Difficulty: Intermediate | Prerequisites: CHM 255 Chapters 7–8, basic understanding of nucleophiles, electrophiles, and leaving groups.
Alkyl halides sit at the crossroads of organic chemistry's four most testable reaction types: SN1, SN2, E1, and E2. This topic covers what happens when a nucleophile or base meets an alkyl halide, and how substrate structure, nucleophile/base strength, and solvent determine which pathway dominates. If you skipped the earlier material on carbocation stability and stereochemistry, revisit that first.
Alkyl halides react with nucleophiles via substitution (SN1 or SN2) or with bases via elimination (E1 or E2). The choice of pathway depends on the substrate (methyl, 1°, 2°, 3°), the strength of the nucleophile/base, and the solvent. Weak nucleophiles with 2° and 3° substrates give mixtures of SN1 and E1 products.
Nucleophilic substitution
A reaction in which a nucleophile replaces a leaving group on a carbon atom. In simple terms, one group swaps out for another at the same carbon.
SN1 (substitution, nucleophilic, unimolecular)
A two-step substitution where the leaving group departs first to form a carbocation, and the nucleophile attacks second. Think of it as: the leaving group leaves on its own, then the nucleophile swoops in.
SN2 (substitution, nucleophilic, bimolecular)
A one-step substitution where the nucleophile attacks at the same time as the leaving group departs, via a backside attack. Think of it as: the nucleophile shoves the leaving group out the back door.
Elimination (E1 and E2)
Reactions where a proton and a leaving group are removed from adjacent carbons, forming a double bond (alkene). In simple terms, instead of swapping groups, the molecule loses two pieces and gains a pi bond.
Leaving group
The atom or group that departs with the bonding electrons. Good leaving groups are weak bases (e.g. Br⁻, Cl⁻, I⁻, H₂O). Think of it as: the group that is willing to walk away.
Nucleophile
An electron-rich species that attacks an electrophilic carbon. Strong nucleophiles (e.g. CN⁻, RS⁻, RO⁻) favour SN2; weak nucleophiles (e.g. H₂O, CH₃OH) favour SN1.
Zaitsev's rule
In elimination reactions, the more substituted alkene (more alkyl groups on the double-bond carbons) is the major product. In simple terms, the bigger, more branched alkene wins.
Anti-coplanar geometry
In E2 elimination, the H and the leaving group must be on opposite sides of the C–C bond and in the same plane (180° dihedral). This is required for the concerted mechanism to work.
Starting from a simple alkyl halide (e.g. CH₃–Br), a nucleophile can replace the bromide to form a range of products:
⁻OH → alcohol (CH₃–OH)
⁻CN → nitrile (CH₃–CN)
⁻OR → ether (CH₃–OR)
⁻SR → thioether (CH₃–SR)
⁻C≡CH (acetylide) → alkyne (CH₃–C≡CH), extends the carbon chain
I⁻ → other alkyl halide (CH₃–I), a Finkelstein-type exchange
The nucleophile donates a pair of electrons to the electrophilic carbon bearing the leaving group. The product depends entirely on which nucleophile you choose.
SN2 conditions: strong nucleophile, polar aprotic solvent (e.g. DMF, DMSO, acetone), methyl or 1° substrate preferred. One concerted step with backside attack, producing inversion of stereochemistry.
SN1 conditions: weak nucleophile, polar protic solvent (e.g. H₂O, CH₃OH), 3° or 2° substrate. Two steps: the leaving group departs first (rate-determining step) to form a carbocation, then the nucleophile attacks. Produces a racemic mixture because the carbocation is planar and the nucleophile can attack from either face.
Instead of substitution, a base can remove a proton from the beta carbon (the carbon next to the one bearing the leaving group). As the proton leaves and the leaving group departs, a double bond forms between the alpha and beta carbons.
E2 (bimolecular elimination):
One concerted step: the base removes the beta-H at the same time as the leaving group departs
Requires anti-coplanar geometry: the H and leaving group must be anti-periplanar (180° dihedral)
Strong, bulky bases favour E2 (e.g. ⁻OH, ⁻OR, tert-butoxide)
Follows Zaitsev's rule: the more substituted alkene is the major product
Works best with 2° and 3° substrates
E1 (unimolecular elimination):
Two steps: the leaving group departs first (rate-determining step) to form a carbocation, then a base removes a beta-H
Weak bases favour E1 (e.g. H₂O, CH₃OH)
Also follows Zaitsev's rule
Occurs alongside SN1 with 2° and 3° substrates in polar protic solvents
Weak bases and nucleophiles such as H₂O and CH₃OH will undergo both SN1 substitution and E1 elimination when reacted with secondary and tertiary alkyl halides. You will typically get a mixture of products (substitution product + alkene). This is a high-frequency exam point: when you see a weak nucleophile with a 2° or 3° substrate, you should expect a mixture, not a single clean product.
Students often think SN1 and SN2 are just "fast vs slow" versions of the same reaction. They are not. SN1 is two steps (carbocation intermediate); SN2 is one concerted step (no intermediate). The mechanism, stereochemistry, and preferred substrates are completely different.
Students often assume that a strong nucleophile always means substitution. A strong nucleophile that is also a strong base (e.g. ⁻OH, ⁻OEt) can just as easily cause elimination, especially with 2° and 3° substrates.
Students frequently forget that SN2 requires backside attack and therefore inverts stereochemistry. If an exam question shows a chiral centre, you must flip the configuration in the product.
Students sometimes think E1 and E2 always give the same product. E2 requires anti-coplanar geometry, which can limit which beta-H is removed. E1 goes through a planar carbocation, so Zaitsev's rule governs without geometric constraints.
⚠️ You will almost certainly be asked to predict whether a given set of conditions favours SN1, SN2, E1, or E2. Know the decision tree: substrate degree + nucleophile/base strength + solvent.
⚠️ Stereochemistry questions are common: SN2 gives inversion, SN1 gives racemisation. If the question shows a wedge-dash structure, you need to draw the correct stereochemical outcome.
⚠️ Zaitsev's rule applies to both E1 and E2. Expect a question asking you to identify the major alkene product.
⚠️ The SN1/E1 mixture with weak nucleophiles and 2°/3° substrates is a classic exam trap. Do not predict only one product.
True or false: SN2 reactions proceed through a carbocation intermediate. (False, SN2 is concerted with no intermediate.)
Fill in the blank: SN1 reactions favour ______ protic solvents, while SN2 reactions favour polar ______ solvents. (Polar protic; aprotic.)
True or false: E2 elimination requires the H and leaving group to be syn-coplanar. (False, they must be anti-coplanar.)
Fill in the blank: In SN2, the stereochemistry at the reactive carbon undergoes ______. (Inversion.)
True or false: A weak nucleophile like H₂O reacting with a 3° alkyl halide gives only substitution products. (False, you get a mixture of SN1 and E1 products.)
Q: A tertiary alkyl bromide is treated with CH₃OH. What reaction pathway(s) will occur, and what product(s) do you expect?
A: Both SN1 and E1. CH₃OH is a weak nucleophile and a weak base. With a 3° substrate, the leaving group departs first to form a carbocation (SN1/E1 share the same first step). The nucleophile attacks the carbocation to give a methyl ether (SN1), and deprotonation of a beta-H gives an alkene (E1). Expect a mixture.
Q: Methyl bromide is treated with NaCN in DMF. What is the product and the mechanism?
A: SN2. CN⁻ is a strong nucleophile, DMF is a polar aprotic solvent, and the substrate is methyl (no steric hindrance). The product is CH₃–CN (acetonitrile). The reaction proceeds in one concerted step with backside attack.
Q: Draw the major elimination product when 2-bromobutane is treated with KOH. Which rule determines the major product?
A: E2 elimination (KOH is a strong base). Zaitsev's rule: the more substituted alkene is the major product, so 2-butene (not 1-butene) is the major product.
Q: In an SN2 reaction at a chiral centre, what happens to the stereochemistry?
A: The configuration is inverted. Backside attack by the nucleophile means the three remaining groups flip to the opposite side, like an umbrella inverting in the wind.
This material connects directly to carbocation stability (3° > 2° > 1° > methyl), which you covered in earlier chapters. Carbocation stability governs whether SN1/E1 pathways are viable. It also ties into stereochemistry: understanding R/S configuration is essential for predicting SN2 inversion outcomes. Later in the course, these same reaction types appear in alcohol chemistry, ether synthesis (Williamson), and alkyne formation via acetylide nucleophiles.
SN1, SN2, E1, E2, nucleophilic substitution, elimination reaction, alkyl halide reactions, Zaitsev's rule, Saytzeff rule, anti-coplanar, anti-periplanar, carbocation intermediate, backside attack, inversion of configuration, racemic mixture, polar protic solvent, polar aprotic solvent, leaving group, nucleophile strength, base strength, CHM 255, organic chemistry exam 3, Purdue organic chemistry