11.3 Characteristics of the SN2 Reaction

Factors Affecting SN2 Reaction Rates

The SN2 reaction is a one-step nucleophilic substitution where the nucleophile attacks at the same time the leaving group departs. Because everything happens in a single concerted step, every factor that influences either the nucleophile's ability to attack or the leaving group's ability to leave directly changes the reaction rate. That makes SN2 reactions highly predictable once you understand four key variables: substrate structure, nucleophile strength, leaving group ability, and solvent choice.

Substrate Structure

Steric hindrance is the single biggest factor determining whether an SN2 reaction will even occur. The nucleophile must attack from the backside of the carbon bearing the leaving group, so anything blocking that approach slows or prevents the reaction.

• Methyl substrates (e.g., $\ce{CH3Br}$) react fastest because there are only hydrogen atoms around the electrophilic carbon.
• Primary substrates (e.g., $\ce{CH3CH2Cl}$) react quickly. One alkyl group creates only minor steric crowding.
• Secondary substrates (e.g., isopropyl bromide) react much more slowly. Two alkyl groups partially block backside attack.
• Tertiary substrates (e.g., tert-butyl chloride) essentially do not undergo SN2 reactions. Three bulky groups make backside attack practically impossible.

Nucleophile Strength

SN2 reactions are first-order in nucleophile concentration, so a stronger nucleophile directly increases the rate. Two properties govern nucleophilic strength:

• Basicity: Stronger bases tend to be better nucleophiles. A negatively charged species like $\ce{OH-}$ is a better nucleophile than its neutral counterpart $\ce{H2O}$.
• Polarizability: Larger, more polarizable atoms donate electron density more easily to the electrophilic carbon. This is why sulfur-based nucleophiles outperform oxygen-based ones of similar basicity.

A rough ranking of common nucleophiles (strongest to weakest):

1. Thiolates ($\ce{RS-}$) — highly polarizable sulfur atom
2. Cyanide ($\ce{CN-}$) — polarizable carbon with a negative charge
3. Iodide ($\ce{I-}$) — large, very polarizable
4. Azide ($\ce{N3-}$) — moderately strong, resonance-stabilized
5. Bromide ($\ce{Br-}$) — moderate nucleophile
6. Chloride ($\ce{Cl-}$) — smaller and less polarizable than bromide
7. Fluoride ($\ce{F-}$) — small, tightly solvated, weak nucleophile in solution
8. Water ($\ce{H2O}$), alcohols ($\ce{ROH}$) — neutral, weak nucleophiles

Leaving Group Ability

A good leaving group departs easily because it forms a stable species once it leaves. Stability of the departing anion (the conjugate base) is the key predictor.

• Good leaving groups (fast SN2): $\ce{I-}$, $\ce{Br-}$, $\ce{OTs-}$ (tosylate)
• Moderate leaving group: $\ce{Cl-}$
• Poor leaving groups (SN2 very slow or doesn't occur): $\ce{F-}$, $\ce{OH-}$, $\ce{OR-}$

This trend tracks with bond strength to carbon: the $\ce{C-I}$ bond is the weakest and easiest to break, while the $\ce{C-F}$ bond is the strongest. Tosylate ($\ce{OTs-}$) is an excellent leaving group because its conjugate base is stabilized by resonance across three oxygen atoms.

Hydroxide ($\ce{OH-}$) and alkoxides ($\ce{OR-}$) are very poor leaving groups because they are strong bases. This is why you can't directly substitute an $\ce{-OH}$ group via SN2 without first converting it to a better leaving group (e.g., by protonation or tosylation).

Solvent Effects

Solvent choice can make or break an SN2 reaction, and the reasoning comes down to how the solvent interacts with the nucleophile.

• Polar aprotic solvents (acetone, DMSO, DMF, acetonitrile) speed up SN2 reactions. These solvents solvate the cation (like $\ce{Na+}$) but leave the nucleophilic anion relatively "naked" and free to attack. This is the preferred solvent type for SN2.
• Polar protic solvents (water, methanol, ethanol) slow down SN2 reactions. Their hydrogen-bonding ability forms a solvation shell around the nucleophile, effectively caging it and reducing its reactivity.

Mechanism and Kinetics of SN2 Reactions

The Concerted Mechanism

The "2" in SN2 stands for bimolecular, meaning two species are involved in the single rate-determining step. Bond formation (nucleophile attacking) and bond breaking (leaving group departing) happen simultaneously. There is no intermediate, only a single transition state.

In the transition state, the carbon is partially bonded to both the incoming nucleophile and the outgoing leaving group, giving it a pentacoordinate geometry:

$\ce{Nu^{\delta-}...C...LG^{\delta-}}$

Rate Law

Because both the nucleophile and the substrate participate in the rate-determining step, the rate law is second-order overall:

$\text{rate} = k[\text{nucleophile}][\text{substrate}]$

This means doubling the concentration of either the nucleophile or the substrate will double the reaction rate. Doubling both will quadruple it.

Stereochemistry: Walden Inversion

Every SN2 reaction proceeds with inversion of configuration at the carbon being attacked. The nucleophile approaches from the back side (180° from the leaving group), which flips the spatial arrangement of the other three groups, like an umbrella inverting in the wind.

If the starting material has R configuration at the stereocenter, the product will have S configuration (or vice versa). This is called Walden inversion and is one of the most reliable stereochemical outcomes in organic chemistry.

Predicting SN2 Reaction Outcomes

When you encounter a substitution problem, work through these four factors in order:

1. Check the substrate. Is it methyl, primary, or secondary? If tertiary, SN2 won't happen.
2. Identify the nucleophile. Is it strong (charged, polarizable) or weak (neutral, small)?
3. Evaluate the leaving group. Is it a good leaving group ($\ce{I-}$, $\ce{Br-}$, $\ce{OTs-}$) or a poor one ($\ce{OH-}$, $\ce{F-}$)?
4. Note the solvent. Polar aprotic favors SN2; polar protic hinders it.

Worked Examples

Fast SN2:

$\ce{CH3CH2Br + NaCN ->[\text{acetone}] CH3CH2CN + NaBr}$

Primary substrate, strong nucleophile ($\ce{CN-}$), good leaving group ($\ce{Br-}$), polar aprotic solvent. All four factors favor SN2, so this reaction is fast.

No SN2 reaction:

$\ce{(CH3)3CBr + NaOH ->[\text{ethanol}] \text{No SN2 product}}$

Tertiary substrate blocks backside attack entirely. Even though $\ce{OH-}$ is a decent nucleophile, steric hindrance prevents SN2. (This substrate would instead undergo E2 elimination with a strong base.)

Slower SN2:

$\ce{CH3CH2CH2Cl + NaI ->[\text{acetone}] CH3CH2CH2I + NaCl}$

Primary substrate and polar aprotic solvent both favor SN2, but $\ce{Cl-}$ is only a moderate leaving group (weaker than $\ce{Br-}$). The reaction proceeds, just not as quickly as it would with a bromide or tosylate leaving group.