at the first step of that same reaction again. The carbon-chlorine bond breaks, and both its electrons go with the chloride. Chemists call that heterolysis. Heterolysis is a bond breaking with the pair going entirely to one atom. Ask where the pair ends up. It ends up on chlorine, which is the electron-rich partner of the two. The same is true with bromide in place of chloride. Now read the sentence you keep against that step. It begins with the words when a new bond forms. This step forms no bond at all, so the sentence makes no claim here. It is silent by scope, not wrong. That silence was designed in. Any sentence about electrons flowing toward the poorer end would get this step backwards.
One step makes the intermediate. Everything after it is fast by comparison. That is why the first peak sets the rate. And the cation itself does not linger. Bench 3's own readout gives its lifetime in water, and it is shorter than a picosecond. That is barely longer than a single bond vibration.
Which step of this reaction does the kept sentence describe, the chloride leaving or the water arriving?
— A peak on the path has no lifetime. A real intermediate has one, and it is tiny.
The clearest case of the kept sentence on this page was the capture step. Water's pair moved into the space the chloride had left empty. SN1 and SN2 are not rivals picked at random. They are two answers to one question. Everything this page has driven is a condition. The substrate, the leaving group, the nucleophile and the solvent all have a say. Change any one of them far enough and the losing route starts winning. PVC piping is the everyday case. Its carbon-chlorine bonds hold for a building's lifetime. Nothing in a wall attacks them, and no solvent there ionises anything.
Take bromomethane, the case bench 1 opens on. To make SN1 the winner there, three things would have to change. The substrate would have to change, in the way section 2's own bench shows. The nucleophile would have to stop attacking. The solvent would have to pull ions apart on its own. Even all three together would not be enough. The cation left behind would be a methyl cation. Bench 3 puts methyl at the unstable end of its ladder. Now take the tertiary halide, whose own row on bench 1 you have already read. Its cation is the steadiest of the four. Give it an ionising solvent and SN1 is the route it takes. The substrate is the decisive choice, not the solvent.
Neopentyl bromide is blocked from SN2 by its own approach cone. Can it go SN1 instead?
Four things leave this module with you. Nucleophile and electrophile now have working definitions, one giving a pair and one accepting it. SN2 and SN1 are two ends of one spectrum, not two unrelated reactions. A transition state and an intermediate are now different things. One is a peak with no lifetime. The other is a real species, though a very short-lived one. Carbocation stability has both a ladder and a warning attached to it. The ladder is bench 3. The warning is that counting alpha-hydrogens orders that ladder without predicting its spacing.
— SN1 and SN2 are not unrelated reactions. They are two answers to one question.
Assume the reaction already has the push module 16 required. The sample starts as a single enantiomer, in module 23's sense. It exchanges its leaving group at {t} arbitrary units per unit time. Give the racemisation rate that would signal complete inversion. Then answer one more thing for yourself, in a sentence the box will not mark. Module 21 had a sodium atom hand one electron to another molecule. Say why that case could never be tested this way.
1.
2. Bench 3's table has the numbers for this. Divide ethyl's alpha-hydrogen count by isopropyl's. Then divide ethyl's formation enthalpy by isopropyl's. Say whether the two results are equal. Enter the second one as your answer.
The two ratios sit side by side on purpose. Counting alpha-hydrogens orders the ladder, and it does not predict how far apart the rungs sit.
When a new bond forms, one pair of electrons moves into a space that was empty.
Sources for this module, in the order the page uses them. The course's own textbook reference is Clayden, Greeves and Warren, Organic Chemistry. The curly arrow's convention is the IUPAC Gold Book. Wikipedia and Chemistry LibreTexts confirm the full head and the fishhook independently. Robinson's 1924 arrow is dated through Rzepa's own examination of the original figures, alongside Kermack and Robinson, 1922. The gate hints are built from documented student errors. Those are Flynn and Featherstone, 2017, in Chemistry Education Research and Practice. The inversion experiment is Hughes, Juliusburger, Masterman, Topley and Weiss, 1935. That paper's own full text could not be reached. The rate ratio is taken from a secondary source. The derivation is worked on the page instead.
The substrate rate ladder is Neuman, Organic Chemistry, Table 7.1. Neopentyl's figure is a separate experiment, Bordwell and Brannen, 1964, via Wikipedia's article on the Finkelstein reaction. The two are never put on one scale. The leaving-group rates are Hoffmann, 1965. The reference transition-state geometry is a teaching-level AM1 calculation, from an ekwan.github.io handout. The angle was computed here from that handout's own coordinates. The bond enthalpies are Blanksby and Ellison, 2003. The bond lengths are the NIST CCCBDB. No carbon-fluorine bond enthalpy is quoted, because two independent compilations disagree and their ranges barely overlap. The conjugate-acid pKa values are the Kutt and Leito compilation. Hydrofluoric acid's is from Petrucci, Harwood and Madura, 2007. Tosylic acid's is a Wikipedia infobox figure with no citation attached.
The protic nucleophilicity scale is Pearson, Sobel and Songstad, 1968. The aprotic reversal is stated as an order only, from three independent textbook sources. No numeric table across all four halides in one solvent was found. The hydration enthalpies are Burgess, Ions in Solution, 1999, Table 4.7. The gas-phase profile is Glukhovtsev, Pross and Radom, Table 5. The disagreeing figure quoted against it is that same paper's running text. The solution-phase figures are Tirado-Rives and Jorgensen, 2019. The experimental value there is McLennan, 1978. The ionisation barrier and rate are a solvolysis kinetics paper in aqueous 2-propanol. The capture rate is Toteva and Richard, 1996. The flat ion-pair surface is Schaefer, Mallavia and Ess, 2025, in Chemical Science, open access. The carbocation ladder is Houle and Beauchamp, 1979. Its four rows are the whole of that source's own table. Every citation here was checked for existence, link and licence at build time, and none was quoted from memory.