Start with a reaction that works. Hydroxide ion, HO⁻, meets bromomethane, CH₃Br. The oxygen uses one of its lone pairs to bond to the carbon, and at the same moment the C–Br bond breaks. Both electrons of that bond go with the bromine, which leaves as bromide ion, Br⁻. The products are methanol and bromide.
Bromide is the leaving group: the atom or group that breaks away from carbon and takes the bonding pair with it. The whole change is a substitution, because one group on the carbon has replaced another. The other two roles come from earlier in this chapter. Hydroxide is the nucleophile, the electron-pair donor that attacks. The δ+ carbon is the electrophile, the atom it attacks. The leaving group completes the set.
Now try the reverse. Mix methanol with bromide and nothing happens, even though bromide is a good nucleophile and the carbon of methanol is δ+. The trouble is that hydroxide would have to leave, and hydroxide will not go.
So the leaving group often decides whether a reaction happens at all. It is also the commonest reason a proposed mechanism is wrong: something departs that cannot. This section explains which groups leave and why, and how chemists turn a group that will not leave into one that will.
Why spreading charge out matters
Compare hydroxide, the group that would not leave, with iodide, bromide's larger neighbor in the halogen column. Hydroxide carries its negative charge on one small oxygen atom. Iodide carries the same single charge on a much bigger atom.
A negative charge is a crowd of electrons, and electrons repel each other. Squeeze the charge into a small space and the repulsion is strong. That repulsion is stored energy, and high energy means unstable. Give the same charge more room and the repulsion eases, so the ion is lower in energy and more stable. A stable ion is content to exist on its own, and existing on its own is exactly what a leaving group has to do.
There are two ways to give a charge room. One is a bigger atom, as in iodide. The other is to share the charge among several atoms, which is called delocalization. The tosylate ion is the standard example of the second way.
The name tosylate is short for p-toluenesulfonate. Chemists write the p-toluenesulfonyl group, CH₃–C₆H₄–SO₂–, as Ts. An alcohol's tosylate ester is then R–OTs, which in full is R–O–SO₂–C₆H₄–CH₃. Its sulfur carries two double-bonded oxygens and a benzene ring, and the ring has a methyl group on the carbon directly opposite the sulfur (the para position).
When the C–O bond breaks, the oxygen keeps the bonding pair and takes on a negative charge. The three resonance forms show the charge moving onto each of the three oxygens in turn (Resonance covers how to read them). The real ion is a blend of the three, with about a third of the charge on each oxygen, so no single oxygen is crowded.
So iodide leaves well, tosylate leaves well, and hydroxide leaves very poorly. Iodide and tosylate get their room in different ways, but the principle is the same: the more room a charge has, the lower its energy.
What makes a good one
Go back to bromide. Its conjugate acid, HBr, is a strong acid that gives up its proton completely in water. As Acids & Bases showed, the stronger an acid, the weaker its conjugate base. So bromide is a weak base: it is content to keep its electron pairs rather than use them. Hydroxide is the opposite case. Its conjugate acid is water, a very weak acid, so hydroxide is a strong base.
That gives the rule. A good leaving group is one that is stable on its own once it holds the bonding pair. That is the same as saying it is a weak base, which is the same as saying it is the conjugate base of a strong acid. Being able to move freely between these three statements is most of the skill.
The poor leaving groups are the strong bases: HO⁻, ⁻NH₂ and ⁻CH₃. Each keeps its whole charge on one small atom. Nitrogen and carbon are also less electronegative than oxygen, so they hold a negative charge even worse, and carbon, the least electronegative of the three, is the worst of all.
Quantifying it: pKaH
The number behind "weak" and "strong" is the pKa of the leaving group's conjugate acid: the species you get by putting a proton back on the leaving group. That number is written pKaH. For bromide the conjugate acid is HBr, so bromide's pKaH is the pKa of HBr, about −9. For hydroxide it is the pKa of water, 15.7.
This is the Acids & Bases scale read for a new purpose. A lower pKaH means a weaker base, and a weaker base is a better leaving group. The table lists the groups that commonly sit on carbon.
| Leaving group | Conjugate acid | pKaH | Verdict |
|---|---|---|---|
| N₂ | HN₂⁺ | very low | Superb |
| TsO⁻ | TsOH | −2.8 | Excellent |
| I⁻ | HI | −10 | Excellent |
| Br⁻ | HBr | −9 | Very good |
| Cl⁻ | HCl | −7 | Good |
| H₂O | H₃O⁺ | −1.7 | Good |
| F⁻ | HF | 3.2 | Poor |
| HO⁻ | H₂O | 15.7 | Terrible |
| ⁻NH₂ | NH₃ | 38 | Never |
| ⁻CH₃ | CH₄ | 50 | Never |
Read the table as a guide rather than a ruler. It gets the big gaps right. More than twenty pKa units separate chloride, bromide or iodide from hydroxide, and that gap is real: nothing with a pKaH above about 10 leaves an sp³ carbon on its own.
The table is less reliable at the top, where the values crowd together and other effects start to matter. It lists HI at −10 and TsOH at −2.8, so it predicts that iodide leaves faster than tosylate. Measured rates put the two close together, and which is faster depends on the reaction. So treat sulfonate ions such as tosylate, and iodide, as one top tier; a course may rank them either way. What the table does well is separate that tier from chloride, and chloride from hydroxide.
Rank CH₃–I, CH₃–OTs, CH₃–OH and CH₃–NH₂ by how fast a nucleophile displaces the group on carbon. The carbon is identical in all four, so the only variable is what leaves.
CH₃–NH₂ leaves as ⁻NH₂, whose conjugate acid is ammonia, pKaH 38. That is off the bottom of the table. This anion, the amide ion (not to be confused with the amide functional group), is one of the strongest bases in the course, and it never departs from carbon. It comes last, by a wide margin.
CH₃–OH leaves as HO⁻, pKaH 15.7. Hydroxide is also a strong base and also refuses to go. It comes third, and in practice it is as unreactive as the amine until the alcohol is activated.
CH₃–I leaves as I⁻, pKaH −10, and CH₃–OTs leaves as TsO⁻, pKaH −2.8. Both are conjugate bases of very strong acids, both are very weak bases, and both leave easily. Treat them as tied (see the paragraph on the table's limits) unless a question tells you otherwise.
Answer: CH₃–OTs ≈ CH₃–I ≫ CH₃–OH > CH₃–NH₂. The gap between the top pair and the bottom pair is many orders of magnitude. Inside the top pair the gap is small. Inside the bottom pair it does not matter, because neither group leaves at all. Getting the tiers right is worth more marks than getting the order inside a tier right.
The table also leaves out a second, smaller factor: the strength of the C–X bond that has to break on the way. The anion's stability matters more, but the bond matters too, and the halides show where each factor wins.
Ranking the halides
Put the four halomethanes side by side: CH₃F, CH₃Cl, CH₃Br and CH₃I. Two things change as you go down the column, and the figure shows both.
The order is I⁻ > Br⁻ > Cl⁻ > F⁻. Iodide is the largest halide and the most polarizable, meaning its electron cloud is the easiest to distort. It spreads its charge over the biggest volume, so it is the weakest base of the four. The bond points the same way: the C–I bond is the longest and weakest of the four, so it is the easiest to break. Both factors agree, which is why the halide order is so clean.
Fluorine is the most electronegative element, so you might expect fluoride to take the electrons happily and leave well. It does not, and that catches people out. Fluoride is still a weak base, since HF has a pKa of 3.2. It is, however, a much stronger base than the other three halides, because its charge sits on a small ion.
The bigger problem for fluoride is the bond. The C–F bond is the strongest bond carbon makes to any halogen: breaking it in CH₃F costs about 110 kcal/mol, against about 57 kcal/mol for CH₃I. Here the two factors disagree, and the bond strength decides. Alkyl fluorides are close to inert toward substitution. The same strong bond is why drug chemists often put a fluorine on a carbon where the body would otherwise oxidize a C–H bond: the C–F bond survives where the C–H bond would not.
Which way a substitution runs
Go back to the pair of reactions at the top of the page. Hydroxide pushes bromide off bromomethane, but bromide cannot push hydroxide off methanol. Hydroxide is the stronger base, and bromide is the weaker one. The reaction runs in the direction that sets the weaker base free.
Leaving as a neutral molecule
Everything so far has left as an anion. A group bonded to carbon through a positively charged atom leaves differently. Take CH₃–OH₂⁺, which is methanol with an extra proton on its oxygen. When the C–O bond breaks, the oxygen takes the bonding pair and becomes a neutral water molecule.
Leaving as a neutral molecule is far easier than leaving as an anion. The pair that the group takes with it cancels the positive charge instead of creating a negative one, and the negative charge was the expensive part. That is why water (pKaH −1.7) is a good leaving group while hydroxide (pKaH 15.7) is a very poor one.
Nitrogen follows the same pattern. An amine group, R–NH₂, never leaves, because it would have to go as ⁻NH₂. Protonate it to R–NH₃⁺ and the group that departs is neutral ammonia.
Ammonia is still fairly basic, though. Its conjugate acid, the ammonium ion NH₄⁺, has a pKa of about 9, so protonated amines rarely leave in practice. The version that gets used is the quaternary ammonium ion, R–NR₃⁺, a nitrogen bonded to four carbon groups. It leaves as a neutral amine, NR₃, and the Hofmann elimination is built on it.
On oxygen or on nitrogen, the idea is the same: put a positive charge on the leaving atom and the group can leave neutral. So when a poor leaving group sits on an atom that could take on a proton or an extra carbon group, look for that step. It is usually what a question is testing.
An alcohol has to be activated before it can leave
Alcohols are everywhere in synthesis, and the step that unlocks their chemistry is turning the OH into a group that will leave. That step is called activating the alcohol, and there are two ways to do it. Both follow from the sections above.
Put a positive charge on the oxygen. Protonate it, and the group that departs is neutral water instead of hydroxide, exactly as in the previous section.
Or replace the oxygen's hydrogen with a group that spreads the charge. Converting R–OH to the tosylate R–OTs, a sulfonate ester (an ester of a sulfur acid), does this. The group that has to leave is now the tosylate ion, pKaH −2.8, with its charge shared over three oxygens. That is about eighteen pKa units better than hydroxide.
Put 1-butanol in a flask with sodium bromide. Bromide is a good nucleophile and the carbon bearing the OH is δ+, so the ingredients of a substitution seem to be there. Nothing happens. Work out why.
What would have to leave? For bromide to bond to that carbon, the carbon must give up a bond, and the only candidate is the C–O bond. Breaking it would push the OH off as hydroxide.
Check the leaving group. Hydroxide's conjugate acid is water, pKaH 15.7. Hydroxide is a strong base and will not depart. The reaction stops here, and no amount of nucleophile fixes it, because the problem is not on the nucleophile's side.
Now run the same reaction with HBr instead of NaBr. HBr supplies the same bromide, but it also supplies a proton, and that changes everything.
Step 1: protonate. An oxygen lone pair takes the proton from HBr, and the H–Br bonding pair goes to the bromine. That is two arrows in a single step. The alcohol becomes R–OH₂⁺, and bromide is released into solution.
Step 2: substitute. The group that has to leave is now water, whose conjugate acid, H₃O⁺, has a pKa of −1.7. That is roughly seventeen units better than hydroxide. Bromide bonds to the carbon from the side opposite the oxygen, and the C–O bond breaks at the same moment, giving 1-bromobutane and water.
Step 2 is one event, not two. This carbon is primary, bonded to only one other carbon. If water left first, it would leave behind a primary carbocation, a carbon with only three bonds and a positive charge, and a primary carbocation is too unstable to form here. Drawing water leaving before bromide arrives invents that cation, and it is the commonest way this mechanism is drawn wrongly.
The lesson is not that acid is magic. The reaction was never held back by a weak nucleophile or an unreactive carbon. It was blocked by one impossible step, and the fix was to change the group that leaves. When a mechanism you have written refuses to work, look there first.
The tosylate route needs a reagent. The alcohol reacts with TsCl, p-toluenesulfonyl chloride, with pyridine present. Pyridine, C₅H₅N, is a benzene-like ring with one nitrogen in place of a CH, and here it serves as a mild base.
Look at which bonds break. The oxygen bonds to sulfur and chloride leaves, and then pyridine takes the proton off the oxygen. The C–O bond is never touched, so the carbon keeps the same oxygen from start to finish. Only the group beyond the oxygen has changed. Methanesulfonyl chloride, MsCl, does the same job and gives the closely related mesylate, R–OMs (Ms is the methanesulfonyl group, CH₃SO₂–).
Two more reagents activate the OH and then replace it with a halogen, all in one flask. Phosphorus tribromide, PBr₃, turns an alcohol into an alkyl bromide, and thionyl chloride, SOCl₂, turns it into an alkyl chloride. Both begin the way tosylation does: the alcohol's oxygen bonds to the reagent's central atom, and a halide ion leaves.
With PBr₃ the oxygen ends up bonded to phosphorus while still carrying its H, so it is positively charged. That lets H–O–PBr₂ leave as a neutral molecule, just as water did. The bromide released in step 1 bonds to the carbon from the side opposite the oxygen, and the C–O bond breaks in the same step.
SOCl₂ follows a similar plan with one extra step. After step 1, pyridine removes the H from the oxygen, leaving R–O–S(=O)–Cl. When chloride bonds to the carbon, this leaving group falls apart into sulfur dioxide gas and a second chloride ion. Both pieces are very stable, so the group leaves easily.
Compare the routes by the bond that breaks. Tosylation breaks O–H and S–Cl, and the C–O bond survives. PBr₃ and SOCl₂ break the C–O bond in their second step, as the halide arrives. What each reagent does to the three-dimensional arrangement at that carbon, why hot H₂SO₄, which protonates the OH but supplies no good nucleophile, gives alkenes instead, and which to reach for when, is the opening business of Alcohols, Ethers & Related Chemistry.
Reading the names SN1, SN2, E1, E2
Two kinds of reaction need a leaving group. Bromoethane with hydroxide can do either one.
In a substitution, the new group bonds to the carbon that held the leaving group, so one group replaces another. In an elimination, the base removes an H from the carbon next door instead. The pair from that C–H bond becomes a second bond between the two carbons, a C=C, and the leaving group departs.
The names SN1, SN2, E1 and E2 are built from these. The letter is the reaction type: S for substitution and E for elimination. The N in SN stands for nucleophilic, because the incoming group is a nucleophile.
The number counts the species that take part in the slowest step, the rate-determining step that sets the reaction's speed. 2 means two: the leaving group departs at the same moment the nucleophile or base attacks, in one step. Hydroxide and bromomethane at the top of this page react that way. 1 means one: the molecule ionizes on its own first, with the leaving group departing unassisted, and the nucleophile or base arrives afterward.
Step 1 leaves a carbocation; Carbocations covers these in detail. All four reactions need a leaving group to depart at some point, so no leaving group means no reaction. Substitution & Elimination covers each one in full.
What carries forward
Leaving-group quality is one of the four variables, along with the substrate, the nucleophile and the solvent, that decide which of SN1, SN2, E1 and E2 you get in Substitution & Elimination. Alcohol activation is a recurring move throughout Alcohols, Ethers & Related Chemistry and Amines. And in Carboxylic Acids & Derivatives, the entire reactivity order of the acid derivatives, acid chloride > anhydride > ester > amide, turns out to be largely a leaving-group ranking in disguise.