Difficulty: Intermediate | Prerequisites: CHEM 237A foundations (alkyl halides, nucleophilic substitution, SN1/SN2 mechanisms, conformational analysis of cyclohexanes, Newman projections)
This material sits at the heart of second-quarter organic chemistry: how and why molecules lose atoms to form carbon-carbon double bonds. Elimination reactions (E2 in particular) compete directly with the substitution reactions you learnt earlier, and the exam tests whether you can predict which pathway wins for a given substrate, base, and set of conditions. You need a solid grasp of Newman projections, stereochemistry (R/S and now E/Z), and the concept of antiperiplanar geometry. If conformational analysis or leaving-group trends still feel shaky, revisit those before diving in here.
E2 elimination requires a strong base, a decent leaving group, and an antiperiplanar arrangement of the H and the leaving group. The geometry of the resulting alkene (E vs Z) is locked in at the transition state and depends on which conformation reacts. Zaitsev's rule predicts the more-substituted alkene as the major product because its transition state is lower in energy.
E2 reaction (bimolecular elimination)
A one-step elimination in which a strong base removes a beta-hydrogen at the same time as the leaving group departs, forming a new pi bond. The rate depends on the concentrations of both the substrate and the base. Think of it as a coordinated, simultaneous event: base grabs H, leaving group leaves, double bond forms, all in one motion.
Antiperiplanar
A dihedral angle of 180° between the C-H bond being broken and the C-LG (leaving group) bond, viewed along the C-C axis in a Newman projection. In simple terms, the H and the leaving group must be on opposite sides and perfectly lined up for E2 to proceed.
Z configuration (zusammen)
The higher-priority substituents on each carbon of the double bond are on the same side. Assigned using Cahn-Ingold-Prelog priority rules. Think of Z as meaning "same side" (the Z in zusammen, German for "together").
E configuration (entgegen)
The higher-priority substituents on each carbon of the double bond are on opposite sides. Think of E as "opposite" (entgegen, German for "opposite").
Zaitsev's rule
In elimination reactions, the major product is the more highly substituted alkene (the one with more alkyl groups on the double bond carbons). This is because the more-substituted alkene is more thermodynamically stable, and its transition state reflects that stability.
Newman projection
A way of viewing a molecule by looking straight down a C-C bond axis. The front carbon is drawn as a dot, the back carbon as a circle. Useful for visualising the dihedral relationship between substituents, which is critical for predicting E2 outcomes.
Transition state
The highest-energy arrangement of atoms along a reaction pathway. For E2, the transition state has partial bonds forming and breaking simultaneously. It is not an intermediate; it cannot be isolated.
t-Butoxide (potassium tert-butoxide, KOtBu)
A strong, bulky base commonly used to promote E2 elimination over substitution. Its size makes it poor at SN2 but excellent at pulling off a proton.
For E2 to occur, four conditions must all be met:
A beta-hydrogen must exist (a hydrogen on the carbon adjacent to the one bearing the leaving group).
A leaving group must be present (halides such as Br, Cl, I; tosylate; mesylate).
A strong base is required (hydroxide, alkoxide, t-butoxide).
The beta-H and the leaving group must be able to adopt an antiperiplanar geometry.
Substrates that cannot undergo E2:
Vinyl halides and aryl halides: the leaving group is on an sp2 carbon, and there is no suitable beta-H in the right geometry.
Substrates with no beta-hydrogens at all (e.g. neopentyl-type structures with no H on the adjacent carbon, or methyl halides).
Bridgehead halides in small bicyclic systems: Bredt's rule forbids a double bond at the bridgehead of small bridged rings because the geometry cannot accommodate a planar alkene.
When you see a question asking which compounds "cannot undergo E2," check each of these conditions. The most common traps are bridgehead positions and the absence of beta-hydrogens.
E/Z naming applies to any alkene where each carbon of the double bond bears two different substituents. If either carbon of the double bond carries two identical groups, E/Z does not apply, and the compound has no geometric isomers.
Assignment method (Cahn-Ingold-Prelog priority rules):
At each carbon of the double bond, rank the two attached groups by atomic number of the atom directly bonded to that carbon.
If there is a tie at the first atom, move outward along the chain until a difference is found.
If the two higher-priority groups are on the same side of the double bond, the configuration is Z.
If the two higher-priority groups are on opposite sides, the configuration is E.
Key points for exam questions:
Lone pairs and hydrogen atoms count: H has the lowest priority of common substituents.
Halogens outrank carbon: Br > Cl > O-containing groups when the first atom is the deciding factor.
Double and triple bonds are treated as if each bond were to a separate atom (a C=O counts as C bonded to O twice).
To check whether a molecule can have E/Z isomers, verify that each doubly bonded carbon has two different substituents. For example, CH3CH2CH=CHCH2CH3 (3-hexene) can: each side has an ethyl group and a hydrogen, and those are different. But (CH3)2C=CHCH3 cannot meaningfully have E and Z because one carbon carries two identical methyl groups: wait, that is wrong. One carbon carries two methyls, meaning both substituents on that carbon are the same, so there are no geometric isomers. H2C=CHCH2CH(CH3)2 cannot either, because one carbon carries two hydrogens.
The stereochemical outcome of an E2 reaction is determined by which conformation reacts. Because E2 requires antiperiplanar geometry (H and leaving group at 180°), you must identify the Newman projection that places the correct H anti to the leaving group.
Worked approach for predicting the alkene product from a Newman projection:
Draw the Newman projection looking down the C2-C3 bond (or whichever bond loses the H and the leaving group).
Identify which H is antiperiplanar (180°) to the leaving group.
Remove that H and the leaving group, and form the double bond.
The groups that were on the same side in the Newman projection end up cis in the product; groups on opposite sides end up trans.
Example from the exam: to get cis-2-butene from 2-bromobutane, you need a conformation where a methyl group and a hydrogen on one carbon, and a methyl and Br on the other, are arranged so that H is anti to Br, and the two methyl groups end up on the same side of the resulting double bond.
The conformation where H3C and CH3 are both "up" (or both on the same side in the Newman projection) when the anti H is removed gives the cis (Z) product. The conformation where they are on opposite sides gives the trans (E) product.
Several factors influence how fast an E2 reaction proceeds.
Leaving group ability:
Better leaving groups accelerate the reaction. The order is I > Br > Cl > F for halides.
Tosylate and mesylate are also excellent leaving groups.
Base strength and bulk:
Stronger bases give faster E2 rates. t-Butoxide is both strong and bulky, making it a classic E2-promoting base.
Bulky bases favour elimination over substitution because they struggle to attack the electrophilic carbon (SN2) but can still reach a peripheral hydrogen.
Substrate structure:
More substituted substrates (3° > 2° > 1°) generally undergo E2 faster because the developing double bond is more stabilised by hyperconjugation and alkyl substitution in the transition state.
However, steric effects also matter. A very crowded substrate might slow E2 slightly if the base cannot reach the beta-H.
For an exam question asking which substrate reacts fastest with t-butoxide:
Compare leaving group quality first (Br > Cl at the same position).
Then consider whether the leaving group is axial or equatorial in a cyclohexane (axial undergoes E2 more readily because it can achieve antiperiplanar geometry with an adjacent axial H).
A substrate with the leaving group locked equatorial in a cyclohexane ring will be very slow for E2.
Zaitsev's rule states that elimination reactions favour the more-substituted alkene product. The thermodynamic basis is straightforward: more-substituted alkenes are more stable due to hyperconjugation (overlap of adjacent C-H sigma bonds with the pi system), and this stability difference is reflected in the transition state.
Reaction coordinate diagram for E2:
The starting material (alkyl halide + base) sits at one energy level on the left.
The transition state is the single energy maximum (E2 is concerted, one step).
The products (alkene + leaving group anion + conjugate acid of the base) sit at a lower energy on the right.
When two products are possible (a more-substituted and a less-substituted alkene), both share the same starting material but have different transition state energies.
The transition state leading to the Zaitsev (more-substituted) product is lower in energy, so its activation energy is smaller.
The Zaitsev product itself is also lower in energy than the Hofmann (less-substituted) product.
On the diagram:
The lower-energy product goes in the lower box on the right.
The higher-energy product goes in the upper box.
The transition state for the Zaitsev product is drawn as a lower hump than the one for the Hofmann product.
The key explanation is that the transition state for E2 has partial double-bond character. Because the more-substituted double bond is more stable, the transition state leading to it is also more stable (Hammond's postulate applied loosely: for an exothermic step, the transition state resembles the reactants, but the product stability still influences the relative energies of competing transition states).
E2 rate law: Rate = k[substrate][base]. Both concentrations matter, which is why it is called bimolecular.
Degree of unsaturation (index of hydrogen deficiency): (2C + 2 + N - H - X) / 2, where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Each degree of unsaturation corresponds to one ring or one double bond. Useful for working backwards from molecular formulae in synthesis problems.
Cahn-Ingold-Prelog priority: rank by atomic number at the point of first difference. Higher atomic number wins. For isotopes, higher mass number wins.
Elimination reactions are a primary method for synthesising alkenes in the lab, and alkenes are the starting point for a huge number of industrial chemical processes, from making polyethylene to pharmaceutical intermediates. Understanding E/Z selectivity matters in drug design: many biologically active molecules have a specific geometric isomer that is active while the other is inactive or even harmful.
Students often think any hydrogen adjacent to the leaving group can be removed in E2. It cannot: only a hydrogen that is antiperiplanar (180°) to the leaving group will react. Gauche (60°) arrangements do not work for E2.
Students confuse cis/trans with E/Z. Cis/trans is based on whether like groups are on the same side; E/Z is based on CIP priority. They often give the same answer, but not always, especially when the substituents include heteroatoms.
Students forget that E2 requires a strong base. Weak bases and polar protic solvents typically promote E1 or SN1 instead.
Students assume that the most stable chair conformation is the one that reacts in E2. In cyclohexanes, the reactive conformation for E2 is the one that places the leaving group axial with an adjacent axial H, even if that chair is less stable overall.
Identifying compounds that cannot undergo E2 (bridgehead halides, no beta-H, vinyl halides) is a recurring question type. Expect negative-marking for wrong answers.
Predicting cis vs trans alkene products from Newman projections is heavily tested. Practice drawing the Newman, finding the anti H, and mapping the remaining groups onto the product alkene.
Reaction coordinate diagrams for Zaitsev vs Hofmann products appear as a 10-point question. Be prepared to sketch transition states and label which product is more stable and why.
E/Z assignment questions test priority rules under time pressure. Drill these until they are automatic.
Questions about elimination rate often hinge on leaving group quality (Br vs Cl) and whether the leaving group can achieve axial orientation in a cyclohexane.
True or False: E2 reactions can occur with a gauche arrangement of H and the leaving group.
False. Antiperiplanar (180°) is required.
True or False: A vinyl bromide can undergo E2 elimination.
False. The leaving group is on an sp2 carbon with no suitable beta-H geometry.
Fill in the blank: In E/Z nomenclature, Z means the higher-priority groups are on the ______ side of the double bond.
Same.
True or False: Zaitsev's rule predicts the less-substituted alkene as the major product.
False. Zaitsev's rule predicts the more-substituted alkene.
Fill in the blank: The E2 rate law is Rate = k[][].
Substrate, base.
Q: A student draws a Newman projection of 2-bromobutane and identifies a conformation where the two methyl groups are on opposite sides when looking down the C2-C3 bond, with H anti to Br. What alkene geometry does E2 give from this conformation?
A: Trans-2-butene (E-2-butene). When the methyl groups are on opposite sides in the reactive conformation, they end up on opposite sides of the double bond.
Q: Why can a bridgehead bromide in bicyclo[2.2.1]heptane not undergo E2?
A: Bredt's rule. Forming a double bond at the bridgehead of this small bridged ring would require extreme strain because the pi bond cannot adopt the necessary planar geometry.
Q: Compound A (a cyclohexane with Br equatorial and H axial at adjacent carbons) and Compound B (Br axial and H axial at adjacent carbons) are treated with NaOEt. Which reacts faster by E2?
A: Compound B. E2 requires a trans-diaxial (antiperiplanar) arrangement. Compound B already has Br and H both axial and anti to each other. Compound A would need to ring-flip first.
Q: Which alkenes from the list I. CH3CH2CH=CHCH2CH3, II. (CH3)2C=CHCH3, III. CH3CH2CH=CHBr, IV. H2C=CHCH2CH(CH3)2 can have E and Z isomers?
A: I and III (answer B). Compound I has two different groups on each doubly bonded carbon (ethyl and H on each side). Compound III has ethyl and H on one carbon, Br and H on the other, so both carbons have two different substituents. Compound II has two identical methyl groups on one carbon, so no E/Z. Compound IV has two hydrogens on one carbon, so no E/Z.
Q: On a reaction coordinate diagram for E2 elimination giving both Zaitsev and Hofmann products, which product has the lower transition state energy?
A: The Zaitsev product. Its transition state has partial double-bond character, and because the more-substituted alkene is more stable, the transition state leading to it is lower in energy.
E2 elimination connects directly to SN2 substitution: the same substrates, bases, and leaving groups appear in both pathways, and exam questions often ask you to predict which one dominates. This connects to alkene addition reactions (the reverse direction: adding HX, H2O, or halogens across a double bond), covered in Part 2 of these notes. The stereochemical principles here (antiperiplanar geometry, E/Z assignment) reappear in the study of alkynes, dienes, and later in pericyclic reactions.
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