Organic Chemistry · Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations
The SN2 Reaction
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In 30 seconds
A substitution reaction swaps one group on a carbon for another: the halide ion departs while a Nucleophile Electron-rich species (often negatively charged) that donates a lone pair to form a new bond Full entry → takes its place, as in CH₃CH₂Br + HO⁻ → CH₃CH₂OH + Br⁻. The SN2 reaction is the simplest mechanism for this exchange. The name is a code: S = substitution, N = nucleophilic, 2 = Bimolecular Two molecules participate in the rate-determining step Full entry → — the alkyl halide and the nucleophile both appear in the rate-determining step. Everything distinctive about SN2 follows from that fact: bond formation and breaking happen in the same step, so there is no intermediate — the nucleophile pushes in from the back side while the halide leaves, flipping the carbon's geometry inside out (Walden inversion Complete flip of configuration at the reacting carbon Full entry →). This topic covers the reaction, its rate law, and the geometry of attack; the next topic examines the substrate, nucleophile, Leaving group The group that departs with the bonding electrons (e.g., Br⁻) Full entry →, and solvent factors.
Why this matters
- Synthesis: SN2 alkylation attaches alkyl groups to oxygen, nitrogen, sulfur, and carbon nucleophiles — the route to countless drugs, dyes, and materials (methyl iodide, methyl tosylate, dimethyl sulfate all react by SN2).
- Biology: Methyl transfers that regulate gene expression and hormones (topic 6) are SN2 reactions run by enzymes.
- Drug metabolism: Many drugs are activated or deactivated by substitution at electrophilic carbons; recognizing SN2 explains metabolic transformations and toxicity.
- Exams: Product prediction, reactivity ranking, and SN2 vs SN1 decisions are heavily tested.
- Safety: Many alkyl halides are volatile and several are toxic or suspected carcinogens; work with them in a fume hood and avoid skin contact.
The college version
Core Concepts
What the name means
SN2 = Substitution, Nucleophilic, bimolecular. "Bimolecular" refers to the rate-determining step: both the nucleophile and the alkyl halide appear in the Transition state The highest-energy, partially bonded arrangement along the reaction path Full entry →. The general equation:
R–X + Nu- → R–Nu + X-
R–X is the alkyl halide, Nu⁻ the nucleophile (electron-rich, often negatively charged), R–Nu the product, and X⁻ the leaving group.
One step, one transition state
SN2 is concerted: bond making and bond breaking occur simultaneously, so no carbocation or other intermediate forms. The nucleophile's lone pair begins bonding to carbon while the C–X bond begins to break; at the midpoint the carbon carries five partial bonds in a pentacoordinate transition state. Because the transition state is the highest-energy point on the path, anything that stabilizes the nucleophile — or destabilizes the crowded transition state — changes the rate.
The rate law
Because two molecules must collide, the rate depends on both:
rate = k[R–X][Nu-]
The reaction is first order in each reactant and second order overall. Doubling either concentration doubles the rate; doubling both quadruples it. This is the experimental fingerprint that distinguishes SN2 from SN1.
Back-side attack and Walden inversion
The nucleophile cannot approach from the same side as the leaving group — that side is blocked. It attacks from the back side, 180° from the C–X bond, overlapping the far lobe of the carbon's σ* orbital. As the nucleophile bonds, the three other groups on carbon swing past one another like an umbrella blown inside out. The product therefore has the opposite configuration at the reacting carbon: (R) reactant gives (S) product, and vice versa. This complete inversion — Walden inversion — is diagnostic of SN2.
Steric hindrance sets the reactivity order
Because the nucleophile must physically reach the carbon, bulky groups slow it:
CH3X (methyl) > 1°> 2° ≫ 3°
Methyl and primary halides react readily; secondary slowly; tertiary essentially not at all (they prefer elimination or SN1). This ordering is the chapter's most important predictor.
How It Works / Step-by-Step Process
Mechanism of CH₃CH₂Br + HO⁻ → CH₃CH₂OH + Br⁻:
- The hydroxide lone pair aims at the back lobe of the C–Br σ* orbital — opposite the bromine.
- As HO⁻ approaches, the C–Br bond stretches; the transition state has partial C···OH and C···Br bonds and a roughly planar carbon.
- The C–Br bond finishes breaking; Br⁻ departs with the electron pair; the C–OH bond finishes forming.
- The three hydrogens swing to the opposite face — the carbon appears to turn inside out. The product, ethanol, carries the nucleophile where the leaving group used to be.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| "Bimolecular" | "Two steps" | Bimolecular = two species in one transition state; SN2 is a single step with no intermediate |
| Inversion of configuration | Retention or racemization | SN2 always inverts the reacting carbon: (R) → (S); SN1 gives mostly racemized product |
| Tertiary halides "react fast" | SN2 reactivity | Tertiary halides are fast in SN1/E1/E2 but essentially inert to SN2 because of crowding |
| Rate depends on the nucleophile | Rate depends only on substrate | Only SN2 puts [Nu⁻] in the rate law; SN1 does not (topics 4–5) |
| Polar protic solvents speed all substitutions | Solvent effects differ | Protic solvents slow SN2 by solvating the nucleophile; they favor SN1 instead |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine a door that opens away from you; to get through, you must run around and push from the back, so it swings open the other way. That is SN2: the incoming molecule sneaks behind the carbon and shoves the leaving group out the front, flipping the molecule's handedness like an umbrella in the wind. It all happens in one quick push, with no stopping in the middle — and big groups guarding the carbon make the push much harder.
Worked example
Example 1: Rate-law arithmetic
Methyl bromide reacts with hydroxide: CH₃Br + HO⁻ → CH₃OH + Br⁻, rate = k[CH₃Br][HO⁻]. Using an illustrative k = 2.0 × 10⁻³ M⁻¹·s⁻¹, [CH₃Br] = 0.10 M, [HO⁻] = 0.20 M, what is the initial rate?
Formula first:
rate = k[CH3Br][HO-]
Substitute with units (dimensional analysis):
rate = (2.0 × 10-3 M-1s-1)(0.10 M)(0.20 M) = 4.0 × 10-5 M · s-1
The units cancel: M⁻¹·s⁻¹ × M × M = M·s⁻¹ (moles per liter per second). If [CH₃Br] doubled to 0.20 M, the rate would double to 8.0 × 10⁻⁵ M·s⁻¹ — direct evidence that both partners matter.
Example 2: Predicting the stereochemical outcome
(R)-2-Bromobutane, CH₃CH₂CH(Br)CH₃, reacts with sodium cyanide in DMSO. Predict the product and its configuration.
Reasoning: Cyanide is a strong nucleophile and DMSO is polar aprotic — classic SN2 conditions. The substrate is secondary, so SN2 is slow but viable with a good nucleophile.
Product: CH₃CH₂CH(CN)CH₃ — 2-cyanobutane — with inverted configuration: (S)-2-cyanobutane. CN⁻ attacks from the back side, flipping the stereocenter while the rest of the molecule is unchanged.
Example 3: Ranking substrates
Rank these bromides from fastest to slowest SN2 with NaN₃ in acetone: (a) CH₃Br, (b) CH₃CH₂Br, (c) (CH₃)₂CHBr, (d) (CH₃)₃CBr.
Answer: a > b > c ≫ d — methyl fastest, primary next, secondary much slower, tertiary essentially unreactive (it would prefer elimination). The trend tracks steric crowding around the reacting carbon.
Key takeaways
- SN2 = substitution, nucleophilic, bimolecular: one concerted step, no intermediate, one transition state.
- Rate law: rate = k[R–X][Nu⁻] — second order overall; both reactants in the rate-determining step.
- Back-side attack always gives complete inversion of configuration: (R) → (S).
- Reactivity order: methyl > 1° > 2° ≫ 3° (steric hindrance, not electronics).
- Requires a strong nucleophile and a good leaving group; polar aprotic solvents speed it up (next topic).
- Tertiary halides do not do SN2 — they eliminate or react by SN1 instead.
- Mechanism in words: nucleophile lone pair → back lobe of C–X σ* orbital while C–X breaks; X⁻ departs with the electron pair.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
What does each letter in "SN2" mean?
Show answer
S = substitution (a group is replaced), N = nucleophilic (an electron-rich species attacks), 2 = bimolecular (both R–X and Nu⁻ appear in the rate-determining step).
Write the rate law for CH₃CH₂Br + I⁻ → CH₃CH₂I + Br⁻. What happens to the rate if both concentrations are doubled?
Show answer
rate = k[CH₃CH₂Br][I⁻]; doubling both concentrations quadruples the rate (2 × 2 = 4).
Which product forms when (S)-2-bromobutane reacts with a strong nucleophile under SN2 conditions — the (R) or (S) product? Why?
Show answer
The (R) product — back-side attack inverts the configuration (Walden inversion), so (S) reactant gives (R) product.
Why is (CH₃)₃CBr essentially unreactive toward SN2?
Show answer
The three methyl groups crowd the carbon and block the back-side approach the nucleophile needs.
How many transition states and how many intermediates does an SN2 reaction have?
Show answer
One transition state and zero intermediates — the reaction is concerted.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Nucleophile
- Electron-rich species (often negatively charged) that donates a lone pair to form a new bond
- Leaving group
- The group that departs with the bonding electrons (e.g., Br⁻)
- Bimolecular
- Two molecules participate in the rate-determining step
- Transition state
- The highest-energy, partially bonded arrangement along the reaction path
- Concerted reaction
- Bond breaking and bond forming in one step, no intermediate
- Walden inversion
- Complete flip of configuration at the reacting carbon
- Back-side attack
- Nucleophile approaches 180° from the leaving group
Sources & references
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