Clutch Prep is now a part of Pearson
Ch. 7 - Substitution ReactionsWorksheetSee all chapters
All Chapters
Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins
Ch. 26 - Transition Metals
Nucleophilic Substitution
Good Leaving Groups
SN2 Reaction
SN1 Reaction
Substitution Comparison
Additional Guides
Alkyl Halide
Johnny Betancourt

The SN2 reaction is a bimolecular nucleophilic substitution reaction that occurs in one step. The nucleophile performs a backside attack on the carbon to which the leaving group is attached. If the carbon is asymmetric, inversion of stereochemistry is observed. 


The SN­­2 mechanism proceeds through a concerted mechanism. All that means is that it proceeds in one step and there’s no intermediate. As the nucleophile (Nu–) performs a “backside attack,” the leaving group (LG) dissociates. In other words, the nucleophile makes a bond and breaks the bond of the leaving group to the carbon holding it. 

SN2 generic mechanismSN2 generic mechanism
 In this generic reaction, the nucleophile is attacking the carbon holding the leaving group, which causes the nucleophile to dissociate. Notice that the net charge of the reaction stays the same. There’s no intermediate in this reaction; it goes straight from reagents to products.

SN2 transition stateSN2 transition state

As the nucleophile forms a bond to the carbon, the leaving group’s bond is broken. This is called the transition state, and it’s indicated from the double dagger (­≠) that’s generally placed at the top right of the box it’s included in. Notice that both the nucleophile and leaving group have partial negative charges. That makes sense because the nucleophile is donating an electron while the leaving group is accepting an electron. 

Now that we know what’s happening in the reaction, let’s look at the reaction-coordinate diagram:

SN2 reaction coordinate diagramSN2 reaction coordinate diagram

In this diagram, there are really only three parts: the reagents, the transition state, and the products. The transition state is the point in the reaction with the highest energy level, and the difference in energy between the reagents and transition state is called the activation energy (often abbreviated as Ea). Remember that the rate-determining step is the step that has the highest activation energy in the reaction.


The nucleophile in SN2 reactions is generally anionic. Sometimes the negative charges aren’t shown explicitly because a cationic metal is used to stabilize the negative. A great example of this is NaCN. CN is negatively charged, and Na is positively charged. It’s an ionic bond that readily dissociates in the presence of a good electrophile or leaving group. Group 1 atoms like Na, Li, and K are a dead giveaway! 

Leaving groups

Atoms or molecules that can easily hold a negative charge are generally good leaving groups. What’s a good way to know if a molecule can be a good leaving group? Check its conjugate acid’s pKa! 

Bromide leaving groupBromide leaving group

When a leaving group dissociates from the substrate, it gains an electron. If it can hold the resulting negative charge “comfortably,” it’s a good leaving group. The bromine in the reaction above is a good leaving group because the pKa of its conjugate acid HBr is -9, which means it can hold a negative charge well.

Not all leaving groups are created equal! Even if you’ve got a bromine like in the example above, its degree greatly influences reactivity toward SN2 reactions.

Degree affects reactivity toward SN2Degree affects reactivity toward SN2

Methyl (0º) leaving groups are very reactive toward SN2 reactions, and tertiary (3º) are not reactive toward them at all. Secondary leaving groups risk competing with E2 reactions. Try to imagine a nucleophile trying to overcome the sterics of even three methyl groups, let alone three phenyl groups. It’s almost impossible for it to squeeze through; SN1 reactions are much more likely to happen in that case. 

Steric hindranceSteric hindrance

See how hard it is for the nucleophile to bypass all the R-groups to get to the carbon holding the tertiary halogen (X)? A tertiary halide with just methyl groups is the best-case scenario, and that’s already enough to stop an SN2 from occurring. 

Let’s look at this oversimplified coordinate diagram of a theoretical tertiary SN2 reaction relative to secondary, primary, and methyl reactions. 

Theoretical tertiary SN2 reaction coordinate diagramTheoretical tertiary SN2 reaction coordinate diagram

The activation energy is so ridiculously high that it’s just not going to happen. Other reactions that have lower activation energies will happen instead. 


SN2 reactions function best in polar aprotic reactions. Basically that means polar solvents that don’t have any acidic protons. Acetone, DMSO, and dimethyl chloride are commonly used polar aprotic solvents. 

Reaction rate

The reaction rate of SN2 reactions depends on the concentrations of both the nucleophile and the leaving group. That’s why it’s called a bimolecular nucleophilic substitution reaction. Here’s the rate law:

SN2 rate lawSN2 rate law

So, what happens to the rate if we double the concentration of the nucleophile? The leaving group? Both?

  • Nucleophile = rate doubles
  • Leaving group = rate doubles
  • Both = rate quadruples


Let’s go ahead and react an achiral, secondary alkyl halide with NaCN. 

Sodium-cyanide-and-isopropyl-bromideSodium cyanide and isopropyl bromide

All that happens is that the –CN attacks the secondary carbon bonded to the Br, and the formation of that bond causes the C-Br bond to break. The NaBr salt that forms is an inorganic product, and it can generally be ignored.

Now, wait a minute. See how I specified that the alkyl halide above was achiral? What happens when the leaving group is at a chiral center like in the following example?

Hydroxide and 2-iodobutaneHydroxide and 2-iodobutane

In this case, the mechanism looks the same, but there’s one key difference. The iodine was on a wedged bond, but the resulting alcohol is on a dashed bond. That’s inversion of stereochemistry! In the very first paragraph, I said that the nucleophile performs a “backside attack,” and this is the result. Remember that the iodine is facing toward us, so that means the hydroxide attacked from the back and forms a bond on dash. 

P.S. In case you’re wondering how we know if an anion will work as a nucleophile vs base, check out the Big Daddy Flowchart! It’ll also help you determine if a mechanism will be SN1, SN2, E1, or E2. 

Johnny Betancourt

Johnny got his start tutoring Organic in 2006 when he was a Teaching Assistant. He graduated in Chemistry from FIU and finished up his UF Doctor of Pharmacy last year. He now enjoys helping thousands of students crush mechanisms, while moonlighting as a clinical pharmacist on weekends.