Difficulty: Intermediate | Prerequisites: a working grasp of Lewis structures, bond polarity, carbon hybridisation, and the basics of reaction mechanisms (arrow pushing). If "nucleophile" and "leaving group" do not yet mean anything to you, review your Organic Chemistry I notes on polar reactions first.
Organic Chemistry II spends a large chunk of its early weeks on four competing reaction pathways: S_N1, S_N2, E1 and E2. Every time a nucleophile or base meets a haloalkane (an alkyl halide), one of these four mechanisms wins out. The trouble is that the "winner" depends on several variables at once: the structure of the substrate, the strength and bulkiness of the nucleophile or base, and sometimes the solvent. This decision map is the cheat sheet that ties those variables together into a single flowchart you can follow from left to right. If you can classify the substrate (methyl, primary, secondary, tertiary) and classify the nucleophile/base (strong, medium, weak, very weak base, very strong base), the map tells you which mechanism dominates.
To predict whether a haloalkane reacts by substitution (S_N1 or S_N2) or elimination (E1 or E2), classify the substrate by carbon type and the nucleophile/base by strength. Methyl and primary substrates almost always go S_N2 unless a bulky, strong base (tBuOK) forces E2. Secondary substrates are the battleground: a very weak base gives S_N1/E1, a very strong base gives E2, and everything in between gives S_N2. Tertiary substrates never do S_N2; they split between S_N1/E1 (very weak base) and E2 (anything stronger).
Nucleophile
A species that donates an electron pair to an electrophilic (electron-poor) carbon. In simple terms, it is the "attacker" that forms a new bond to carbon.
Electrophile
A species that accepts an electron pair. The carbon bearing the leaving group is the electrophilic centre in these reactions.
Base
A species that abstracts (removes) a proton. Many nucleophiles are also bases, but the two roles are distinct: a nucleophile attacks carbon, a base attacks hydrogen.
Leaving group
The group that departs with the bonding electrons (e.g. Br⁻, Cl⁻, I⁻, tosylate). Think of it as the piece that gets kicked off the substrate.
Haloalkane (alkyl halide)
An organic compound in which a halogen (F, Cl, Br, I) is bonded to an sp³ carbon. The substrate in all four mechanisms discussed here.
Methyl substrate
CH₃–X. Carbon bears three hydrogens and the leaving group. No steric hindrance at all.
Primary (1°) substrate
The carbon bearing the leaving group is bonded to one other carbon. Low steric hindrance.
Secondary (2°) substrate
The carbon bearing the leaving group is bonded to two other carbons. Moderate steric hindrance; this is where mechanism prediction gets tricky.
Tertiary (3°) substrate
The carbon bearing the leaving group is bonded to three other carbons. High steric hindrance; backside attack (S_N2) is blocked.
S_N2 (bimolecular nucleophilic substitution)
A one-step mechanism: the nucleophile attacks the electrophilic carbon at the same time the leaving group departs, in a single concerted step. The rate depends on both substrate and nucleophile concentration. Think of it as a simultaneous "in the front door, out the back door" process. Results in inversion of configuration at the chiral centre (Walden inversion).
S_N1 (unimolecular nucleophilic substitution)
A two-step mechanism: the leaving group departs first to form a carbocation intermediate, then the nucleophile attacks. The rate depends only on substrate concentration. Because the carbocation is planar, attack can come from either face, so you get scrambling (racemisation) at the chiral centre.
E2 (bimolecular elimination)
A one-step mechanism: the base removes a β-hydrogen at the same time the leaving group departs, forming a double bond in a single concerted step. Requires anti-periplanar geometry between the β-hydrogen and the leaving group. Typically follows Zaitsev's rule (the more substituted alkene is the major product), unless the base is bulky (tBuOK), in which case the less substituted (Hofmann) product can form in appreciable amounts.
E1 (unimolecular elimination)
A two-step mechanism: the leaving group departs first to form a carbocation, then a base removes a β-hydrogen to form the alkene. Follows Zaitsev's rule. Often competes with S_N1 under the same conditions.
Strong nucleophiles
Br⁻, I⁻, RS⁻, HS⁻, N≡C⁻ (cyanide), N₃⁻ (azide). Also RC≡C⁻, RO⁻, HO⁻, which double as strong bases.
Medium nucleophiles
RCO₂⁻ (carboxylate), RSH, R₂S, NH₃, RNH₂, R₂NH, NR₃. Decent nucleophiles but not particularly strong bases.
Weak nucleophiles / very weak bases
RCO₂H (carboxylic acids), ROH (alcohols), H₂O (water). These are too weak to force S_N2 on anything but a methyl or unhindered primary substrate, and too weak to force E2.
tBuOK (potassium tert-butoxide)
A strong base but a poor nucleophile because the tert-butyl group creates so much steric bulk around the oxygen. When you see tBuOK with a primary substrate, expect E2 instead of S_N2.
Zaitsev's rule
In elimination reactions, the more substituted alkene (the one with more alkyl groups on the double-bond carbons) is the major product, because it is thermodynamically more stable.
Anti-periplanar geometry
The spatial arrangement required for E2: the β-hydrogen and the leaving group must be on opposite sides of the C–C bond being rotated, at a dihedral angle of 180°. In simple terms, they point in opposite directions when you look down the bond.
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The map works in two steps. First, classify the substrate. Then ask about the nucleophile/base. The flowchart above covers methyl, primary and secondary substrates. Tertiary substrates follow a simpler rule covered below.
Methyl halides (CH₃–X)
Always S_N2. No exceptions in a standard Organic Chemistry II course.
The carbon is completely unhindered, so backside attack by a nucleophile is easy.
There is no β-hydrogen available for E2 (no adjacent carbon to lose a proton from), so elimination is not possible.
Primary halides (RCH₂–X)
Default outcome is S_N2.
The one exception to watch for: if the base is tBuOK (potassium tert-butoxide), predict E2. tBuOK is too bulky to act as a nucleophile, so it acts exclusively as a base and forces elimination.
With any other strong or medium nucleophile, S_N2 wins.
Secondary halides (R₂CH–X), including allylic and benzylic halides
This is the contested territory. Three questions, asked in order:
Is the nucleophile/base a very weak base (H₂O, ROH, RCO₂H)? If yes, the reaction is slow and goes through the carbocation pathway: S_N1/E1.
Is the nucleophile/base a very strong base (e.g. tBuOK, LDA)? If yes, E2.
If neither extreme applies (the nucleophile is strong or medium but not a very strong base), the answer is S_N2.
Note: with very weak bases, S_N2 can technically compete, but for course purposes assume S_N1/E1 predominate.
Tertiary halides (R₃C–X)
S_N2 never happens. Three bulky groups around the electrophilic carbon completely block backside attack.
If the nucleophile/base is a very weak base: S_N1/E1.
If the nucleophile/base is anything stronger: E2.
Knowing which category a given species falls into is half the battle. Here is the classification:
Category | Examples | Also classified as |
|---|---|---|
Strong nucleophiles | Br⁻, I⁻, RS⁻, HS⁻, CN⁻, N₃⁻ | Nucleophiles (not strong bases) |
Strong nucleophiles AND strong bases | RC≡C⁻, RO⁻, HO⁻ | Both roles |
Medium nucleophiles | RCO₂⁻, RSH, R₂S, NH₃, RNH₂, R₂NH, NR₃ | Moderate bases |
Weak nucleophiles / very weak bases | RCO₂H, ROH, H₂O | Too weak to force S_N2 or E2 |
The key dividing line: species in the "very weak base" row push reactions towards S_N1/E1 (they wait for the substrate to ionise on its own). Species in the other rows are strong enough to actively attack or abstract.
tBuOK: the base that is not a nucleophile
tert-Butoxide (tBuO⁻) is thermodynamically a strong base, but the three methyl groups around the oxygen make it far too bulky to attack carbon effectively. Wherever you see tBuOK, think "E2" rather than "S_N2."
With primary substrates, tBuOK overrides the usual S_N2 default and gives E2.
With secondary and tertiary substrates, tBuOK drives E2 as well. Because of its bulk, it tends to abstract the most accessible (least substituted) β-hydrogen, which can produce an appreciable amount of the Hofmann (less substituted) product alongside the Zaitsev product.
Stereochemistry checklist by mechanism
S_N2: inversion of configuration at the chiral centre (the nucleophile enters from the opposite side of the leaving group).
S_N1: racemisation (scrambling) at the chiral centre, because the planar carbocation intermediate can be attacked from either face.
E2: requires anti-periplanar arrangement of the β-H and the leaving group. Product regiochemistry usually follows Zaitsev's rule (more substituted alkene).
E1: Zaitsev's rule applies. No geometric constraint on the elimination step itself (carbocation forms first).
Students often think a "strong nucleophile" and a "strong base" are the same thing. They overlap (RO⁻ and HO⁻ are both), but many strong nucleophiles (I⁻, RS⁻, CN⁻) are weak bases. Nucleophilicity is about attacking carbon; basicity is about attacking hydrogen.
Students frequently try to run S_N2 on tertiary substrates. It does not happen. Three bulky groups block the backside attack entirely.
Seeing "NaOH" or "KOH" and defaulting to E2 regardless of substrate is a common error. With a methyl or primary substrate, HO⁻ is a strong nucleophile, and S_N2 wins. E2 dominates only when the substrate is secondary or tertiary, or when the base is specifically bulky (tBuOK).
Confusing Zaitsev and Hofmann products. Zaitsev = more substituted alkene (the usual product). Hofmann = less substituted alkene (favoured by bulky bases such as tBuOK). If the exam says "bulky base," think Hofmann product.
⚠️ Predicting the correct mechanism (S_N1, S_N2, E1, E2) is one of the most frequently tested skills in Organic Chemistry II exams. Expect multiple questions requiring you to identify substrate class, nucleophile/base strength, and the resulting mechanism.
⚠️ Stereochemistry of the product is almost always tested alongside the mechanism. If the question asks you to predict the product of an S_N2 reaction, you must show inversion. For S_N1, show racemisation.
⚠️ E2 regiochemistry (Zaitsev vs. Hofmann) comes up whenever a bulky base is involved. If the exam gives you tBuOK as the base, consider whether the less substituted alkene is the major product.
⚠️ Secondary substrates are the most common exam trap, precisely because they can go S_N1, S_N2, E1 or E2 depending on conditions. Work through the decision tree carefully for these.
Q: True or false: S_N2 can occur on a tertiary substrate if the nucleophile is strong enough.
A: False. Steric hindrance blocks backside attack regardless of nucleophile strength.
Q: Fill in the blank: tBuOK is a strong ______ but a poor ______ due to steric bulk.
A: Base; nucleophile.
Q: True or false: S_N1 and E1 typically compete under the same conditions.
A: True. Both proceed through a carbocation intermediate and are favoured by weak nucleophiles/bases.
Q: Fill in the blank: E2 requires ______ geometry between the β-hydrogen and the leaving group.
A: Anti-periplanar (180° dihedral angle).
Q: True or false: a methyl halide can undergo E2 elimination.
A: False. Methyl halides have no β-hydrogen, so elimination is not possible.
Q: You react 2-bromobutane with NaOH in DMSO. Predict the mechanism and major product.
A: 2-bromobutane is a secondary substrate. NaOH provides HO⁻, which is a strong nucleophile and strong base but not a very weak base and not a very strong/bulky base. Following the decision map: not a very weak base (rules out S_N1/E1), not a very strong base like tBuOK (rules out E2). The answer is S_N2, giving 2-butanol with inversion of configuration.
Q: You react 2-bromo-2-methylpropane (tert-butyl bromide) with water. Predict the mechanism.
A: Tertiary substrate. Water is a very weak base/nucleophile. The map says: tertiary + very weak base = S_N1/E1. Expect a mixture of 2-methyl-2-propanol (substitution) and 2-methylpropene (elimination), with the carbocation intermediate forming first.
Q: You react 1-bromobutane with tBuOK in tBuOH. Predict the mechanism and major product.
A: Primary substrate + tBuOK = E2 (the one exception for primary substrates). The major product is but-1-ene. Because tBuOK is bulky, it preferentially abstracts the most accessible hydrogen, which may favour the less substituted alkene (Hofmann product) in more complex primary substrates, though for 1-bromobutane the only β-hydrogens are on the same carbon.
Q: You react (R)-2-chlorobutane with NaCN in DMSO. Predict the mechanism, product and stereochemistry.
A: Secondary substrate. CN⁻ is a strong nucleophile but not a very weak base, and not a very strong base. The map gives S_N2. The product is (S)-2-cyanobutane (or 2-methylbutanenitrile), with inversion of configuration at the chiral centre.
Q: You react 3-bromo-3-methylpentane with NaOEt (sodium ethoxide) in ethanol. Predict the mechanism and major product.
A: Tertiary substrate. Ethoxide (EtO⁻) is a strong base, which is not a very weak base. The map says: tertiary + not a very weak base = E2. The major product is the more substituted alkene (Zaitsev product): 3-methylpent-2-ene.
This material connects directly to alkene synthesis (Chapters on elimination reactions produce alkenes, which are the starting materials for addition reactions in subsequent chapters). Understanding when E2 vs. E1 dominates determines which alkene isomer you start with.
It also connects to alcohol chemistry: alcohols can be converted to alkyl halides, and the reverse reactions (nucleophilic substitution on alkyl halides with hydroxide) produce alcohols. The mechanism governs stereochemical outcomes that carry through multi-step synthesis problems.
The carbocation intermediates in S_N1 and E1 tie into carbocation stability and rearrangements. If you are shaky on why tertiary carbocations are more stable than primary ones, review that material, because it underpins every S_N1/E1 prediction.
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