Difficulty: Intermediate | Prerequisites: CHEM 211 (Organic Chemistry I), familiarity with Lewis structures, acid-base chemistry, nucleophiles and electrophiles.
Organic chemistry is largely about understanding how and why molecules react. This worksheet sits at the crossroads of two of the most important reaction families in the course: substitution/elimination (what happens when a leaving group departs from a carbon) and addition (what happens when a pi bond opens up to accept new atoms). If you can reliably predict which pathway wins for a given set of conditions, you can predict the product of almost any reaction in CHEM 212.
You should already be comfortable drawing Lewis structures, identifying nucleophiles and electrophiles, and understanding pKa as a measure of acidity. The concepts here build directly on the SN1, SN2, E1 and E2 frameworks from Chapter 9, and extend them to alcohols (Chapter 10) and alkene additions (Chapter 6).
When an organic molecule has a leaving group (halide, protonated OH) and you add a nucleophile or base, it can undergo substitution (replace the leaving group) or elimination (lose the leaving group plus a proton to form a double bond). Which pathway wins depends on the substrate's degree (1°, 2°, 3°), the strength and size of the nucleophile/base, and the solvent. Alkene addition reactions run the reverse logic: a double bond reacts with HX, water, or borane to add atoms across the pi bond, with regio- and stereoselectivity controlled by the mechanism.
SN2 (bimolecular nucleophilic substitution)
A one-step mechanism where the nucleophile attacks the electrophilic carbon at the same time the leaving group departs. The rate depends on the concentration of both the substrate and the nucleophile. Always produces inversion of configuration (backside attack). In simple terms, think of it as the nucleophile kicking out the leaving group in a single move, like an umbrella flipping inside out.
SN1 (unimolecular nucleophilic substitution)
A two-step mechanism where the leaving group departs first to form a carbocation intermediate, then the nucleophile attacks. The rate depends only on the substrate concentration. Produces a racemic mixture because the carbocation is planar, so the nucleophile can attack from either face. Think of it as the leaving group walking away first, leaving a flat, open spot that can be hit from either side.
E2 (bimolecular elimination)
A one-step mechanism where a strong base removes a proton from a carbon adjacent to the leaving group (the beta carbon) at the same time the leaving group departs, forming a double bond. Requires anti-periplanar geometry between the departing H and the leaving group. In simple terms, the base grabs a proton and the leaving group exits simultaneously, like two doors opening at once.
E1 (unimolecular elimination)
A two-step mechanism where the leaving group departs first to form a carbocation, then a base removes a proton from the beta carbon to form the alkene. Often competes with SN1 and typically favoured by heat. Think of it as the leaving group departing first (just like SN1), but instead of a nucleophile attacking, a base pulls off a proton to make a double bond.
Nucleophile
A Lewis base that donates an electron pair to an electrophilic carbon. Good nucleophiles are negatively charged or have lone pairs on atoms with low electronegativity. In simple terms, it is the electron-rich species that does the attacking.
Leaving group (LG)
An atom or group that departs with the bonding electrons. Good leaving groups are stable once they leave (weak bases). Halides (Cl⁻, Br⁻, I⁻) and water (from protonated alcohols) are common examples. Think of it as the group that can comfortably walk away carrying the electrons.
Carbocation
A positively charged carbon intermediate with an empty p-orbital. Stability order: 3° > 2° > 1° > methyl. Benzylic and allylic carbocations gain extra stability through resonance. In simple terms, it is a carbon that has lost its leaving group and is electron-hungry.
Substrate degree (1°, 2°, 3°)
Refers to how many other carbon atoms are bonded to the carbon bearing the leaving group. A primary (1°) carbon is bonded to one other carbon, secondary (2°) to two, tertiary (3°) to three. This is the single most important variable in predicting SN1/SN2/E1/E2.
Polar protic solvent
A solvent that can donate hydrogen bonds (e.g. water, ethanol, methanol). Stabilises carbocations and anions through solvation. Favours SN1 and E1 pathways.
Polar aprotic solvent
A solvent that cannot donate hydrogen bonds (e.g. acetone, DMSO, DMF). Does not stabilise anions well, leaving nucleophiles "naked" and more reactive. Favours SN2.
Zaitsev's rule
In elimination reactions, the more substituted alkene (more R groups on the double-bond carbons) is the major product. This is the thermodynamically more stable product.
Anti-periplanar geometry
A spatial arrangement where the departing H and the leaving group are on opposite sides of the C-C bond and 180° apart (anti to each other). Required for the E2 mechanism. Best visualised using Newman projections.
The worksheet's central skill is reading a set of reaction conditions and predicting which mechanism dominates. Five questions guide every prediction:
What is the substrate's degree? (methyl, 1°, 2°, 3°)
Is the reagent a good nucleophile, a strong base, or both?
Is the solvent polar protic or polar aprotic?
Are the conditions acidic, basic, or neutral?
Is heat applied? (heat generally favours elimination)
Methyl and primary (1°) substrates
SN2 is strongly favoured. The carbon bearing the leaving group is relatively unhindered, so backside attack by the nucleophile is easy. E2 can compete if a strong, bulky base is used (e.g. tert-butoxide), but substitution is the default.
Secondary (2°) substrates
This is where it gets tricky, because all four pathways are possible. The reagent and solvent tip the balance:
Strong base + polar protic solvent: E2 dominates (as in Puzzle 4, where NaOEt/EtOH with a 2° alkyl bromide gives the Zaitsev alkene as the major product)
Good nucleophile + polar aprotic solvent: SN2 dominates (as in Puzzle 3, where NaI/acetone converts a 2° alkyl bromide to the alkyl iodide with inversion)
Weak nucleophile/base + polar protic solvent + heat: SN1/E1 mixture
Tertiary (3°) substrates
SN2 is essentially impossible due to steric hindrance. In basic conditions, E2 dominates. In neutral or weakly nucleophilic conditions (polar protic solvent), SN1 and E1 compete, with E1 favoured by heat.
Puzzle 6 illustrates this well: a tertiary alkyl chloride with ethanol (a weak nucleophile, poor base, polar protic solvent) proceeds through SN1. The carbocation intermediate is planar, so nucleophilic attack from either face produces a racemic mixture of ethers.
SN2: one step, inversion, no intermediate
The nucleophile attacks the carbon from the side opposite the leaving group (backside attack). The transition state has five groups around the central carbon. The C-Nu bond forms as the C-LG bond breaks, and the three remaining groups "flip" like an umbrella in wind. The product always has inverted stereochemistry relative to the starting material.
Key markers for SN2:
Good nucleophile present (I⁻, CN⁻, RS⁻, RO⁻)
Polar aprotic solvent (acetone, DMSO, DMF)
Methyl or primary substrate (secondary is possible if the nucleophile is strong and unhindered)
Product shows inversion of configuration
Puzzle 3 is a clean SN2 example. A secondary alkyl bromide reacts with NaI in acetone. Iodide is a strong nucleophile, acetone is polar aprotic, and the pKa of HI is about -10, making iodide a negligible base. The product is the alkyl iodide with inversion, plus NaBr as a byproduct. Acetone is a particularly clever solvent choice here because NaBr precipitates out of acetone, driving the reaction forward (the Finkelstein reaction).
Puzzle 11 follows the same logic: NaCN in acetone with a primary substrate. Cyanide is a good nucleophile, acetone is polar aprotic, and the product is the nitrile with inversion at the reacting carbon.
SN1: two steps, racemisation, carbocation intermediate
Step 1: the leaving group departs to form a carbocation. Step 2: the nucleophile attacks the carbocation. Because the carbocation is sp2-hybridised (planar), the nucleophile can approach from either face, producing a roughly equal mixture of both enantiomers (racemic mixture).
Key markers for SN1:
Tertiary (or benzylic/allylic) substrate
Weak nucleophile (often the solvent itself, called solvolysis)
Polar protic solvent
Neutral or acidic conditions
Product is racemic (if the starting material had a stereocentre)
Puzzle 6 is a textbook SN1. A tertiary alkyl chloride reacts with ethanol. Ethanol is a weak nucleophile and a poor base (pKa of protonated ethanol is about -2.4), so it cannot do SN2 or E2. The chloride leaves, forming a tertiary carbocation. Ethanol attacks from both faces. The resulting oxonium ion is deprotonated by another molecule of ethanol to give the ether product as a racemic mixture.
E2: one step, anti-periplanar requirement, strong base
The base removes a proton from the beta carbon while the leaving group simultaneously departs. The H and the leaving group must be anti-periplanar (180° dihedral angle) for the orbitals to align properly. This requirement controls which alkene isomer forms, and is best analysed using Newman projections.
Key markers for E2:
Strong base present (NaOEt, NaOH, KOtBu, NaNH2)
Secondary or tertiary substrate
Product follows Zaitsev's rule (more substituted alkene is major) unless a bulky base is used, in which case the less substituted (Hofmann) product may dominate
Anti-periplanar geometry is required
Puzzle 4 is a straightforward E2. A secondary alkyl bromide is treated with NaOEt in ethanol. Ethoxide is a strong base. The mechanism is concerted: the base pulls the beta hydrogen, the electrons form the new pi bond, and bromide departs. The major product is the more substituted alkene (Zaitsev's rule). The worksheet notes that E2 dominates over SN2 here because the substrate is secondary and a strong base is present.
Puzzle 8 extends E2 to stereochemistry. The same conditions (NaOEt/EtOH) are used, but the substrate has specific stereochemistry. To predict the correct alkene geometry (E vs Z), you must draw the Newman projection, rotate to place H and Br anti-periplanar, and then determine which groups end up on which side of the resulting double bond. The worksheet asks for this Newman projection to justify the stereochemical outcome.
Puzzle 9 applies E2 to a cyclohexane system. A secondary chloride on a cyclohexyl ring is treated with NaOH in water. For E2 in cyclohexane, the H and the leaving group must both be axial and anti-periplanar (trans-diaxial). Drawing the chair conformation reveals which beta hydrogen can achieve this geometry, which in turn determines the regiochemistry of the product alkene.
E1: two steps, carbocation intermediate, often competes with SN1
Step 1: the leaving group departs to form a carbocation (identical to SN1). Step 2: a base (often the solvent) removes a proton from the beta carbon, forming the alkene. The major product usually follows Zaitsev's rule.
Puzzle 1 illustrates elimination from an alcohol under acidic conditions. A secondary alcohol is treated with H2SO4 and heat. The acid protonates the hydroxyl group, converting it to water (a good leaving group). Water departs to form a secondary carbocation. Then a base (often HSO4⁻ or water) removes a beta proton to form the more substituted alkene (major) and the less substituted alkene (minor).
Alcohols are poor substrates for substitution and elimination because hydroxide (OH⁻) is a terrible leaving group. The solution is to convert -OH into a better leaving group. Under acidic conditions, the oxygen is protonated to give -OH2⁺, and water is an excellent leaving group.
Dehydration with strong acid and heat (Puzzle 1)
A secondary alcohol treated with H2SO4 and heat undergoes acid-catalysed dehydration. The mechanism is E1:
Proton transfer: H2SO4 protonates the -OH to give -OH2⁺
Loss of leaving group: water departs, forming a carbocation
Proton transfer: a base removes a beta-H to form the alkene
The major product is the more substituted alkene (Zaitsev's rule). Heat drives the equilibrium toward elimination over substitution.
Substitution with HBr (Puzzle 5)
A secondary alcohol treated with HBr undergoes substitution. HBr is acidic, so it protonates the alcohol first, converting -OH to -OH2⁺. Water leaves to form a carbocation. Bromide then attacks the carbocation. The overall result is replacement of -OH with -Br.
The acidic conditions here favour formation of cationic intermediates (carbocations), which is the hallmark of SN1-type reactivity. Note the contrast with Puzzle 1: same type of substrate, but HBr provides a nucleophile (Br⁻) that traps the carbocation before elimination can occur, whereas H2SO4/heat provides no good nucleophile and the conditions push toward alkene formation.
Converting alcohols to good leaving groups with TsCl (Puzzle 12)
Another strategy is to convert the alcohol to a tosylate (OTs) using TsCl and pyridine. The tosylate is an excellent leaving group, and the conversion happens with retention of configuration at the reacting carbon (the C-O bond is not broken in this step). Once the tosylate is in place, a nucleophile such as NaOAc in DMSO (polar aprotic) can carry out an SN2 reaction with inversion.
The Williamson ether synthesis makes an ether by reacting an alkoxide (RO⁻) with an alkyl halide via SN2. Because it relies on SN2, the alkyl halide must be methyl or primary to avoid elimination.
Puzzle 14 asks you to consider two possible disconnections of the same ether target. Each route uses a different combination of alkoxide and alkyl halide:
Route A: use a primary alkyl halide + the other fragment as the alkoxide
Route B: use a tertiary or bulky alkyl halide + the other fragment as the alkoxide
Route A is feasible because SN2 works well on primary substrates. Route B fails because a tertiary alkyl halide cannot undergo SN2 (too sterically hindered), and the strong alkoxide base would instead cause E2 elimination.
The general rule: in Williamson ether synthesis, always make the less hindered fragment the alkyl halide and the more hindered fragment the alkoxide. This ensures SN2 proceeds cleanly without competing elimination.
Students often think SN1 and SN2 differ only in speed. They do not. SN1 involves a carbocation intermediate and produces racemisation; SN2 is concerted and always inverts stereochemistry. The number refers to molecularity (how many species are in the rate-determining step), not the number of steps.
Students often assume that a strong nucleophile always means substitution wins. It does not. If the nucleophile is also a strong base and the substrate is secondary or tertiary, E2 elimination often dominates instead.
Students forget that -OH is a terrible leaving group. Under basic conditions, alcohols do not undergo substitution or elimination. You need acidic conditions (to protonate -OH to -OH2⁺) or conversion to a tosylate/mesylate first.
Students commonly ignore solvent effects. Polar aprotic solvents (acetone, DMSO) dramatically favour SN2 because they do not solvate nucleophiles. Polar protic solvents (water, ethanol) stabilise carbocations and favour SN1/E1. Changing the solvent can change the entire product distribution.
⚠️ The SN1/SN2/E1/E2 decision tree is the most commonly tested topic in Organic Chemistry II. Expect a reaction with conditions given and be ready to name the mechanism, predict the product, and draw the curved-arrow mechanism.
⚠️ Stereochemistry is frequently tested alongside mechanism. SN2 gives inversion. SN1 gives racemisation. E2 requires anti-periplanar geometry. Know how to use Newman projections and chair conformations to justify your stereochemical predictions.
⚠️ Alcohol conversions (protonation to make -OH2⁺, tosylation with TsCl) are a favourite exam topic because they test whether you understand leaving group ability.
⚠️ Williamson ether synthesis questions almost always test whether you can identify the correct disconnection. The alkyl halide must be primary or methyl for SN2 to work.
⚠️ Watch for carbocation rearrangements (hydride shifts, methyl shifts) when SN1 or E1 is involved. If a more stable carbocation can form by rearrangement, it will.
True or False: SN2 reactions always produce inversion of stereochemistry at the reacting carbon.
True or False: A tertiary alkyl halide readily undergoes SN2 with a strong nucleophile.
Fill in the blank: In E2 elimination, the hydrogen and the leaving group must be in a(n) __________ arrangement.
True or False: Ethanol is a polar aprotic solvent.
Fill in the blank: In Williamson ether synthesis, the alkyl halide should be __________ (primary/tertiary) to avoid elimination.
Q: A secondary alkyl bromide is treated with NaI in acetone. What is the major product and by what mechanism does the reaction proceed?
A: The product is the corresponding alkyl iodide with inversion of configuration, formed by SN2. Iodide is a good nucleophile, acetone is polar aprotic (favouring SN2), and iodide is a negligible base (pKa of HI ≈ -10).
Q: A tertiary alkyl chloride is dissolved in ethanol. What products do you expect, and what mechanism operates?
A: The major product is the ethyl ether, formed by SN1. Ethanol is a weak nucleophile and a polar protic solvent. The tertiary substrate cannot undergo SN2. The carbocation forms first, then ethanol attacks from both faces, giving a racemic mixture. E1 elimination to give the alkene is a minor pathway.
Q: Why does E2 dominate over SN2 when NaOEt is used with a secondary alkyl bromide?
A: Ethoxide is both a strong nucleophile and a strong base. With a secondary substrate, the carbon is moderately hindered, slowing SN2. Meanwhile, E2 only requires the base to reach a beta hydrogen, which is less sterically demanding. The combination of a strong base with a secondary substrate tips the balance toward E2.
Q: In a Williamson ether synthesis to make methyl tert-butyl ether, which fragment should serve as the alkyl halide and which as the alkoxide?
A: Methyl bromide (or methyl iodide) should be the alkyl halide, and potassium tert-butoxide should be the alkoxide. Methyl halides are the best SN2 substrates. Using tert-butyl bromide with methoxide would give elimination (E2) because tert-butyl halides are too hindered for SN2 and methoxide is a strong base.
Q: An optically active secondary alcohol is treated with HBr. The product is optically inactive. Explain.
A: HBr protonates the -OH to form -OH2⁺, which leaves as water to form a planar carbocation (SN1 pathway). Bromide attacks the carbocation from both faces equally, giving a racemic (optically inactive) mixture.
Substitution and elimination reactions connect directly to alkene chemistry (Chapter 6): elimination creates alkenes, and alkene additions (covered in the second set of notes) effectively reverse that process. Understanding when elimination wins over substitution helps you design synthetic routes that build molecular complexity in the right direction.
The leaving group concept here extends to carbonyl chemistry later in the course. Acyl substitution reactions (Chapter 12+) follow the same logic of a nucleophile displacing a leaving group, but at an sp2 carbon rather than sp3.
Stereochemistry from SN2 inversion and E2 anti-periplanar requirements will reappear in every reaction type for the rest of the course. If you can draw Newman projections and chair conformations fluently now, the later material becomes considerably more manageable.
SN1, SN2, E1, E2, nucleophilic substitution, elimination reaction, alkyl halide, leaving group, carbocation, carbocation rearrangement, hydride shift, methyl shift, backside attack, inversion of configuration, racemic mixture, racemisation, Zaitsev's rule, Saytzeff rule, anti-periplanar, Newman projection, chair conformation, trans-diaxial, polar protic solvent, polar aprotic solvent, solvolysis, Williamson ether synthesis, tosylate, TsCl, Finkelstein reaction, dehydration, acid-catalysed elimination, substrate degree, primary secondary tertiary, nucleophile vs base, strong base elimination, CHEM 212, organic chemistry II, predict the products, reaction mechanisms