Difficulty: Intermediate
Prerequisites: Chapters 1 to 4 (bonding, Lewis structures, polarity, functional groups). You should be comfortable with curved-arrow notation and the concept of leaving groups before starting here.
Nucleophilic substitution is one of the foundational reaction types in organic chemistry. In Chapter 5 of CHEM 2301 you learn the two mechanisms by which a nucleophile can replace a leaving group on a carbon: SN2 (bimolecular) and SN1 (unimolecular). This document covers SN2. Understanding SN2 is essential because it governs how primary and secondary substrates react with strong nucleophiles, and it is the basis for predicting stereochemical outcomes. If you are not yet confident with leaving-group ability and nucleophile strength from Chapter 4, review those first.
In an SN2 reaction, a nucleophile attacks the electrophilic carbon at the same time the leaving group departs, all in one concerted step. The nucleophile always approaches from the side opposite the leaving group (backside attack), which flips the stereochemistry at that carbon (inversion). SN2 works best on unhindered (methyl and primary) substrates with strong nucleophiles.
Substitution reaction
A reaction following the general pattern A + B → C + D, where a nucleophile replaces a leaving group on a carbon. Think of it as swapping one group for another on the same carbon atom.
SN2 (Substitution Nucleophilic Bimolecular)
A one-step substitution mechanism in which both the nucleophile and the substrate participate in the rate-determining step. In simple terms, this means two molecules must collide for the reaction to proceed.
Nucleophile (Nu)
An electron-rich species that donates a pair of electrons to an electrophilic carbon. Think of it as the "attacker" in the reaction, bringing electrons to form a new bond.
Leaving group (LG)
The atom or group that departs with the bonding electrons when the nucleophile attacks. Good leaving groups are stable once they leave (weak bases such as I⁻, Br⁻, Cl⁻, and tosylate).
Electrophilic carbon
The carbon atom bonded to the leaving group. It carries a partial positive charge (δ+) because the leaving group is electronegative, making it the target for nucleophilic attack.
Backside attack
The nucleophile approaches the electrophilic carbon from the side directly opposite the leaving group. This geometry is required in SN2 because the leaving group's electron cloud blocks the front side.
Transition state
The highest-energy arrangement along the reaction pathway, where the nucleophile is partially bonded to the carbon and the leaving group is partially departed. It cannot be isolated. In simple terms, it is the "halfway point" of the single step.
Inversion of stereochemistry (Walden inversion)
The configuration at the electrophilic carbon flips during the reaction, like an umbrella turning inside out. If the starting material is R, the product is S, and vice versa.
Concerted reaction
A reaction in which bond-making and bond-breaking happen simultaneously in a single step, with no intermediate. SN2 is the textbook example.
Rate-determining step
The slowest step in a reaction mechanism, which controls the overall rate. In SN2 the only step is the rate-determining step, and it involves both the nucleophile and the substrate.
The nucleophile attacks the electrophilic carbon from the back side (the side opposite the leaving group)
Bond formation (Nu to C) and bond breaking (C to LG) happen at the same time, in one concerted step
The reaction passes through a single transition state in which the carbon is partially bonded to both the nucleophile and the leaving group
The transition state is drawn in square brackets with a double-dagger symbol (‡)
Five groups are arranged around the central carbon in a trigonal bipyramidal geometry at the transition state
The leaving group departs with the bonding electrons
SN2 always produces inversion of configuration at the electrophilic carbon
If the starting material has R configuration, the product has S configuration, and vice versa
This is sometimes called Walden inversion
You get a single enantiomer of product (not a mixture), because the backside attack is the only possible approach
The energy diagram shows a single energy maximum (one "hill")
Starting materials on the left, products on the right, with one transition state at the peak
There is no intermediate, because the reaction is concerted
The height of the peak represents the activation energy (Ea)
Strong nucleophile: the nucleophile is directly involved in the rate-determining step, so a stronger nucleophile speeds the reaction
Methyl or primary substrate: less steric hindrance around the electrophilic carbon allows the nucleophile to reach it. Secondary substrates can sometimes undergo SN2 but much more slowly. Tertiary substrates essentially do not undergo SN2 because the three bulky groups block backside attack
Good leaving group: a more stable leaving group (weaker base) departs more easily
Polar aprotic solvent: these solvents (acetone, DMSO, DMF, acetonitrile) do not hydrogen-bond to the nucleophile, leaving it "naked" and reactive. Polar protic solvents (water, alcohols) solvate the nucleophile and slow SN2
Pyridine (a nitrogen-containing aromatic ring acting as the nucleophile) reacts with an alkyl iodide (the substrate with I as the leaving group) via SN2
The nitrogen lone pair attacks from the back side of the C-I bond
I⁻ departs, and the product is an N-alkylpyridinium salt with inverted stereochemistry at the carbon
A benzylic substrate with Cl as the leaving group reacts with methoxide (⁻OCH₃) as the nucleophile
Backside attack by ⁻OCH₃ displaces Cl⁻
The product shows inversion of stereochemistry: a single enantiomer is formed
SN2 Rate Law
\text{Rate} = k[\text{Nu}][\text{Substrate}]The rate depends on the concentration of both the nucleophile and the substrate. Double either concentration and the rate doubles. This is a second-order rate law, which is where the "2" in SN2 comes from (bimolecular).
Energy Diagram (description)
A single smooth curve rising from starting materials to a single transition-state peak, then falling to products. No dip in the middle (no intermediate). Label the peak "T.S." or "‡". The difference in height between starting materials and the peak is the activation energy.
SN2 reactions are the workhorse of synthetic organic chemistry. Pharmaceutical chemists use SN2 to install specific functional groups on drug molecules with precise stereochemical control, because the inversion is predictable. Biological systems rely on SN2-like chemistry too: methyltransferase enzymes transfer a methyl group from S-adenosylmethionine (SAM) to substrates via a backside-attack mechanism.
Students often think SN2 can happen on tertiary carbons if the nucleophile is strong enough. It cannot. Steric hindrance physically prevents backside attack on a tertiary carbon regardless of nucleophile strength.
Students sometimes draw the nucleophile attacking from the same side as the leaving group (frontside attack). This does not occur in SN2. The nucleophile must approach from the opposite side.
A common error is drawing an intermediate (a distinct species between steps) for SN2. There is no intermediate. The transition state is not an intermediate: it is a fleeting, non-isolable maximum on the energy curve.
Students occasionally confuse "inversion of stereochemistry" with racemisation. Inversion means the configuration flips to the opposite enantiomer (one product). Racemisation means a 50:50 mixture of both enantiomers, which is what happens in SN1, not SN2.
⚠️ You will almost certainly be asked to draw the transition state for an SN2 reaction. Remember: square brackets, partial bonds (dashed lines) to both Nu and LG, five groups around carbon, double-dagger symbol.
⚠️ Expect a question asking you to predict the stereochemical outcome. If the substrate is chiral, assign R/S to the starting material and show that the product has the opposite configuration.
⚠️ Know the difference between a transition state and an intermediate. Transition states appear on energy-diagram peaks and cannot be isolated. Intermediates sit in energy-diagram valleys and can (in principle) be detected.
⚠️ Rate-law problems: if you are given the rate law Rate = k[Nu][Substrate], you should recognise this as SN2. If only the substrate appears in the rate law, that points to SN1.
True or false: SN2 reactions proceed through a carbocation intermediate. ___
In SN2, the nucleophile attacks from the ___ side relative to the leaving group.
True or false: doubling the nucleophile concentration doubles the rate of an SN2 reaction. ___
An SN2 reaction at a chiral centre produces (inversion / retention / racemisation) of configuration.
True or false: tertiary substrates are the best candidates for SN2 reactions. ___
Answers: 1. False (no intermediate at all). 2. Back (opposite). 3. True (second-order kinetics). 4. Inversion. 5. False (methyl and primary are best; tertiary substrates do not undergo SN2).
Q: A primary alkyl bromide reacts with NaCN in DMSO. What mechanism is operating, and what is the stereochemical outcome at the carbon bearing the leaving group?
A: SN2. Strong nucleophile (CN⁻), primary substrate, polar aprotic solvent (DMSO). The product shows inversion of configuration at the electrophilic carbon.
Q: Draw the transition state for the reaction of hydroxide (OH⁻) with 2-iodopropane via SN2.
A: The transition state has HO partially bonded to the central carbon from one side (dashed bond) and I partially bonded from the opposite side (dashed bond). The three remaining groups (H, CH₃, CH₃) are arranged in a plane perpendicular to the HO...C...I axis. The whole structure is enclosed in square brackets with a double-dagger symbol.
Q: Why does SN2 not occur at tertiary carbons?
A: Three bulky alkyl groups surround the electrophilic carbon and block the nucleophile from approaching the back side. The steric hindrance raises the activation energy so much that the reaction does not proceed at a measurable rate.
Q: If (R)-2-bromobutane undergoes SN2 with NaOH, what is the configuration of the product?
A: (S)-2-butanol. SN2 always inverts the configuration, so R becomes S.
Q: How does the rate law distinguish SN2 from SN1?
A: SN2 is second order: Rate = k[Nu][Substrate]. SN1 is first order: Rate = k[Substrate]. If changing the nucleophile concentration changes the rate, the mechanism is SN2.
SN2 connects directly to SN1 (covered in the companion study notes): the two mechanisms compete, and predicting which one dominates for a given substrate, nucleophile, and solvent is a central exam skill in CHEM 2301. SN2 also connects to elimination reactions (E2), which share the same preference for strong bases and unhindered substrates, so you will need to distinguish SN2 from E2 in later chapters. The stereochemical reasoning here (R/S assignment, inversion) builds on the Chapter 3 material on chirality and optical activity.
SN2 reaction, substitution nucleophilic bimolecular, backside attack, Walden inversion, inversion of configuration, concerted mechanism, transition state, nucleophilic substitution, leaving group, nucleophile, electrophilic carbon, second-order kinetics, bimolecular rate law, polar aprotic solvent, steric hindrance, primary substrate, methyl substrate, organic chemistry 1, CHEM 2301 Chapter 5, Hoover, University of Minnesota