\(\text{S}_\text{N}\text{2}\) Reactions
Conditions, Stereochemistry, Rate
Master the backside attack mechanism, understand Walden inversion, and predict reaction rates based on substrate, nucleophile, and solvent.
Organic Chemistry 1
What Is an \(\text{S}_\text{N}\text{2}\) Reaction?
\(\text{S}_\text{N}\text{2}\) stands for Substitution, Nucleophilic, Bimolecular. A nucleophile attacks a carbon bearing a leaving group, and the bond-breaking and bond-forming happen simultaneously in a single concerted step.
S
Substitution: one group replaces another
N
Nucleophilic: electron-rich species attacks
2
Bimolecular: rate depends on both reactants
The Concerted Mechanism
Why \(\text{S}_\text{N}\text{2}\) Matters
\(\text{S}_\text{N}\text{2}\) reactions are among the most predictable and stereospecific reactions in organic chemistry. They let chemists:
1
Build new bonds selectively: Replace one functional group with another at a specific carbon.
2
Control stereochemistry: Predict inversion of configuration every time.
3
Synthesize pharmaceuticals: Many drug syntheses rely on \(\text{S}_\text{N}\text{2}\) to set chirality.
The Rate Law
Because the nucleophile and substrate are both involved in the rate-determining step, the rate law is second-order overall:
Double [Sub]
Rate doubles: first-order in substrate
Double [Nu]
Rate doubles: first-order in nucleophile
Overall
Second-order: \(1 + 1 = 2\)
Walden Inversion: Stereochemistry
Factors Affecting \(\text{S}_\text{N}\text{2}\) Rate
Substrate
\(\text{Methyl} > 1^\circ > 2^\circ \gg 3^\circ\) (steric hindrance slows reaction)
Nucleophile
Strong nucleophiles required: \(\text{HO}^-\), \(\text{CN}^-\), \(\text{I}^-\), \(\text{RS}^-\)
Solvent
Polar aprotic solvents: DMSO, DMF, acetone
Leaving Group
Better LG = faster: \(\text{I}^- > \text{Br}^- > \text{Cl}^- \gg \text{F}^-\)
Steric Hindrance and Substrate
\(\text{S}_\text{N}\text{2}\) vs \(\text{S}_\text{N}\text{1}\)
| \(\text{S}_\text{N}\text{2}\) | \(\text{S}_\text{N}\text{1}\) | |
|---|---|---|
| Mechanism | One step (concerted) | Two steps (carbocation) |
| Rate Law | \(k[\text{Sub}][\text{Nu}]\) | \(k[\text{Sub}]\) |
| Substrate | \(\text{Methyl}, 1^\circ, (2^\circ)\) | \(3^\circ, (2^\circ)\) |
| Nucleophile | Strong required | Weak is fine |
| Solvent | Polar aprotic | Polar protic |
| Stereochemistry | Inversion | Racemization |
Solvent: Aprotic vs Protic
Polar Aprotic
DMSO, DMF, acetone, acetonitrile. No N-H or O-H bonds. Cannot hydrogen-bond to nucleophile, so the nucleophile stays "naked" and reactive.
Polar Protic
Water, methanol, ethanol. Have N-H or O-H bonds. They solvate (surround) the nucleophile, creating a cage that slows the attack.
How to Predict an \(\text{S}_\text{N}\text{2}\) Reaction
1
Identify the substrate: Is it methyl, \(1^\circ\), or \(2^\circ\)? If \(3^\circ\), stop: \(\text{S}_\text{N}\text{2}\) is not possible.
2
Check the nucleophile: Is it a strong nucleophile (charged, unhindered)? Examples: \(\text{HO}^-\), \(\text{CN}^-\), \(\text{I}^-\), \(\text{N}_3^-\)
3
Check the solvent: Is it polar aprotic (DMSO, DMF, acetone)? This favors \(\text{S}_\text{N}\text{2}\).
4
Identify the leaving group: Good LGs are weak bases: \(\text{I}^-\), \(\text{Br}^-\), \(\text{TsO}^-\)
5
Assign stereochemistry: The product has inverted configuration at the carbon bearing the leaving group.
Example 1: Simple \(\text{S}_\text{N}\text{2}\)
Example 2: Stereochemical Outcome
Example 3: Will It Go \(\text{S}_\text{N}\text{2}\)?
Predict whether this reaction proceeds via \(\text{S}_\text{N}\text{2}\):
1
Substrate: tert-Butyl = \(3^\circ\) carbon. Three bulky methyl groups surround the electrophilic carbon.
2
Nucleophile: \(\text{I}^-\) is a strong nucleophile.
3
Solvent: Acetone is polar aprotic: favors \(\text{S}_\text{N}\text{2}\).
Common Mistakes
✕ Forcing \(3^\circ\) \(\text{S}_\text{N}\text{2}\)
Tertiary substrates never undergo \(\text{S}_\text{N}\text{2}\). Always check the substrate first.
✕ Wrong Stereochemistry
Forgetting inversion: the product must have opposite configuration, not retention.
✕ Protic Solvent
Using water or ethanol and expecting fast \(\text{S}_\text{N}\text{2}\). Protic solvents cage the nucleophile.
\(\text{S}_\text{N}\text{2}\) Quick Reference
| Factor | Favors \(\text{S}_\text{N}\text{2}\) | Disfavors \(\text{S}_\text{N}\text{2}\) |
|---|---|---|
| Substrate | \(\text{Methyl}, 1^\circ\) | \(3^\circ\) |
| Nucleophile | Strong (charged) | Weak (neutral) |
| Solvent | Polar aprotic | Polar protic |
| Leaving Group | \(\text{I}^-\), \(\text{TsO}^-\) | \(\text{F}^-\), \(\text{HO}^-\) |
| Stereochemistry | Always inversion (Walden) | |
| Rate Law | \(k[\text{Sub}][\text{Nu}]\) | |
Question 1 of 5
What does the "2" in \(\text{S}_\text{N}\text{2}\) refer to?
The "2" means bimolecular: the rate depends on the concentration of both the substrate and the nucleophile: \(\text{Rate} = k[\text{Sub}][\text{Nu}]\).
Question 2 of 5
Which substrate reacts fastest in an \(\text{S}_\text{N}\text{2}\) reaction?
Methyl bromide has the least steric hindrance around the electrophilic carbon, making backside attack easiest. \(3^\circ\) substrates do not undergo \(\text{S}_\text{N}\text{2}\) at all.
Question 3 of 5
An \(\text{S}_\text{N}\text{2}\) reaction on an (R)-configured substrate produces a product with which configuration?
\(\text{S}_\text{N}\text{2}\) always produces inversion (Walden inversion). The nucleophile attacks from \(180^\circ\) opposite the leaving group, flipping the configuration from (R) to (S).
Question 4 of 5
Which solvent would best promote an \(\text{S}_\text{N}\text{2}\) reaction?
DMSO is a polar aprotic solvent. It dissolves ionic nucleophiles without hydrogen-bonding to them, keeping the nucleophile "naked" and highly reactive. Water, ethanol, and acetic acid are all protic and would solvate the nucleophile.
Question 5 of 5
If the concentration of the nucleophile is tripled while all other conditions remain constant, the \(\text{S}_\text{N}\text{2}\) rate will:
The rate law is \(\text{Rate} = k[\text{Sub}][\text{Nu}]\). Since the rate is first-order in nucleophile, tripling \([\text{Nu}]\) triples the rate. It would only increase by 9x if the rate were second-order in nucleophile.