Warm tert-butyl bromide, (CH3)3C–Br, in water and it slowly turns into tert-butanol, (CH3)3C–OH. The bromine has been swapped for an OH group, just as in an SN2 reaction, but the route is different. Here the bromide leaves first, on its own, and water bonds to the carbon only afterward. This two-step route is SN1. Its rate law, its stereochemistry, the substrates that react and the solvents that help all differ from SN2, and each difference follows from that order of events.
Two steps instead of one
Follow tert-butyl bromide through the reaction. In the first step the C–Br bond breaks without any nucleophile attacking the carbon. Bromine leaves with both electrons of the bond and becomes bromide ion, Br−, with four lone pairs. The carbon it leaves behind keeps three bonds, no lone pair and a positive charge. That species, (CH3)3C+, is the tert-butyl carbocation: a flat, sp2 carbon with six valence electrons and an empty p orbital, as the Carbocations section of How Reactions Happen showed. A step like this one, in which a neutral molecule comes apart into two ions, is called ionization.
In the second step, water attacks the cation. A lone pair on the water oxygen moves into the empty p orbital and forms a new C–O bond. The cation acts as a Lewis acid, an electron-pair acceptor, and water as a Lewis base, just as the Lewis acids/bases section of Acids & Bases described. The oxygen now has three bonds and one lone pair, so it carries a positive charge. The species formed, (CH3)3C–OH2+, is an oxonium ion: an alcohol with an extra proton on its oxygen.
A third step is needed whenever the nucleophile is neutral, as water is. A second water molecule uses a lone pair to take the extra proton, and the electrons of the O–H bond move back onto the oxonium oxygen. The products are the neutral alcohol and H3O+.
When the solvent is also the nucleophile, as water is here, the reaction is a solvolysis. Most SN1 reactions are run this way. The nucleophile only has to capture a cation that already exists, so a weak one such as water or an alcohol is enough, and the solvent supplies plenty of it.
The rate law: only the substrate counts
Double the concentration of tert-butyl bromide and the reaction goes twice as fast. Add more nucleophile and the bromide disappears no faster. That second result is the clearest experimental sign of SN1 rather than SN2.
The order of the steps explains it. Ionization is slow, while the capture of the cation and the proton transfer are fast. The slow step is the rate-determining step, the step that holds the whole reaction back (see the Energy diagrams section of How Reactions Happen). Only the substrate takes part in it, so only the substrate's concentration appears in the rate law: rate = k[substrate]. A rate-determining step with one species in it is unimolecular, and that is the 1 in SN1. SN2 is bimolecular, with rate = k[Nu][substrate].
The nucleophile still decides which product forms. Warm tert-butyl bromide in a mixture of water and ethanol and both solvents compete for the same cation, giving tert-butanol and tert-butyl ethyl ether. Changing the mixture changes the ratio of those two products, but not how fast the bromide is used up.
Stereochemistry: the flat cation loses the configuration
Start with (R)-3-bromo-3-methylhexane and warm it in aqueous acetone, a mixture of water and acetone that dissolves the bromide and supplies water as the nucleophile. Its C3 carries a bromine, a methyl, an ethyl and a propyl group. Those are four different groups, so C3 is a stereocenter and the sample rotates plane-polarized light. What is the product?
Ionize. Bromide leaves, and C3 becomes a tertiary cation. Its three remaining groups now lie in one plane, and the empty p orbital has one lobe above that plane and one below it. This flat cation is not chiral, so the configuration of the starting material is lost at this point. It is not scrambled at some later step.
Attack. Nothing blocks either lobe, so water bonds through the top lobe in about half of the cations and through the bottom lobe in the other half. Attack on the face the bromine left puts the OH where the bromine was, which is retention of configuration. Attack on the opposite face gives the mirror image, which is inversion.
Assign the letters. After deprotonation, OH ranks first in the product just as Br did in the starting material, followed by propyl, ethyl and methyl. So retention gives the (R) alcohol and inversion gives the (S) alcohol.
Answer: (R)- and (S)-3-methylhexan-3-ol in nearly equal amounts: a largely racemic product.
A 50:50 mixture of two enantiomers is a racemic mixture, and it does not rotate plane-polarized light (see the Enantiomers section of Stereochemistry). So an optically active starting material has given a product with little or no optical activity. SN2 behaves differently. Put a single enantiomer of a secondary bromide, such as (R)-2-bromobutane, through SN2 with a good nucleophile, and every molecule inverts, giving one enantiomer of the product. That contrast is the second classic test for telling the two mechanisms apart.
Real SN1 products are rarely exactly 50:50. Most carry a modest excess of the inverted enantiomer, anywhere from about 55:45 to 70:30. The reason is timing. Right after the C–Br bond breaks, the bromide is still beside the cation, inside the same shell of solvent molecules. The two ions, held together only by the attraction between their charges, form an ion pair. While the bromide stays there, it partly blocks the face it just left, so water reaches the other face more easily and inversion wins by a little. Once solvent molecules slip between the two ions, both faces are equally open again.
The skew is evidence for the mechanism, not an exception to it. A cation that forms beside its leaving group should give a little extra inversion, and the longer a cation lives, the closer its product comes to 50:50. When a question asks for the outcome, call the SN1 product at a stereocenter racemic, or largely racemic.
Which substrates react: the more stable the cation, the faster
Warm four bromides in aqueous ethanol: methyl, ethyl (primary), isopropyl (secondary) and tert-butyl (tertiary). tert-Butyl bromide ionizes readily and isopropyl bromide slowly. Methyl and ethyl bromide hardly react, and the little reaction they do show goes mostly by SN2, with the solvent as the nucleophile. In this comparison a tertiary bromide reacts a million or more times faster than methyl bromide, and in formic acid the gap is larger still.
This order follows from the rate law. Ionization is the rate-determining step, and its product is the carbocation. Whatever makes the cation more stable lowers the barrier to making it, so the SN1 rate follows carbocation stability: 3° > 2° > 1° > methyl. The reasons for that order, hyperconjugation and induction from the neighboring alkyl groups, are set out in the Carbocations section of How Reactions Happen.
SN2 runs the other way. Its nucleophile has to reach the back of the carbon, so each alkyl group slows it, and a tertiary carbon does not react by SN2 at all. The same feature, alkyl groups on the carbon that carries the leaving group, speeds one mechanism and blocks the other. That is why substrate class is the first thing to check when you decide between the two.
Exceptions to the ladder: charge spread by resonance
Benzyl chloride, C6H5CH2Cl, is a primary halide, yet it solvolyzes readily. Its cation spreads the positive charge into the benzene ring by resonance, which makes a primary benzylic cation about as stable as a tertiary alkyl one. A primary allylic cation, next to a C=C, gets the same kind of help, and is about as stable as a secondary alkyl cation. So the rule that primary substrates do not react by SN1 covers only simple primary carbons, with no C=C or benzene ring next door.
Sharing the charge has a second consequence. Ionize 3-chlorobut-1-ene, CH2=CH–CHCl–CH3, in water. The cation is allylic, and its two resonance structures put the positive charge on C3 in one and on C1 in the other. Water can bond to either carbon. Attack at C3, the carbon that held the chlorine, gives but-3-en-2-ol. Attack at C1 gives but-2-en-1-ol, in which the double bond has moved over to C2–C3. Both alcohols come from the same cation, so the reaction gives a mixture of the two.
This shift of the double bond is called an allylic rearrangement. No atom moves from one carbon to another, unlike the hydride shift later on this page. The charge was shared by two carbons all along, and water bonded to each of them.
Vinyl and aryl halides sit at the other extreme. In a vinyl cation the charge would sit on a carbon of a C=C, and in an aryl cation on a carbon of a benzene ring. Neither can use the nearby π electrons, because its empty orbital lies at right angles to them, as the Carbocations section showed. So an ordinary vinyl or aryl halide does not ionize, and it does not react by SN2 either. Replacing a group on a benzene ring therefore takes different mechanisms, which Aromatic Chemistry covers.
Rearrangement: check the cation before the nucleophile arrives
A carbocation lives only briefly, but usually long enough for a 1,2-shift. In a 1,2-shift, a hydrogen or an alkyl group on the neighboring carbon moves across to the positive carbon together with its bonding pair. The Carbocations section taught the rule: a shift happens whenever it gives a more stable cation. Every SN1 reaction makes a cation, so a shift can happen before the nucleophile arrives, and the product then has a different carbon skeleton from the starting material.
The Carbocations section drew this same cation; here it appears in a full reaction. The substrate is CH3–CH(OH)–CH(CH3)–CH3. Number the chain from the end nearer the OH. The OH is on C2, and C3 carries one hydrogen and a methyl branch. Simply swapping the OH for Br would give 2-bromo-3-methylbutane.
Step 1 — activate the OH. Hydroxide is a very poor leaving group, so the alcohol cannot ionize as it is. HBr puts a proton on the oxygen and turns the OH into –OH2+, which leaves as a neutral water molecule, a good leaving group (see the Leaving groups section of How Reactions Happen).
Step 2 — ionize (slow). Water leaves and takes the C–O bonding pair with it. C2 keeps three bonds, to C1, C3 and its hydrogen, so the cation is secondary.
Step 3 — check each neighbor. Ask what each possible shift would leave behind. A hydrogen moving from C1 would put the charge on C1, a primary carbon, so that shift does not happen. The hydrogen on C3 is different. If it moves to C2, the charge lands on C3, which is then bonded to two methyl groups and C2. That is a tertiary cation, so this shift happens. Moving a methyl group from C3 instead would leave a secondary cation on C3, which is no gain. The hydrogens on C4 are two carbons away from C2 and out of reach.
Step 4 — the shift. The arrow's tail sits on the C3–H bond and its head on C2. The hydrogen moves as hydride, H−, because it takes both electrons of the bond with it.
Step 5 — capture. Bromide bonds to C3. Numbered afresh from the end nearer the Br, that carbon is C2, so the product is 2-bromo-2-methylbutane, not 2-bromo-3-methylbutane.
Three pieces of evidence that the cation is real
Three observations show that SN1 passes through a free carbocation, and exam questions ask about each one.
Rearranged products. SN2 happens in one step with no intermediate, so its carbon skeleton never changes. A shifted skeleton, as in 3-methylbutan-2-ol with HBr, means a cation lasted long enough to rearrange.
Loss of configuration. A single enantiomer gives a largely racemic product, as in Worked example 1.
The common-ion effect. Run the solvolysis of a bromide, R–Br, in water that also contains dissolved sodium bromide, and the reaction slows down. Ionization is reversible: bromide can bond back to the cation and re-form R–Br. Extra bromide makes that return more frequent, so fewer cations go on to the alcohol. This slowdown caused by adding the leaving group’s own ion is the common-ion effect, named because the added salt shares an ion with the reaction. A concerted SN2 has no intermediate for the added ion to capture, so it shows no such slowdown.
Do not confuse this with a salt that shares no ion with the reaction. Sodium perchlorate, NaClO4, speeds an SN1 reaction up slightly, because the extra ions in solution help the solvent support the charges that form in the rate-determining step.
Solvent: polar protic solvents help
tert-Butyl chloride ionizes far faster in water than in ethanol, and it does not ionize at all in a nonpolar solvent such as hexane. SN1 is favored by polar protic solvents, those with an O–H or N–H bond, such as water, alcohols and acetic acid. They help both ions that form in the rate-determining step. Their molecules cluster around the growing positive charge on carbon, oxygen lone pairs toward it, and their O–H hydrogens hydrogen-bond to the departing anion. Both effects lower the barrier to ionization.
This is the opposite of what SN2 wants, and it gives you a practical lever. The same secondary substrate can be pushed toward SN2 by running it in DMSO with a strong nucleophile, or toward SN1 by running it in aqueous ethanol with no added nucleophile.
The leaving group matters even more than in SN2
Compare a tosylate and a chloride on the same carbon. The tosylate ionizes orders of magnitude faster. A tosylate ion spreads its charge over three oxygens, as the Leaving groups section of How Reactions Happen showed, so it is a far more stable anion than chloride.
Leaving-group quality counts for more in SN1 than in SN2 because no nucleophile pushes the leaving group out. In SN2 the nucleophile does push, so a mediocre leaving group can be helped along. In SN1 the bond breaks without that push, and the barrier depends heavily on how stable the leaving group is once it has gone. That is also why an alcohol does not ionize until its OH is protonated, as in Worked example 2, or turned into a tosylate.
The four variables, summarized
SN1 is favored by:
- a substituted substrate: 3° > 2°, plus allylic and benzylic carbons, but never a simple 1° carbon or methyl;
- a weak, usually neutral nucleophile, often the solvent itself;
- a good leaving group, which matters even more than in SN2 because no nucleophile pushes it out;
- a polar protic solvent.
What carries forward
As a preview: carbocation stability and rearrangement come back throughout Alkenes & Alkynes and Alcohols, Ethers & Related Chemistry. They explain which way an acid adds across a C=C, how alcohols lose water to give alkenes, and several surprising products along the way. Closer to hand, E1, a later section of this chapter, begins with exactly the same ionization step as SN1, so the two mechanisms compete whenever a cation forms. Choosing among SN1, SN2 and the eliminations is the job of the Substrate & solvent effects section at the end of this chapter.