Difficulty: Intermediate | Prerequisites: Functional groups and nomenclature notes, basic understanding of Lewis acid/base chemistry and electronegativity.
Nucleophilic substitution is one of the two main reaction families for haloalkanes and similar substrates (the other being elimination). Every substitution swaps a leaving group for a nucleophile, but the pathway it takes, SN2 (one concerted step) or SN1 (two steps via a carbocation), controls the stereochemical outcome, the rate law, and which substrates and reagents favour it. This is the heart of predicting what happens when you mix a haloalkane with a nucleophile, and exam questions test it from every angle: drawing products, comparing rates, choosing reagents, and explaining stereochemistry.
SN2 is a one-step backside attack: the nucleophile strikes while the leaving group departs, giving inversion of configuration and a rate that depends on both nucleophile and substrate concentrations. SN1 is a two-step process: the leaving group departs first to form a carbocation, then the nucleophile attacks the flat carbocation from either face, typically giving a racemic (or partially racemised) mixture. Primary substrates favour SN2; tertiary substrates favour SN1; secondary substrates can go either way depending on nucleophile strength and solvent.
Nucleophile
A species that donates an electron pair to an electrophilic carbon. Think of it as the "attacker" in a substitution, the Lewis base that wants to bond to a positive or partially positive carbon.
Leaving group
The atom or group that departs with the bonding electrons. In simple terms, the piece that gets kicked off the substrate. Good leaving groups are weak bases (e.g. Br⁻, I⁻, TsO⁻).
SN2 (bimolecular nucleophilic substitution)
A concerted, one-step mechanism where the nucleophile attacks the electrophilic carbon at 180° to the leaving group (backside attack), causing simultaneous bond formation and bond breaking. Rate = k[substrate][nucleophile].
SN1 (unimolecular nucleophilic substitution)
A two-step mechanism: (1) the leaving group departs to form a carbocation intermediate, then (2) the nucleophile attacks the carbocation. Rate = k[substrate]. The nucleophile concentration does not appear in the rate law.
Inversion of configuration (Walden inversion)
The stereochemical outcome of SN2: the nucleophile enters from the back, flipping the configuration at the carbon like an umbrella turning inside out.
Racemisation
The formation of equal amounts of both enantiomers. In SN1, the planar carbocation intermediate can be attacked from either face, so the product is typically racemic (or close to it).
Carbocation
A positively charged carbon with an empty p orbital. Stability order: 3° > 2° > 1° > methyl, which is why tertiary substrates favour SN1.
Steric hindrance
The physical crowding around the electrophilic carbon by bulky groups. More hindrance slows SN2 (which needs backside access) but does not affect SN1 (which forms the carbocation first).
Nucleophilicity vs. basicity
Nucleophilicity is the rate of attack on carbon (kinetic). Basicity is the equilibrium tendency to accept a proton (thermodynamic). They often track together, but steric effects and solvent can decouple them.
SN2 is a single concerted step. The nucleophile approaches from 180° opposite the leaving group (backside attack). As the new bond forms, the old bond breaks simultaneously, passing through a pentacoordinate transition state.
Stereochemistry: always inversion (Walden inversion). If the starting material is (R), the product is (S), and vice versa.
Rate law: Rate = k[nucleophile][substrate]. Doubling the nucleophile concentration doubles the rate.
Substrate preference: methyl > primary > secondary. Tertiary substrates essentially do not undergo SN2 because the three bulky groups block backside approach.
Solvent: polar aprotic solvents (DMF, DMSO, acetone) favour SN2. They dissolve the salt but do not solvate the nucleophile's lone pairs, leaving it "naked" and reactive.
Exam key example (Problem 2a): a secondary benzylic bromide treated with NaCN in DMF. CN⁻ is a strong nucleophile, DMF is polar aprotic. The product is a nitrile with inverted configuration at the carbon that bore the Br.
SN1 proceeds in two discrete steps:
Step 1 (rate-determining): the leaving group departs, forming a planar carbocation.
Step 2 (fast): the nucleophile attacks the carbocation from either face.
Key features:
Stereochemistry: the planar carbocation can be attacked from both sides, so the product is typically a racemic mixture (equal R and S). In practice, slight preference for attack from the side opposite the departing group can give partial inversion, but for exam purposes expect racemisation.
Rate law: Rate = k[substrate]. The nucleophile is not involved in the rate-determining step, so its concentration does not affect the rate.
Substrate preference: tertiary > secondary > primary > methyl, because more substituted carbocations are more stable.
Solvent: polar protic solvents (water, methanol, ethanol, acetic acid) favour SN1. They stabilise the carbocation intermediate and solvate the departing leaving group.
Exam key example (Problem 2b): a secondary cyclohexyl chloride treated with CH₃OH (a weak nucleophile, polar protic solvent). SN1 conditions. Products are both enantiomers of the methyl ether, showing racemisation.
Factors that make a species a better nucleophile:
Charge: a negatively charged species is a better nucleophile than its neutral conjugate acid (e.g. RO⁻ > ROH).
Electronegativity (same row): less electronegative atoms are more nucleophilic (e.g. N > O > F within the same period).
Size / polarisability (same column, in protic solvents): larger atoms are more nucleophilic in protic solvents because they are less tightly solvated (e.g. I⁻ > Br⁻ > Cl⁻ > F⁻ in water or alcohol). In polar aprotic solvents, this trend can reverse.
Steric effects: less hindered nucleophiles react faster in SN2.
Exam key examples (Problem 5a,b):
5a: a phenoxide (ArO⁻) without an adjacent carbonyl vs. a carboxylate (RCO₂⁻, where the charge is delocalised over two oxygens). The phenoxide, whose charge is localised on one oxygen, is the better nucleophile.
5b: a small alkoxide vs. a bulkier alkoxide with delocalisation. The bulkier, resonance-stabilised anion is weaker. Here, the answer was B (the less resonance-stabilised species).
A good leaving group is a weak base, the conjugate base of a strong acid:
I⁻ (pKa of HI ≈ -10) > Br⁻ (pKa of HBr ≈ -9) > Cl⁻ (pKa of HCl ≈ -7) > F⁻ (poor leaving group)
TsO⁻ (tosylate) is an excellent leaving group, comparable to I⁻.
HO⁻ and H₂N⁻ are terrible leaving groups. Alcohols must be protonated (to H₂O) or converted to tosylates before they can leave.
Exam key example (Problem 5f): two secondary substrates, one with Br and one with I. Both are SN1 candidates. The iodide reacts faster because I⁻ is a better leaving group than Br⁻.
SN1 rate depends on carbocation stability:
Tertiary > secondary > primary > methyl.
Allylic and benzylic carbocations are stabilised by resonance and react faster than their non-conjugated equivalents.
Exam key example (Problem 5d): three bromides ranked by SN1 rate. A primary bromide (slowest, least stable carbocation), an allylic/secondary bromide (middle), and a tertiary or benzylic bromide (fastest). The ranking was A < C < B.
Students often think SN1 always gives a perfect 50:50 racemic mixture. In practice, the leaving group may partially shield one face of the carbocation, giving slight excess inversion. But for most exam purposes, treat SN1 as racemisation.
Students confuse nucleophilicity with basicity. A bulky, strong base (like t-BuO⁻) is an excellent base but a poor nucleophile for SN2 because it cannot reach the electrophilic carbon. It will favour elimination instead.
A common error is predicting SN2 for a tertiary substrate. Tertiary carbons are far too crowded for backside attack. If the substrate is tertiary, SN2 is effectively ruled out.
Students sometimes forget that the solvent matters. The same substrate and nucleophile can go SN2 in DMSO and SN1 in methanol.
⚠️ Drawing the stereochemical product is essential. For SN2, show inversion using wedge/dash notation. For SN1, draw both enantiomers.
⚠️ Know the SN2 rate law: Rate = k[Nu][substrate]. If asked how doubling the nucleophile concentration affects the rate, the answer is the rate doubles.
⚠️ The SN1 rate law is Rate = k[substrate]. Nucleophile concentration has no effect on the rate.
⚠️ Reagent-choice questions test whether you can pick the right nucleophile/solvent combination. NaOAc in DMF (strong nucleophile, polar aprotic) signals SN2. CH₃OH alone (weak nucleophile, polar protic) signals SN1.
⚠️ Leaving group comparisons are commonly tested. Remember the pKa trend: lower pKa of the conjugate acid = better leaving group (I⁻ > Br⁻ > Cl⁻).
True or false: SN2 gives retention of configuration. False. SN2 gives inversion.
Fill in the blank: the rate-determining step in SN1 is the departure of the ____. leaving group
True or false: a tertiary substrate can readily undergo SN2. False. Too sterically hindered.
Fill in the blank: in the SN2 rate law, Rate = k[____][____]. nucleophile, substrate
True or false: I⁻ is a better leaving group than Br⁻. True. (pKa of HI is lower than pKa of HBr.)
Q: A chiral secondary bromide is treated with NaCN in DMF. What mechanism operates, and what is the stereochemical outcome?
A: SN2. CN⁻ is a strong nucleophile, DMF is a polar aprotic solvent, and secondary substrates can undergo SN2 with a good nucleophile. The product is the nitrile with inversion of configuration at the carbon that bore the Br.
Q: A secondary alkyl chloride is dissolved in methanol with no added nucleophile. What mechanism operates, and what products form?
A: SN1. CH₃OH is a weak nucleophile and a polar protic solvent. The chloride leaves to form a carbocation, then methanol attacks from either face. The products are both enantiomers of the methyl ether.
Q: Rank the following in order of increasing SN1 rate: primary bromide (A), allylic secondary bromide (C), tertiary bromide (B).
A: A < C < B. SN1 rate depends on carbocation stability: 1° < allylic 2° < 3°.
Q: In an SN2 reaction, how does doubling the nucleophile concentration affect the rate?
A: The rate doubles. Rate = k[Nu][substrate], so doubling [Nu] doubles the rate.
Q: Two secondary substrates, one bearing Br and the other bearing I, are compared under SN1 conditions. Which reacts faster, and why?
A: The iodide reacts faster. I⁻ is a better leaving group than Br⁻ because HI is a stronger acid (lower pKa) than HBr.
Q: A primary alkyl tosylate is treated with NaOAc in DMF. What mechanism operates?
A: SN2. Primary substrate, strong nucleophile (acetate), polar aprotic solvent (DMF). The acetate displaces the tosylate with inversion.
Substitution always competes with elimination. The same substrate, nucleophile, and conditions that could give SN2 might also give E2, and conditions favouring SN1 often also produce E1 products. The next set of notes covers elimination in detail and includes the decision framework for predicting which pathway dominates.
Carbocation stability, covered in the elimination and miscellaneous notes, is also critical here: every SN1 prediction relies on knowing whether the intermediate carbocation is stable enough to form. Rearrangement of carbocations (hydride and methyl shifts) can also occur during SN1 reactions, giving unexpected products.
SN1, SN2, nucleophilic substitution, backside attack, Walden inversion, racemisation, carbocation, leaving group, nucleophile, electrophile, polar aprotic solvent, polar protic solvent, DMF, DMSO, methanol, steric hindrance, primary secondary tertiary substrate, rate law, bimolecular, unimolecular, organic chemistry I, OChem 1, UMN, Exam 3 review