Difficulty: Intermediate | Prerequisites: Lewis structures, electronegativity, polar bonds, functional groups (Ch. 1-7 material)
Nucleophilic substitution is one of the most fundamental reaction types in organic chemistry. When a nucleophile (an electron-rich species) attacks an electrophilic carbon bearing a leaving group, the leaving group departs and a new bond forms. This single idea underpins drug synthesis, polymer chemistry, and countless lab transformations you will meet throughout CHEM 237-239.
There are two competing pathways, SN1 and SN2, and the outcome depends on substrate structure, nucleophile strength, solvent, and leaving group. If you can predict which pathway dominates, you can predict the product, its stereochemistry, and the rate law. That prediction is exactly what exam questions test.
SN2 reactions are one-step, backside-attack processes that invert stereochemistry and work best on unhindered (primary, methyl) substrates with strong nucleophiles. SN1 reactions are two-step processes that go through a carbocation intermediate, produce racemisation, and favour tertiary substrates with weak nucleophiles in polar protic solvents. Choosing between the two comes down to substrate class, nucleophile strength, and solvent.
Nucleophile
An electron-rich species that donates a pair of electrons to an electrophilic carbon. Think of it as the attacker: it has electrons to share and seeks a positive or partially positive site.
Electrophile
An electron-poor species that accepts a pair of electrons. In substitution reactions the electrophilic site is the carbon bonded to the leaving group.
Leaving group
The atom or group that departs with the bonding electrons when the nucleophile attacks. Good leaving groups are weak bases (e.g. I⁻, Br⁻, Cl⁻, tosylate). In simple terms, the better the leaving group, the more stable it is once it leaves.
SN2 (substitution, nucleophilic, bimolecular)
A one-step mechanism: the nucleophile attacks the electrophilic carbon at the same time the leaving group departs. "Bimolecular" means the rate depends on the concentration of both the substrate and the nucleophile. Think of it as a door that only opens when you push from behind while someone pulls from the front.
SN1 (substitution, nucleophilic, unimolecular)
A two-step mechanism: the leaving group departs first to form a carbocation, then the nucleophile attacks. "Unimolecular" means only the substrate concentration appears in the rate law. Think of it as the leaving group walking away on its own, and then the nucleophile arriving to fill the vacancy.
Carbocation
A positively charged carbon with an empty p-orbital. Stability follows the order: tertiary > secondary > primary > methyl. In simple terms, more alkyl groups around the positive carbon stabilise it through hyperconjugation and inductive effects.
Inversion of configuration (Walden inversion)
The stereochemical outcome of an SN2 reaction. The nucleophile attacks from the back side of the leaving group, flipping the spatial arrangement at the carbon, much like an umbrella inverting in the wind.
Racemisation
Formation of equal amounts of both enantiomers, producing a racemic mixture. This is the typical stereochemical outcome of SN1 reactions, because the planar carbocation intermediate can be attacked from either face.
Rate law
The mathematical expression showing how reaction rate depends on reactant concentrations. SN2: rate = k[substrate][nucleophile]. SN1: rate = k[substrate].
How it works
The nucleophile attacks the electrophilic carbon from the side opposite the leaving group (backside attack).
Bond formation and bond breaking happen simultaneously in a single concerted step.
The transition state is a pentacoordinate carbon with partial bonds to both the incoming nucleophile and the departing leaving group.
Substrate effects
Methyl > primary > secondary >> tertiary (tertiary substrates essentially do not undergo SN2).
Steric hindrance at the electrophilic carbon slows the reaction. Bulky groups block the nucleophile's approach.
Example from the exam: a cyclopentane ring with a bromine and a methyl group. The nucleophile (NH₃) attacks via SN2 at the less hindered carbon, producing inversion.
Nucleophile effects
SN2 requires a strong nucleophile. The nucleophile appears in the rate law: rate = k[substrate][nucleophile].
Doubling the nucleophile concentration doubles the rate.
Strong nucleophiles include: OH⁻, CN⁻, I⁻, RS⁻, NH₃ (when used with a primary or secondary substrate).
Weak or bulky nucleophiles (e.g. H₂O, tert-butoxide) are poor for SN2.
Leaving group effects
Better leaving groups accelerate both SN1 and SN2. The order is: I⁻ > Br⁻ > Cl⁻ >> F⁻.
Iodide is both a good leaving group and a good nucleophile. This means an SN2 reaction with NaI can run in both directions, which is why (R)-2-bromobutane treated with NaI eventually gives racemic 2-iodobutane: each SN2 step inverts, but repeated back-and-forth inversions scramble the stereochemistry.
Solvent effects
Polar aprotic solvents (DMSO, DMF, acetone) favour SN2. They dissolve ionic nucleophiles without solvating them heavily, leaving the nucleophile free to attack.
Polar protic solvents (water, alcohols) slow SN2 by hydrogen-bonding to the nucleophile and reducing its reactivity.
Stereochemistry
Always inversion of configuration at the electrophilic carbon (Walden inversion).
If the starting material is (R), the product at that centre is (S), and vice versa.
Example: a secondary alkyl chloride treated with NaCN undergoes backside attack, giving a nitrile product with inverted configuration.
How it works
Step 1 (rate-determining): the leaving group departs on its own, forming a carbocation.
Step 2: the nucleophile attacks the carbocation.
Because the rate-determining step involves only the substrate, rate = k[substrate]. Doubling the nucleophile concentration has no effect on the rate.
Substrate effects
Tertiary > secondary >> primary > methyl. The more substituted the carbon, the more stable the carbocation intermediate.
Primary and methyl substrates essentially never react by SN1 because their carbocations are far too unstable.
Example from the exam: 2-methyl-2-bromopropane (a tertiary substrate) reacts with NH₃. Doubling [NH₃] has no effect on the rate, confirming SN1. Compare this with 2-bromopropane (secondary), where doubling [NH₃] doubles the rate, confirming SN2.
Nucleophile effects
The nucleophile does not participate in the rate-determining step, so its strength and concentration do not affect the rate.
Weak nucleophiles (H₂O, alcohols) work perfectly well for SN1 because they only need to attack the already-formed carbocation.
Solvent effects
Polar protic solvents (water, alcohols) favour SN1. They stabilise the carbocation intermediate and the departing anion through solvation.
Stereochemistry
The carbocation intermediate is sp² hybridised and planar. The nucleophile can attack from either face.
The theoretical outcome is a 50:50 racemic mixture, though in practice there is often a slight preference for inversion (the leaving group may partially shield one face as it departs).
Carbocation rearrangements
Carbocations can rearrange to form a more stable cation via 1,2-hydride shifts or 1,2-methyl shifts.
A secondary carbocation adjacent to a quaternary carbon may rearrange to a tertiary carbocation, which changes the carbon skeleton of the product.
Example from the exam: a neopentyl-type bromide reacting with H₂O. The initially formed secondary carbocation rearranges (methyl shift) to a more stable tertiary carbocation, leading to a product with a different carbon skeleton from the starting material. The mechanism must show the skeletal change via the rearrangement step.
Factor | SN2 | SN1 |
|---|---|---|
Steps | One (concerted) | Two (carbocation intermediate) |
Rate law | rate = k[sub][nuc] | rate = k[sub] |
Best substrate | Methyl, primary | Tertiary |
Nucleophile | Strong, required | Weak is fine |
Solvent | Polar aprotic | Polar protic |
Stereochemistry | Inversion | Racemisation |
Rearrangements | Never | Yes (hydride/methyl shifts) |
Secondary substrates sit in the middle and can react by either pathway. The dominant mechanism depends on nucleophile strength, solvent, and temperature. A strong nucleophile in a polar aprotic solvent pushes secondary substrates toward SN2. A weak nucleophile in a polar protic solvent pushes them toward SN1.
Why acetate substitutes but does not eliminate, while ethoxide does both
Both acetate (CH₃COO⁻) and ethoxide (CH₃CH₂O⁻) carry a negative charge on oxygen and act as nucleophiles in SN2 with 1-chloropropane. However, ethoxide is also a strong base (the conjugate base of ethanol, pKa ~16), so it can abstract a beta-hydrogen and cause E2 elimination to form propene. Acetate is a much weaker base (the conjugate base of acetic acid, pKa ~5), with its charge delocalised across two oxygens. That resonance stabilisation makes it a poor base, so it substitutes but does not eliminate.
SN2 rate law
rate = k [substrate] [nucleophile]
SN1 rate law
rate = k [substrate]
Leaving group ability (halides)
I⁻ > Br⁻ > Cl⁻ >> F⁻
This tracks with base strength in reverse: the weaker the base, the better the leaving group.
Carbocation stability order
3° > 2° > 1° > methyl
Nucleophilicity trends
Within the same row of the periodic table, nucleophilicity roughly tracks basicity. In polar protic solvents, nucleophilicity increases going down a column (I⁻ > Br⁻ > Cl⁻ > F⁻) because larger atoms are less tightly solvated.
SN2 reactions are used extensively in pharmaceutical synthesis to introduce functional groups with precise stereochemistry. For instance, converting an alkyl halide to a nitrile (using NaCN) is a common way to extend a carbon chain by one carbon, and the inversion of configuration is factored into the synthetic plan.
SN1 reactions are relevant in biological chemistry, where solvolysis in aqueous environments (the body's solvent) often proceeds through carbocation intermediates. Understanding carbocation rearrangements also explains why certain terpene biosynthesis pathways produce unexpected carbon skeletons.
Students often assume that SN2 always gives one pure enantiomer. It does invert at the reacting centre, but if repeated SN2 reactions can occur (as with NaI on 2-bromobutane, where I⁻ is both a good nucleophile and a good leaving group), the product eventually racemises through successive inversions.
Students often confuse nucleophilicity and basicity. A species can be a strong base but a poor nucleophile if it is too bulky (e.g. potassium tert-butoxide is a strong base but a poor nucleophile for SN2 due to steric bulk, so it favours E2 instead).
Students sometimes think SN1 produces only one enantiomer because "the nucleophile attacks." The carbocation is flat, so attack from both faces is equally likely, giving a racemic mixture.
Students often forget that carbocation rearrangements change the carbon skeleton. If you draw the product without accounting for the rearrangement, you will get the wrong structure entirely.
⚠️ Predicting SN1 vs SN2 from substrate, nucleophile, and solvent is the single most tested skill in Ch. 8-9.
⚠️ Drawing the correct stereochemistry (inversion for SN2, racemisation for SN1) is always required. "Include stereochemistry where appropriate" appears in exam instructions.
⚠️ Carbocation rearrangements: if the product skeleton differs from the starting material, show the rearrangement step in your mechanism. The exam hint "your mechanisms must show how the skeleton changes" signals this.
⚠️ Explaining rate law differences (why doubling [nucleophile] affects SN2 but not SN1) is a common short-answer question.
⚠️ The acetate vs ethoxide distinction (nucleophile vs base) tests whether you understand the relationship between basicity and elimination.
True or False: SN2 reactions proceed faster with tertiary substrates than with primary substrates.
Fill in the blank: The rate law for an SN1 reaction is rate = k[______].
True or False: A carbocation formed during an SN1 reaction is sp³ hybridised.
Fill in the blank: The stereochemical outcome of an SN2 reaction is ______ of configuration.
True or False: Polar protic solvents favour SN2 reactions.
Answers: 1. False (SN2 is fastest with methyl/primary). 2. substrate. 3. False (it is sp²). 4. inversion. 5. False (polar protic favours SN1; polar aprotic favours SN2).
Q: Acetate and ethoxide both carry a negative charge on oxygen. Both will undergo substitution with 1-chloropropane. Why does only ethoxide also produce propene?
A: Ethoxide is a strong, unhindered base (conjugate base of ethanol, pKa ~16). It can abstract a beta-hydrogen to cause E2 elimination, forming propene. Acetate is a much weaker base because its negative charge is delocalised by resonance across two oxygens (conjugate base of acetic acid, pKa ~5). It is too weak a base to perform elimination.
Q: 2-Bromobutane reacts with NaI via SN2 to form 2-iodobutane. SN2 gives inversion, yet the product is racemic. Explain why (this is not SN1).
A: Iodide is both a good nucleophile and a good leaving group. The initial SN2 reaction inverts the configuration to give one enantiomer. But then I⁻ (released from NaI still in solution) can perform a second SN2 on the product, inverting it back. Repeated rounds of SN2 in both directions eventually produce equal amounts of both enantiomers, giving a racemic mixture.
Q: When 2-bromopropane reacts with NH₃, doubling [NH₃] doubles the rate. When 2-methyl-2-bromopropane reacts with NH₃, doubling [NH₃] has no effect on the rate. Explain.
A: 2-Bromopropane is a secondary substrate reacting with NH₃ (a moderate nucleophile) via SN2, so the rate depends on both [substrate] and [NH₃]. 2-Methyl-2-bromopropane is tertiary, so it cannot undergo SN2 (too much steric hindrance). It reacts via SN1, where the rate-determining step is ionisation of the substrate alone. The nucleophile only enters after the carbocation has formed, so its concentration does not affect the rate.
Q: Draw the major product when a cyclopentane bearing a bromine (trans to a methyl group) reacts with NH₃.
A: NH₃ acts as a nucleophile in an SN2 reaction. It attacks from the back side of the bromine, giving an amine product with inverted configuration at the carbon that bore the bromine. The methyl group's stereochemistry is unchanged.
Substitution reactions connect directly to elimination reactions (E1 and E2), which compete with SN1 and SN2 under many conditions. Understanding when substitution wins over elimination (and vice versa) requires the same set of factors: substrate class, base/nucleophile strength, and solvent.
The stereochemistry concepts here (inversion, racemisation, R/S assignment) build on the foundational stereochemistry from Ch. 5. Carbocation stability connects to the stability of intermediates in electrophilic addition to alkenes (Ch. 9), where Markovnikov's rule is explained by the preference for the more stable carbocation.
SN1, SN2, nucleophilic substitution, backside attack, Walden inversion, inversion of configuration, racemisation, racemic mixture, carbocation, carbocation rearrangement, 1,2-hydride shift, 1,2-methyl shift, leaving group, nucleophile, electrophile, rate law, bimolecular, unimolecular, polar aprotic solvent, polar protic solvent, DMSO, DMF, steric hindrance, primary substrate, secondary substrate, tertiary substrate, methyl substrate, acetate vs ethoxide, nucleophilicity vs basicity, CHEM 237 substitution reactions, organic chemistry substitution