Difficulty: Intermediate | Prerequisites: Acidity and resonance notes, Lewis structures, stereochemistry basics (R/S, E/Z)
Substitution and elimination are the two main reaction families for haloalkanes (and alcohols after activation). Nearly every organic synthesis problem boils down to deciding which of four pathways applies: SN1, SN2, E1, or E2. The variables that determine the pathway are the substrate structure (1°, 2°, 3°), the nucleophile/base strength, and the solvent. This is tested heavily, often as mechanism questions, product prediction, and reagent matching.
Substitution replaces a leaving group with a nucleophile; elimination removes the leaving group and a neighbouring H to form a double bond. SN2 and E2 are concerted (one step, strong reagents, anti stereochemistry for E2). SN1 and E1 go through a carbocation intermediate (weak reagents, rearrangements possible, both products often seen together). Substrate degree and reagent strength determine which pathway dominates.
Nucleophile
A species that donates an electron pair to an electrophilic carbon. Think of it as the attacker in substitution.
Leaving group
The group that departs with the bonding electrons. Good leaving groups are weak bases (e.g. halides, tosylate, water after protonation).
SN2 (substitution, nucleophilic, bimolecular)
A one-step mechanism where the nucleophile attacks the carbon at the same time the leaving group departs. Rate depends on both substrate and nucleophile concentrations.
SN1 (substitution, nucleophilic, unimolecular)
A two-step mechanism: the leaving group departs first to form a carbocation, then the nucleophile attacks. Rate depends only on the substrate concentration.
E2 (elimination, bimolecular)
A one-step mechanism where a base removes a neighbouring H while the leaving group departs simultaneously, forming a double bond. Requires anti-coplanar geometry between the H and the leaving group.
E1 (elimination, unimolecular)
A two-step mechanism: the leaving group departs to form a carbocation, then a base (often the solvent) removes a neighbouring H to form the alkene.
Zaitsev product (Zaitzev product)
The more substituted alkene, which is the major elimination product in most E1 and E2 reactions.
Hofmann product
The less substituted alkene, formed as the major product when a bulky base is used in E2.
Inversion of configuration
Flipping the stereochemistry at the carbon being attacked. SN2 always gives inversion (backside attack). In simple terms, the groups swap sides like an umbrella flipping in the wind.
Carbocation rearrangement
A hydride shift or methyl shift that converts a less stable carbocation into a more stable one. Only possible in SN1/E1 pathways where a free carbocation intermediate exists.
Reactivity order: 1° > 2° > 3° (methyl is best). The reaction is concerted, so steric hindrance at the carbon slows it down.
Reagents: Strong nucleophiles. Common ones include R₃P, NH₂⁻, RS⁻, HS⁻, I⁻, CN⁻, R₃N, OH⁻, RO⁻, Br⁻, RNH₂, NH₃, RCOO⁻, Cl⁻.
Solvent: Polar aprotic solvents work best (DMSO, DMF, acetone). These solvents do not hydrogen-bond to the nucleophile, leaving it free to attack.
Stereochemistry: Only inversion products. Backside attack flips the configuration at the carbon.
Reactivity order: 3° > 2° > 1°. More substituted substrates eliminate more readily because the product alkene is more stable.
Reagents: Strong bases that are weak nucleophiles. Common choices include OH⁻, RO⁻ (methoxide, ethoxide). For exclusively E2 products, use bulky bases such as tert-butoxide, DBU, or DBN (Hofmann elimination).
Stereochemistry: Anti-coplanar arrangement is required. The H being removed and the leaving group must be anti (180°) to each other.
On a Newman projection, orient the molecule so H and X are anti before drawing the product.
On a cyclohexane ring, both H and the leaving group must be axial and trans-diaxial. Draw the chair conformation to confirm.
Product regiochemistry: Zaitsev product (more substituted alkene) is major, unless a bulky base is used, which gives the Hofmann product (less substituted alkene).
They compete when a strong nucleophile/base reacts with a 1° or 2° haloalkane.
SN2 has many more compatible reagents. E2-exclusive reagents are the bulky bases (tert-butoxide, DBU, DBN) that are too hindered to act as nucleophiles.
With a primary substrate and a strong, unhindered nucleophile, SN2 dominates. With a tertiary substrate and a strong base, E2 dominates.
SN1 and E1 typically occur together whenever a carbocation intermediate forms. Expect both substitution and elimination products in the same reaction.
Reactivity order: 3° > 2° > 1°. The substrate must be able to form a reasonably stable carbocation.
Reagents: Weak nucleophiles and weak bases. Common reagents include H₂O, HOR, RCOOH, HSR. Notice these are all neutral molecules that contain an H.
Carbocation rearrangements can occur. If a hydride or methyl shift converts a 2° carbocation to a 3° carbocation, it will happen. Always check for rearrangement before drawing products.
The carbocation is sp²-hybridised and planar, so the nucleophile can attack from either face. This gives a mixture of inversion and retention products. If the starting material is a single enantiomer, the SN1 product is a mixture of both enantiomers (partial or full racemisation).
The Zaitsev product (more substituted alkene) is the major product because the more substituted double bond is more thermodynamically stable.
Condition | SN2 | SN1 | E2 | E1 |
|---|---|---|---|---|
Best substrate | 1° (methyl) | 3° | 3° | 3° |
Reagent | Strong nucleophile | Weak Nu / weak base | Strong base | Weak base |
Solvent | Polar aprotic | Polar protic | Either | Polar protic |
Stereochemistry | Inversion only | Racemisation (mix) | Anti-coplanar required | N/A |
Rearrangements? | No | Yes | No | Yes |
Regiochemistry | N/A | N/A | Zaitsev (or Hofmann with bulky base) | Zaitsev |
Look at the substrate. Is it 1°/methyl, 2°, or 3°?
Look at the reagent. Is it a strong nucleophile, a strong base, or a weak/neutral species?
1° or methyl + strong nucleophile = SN2.
3° + strong base = E2.
3° + weak/neutral reagent = SN1 + E1 (both products).
2° substrates are the tricky ones: strong nucleophile favours SN2; strong base favours E2; weak/neutral reagent gives SN1/E1 mixture.
Bulky base (tert-butoxide, DBU, DBN) always means E2, regardless of substrate degree.
Students often assume SN1 gives complete racemisation. In practice, it gives partial racemisation (the leaving group can shield one face briefly), but for exam purposes, expect both inversion and retention products.
A frequent error is forgetting anti-coplanar geometry in E2. On a cyclohexane, the H and the leaving group must both be axial and on opposite sides (trans-diaxial). Drawing the chair is not optional.
Students sometimes classify OH⁻ as an E2-only reagent. OH⁻ is both a strong base and a strong nucleophile; it can do both SN2 and E2. Only bulky bases (tert-butoxide) are E2-exclusive.
Many students forget to check for carbocation rearrangements in SN1/E1. If a hydride or methyl shift gives a more stable carbocation, it will happen.
⚠️ Mechanism questions are flagged as "most likely tested" in the source material. Expect to write out full arrow-pushing mechanisms for at least one substitution and one elimination.
⚠️ Product prediction problems require you to identify the correct pathway first, then draw all possible products (including stereochemistry). If the pathway is SN1, always draw both enantiomers.
⚠️ Hofmann elimination with a bulky base: only apply this for E2 on tertiary substrates. Using Hofmann logic on a primary substrate is a common exam error.
⚠️ When an alcohol is the starting material, it cannot leave directly. You must protonate the OH first (to convert it into water, a good leaving group) before any substitution or elimination pathway can begin.
True or false: SN2 reactions work best on tertiary substrates. (False. SN2 works best on primary and methyl substrates.)
Fill in the blank: E2 requires ______ geometry between the H and the leaving group. (anti-coplanar)
True or false: SN1 and E1 products are often observed together in the same reaction. (True.)
Fill in the blank: Using tert-butoxide as a base gives the ______ product in E2. (Hofmann / less substituted)
True or false: Carbocation rearrangements can occur in SN2 reactions. (False. Only in SN1/E1, which proceed through a carbocation intermediate.)
Q: 2-bromobutane is treated with NaOCH₃ in DMSO. Predict the major product and mechanism.
A: SN2. NaOCH₃ is a strong nucleophile, DMSO is polar aprotic, and 2-bromobutane is secondary. The major product is 2-methoxybutane with inversion of configuration at the carbon bearing bromine.
Q: 2-bromo-2-methylpropane is dissolved in water. What products form and by which mechanism?
A: SN1 and E1. The substrate is tertiary, and water is a weak nucleophile/base. SN1 gives 2-methyl-2-propanol (with both retention and inversion). E1 gives 2-methylpropene (Zaitsev product). Both products appear.
Q: Explain why tert-butoxide with a secondary haloalkane gives exclusively the Hofmann elimination product.
A: Tert-butoxide is extremely bulky: it cannot approach the carbon for backside attack (ruling out SN2), and when it acts as a base in E2, it preferentially removes the less hindered H, which leads to the less substituted (Hofmann) alkene.
Q: A student treats (R)-2-bromobutane with NaCN in DMF and obtains a product with (S) configuration. Is this expected?
A: Yes. CN⁻ is a strong nucleophile, DMF is polar aprotic, and the substrate is secondary, so SN2 dominates. SN2 always gives inversion of configuration, so (R) starting material produces (S) product.
Q: Why must you protonate an alcohol before it can undergo substitution or elimination?
A: The hydroxyl group (OH⁻) is a poor leaving group because hydroxide is a strong base. Protonation converts OH into water (H₂O), which is a much better leaving group because water is a weak, stable base.
The acid-base framework from Exam 1 determines whether a reagent acts as a nucleophile or a base. A species with a low pKa conjugate acid (strong base) favours E2; a good nucleophile with moderate basicity favours SN2.
Allylic substitution (covered in the Aromaticity and Reactions notes) follows the same SN1/SN2 logic, but proceeds faster because the allylic carbocation or transition state is stabilised by the adjacent double bond.
Alcohol starting materials connect back to protonation (acid-base) before any pathway can begin, and the dehydration of alcohols via E1 is a key mechanism question on the final.
Related Terms / Search Tags
SN1, SN2, E1, E2, nucleophilic substitution, elimination reaction, haloalkane, leaving group, nucleophile, base, carbocation, inversion, retention, racemisation, Zaitsev rule, Hofmann elimination, anti-coplanar, trans-diaxial, polar aprotic solvent, polar protic solvent, backside attack, concerted mechanism, tert-butoxide, DBU, DBN, bulky base, rearrangement, hydride shift, methyl shift