Difficulty: Intermediate | Prerequisites: CHEM 225 Exam 1 and 2 material (functional groups, stereochemistry, acid-base chemistry, conformational analysis).
Nucleophilic substitution is one of the most fundamental reaction types in organic chemistry. It describes how a nucleophile (an electron-rich species) displaces a leaving group on a carbon centre, forming a new bond. This material sits at the heart of CHEM 225 and connects directly to elimination reactions (E1, E2), which often compete with substitution under the same conditions. If you are comfortable with Lewis structures, electronegativity, and basic stereochemistry (R/S assignments, wedge-dash notation), you have what you need to follow along.
SN2 reactions are one-step, backside-attack processes whose rate depends on both substrate and nucleophile concentration. SN1 reactions proceed through a carbocation intermediate, depend only on substrate concentration, and can involve rearrangements. Choosing between SN1, SN2, E1, and E2 comes down to four variables: substrate structure, nucleophile/base strength, solvent polarity, and temperature.
SN2 (bimolecular nucleophilic substitution)
A one-step mechanism in which the nucleophile attacks the electrophilic carbon at the same time the leaving group departs. The rate depends on the concentration of both reactants. In simple terms, think of it as a simultaneous swap: the new group pushes in from the back while the old group leaves from the front.
SN1 (unimolecular nucleophilic substitution)
A two-step mechanism in which the leaving group departs first to form a carbocation intermediate, and the nucleophile attacks second. The rate depends only on the substrate concentration. Think of it as: the leaving group walks away on its own, leaving a positively charged carbon behind, and then any nucleophile in the neighbourhood can step in.
Nucleophile
An electron-rich species that donates a pair of electrons to an electrophilic carbon. In simple terms, it is the "attacker" in a substitution reaction. Examples: CH₃NH₂, HO⁻, HS⁻, CH₃O⁻.
Leaving group
The atom or group that departs with the bonding electrons during substitution. Good leaving groups are weak bases (stable once they leave). Think of it as: the better something is at being on its own, the better a leaving group it is. Examples: Cl⁻, Br⁻, I⁻, H₂O.
Electrophile
An electron-poor species that accepts a pair of electrons. In substitution reactions, the electrophilic carbon bearing the leaving group is the site of attack.
Carbocation
A positively charged carbon intermediate formed in SN1 (and E1) reactions. Stability order: 3° > 2° > 1° > methyl. Think of it as a carbon that has lost its leaving group and is desperate for electrons.
Backside attack
In SN2, the nucleophile must approach from the side opposite the leaving group. This causes inversion of stereochemistry at the carbon centre (Walden inversion).
Williamson ether synthesis
An SN2 reaction between an alkoxide ion (RO⁻) and a primary alkyl halide to form an ether (R-O-R'). Think of it as one of the most reliable ways to stitch two organic fragments together through an oxygen atom.
Elimination (E1 and E2)
Competing pathways where, instead of substitution, a base removes a proton and a leaving group departs to form a double bond. E2 is concerted (one step); E1 goes through a carbocation intermediate, just like SN1.
The SN2 rate law is second-order overall:
\text{Rate} = k[\text{substrate}][\text{nucleophile}]Because both concentrations appear in the rate expression, doubling the nucleophile concentration doubles the rate, and halving the substrate concentration halves it. If you double one and halve the other, the effects cancel and the rate stays the same.
Methyl halides react fastest by SN2 (least steric hindrance)
Primary halides react readily
Secondary halides react slowly
Tertiary halides essentially do not undergo SN2 (too much steric crowding around the electrophilic carbon for backside attack)
The first reaction on the exam (cyclohexyl-CH₂-Cl vs. neopentyl-type Cl) illustrates this directly. A primary substrate with no branching at the beta carbon reacts faster than one with branching, because the incoming nucleophile has a clearer path to the carbon.
Strong nucleophiles (charged species: CH₃O⁻, HO⁻, HS⁻, CN⁻, NH₂⁻) favour SN2
Weak nucleophiles (neutral species: H₂O, CH₃OH, CH₃NH₂) tend to favour SN1
Within a group going down the periodic table, nucleophilicity increases in polar protic solvents (I⁻ > Br⁻ > Cl⁻ > F⁻) because larger atoms are less tightly solvated
Identifying the nucleophile: in the exam reaction of an alkyl chloride with excess CH₃NH₂, the nucleophile is CH₃NH₂ (the neutral amine), not NH₂⁻
Polar aprotic solvents (DMF, DMSO, acetone) favour SN2. They dissolve the salt but do not solvate the nucleophile heavily, leaving it "naked" and reactive.
Polar protic solvents (H₂O, CH₃OH, ethanol) favour SN1. They stabilise the carbocation intermediate through solvation and hydrogen bonding.
The exam tests this with a reaction converting an alkyl bromide to a methyl ether with inversion of configuration. CH₃ONa in DMF (a strong nucleophile in a polar aprotic solvent) is the SN2-favouring combination.
SN2 always gives inversion of configuration at the electrophilic carbon (Walden inversion)
This is a direct consequence of backside attack: the nucleophile approaches 180° from the leaving group
\text{Rate} = k[\text{substrate}]Only the substrate concentration matters. Doubling the nucleophile has no effect on the rate.
Tertiary substrates (3° carbocations are the most stable)
Weak nucleophiles (H₂O, alcohols)
Polar protic solvents (stabilise the carbocation)
The substrate must be able to form a reasonably stable carbocation on its own
Carbocations rearrange to become more stable. A secondary carbocation will undergo a 1,2-hydride shift or 1,2-methyl shift to become tertiary if the geometry allows it.
The exam tests this with KOtBu/tBuOH acting on a secondary substrate. The product shows a rearranged skeleton, which tells you a carbocation intermediate formed (SN1/E1 territory). The explanation: a carbocation intermediate undergoes rearrangement, the substrate undergoes E1 elimination, and the bulky base (tert-butoxide) cannot easily do SN2 on this substrate. All three statements are correct.
E1 shares the same first step as SN1: ionisation to a carbocation
The carbocation then loses a proton from an adjacent carbon to form an alkene
E1 competes with SN1 whenever a carbocation intermediate forms
Higher temperatures favour elimination over substitution
The Zaitsev product (more substituted alkene) is typically the major E1 product
E2 is concerted (one step): a strong base removes a proton while the leaving group departs simultaneously
Requires anti-periplanar geometry (the H and the leaving group must be 180° apart)
Favoured by strong, bulky bases (KOtBu) and higher temperatures
Also gives the Zaitsev product unless a bulky base forces the Hofmann product
The Williamson synthesis uses an alkoxide nucleophile (RO⁻) in an SN2 reaction with a primary (or methyl) alkyl halide to form an ether.
When two routes to the same ether are possible, choose the one that places the SN2 reaction on the less hindered (primary) carbon. This minimises competing E2 elimination.
The exam gives two routes to butyl propyl ether. Using 2-butanol (as the alkoxide) reacting with a primary alkyl chloride is preferred over the reverse arrangement, because 2-butanol's alkoxide attacking a primary carbon avoids the elimination side reactions that would occur if a primary alkoxide attacked a secondary substrate. The answer: 2-butanol, because it is least likely to give undesired elimination products.
When a molecule has both a nucleophile and a leaving group, intramolecular SN2 can form a ring. Ring size matters enormously:
3-membered rings (epoxides, thiiranes): form readily because the two reactive centres are close together
5-membered rings: the most kinetically favoured ring size for intramolecular reactions
6-membered rings: also form readily
4-membered and medium rings (7-12): much harder to form
The exam's extra credit question compares two intramolecular cyclisations with NaOH:
HS-CH₂CH₂CH₂-Br forms a 3-membered thiirane ring (sulfur is a better nucleophile than oxygen, and 3-membered ring closure from a 1,3-arrangement is favourable)
Br-CH₂(CH₂)₄CH₂-Br with NaOH forms a 6-membered tetrahydropyran (oxygen from hydroxide displaces bromide in an intramolecular SN2, and the 6-membered ring is thermodynamically and kinetically accessible)
The key insight: sulfur is a far better nucleophile than oxygen (it is larger, more polarisable, and less tightly solvated in protic solvents), so the thiol cyclises by direct intramolecular SN2 to the smaller ring, while the oxygen-based closure requires a longer chain to reach the preferred ring size.
SN2 rate law:
\text{Rate}_{\text{SN2}} = k[\text{R-LG}][\text{Nu}^-]SN1 rate law:
\text{Rate}_{\text{SN1}} = k[\text{R-LG}]Carbocation stability order:
3^\circ > 2^\circ > 1^\circ > \text{methyl}Leaving group ability (in a given row of the periodic table): correlates with the conjugate acid strength of the leaving group. Weaker bases are better leaving groups.
Nucleophilicity vs. basicity: Nucleophilicity is a kinetic property (how fast the species attacks carbon). Basicity is a thermodynamic property (how strongly it holds a proton). In polar protic solvents, nucleophilicity increases going down a column (I⁻ > Br⁻ > Cl⁻ > F⁻) while basicity increases going up.
SN2 chemistry is the basis of the Williamson ether synthesis, which is used industrially to produce ethers found in fragrances, pharmaceuticals, and fuel additives (MTBE, for instance, was once a common petrol additive). Understanding how nucleophilic substitution and elimination compete is essential in medicinal chemistry, where a drug synthesis route that accidentally favours elimination over substitution can waste millions of pounds in development costs.
Students often think that doubling the nucleophile in an SN1 reaction will double the rate. It will not. SN1 rate depends only on the substrate concentration.
Students frequently confuse nucleophilicity with basicity. A good nucleophile is not always a strong base (HS⁻ is a superb nucleophile but a weak base), and a strong base is not always a good nucleophile (tert-butoxide is a strong base but a poor nucleophile for SN2 due to steric bulk).
Students assume polar aprotic solvents favour SN1 because "polar" sounds like it stabilises ions. Polar aprotic solvents actually favour SN2 by leaving the nucleophile unsolvated and reactive. SN1 needs polar protic solvents.
In Williamson ether synthesis, students sometimes put the SN2 reaction on the more hindered carbon. Always place the alkoxide on the bulkier fragment and the halide on the less hindered (primary) fragment.
⚠️ The SN2 rate law (Rate = k[substrate][nucleophile]) is tested directly. Know how to predict rate changes when concentrations are altered.
⚠️ Choosing the correct solvent/nucleophile combination for SN2 vs. SN1 is a recurring question format. Polar aprotic + strong nucleophile = SN2.
⚠️ Identifying the nucleophile in a reaction (as opposed to the leaving group, substrate, or by-product) appears as a standalone question.
⚠️ Carbocation rearrangements are tested through product prediction: if the product has a different carbon skeleton from the substrate, a rearrangement occurred, pointing to an SN1/E1 pathway.
⚠️ Williamson ether synthesis route selection (which alcohol, which halide) is tested by asking which combination avoids elimination.
True or False: SN2 reactions proceed faster on tertiary substrates than on primary substrates. (False: the opposite. Steric hindrance blocks backside attack on tertiary carbons.)
Fill in the blank: The SN2 rate law is Rate = k × ______ × ______. (Rate = k[substrate][nucleophile])
True or False: Polar aprotic solvents favour SN1 reactions. (False: they favour SN2.)
Fill in the blank: In the Williamson ether synthesis, the halide should be on the ______ carbon to avoid elimination. (primary / least hindered)
True or False: A carbocation rearrangement can occur during an SN2 reaction. (False: SN2 is concerted with no intermediate. Rearrangements require a carbocation intermediate, which only forms in SN1/E1.)
Q: If the concentration of NaOH is doubled and the concentration of propyl chloride is halved in an SN2 reaction, what happens to the rate?
A: The rate stays the same. Doubling one reactant and halving the other cancels out: (2)(0.5) = 1.
Q: What reagent/solvent combination favours SN2 to convert an alkyl bromide to a methyl ether with inversion of stereochemistry?
A: CH₃ONa / DMF. Sodium methoxide is a strong nucleophile, and DMF is a polar aprotic solvent that leaves the nucleophile unsolvated and reactive.
Q: In the reaction of (CH₃)₂CHCl with excess CH₃NH₂, identify the nucleophile.
A: CH₃NH₂. The amine acts as the nucleophile, donating its lone pair to the electrophilic carbon. It is not NH₂⁻ (which would require deprotonation) and not Cl⁻ (which is the leaving group).
Q: Two Williamson ether synthesis routes can produce the same ether. One uses 2-butanol (as its alkoxide) with a primary alkyl chloride; the other uses a primary alcohol (as its alkoxide) with a secondary alkyl chloride. Which route is preferred and why?
A: The route using 2-butanol (as its alkoxide) with the primary alkyl chloride. SN2 works best on primary substrates, and a secondary substrate would give significant elimination side products.
Q: A secondary alkyl chloride is treated with KOtBu/tBuOH. The major product has a rearranged carbon skeleton. Explain why.
A: KOtBu is a strong, bulky base that cannot easily perform SN2 on a secondary substrate. Instead, ionisation to a carbocation occurs (E1/SN1 pathway). The secondary carbocation rearranges via a 1,2-hydride or methyl shift to a more stable tertiary carbocation, and elimination then produces the observed alkene.
This material connects directly to alkene chemistry: elimination reactions (E1 and E2) produce alkenes, and understanding when substitution wins over elimination is a core skill for both Exam 3 and the final. The stereochemistry of SN2 (inversion) ties back to R/S configuration assignments from earlier in the course. Carbocation stability, first encountered in acid-base chemistry and Markovnikov's rule, reappears here as the driving force behind SN1 reactivity and rearrangements.
SN2 reaction, SN1 reaction, nucleophilic substitution, bimolecular substitution, unimolecular substitution, backside attack, Walden inversion, leaving group, nucleophile, electrophile, carbocation rearrangement, 1,2-hydride shift, 1,2-methyl shift, Williamson ether synthesis, alkoxide, polar aprotic solvent, polar protic solvent, DMF, DMSO, acetone, E1 elimination, E2 elimination, Zaitsev rule, anti-periplanar, KOtBu, tert-butoxide, intramolecular SN2, ring closure, thiirane, tetrahydropyran, CHEM 225, organic chemistry II, University of Michigan