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Organic Chemistry 1 · Substitution and Elimination

The SN2 Mechanism

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Worked example
  6. Key takeaway
  7. Study tools

In 30 seconds

stands for bimolecular nucleophilic substitution: the rate depends on both the substrate and the nucleophile. In one step, the nucleophile attacks the backside of the carbon, 180° away from the leaving group, and the new bond forms as the old bond breaks. The result is inversion of configuration at the reacting carbon. SN2 works best on methyl and primary substrates with strong nucleophiles, good leaving groups, and polar aprotic solvents.

Why this matters

SN2 reactions are a workhorse for building carbon–heteroatom and carbon–carbon bonds in drug synthesis — for example, installing a nitrile, azide, or amine handle that later becomes part of a drug candidate. The strict stereospecificity of SN2 is clinically relevant because drug activity often depends on a single enantiomer; an SN2 step can invert a stereocenter cleanly rather than scrambling it. As always, preparative procedures, reagent quantities, and safety measures must follow institutional documentation and are not reproduced here.

The college version

1. The One-Step, Concerted Mechanism and Rate Law

SN2 is a single elementary step. The nucleophile donates its lone pair into the σ* antibonding orbital of the C–X bond while the leaving group departs with the bonding electrons. Because both the substrate and the nucleophile appear in the rate-determining (and only) step, the is:

rate = k [RX] [Nu-]

Doubling either the substrate or the nucleophile doubles the rate (second order overall).

2. Backside Attack and Walden Inversion

The nucleophile approaches along the axis of the C–X bond, from the side opposite the leaving group (180°). This backside approach minimizes repulsion between the incoming electrons and the departing leaving group. The three remaining substituents "flip" through a planar transition state, inverting configuration — the . SN2 is therefore stereospecific: a single enantiomeric starting material gives a single enantiomeric product with opposite configuration.

3. Factors Governing SN2 Reactivity

  • Substrate (): methyl > primary > secondary >> tertiary. Tertiary substrates do not undergo SN2 because three alkyl groups block backside approach.
  • Nucleophile strength: strong, concentrated nucleophiles (for example, I⁻, Br⁻, CN⁻, N₃⁻, RS⁻) accelerate the reaction.
  • Leaving group: good leaving groups (weak bases) are required; I⁻ > Br⁻ > Cl⁻ > F⁻.
  • Solvent: polar aprotic solvents (DMSO, DMF, acetone, acetonitrile) solvate cations but leave the nucleophile's anion "naked" and highly reactive. Polar protic solvents slow SN2 by hydrogen-bonding to the nucleophile.

How it works

  1. A strong nucleophile with a lone pair approaches the backside of the C–X bond.
  2. The lone pair overlaps the σ* orbital of C–X, weakening that bond.
  3. A single transition state forms with partial bonds to both the nucleophile and the leaving group.
  4. The C–X bond fully breaks as the leaving group departs with both electrons.
  5. The three substituents invert, giving the substitution product with inverted configuration.

Common confusions

Do not confuseWithDifference
Transition stateIntermediateA transition state is a fleeting, unisolable maximum; an intermediate is a real species at an energy minimum
NucleophilicityBasicityNucleophilicity is a kinetic rate factor; basicity is a thermodynamic proton-affinity
InversionRacemizationSN2 inverts configuration (one enantiomer in, one out); SN1 gives racemic mixtures
Polar aproticPolar proticAprotic has no acidic O–H/N–H to solvate the anion; protic does and slows SN2

Memory aids

"Backside, Both, Big-bulky-blocked" — SN2 is , depends on Both reactants, and is Blocked by bulky (tertiary) substrates.

Quick review

Topic Recap

SN2 is a concerted, second-order nucleophilic substitution with backside attack and inversion of configuration. It is fastest for methyl and primary substrates, favors strong nucleophiles and good leaving groups, and is accelerated by polar aprotic solvents. Its clean stereospecificity makes it invaluable in synthesis and a favorite exam target.

Knowledge Check

  1. Write the rate law for an SN2 reaction and state the overall kinetic order.
  2. Why does (CH₃)₃CBr not undergo SN2?
  3. What stereochemical change occurs at the reacting carbon in an SN2 reaction?
  4. Why do polar aprotic solvents speed up SN2?
  5. Which substrate reacts fastest with NaCN by SN2: CH₃Br, CH₃CH₂Br, or (CH₃)₂CHBr?

Answers and Rationales

  1. k [RX] [Nu-], second order overall — both species are in the single rate-determining step.
  2. Three methyl groups sterically block backside approach; the bulky tertiary carbon also does not readily form a bond to the incoming nucleophile.
  3. Inversion of configuration (Walden inversion) — attack from the backside flips the substituents.
  4. They solvate cations tightly but leave the anionic nucleophile unsolvated and "naked," which raises its reactivity.
  5. CH₃Br — methyl has the least steric hindrance, so backside attack is fastest.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Picture a crowded four-person seat on a bus where one passenger (the leaving group) is about to get off. The incoming passenger (the nucleophile) cannot push in from the same door the leaving passenger is using; instead, the new passenger slides in from the opposite side, and in the same smooth motion the old passenger is pushed out the far side. That is exactly how SN2 works: attack from the backside while the leaving group departs from the front.

The "umbrella flip" is the classic comparison: if you push an umbrella open in a strong wind, the canopy can invert and blow inside-out. The carbon's three substituents "flip" like an umbrella as the nucleophile attacks the backside, so the product has the opposite three-dimensional arrangement — inversion of configuration.

Where this stops being exact: the umbrella image makes it look like two separate events, but SN2 is concerted — bond-making and bond-breaking happen at the same time through a single transition state, with no intermediate. Also, the "flip" only inverts configuration when the reacting carbon is a stereocenter; if the substrate is achiral, no configuration change is observable even though the geometry still inverts.

Simple Example

Hydroxide attacks methyl bromide. OH⁻ approaches from the backside of the C–Br bond; as the C–O bond forms, the C–Br bond breaks, and Br⁻ leaves. The single transition state has a partially formed C–O bond and a partially broken C–Br bond, and methanol (CH₃OH) is the product.

Worked example

Walk through the SN2 reaction of cyanide with (R)-2-bromobutane.

  1. Identify the electrophile and nucleophile. The δ+ carbon bearing bromine is the electrophilic center; the cyanide ion (CN⁻) is the nucleophile, donating a lone pair.
  2. Draw electron movement before products. A double-headed curved arrow starts on the CN⁻ lone pair and points to the backside carbon. A second double-headed arrow starts at the C–Br bond and points to bromine, showing Br⁻ leaving with its bonding pair.
  3. Show the transition state. The carbon is trigonal planar with three groups in a plane; the incoming CN and the departing Br are on opposite sides, each with partial bonds (C···CN, C···Br).
  4. Account for electrons and charge. CN⁻ loses its lone pair into a new bond (neutral contribution) and Br⁻ gains the old bond's pair, becoming Br⁻. Net charge is conserved: the (−1) reactant nucleophile becomes the (−1) product bromide.
  5. State the stereochemistry. The (R) starting material inverts to the (S) product — 2-methylbutanenitrile with inverted configuration.
  6. Verify atom balance. C₄H₉Br + CN⁻ → C₄H₉CN + Br⁻; all atoms are conserved.

Key takeaways

  • High yield: SN2 = one step, second order, inversion of configuration.
  • High yield: Reactivity order by substrate: methyl > 1° > 2° >> 3° (no SN2 at tertiary).
  • High yield: Strong nucleophiles and good leaving groups speed up SN2.
  • High yield: Polar aprotic solvents accelerate SN2; polar protic solvents slow it.
  • High yield: SN2 is stereospecific — configuration inverts at the reacting carbon.
  • The nucleophile can be neutral (for example, NH₃, H₂O) or anionic, but anionic nucleophiles are generally stronger.
  • High yield: A classic exam trap is writing SN2 on a tertiary substrate, or forgetting that the product is the inverted enantiomer.

Keep learning

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Practice Organic Chemistry 1

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Key vocabulary · Related

You’ll learn to

  • State the meaning of "SN2" and the rate law k [RX] [Nu-].
  • Describe the one-step, concerted mechanism with backside attack and Walden inversion.
  • Explain how substrate structure, nucleophile strength, leaving-group ability, and solvent influence SN2 rates.
  • Draw and interpret SN2 energy diagrams and predict the stereochemical outcome.

Key vocabulary

SN2
Bimolecular nucleophilic substitution
Concerted
Bond-making and bond-breaking in one step
Rate law
k [RX] [Nu-]
Backside attack
Nucleophile approaches 180° from the leaving group
Walden inversion
Stereochemical inversion at the reacting carbon
Steric hindrance
Bulk of alkyl groups blocking approach
Polar aprotic solvent
Polar solvent with no O–H or N–H bonds (for example, DMSO)

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