Difficulty: Intermediate to Advanced | Prerequisites: IUPAC nomenclature, Lewis structures, electronegativity, acid-base chemistry, stereochemistry basics (R/S, E/Z).
Substitution and elimination reactions are the core of Organic Chemistry I. Nearly every transformation you study in this course involves a nucleophile attacking an electrophile (substitution) or a base removing a proton to form a double bond (elimination). The challenge is learning when each mechanism operates and what products to expect. This material builds directly on your knowledge of Lewis structures, electronegativity, and stereochemistry. If you can reliably distinguish SN1 from SN2 and E1 from E2, you can predict the outcome of most reactions on the exam.
SN2 is a one-step, backside-attack mechanism favoured by strong nucleophiles, primary/methyl substrates, and polar aprotic solvents. SN1 is a two-step mechanism (carbocation intermediate) favoured by tertiary substrates and polar protic solvents. E2 is a one-step elimination requiring a strong base and anti-periplanar geometry. E1 shares the carbocation intermediate of SN1 and competes with it at high temperature. Knowing the substrate, nucleophile/base strength, and solvent lets you predict which pathway dominates.
Nucleophile
A species that donates an electron pair to an electrophilic carbon. Think of it as the "attacker" in a substitution reaction. Stronger nucleophiles have a negative charge or high electron density.
Electrophile
A species that accepts an electron pair, typically a carbon bearing a good leaving group. In simple terms, it is the carbon that gets attacked.
Leaving group
The atom or group that departs with the bonding electrons during substitution or elimination. Better leaving groups are weaker bases (the conjugate bases of strong acids). Iodide is better than chloride, and both are better than hydroxide.
SN2 (bimolecular nucleophilic substitution)
A one-step mechanism where the nucleophile attacks the electrophilic carbon at 180 degrees to the leaving group (backside attack), causing inversion of stereochemistry. Rate depends on both substrate and nucleophile concentrations.
SN1 (unimolecular nucleophilic substitution)
A two-step mechanism: the leaving group departs to form a carbocation, then the nucleophile attacks. Rate depends only on substrate concentration. Produces racemisation (a mix of R and S products).
E2 (bimolecular elimination)
A one-step mechanism where a strong base removes a proton anti-periplanar to the leaving group, forming an alkene. Rate depends on both substrate and base.
E1 (unimolecular elimination)
A two-step mechanism sharing the same carbocation intermediate as SN1. After the leaving group departs, a base (often the solvent) removes a proton to form an alkene. Favoured by heat.
Solvolysis
A reaction in which the solvent itself acts as the nucleophile or base (e.g. methanol in an SN1/E1 reaction).
Zaitsev's rule
The more substituted alkene is the major product of an elimination reaction, unless a bulky base is used.
Hofmann product
The less substituted alkene, favoured when a sterically bulky base (such as potassium tert-butoxide) removes the more accessible proton instead of the one that would give the Zaitsev product.
Anti-periplanar geometry
The spatial arrangement in which the proton being removed and the leaving group are on opposite sides of the C-C bond (180-degree dihedral angle). Required for E2 elimination.
Mesylate (OMs) / Tosylate (OTs)
Sulfonate ester leaving groups formed by reacting an alcohol with mesyl chloride (MsCl) or tosyl chloride (TsCl) in the presence of pyridine. They convert a poor leaving group (OH) into an excellent one without changing the carbon skeleton or stereochemistry at that carbon.
A stronger nucleophile drives SN2 faster.
Negatively charged species are stronger nucleophiles than their neutral counterparts (e.g. CH3O- > CH3OH).
In a given row of the periodic table, nucleophilicity roughly follows basicity.
In polar aprotic solvents, nucleophilicity follows the periodic table trend going up and left (smaller, more electronegative atoms with concentrated charge are better nucleophiles).
In polar protic solvents, the trend reverses for atoms in the same group: larger atoms are better nucleophiles because they are less well-solvated (e.g. S- > O- in protic solvents).
Exam example: CH3S- was the strongest nucleophile (answer d) because sulfur is large, polarisable, and carries a negative charge.
SN2 requires backside attack, so steric hindrance at the electrophilic carbon is critical.
Methyl > primary > secondary >> tertiary (tertiary substrates essentially do not undergo SN2).
Better leaving groups also increase SN2 rate: I- > Br- > Cl- >> OH- (hydroxide is not a leaving group without activation).
Exam example: CH3CH2Cl (primary, Cl leaving group, answer a) was the most reactive electrophile. The tertiary neopentyl-type iodide (answer c) was too hindered despite having a good leaving group. The alcohol (answer d) has no leaving group at all.
E2 requires a strong base. The strongest bases are small, negatively charged alkoxides.
Methoxide (CH3O-) is a strong, non-bulky base that gives Zaitsev products.
Potassium tert-butoxide (KOC(CH3)3) is a strong, bulky base that gives Hofmann (less substituted) products.
Exam example: CH3O- (answer b) was the strongest base for E2, stronger than Br- (weak base), CH3S- (weaker base because S is less electronegative than O), or an acetate/carboxylate (resonance-stabilised, weaker base).
Triple bonds are shorter than double bonds, which are shorter than single bonds.
In a conjugated or simple alkene, rank bond lengths: C-C single (longest) > C=C double (shorter) > C-H (shortest in many contexts).
Exam example: in propene, the bonds were ranked c (C-C single) > b (C=C double) > a (C-H, shortest).
One-step, concerted: nucleophile attacks as leaving group departs.
Backside attack causes inversion of configuration (Walden inversion).
Polar aprotic solvents (acetone, DMSO, DMF) are best because they do not solvate the nucleophile.
Exam example (Q3-i): a secondary substrate with CH3SNa in acetone undergoes SN2. The thiolate attacks from the back, inverting stereochemistry.
Exam example (Q3-vii): conversion of a mesylate to a nitrile using NaCN in DMSO. The cyanide ion displaces the mesylate leaving group via SN2 with inversion.
Both proceed through a carbocation intermediate. Step 1: leaving group departs. Step 2: nucleophile attacks (SN1) or base removes a proton (E1).
Favoured by tertiary substrates (stable carbocations), weak nucleophiles/bases, and polar protic solvents.
SN1 gives racemisation at the stereogenic centre.
E1 is favoured over SN1 at higher temperatures.
Exam example (Q3-ii): a tertiary alkyl chloride in methanol. At 25 degrees C, SN1 dominates, giving a methyl ether (OCH3 product). With heat, E1 dominates, giving the alkene.
Exam example (Q5-i): tert-butyl iodide reacts faster than tert-butyl chloride in solvolysis (E1) because the C-I bond is weaker than C-I, making the rate-determining ionisation step faster. Iodide is also a better leaving group.
Requires anti-periplanar geometry: the H and the leaving group must be on opposite sides of the C-C bond in a 180-degree arrangement.
Strong base, typically an alkoxide.
Zaitsev product (more substituted alkene) is the default major product, unless a bulky base is used.
Exam example (Q5-ii): the trans isomer of a bromocyclohexane reacts fastest via E2 because the axial bromine and an axial hydrogen are already anti-periplanar in the favoured chair conformation. The cis isomer cannot achieve this geometry as easily.
Exam example (Q5-iv): sodium tert-butoxide (a bulky base) gives a non-Zaitsev (Hofmann) major product because it preferentially removes the most accessible (least sterically hindered) proton.
Converting an alcohol to an alkene often requires two steps: first activate the OH (make it a better leaving group), then eliminate.
Exam example (Q4): cyclopentanol to methylenecyclopentane. Step 1: SOCl2 with pyridine converts the alcohol to a chloride (retaining configuration, good leaving group). Step 2: potassium tert-butoxide with tert-butanol promotes E2 elimination to give the less substituted (exocyclic) alkene as the major product.
Alcohols are poor substrates for substitution/elimination because OH- is a terrible leaving group.
Three common activation methods:
Protonation with strong acid (H2SO4, HI, HBr) converts OH to OH2+, which leaves as water.
SOCl2 with pyridine converts OH to Cl with inversion of configuration.
Mesyl chloride (MsCl) or tosyl chloride (TsCl) with pyridine converts OH to OMs or OTs without breaking the C-O bond (retention of configuration at carbon).
Exam example (Q3-vi): an alcohol is treated with MsCl/pyridine to form a mesylate, then NaCN in DMSO displaces the mesylate via SN2 with inversion.
Students often assume that the best nucleophile is also the best base. They overlap but are not the same. Nucleophilicity is a kinetic property (how fast it attacks carbon). Basicity is a thermodynamic property (how strongly it binds a proton). Sulfur anions are great nucleophiles but mediocre bases.
Thinking SN1 always gives a 50:50 racemic mixture. In practice the leaving group often partially shields one face of the carbocation, so the product is not perfectly racemic, though it does lose optical purity.
Confusing the effect of solvent on SN1 vs SN2. Polar protic solvents favour SN1 (they stabilise the carbocation and the leaving group). Polar aprotic solvents favour SN2 (they do not solvate the nucleophile, leaving it free to attack).
Assuming heat always means E1. Heat favours elimination over substitution in general, but E2 can also be the elimination pathway if a strong base is present. Heat alone, with a weak base, points to E1.
Predicting the correct mechanism (SN1, SN2, E1, E2) from given conditions is the single most important skill tested. Expect multiple questions on this.
Stereochemistry of products: SN2 gives inversion, SN1 gives racemisation, E2 requires anti-periplanar H and leaving group. The exam tested all three.
Zaitsev vs Hofmann products: know that bulky bases (tert-butoxide) flip the regiochemistry of elimination.
Leaving group quality is tested both directly (ranking leaving groups) and indirectly (why tert-butyl iodide reacts faster than tert-butyl chloride).
Multistep synthesis questions test your ability to string together activation of an alcohol (SOCl2, MsCl, acid) followed by substitution or elimination.
True or false: SN2 reactions proceed faster at tertiary carbons than primary carbons. (False, the opposite.)
Fill in the blank: In an SN2 reaction, the nucleophile attacks from the ______ side, causing ______ of configuration. (back; inversion)
True or false: Polar protic solvents favour SN2 reactions. (False, they favour SN1.)
Fill in the blank: E2 elimination requires the hydrogen and leaving group to be in an ______ arrangement. (anti-periplanar)
True or false: Heating an SN1 reaction mixture shifts the product ratio towards elimination (E1). (True)
Q: A secondary alkyl bromide is treated with CH3SNa in acetone. What mechanism operates, and what happens to the stereochemistry at the carbon bearing the leaving group?
A: SN2. The thiolate is a strong nucleophile, acetone is a polar aprotic solvent, and the substrate is secondary (borderline, but the strong nucleophile pushes SN2). The stereochemistry is inverted.
Q: Tert-butyl chloride is dissolved in methanol at 25 degrees C. What is the major product and mechanism?
A: SN1. Methanol is a weak nucleophile and polar protic solvent; the substrate is tertiary. The major product is tert-butyl methyl ether (methanol replaces chloride). At higher temperature, E1 would dominate and the alkene would be the major product.
Q: Why does tert-butyl iodide undergo solvolysis faster than tert-butyl chloride?
A: The C-I bond is weaker than the C-Cl bond, so the rate-determining ionisation step (forming the carbocation) is faster. Iodide is also a better leaving group because it is a weaker base.
Q: A cyclohexane ring has a bromine substituent. The trans isomer undergoes E2 elimination faster than the cis isomer. Explain why.
A: In the trans isomer, the bromine can adopt an axial position in one chair conformation, placing it anti-periplanar to an axial hydrogen on the adjacent carbon. This geometry is required for E2. The cis isomer cannot achieve anti-periplanar geometry as readily.
Q: Sodium tert-butoxide reacts with a secondary alkyl bromide bearing both a more substituted and a less substituted beta-carbon. Which alkene is the major product?
A: The less substituted (Hofmann) alkene. Tert-butoxide is a bulky base that preferentially removes the most accessible proton, which is on the less substituted side.
Q: An alcohol is treated with mesyl chloride and pyridine, then with NaCN in DMSO. What is the overall transformation and the stereochemical outcome?
A: The alcohol is converted to a mesylate (retention of configuration at carbon), then the cyanide displaces the mesylate via SN2 (inversion). The net result is substitution of OH by CN with inversion of configuration.
Substitution and elimination reactions build on nomenclature (you need to name the products) and stereochemistry (you need to assign R/S and E/Z to products). These reactions also connect forward to alcohol chemistry, ether synthesis, and organometallic reactions in Organic Chemistry II. The concept of leaving group activation (mesylates, tosylates) returns in carbohydrate chemistry and biochemistry.
SN1, SN2, E1, E2, nucleophilic substitution, elimination reaction, nucleophile strength, electrophile reactivity, leaving group, backside attack, Walden inversion, racemisation, carbocation, solvolysis, polar aprotic solvent, polar protic solvent, Zaitsev rule, Hofmann elimination, anti-periplanar, mesylate, tosylate, SOCl2 pyridine, tert-butoxide, bulky base, organic chemistry I exam, OChem reactions review