Take 2-bromobutane. Heat it with sodium ethoxide in ethanol and the main product is but-2-ene. Stir it with sodium azide in acetone and you get 2-azidobutane instead. Warm it in water with nothing added and you get butan-2-ol along with some alkene. The substrate never changed. The reagent and the solvent picked a different mechanism each time.
You have met the four mechanisms one at a time: SN2, SN1, E2 and E1. This section is about choosing among them. A problem gives you a substrate, a reagent, a solvent and often a temperature, and you work out which mechanism wins or which ones share the flask. Three checks, taken in order, settle almost every case: the substrate first, then the reagent, then the solvent and temperature. The leaving group matters too, but most questions keep it fixed. Exams test this skill heavily.
Step 1: substrate class rules mechanisms in or out
Start with the carbon that carries the leaving group, called the α carbon (alpha carbon). A carbon bonded to it is a β carbon (beta carbon), and E2 always takes its hydrogen from a β carbon. The substrate’s class is the number of carbons bonded to the α carbon: none for methyl, one for primary (1°), two for secondary (2°) and three for tertiary (3°).
The class matters because the two substitution mechanisms need opposite things. SN2 needs room, since the nucleophile has to reach the back of the α carbon, so every group on that carbon slows it down. SN1 needs a stable carbocation, and every alkyl group on the α carbon makes the cation more stable (Carbocations). So in SN2 crowding decides, and in SN1 cation stability decides. E1 starts from the same cation as SN1, so it needs the same stable cation. E2 needs a β hydrogen and a strong base, and a crowded α carbon does not stop it.
Methyl and primary. SN2 is the pathway. Nothing much crowds the back of the carbon, and a methyl or primary carbocation is too unstable to form, which rules out both SN1 and E1. A strong bulky base is the one exception on a primary substrate. It cannot reach the carbon, so it takes a β hydrogen instead and gives E2. A methyl substrate has no β carbon at all, so it can never eliminate.
Tertiary. SN2 is impossible, because three groups block the back of the carbon. What remains is SN1 and E1 with a weak reagent, or E2 with a strong base. E2 is the only bimolecular pathway a tertiary substrate has.
Secondary. This is the hard case. All four mechanisms are possible, so the substrate rules nothing out and the reagent and solvent decide. Most exam questions live here.
Two kinds of substrate do not follow these class rules. In an allylic substrate the α carbon sits next to a C=C, and in a benzylic one it sits next to a benzene ring. Both react fast by SN1 and by SN2. Their cations are stabilized by resonance, as Carbocations showed, and the neighboring π bond also stabilizes the SN2 transition state.
In a vinyl halide the leaving group sits on a carbon of the C=C itself, and in an aryl halide it sits on a ring carbon. Neither one undergoes SN2 or SN1. The flat sp² geometry blocks SN2: in bromobenzene the back of the C–Br bond points straight into the ring, and in a vinyl halide it lies in the crowded plane of the double bond. SN1 fails because vinyl and aryl cations are far too unstable to form. (A very strong base can still pull HBr out of a vinyl bromide to make an alkyne; Alkynes covers that reaction.)
Step 2: the reagent decides bimolecular versus unimolecular
Compare 2-bromopropane with sodium ethoxide and with ethanol alone. Ethoxide reacts with it in a single step, long before the C–Br bond would break on its own. Ethanol is too weak to do that. With ethanol alone, nothing happens until the substrate ionizes.
That is the general rule. A strong nucleophile or base forces a bimolecular pathway, SN2 or E2, whenever the substrate allows one. Strong reagents usually carry a negative charge: hydroxide, alkoxides, cyanide, azide, thiolates, acetylides and amide ion, H₂N⁻.
A weak nucleophile or base is usually neutral: water, alcohols and carboxylic acids. It cannot force a bimolecular step at a useful rate, so the reaction waits for the substrate to ionize, and the weak reagent then reacts with the carbocation. A weak reagent and a substrate that can ionize therefore give SN1 and E1 together.
Once you know the reaction is bimolecular, ask whether the reagent is better as a nucleophile, attacking carbon, or as a base, taking a proton. Nucleophiles showed that these are different properties. The table sorts the common reagents.
| Reagent type | Examples | Drives |
|---|---|---|
| Strong Nu, weak base | I⁻, RS⁻, N₃⁻, ⁻CN | SN2 |
| Strong Nu and base | HO⁻, CH₃O⁻, EtO⁻ | SN2 on 1°, E2 on 2°/3° |
| Strong bulky base | t-BuO⁻, DBU | E2 |
| Weak Nu and base | H₂O, ROH | SN1 + E1 |
DBU is short for 1,8-diazabicyclo[5.4.0]undec-7-ene. It is an amidine, a neutral nitrogen base in which one carbon is double-bonded to one nitrogen and single-bonded to another. In DBU this N–C=N unit sits inside two fused rings. The basic nitrogen is the one in the C=N bond. When that nitrogen takes a proton, resonance spreads the positive charge over both nitrogens. That stable conjugate acid is what makes DBU a strong base.
The basic nitrogen sits right beside the ring fusion, where the rings crowd it. Chemists use DBU mainly as a base for elimination, although it can still attack very reactive electrophiles.
Acetylide, ⁻C≡CR, is the anion left when a terminal alkyne loses its C–H proton. You met it as a conjugate base in Acids & Bases, and Alkenes & Alkynes uses it as a carbon nucleophile that builds longer chains. For now, file it with the reagents that are strong both as a nucleophile and as a base.
Size can override strength. Sodium ethoxide and potassium tert-butoxide are both strong bases, but with 2-bromopropane ethoxide gives some ether along with propene, while tert-butoxide gives propene and almost no ether. The three methyl groups of tert-butoxide keep its oxygen away from a crowded carbon, so SN2 fails. A β hydrogen sits on the outside of the molecule, so the oxygen can still reach it, and E2 goes ahead.
A bulky base therefore pushes secondary and tertiary substrates, and even primary ones, toward elimination. It also changes which alkene forms, favoring the less substituted Hofmann product, as E2 showed. This is the same pair Nucleophiles used to show that one molecule can be a strong base and a poor nucleophile.
Step 3: solvent and temperature reinforce the call
Go back to 2-bromobutane and sodium azide. In acetone the pair gives clean SN2. In water the same pair reacts more slowly and gives a mixture, as worked example 4 below explains. Only the solvent changed.
The split that matters is protic versus aprotic. A protic solvent has an O–H or N–H bond, so it can donate hydrogen bonds. An aprotic solvent has neither (Nucleophiles). All the solvents in this section are polar: each has a strong dipole and dissolves ionic reagents.
Polar aprotic solvents include DMSO, DMF, acetone and acetonitrile. They cannot hydrogen-bond to an anion, so a strong nucleophile stays bare and reactive. That favors SN2 and E2.
Polar protic solvents include water, alcohols and acetic acid. They hydrogen-bond to anions and cluster around cations, so they stabilize both pieces that an ionization makes: the carbocation and the departing anion. That favors SN1 and E1. The same hydrogen bonds wrap around a strong nucleophile and slow it down.
Heat favors elimination: E2 over SN2, and E1 over SN1. Count the particles to see why. Ethoxide and 2-bromopropane are two particles. SN2 turns them into two, the ether and bromide ion. E2 turns them into three: propene, ethanol and bromide ion. Elimination makes one more particle than the matching substitution.
E2 explained that more free particles means more entropy, and E1 showed that entropy enters the free energy through ΔG = ΔH − TΔS. So the extra particle counts for more as the flask gets hotter. That is why “heat” or “reflux” (boiling the mixture under a condenser) in a set of conditions usually signals elimination.
Worked examples
Substrate. Primary. That rules out SN1 and E1 at once, because a primary cation does not form.
Reagent. Cyanide is a strong nucleophile and a weak base. HCN’s pKa is about 9, so ⁻CN is a far weaker base than an alkoxide. A strong reagent means a bimolecular pathway, and a weak base means substitution rather than elimination.
Solvent. DMSO is polar aprotic, so it leaves cyanide bare and at its most reactive. Everything points the same way.
Products. Pentanenitrile, CH₃CH₂CH₂CH₂CN, and almost nothing else. The competing E2 product would be but-1-ene, and it barely appears because cyanide is too weak a base to pull off a proton.
Substrate. Secondary, the ambiguous class. Nothing is ruled out.
Reagent. Ethoxide is strong on both counts: a good nucleophile and a strong base (ethanol’s pKa is 16). The pathway is bimolecular for certain, but both SN2 and E2 are still possible.
Solvent and temperature. Ethanol is protic, but ethoxide stays a strong base in it, so the reaction does not switch to SN1 and E1. Heat favors elimination.
Products, major and minor. Elimination wins. The major product is but-2-ene, the Zaitsev alkene, mostly as its E isomer because the trans double bond is less crowded. The minor alkene is but-1-ene, from the other β carbon. The flask also holds a third product that students often leave out: 2-ethoxybutane (sec-butyl ethyl ether), from the SN2 that lost the race. A secondary substrate with a strong base gives a mixture, and naming only the major product is a partial answer.
Substrate. Tertiary, so SN2 is impossible.
Reagent. Water and ethanol are weak nucleophiles and weak bases, so nothing forces a bimolecular step. This is a solvolysis, a reaction in which the solvent is the nucleophile. SN1 and E1 both run from the same cation.
Solvent and temperature. The solvent is protic, which is exactly what ionization needs, and the warmth adds elimination.
Products. SN1 gives tert-butanol and tert-butyl ethyl ether, one from each nucleophile in the solvent. E1 gives 2-methylpropene. All three form together. That mixture is the trouble with unimolecular conditions, and it is why chemists rarely use them to make a single product.
The substrate and the reagent are the same in both flasks, and the answers differ. The figure in Step 3 draws both flasks.
In acetone (polar aprotic). Azide is a strong nucleophile and a very weak base (HN₃’s pKa is about 4.6). With nothing holding it, it attacks fast and gives clean SN2 to 2-azidobutane, with inversion at C2. Almost no alkene forms.
In water. Water hydrogen-bonds to azide and slows it down. Water is also a good ionizing solvent, so the substrate’s own slow ionization starts to compete. The product becomes a mixture: some inverted 2-azidobutane from SN2, some racemic 2-azidobutane and butan-2-ol from SN1, and a little but-2-ene. The right answer to a borderline secondary question is a mixture, with a reason for each piece, not a single product.
What to say in an exam. Name the major pathway, name its competitor, and give the variable that decides between them. A question with a secondary substrate is usually testing exactly that.
Exam questions rarely change the leaving group, so it is worth seeing once on its own. Warm 2-chlorobutane and butan-2-yl tosylate in ethanol, each with no added reagent.
The substrate class, the reagent and the solvent are the same. The chloride barely reacts. Chloride is only a fair leaving group, and with nothing to push it out, ionization is too slow to see. The tosylate solvolyzes smoothly, because ⁻OTs spreads its charge over three oxygens and is a far more stable anion (Leaving groups). The products are 2-ethoxybutane and some butene, from SN1 and E1.
This is the point SN1 made. A leaving group that is good enough when a strong nucleophile pushes it out can be hopeless when it has to leave on its own.
Treat neopentyl bromide, (CH₃)₃C–CH₂Br (1-bromo-2,2-dimethylpropane), with sodium ethoxide, and run the checks in order.
Substrate. Primary, so SN1 and E1 are out. SN2 is allowed on paper, but the three methyl groups on the next carbon block the path to the back of the α carbon. The reaction runs about 10⁵ times slower than it does with ethyl bromide.
What about E2? Find the β carbon. There is exactly one, and it is the quaternary carbon, which carries three methyls and no hydrogen at all. E2 needs a β hydrogen, and there is none, so E2 is not merely slow here. It is impossible.
So what happens? A very slow SN2 gives the ether, and nothing else competes. Count the β hydrogens before you propose an elimination. A substrate with none cannot eliminate, however strong the base.
Quick reference
| Substrate | Strong Nu, weak base | Strong base | Bulky base | Weak Nu/base |
|---|---|---|---|---|
| methyl | SN2 | SN2 | SN2 (slow) | no reaction |
| 1° | SN2 | SN2 > E2 | E2 | no reaction |
| 2° | SN2 | E2 > SN2 | E2 | SN1 + E1 |
| 3° | SN1 (slow) | E2 | E2 | SN1 + E1 |
Reading the evidence backwards
Exam questions often run the other way: here is what happened, so which mechanism was it? Each mechanism leaves its own evidence. Chemists first worked out the mechanisms from these same observations.
- A rate that depends on the nucleophile’s concentration means a bimolecular mechanism. A rate that does not means a unimolecular one.
- Clean inversion at a stereocenter means SN2. Racemization means SN1.
- A product with the new group on a different carbon, or with a rearranged carbon skeleton, means a carbocation that shifted, so SN1 or E1.
- An alkene whose geometry follows from which diastereomer you started with means E2.
Worked diagnosis. An optically active secondary bromide is warmed in methanol, and three observations come back. First, adding sodium azide, a far better nucleophile, does not speed up the loss of the bromide at all. Second, the product is optically inactive. Third, a second, isomeric product carries its methoxy group on a different carbon from the one that held the bromine.
Take them in turn. A rate that ignores the nucleophile means the nucleophile is not in the rate-determining step, so the mechanism is unimolecular. Loss of optical activity means the stereocenter passed through a flat intermediate. A group that ends up on a new carbon means an intermediate lived long enough to shift. All three point to a carbocation, so the answer is SN1, with the second product coming from the same cation after a hydride or methyl shift. Any one observation alone would be suggestive. Together they are conclusive, and that is how these questions are built.
What carries forward
This four-way choice is the model for every later competition in the course: 1,2- versus 1,4-addition in Conjugation & Pericyclic Reactions, substitution versus addition on aromatic rings in Aromatic Chemistry, and which carbonyl derivative reacts fastest in Carboxylic Acids & Derivatives. The habit of asking what the substrate allows, what the reagent wants and what the solvent rewards carries straight over.