Predicting the products of SN1 and SN2 reactions

S_N1 vs S_N2 Reactions

Alkyl halides commonly undergo nucleophilic substitution reactions, where a nucleophile replaces the halogen.

But not all substitution reactions happen the same way—there are two main pathways: S_N1 and S_N2. Understanding these mechanisms will help you predict reaction outcomes quickly and accurately.

S_N2 – Substitution Nucleophilic Bimolecular

The S_N2 reaction happens in a single step—the nucleophile attacks causing the leaving group to exit.

Since both events occur at the same time, the reaction follows a concerted mechanism and its rate depends on both the nucleophile and the alkyl halide:

Rate = k [Nucleophile] [Alkyl halide]

Stereochemistry of S_N2 reactions

Because the nucleophile approaches from the opposite side of the leaving group (a backside attack), the reaction causes an inversion of configuration at the chiral center.

This means that if the starting material was R, the product will be S, and vice versa.

Since the configuration of the product directly depends on the structure of the starting material, S_N2 reactions are stereospecific.

S_N1 – Substitution Nucleophilic Unimolecular

Unlike S_N2, the S_N1 reaction occurs in two main steps.

The leaving group exits first, forming a carbocation intermediate.

The nucleophile attacks the carbocation in a separate step.

The oxonium ion can then lose its proton to water to give the observed substitution product (since it’s both the nucleophile and solvent in this reaction).

Because the rate of the reaction only depends on the first step (formation of the carbocation), the rate law is:

Rate = k [Alkyl halide]

Stereochemistry of S_N1 reactions

Since the carbocation intermediate is planar (flat), the nucleophile can attack from either side. This leads to both inversion and retention of configuration.

Carbocation rearrangements

A key feature that sets S_N1 reactions apart from S_N2 is the formation of a carbocation intermediate. These carbocations can rearrange to become more stable through:

• Hydride shifts
• Methyl shifts
• Ring expansions

Recognizing when a rearrangement will occur is a key skill, which we cover in another guide.

Let’s apply what we’ve learned!

Now that we understand S_N1 and S_N2 mechanisms and how they affect stereochemistry, let’s put that knowledge into practice by predicting the products of a given reaction.

Challenge

Question:

Predict the product(s) of the following reactions and identify whether the mechanism follows S_N1 or S_N2.

Before we jump in, remember that the mechanism depends on the:

• The structure of the alkyl halide
• The nature of the nucleophile/base.

If you need a refresher, we’ve put together a detailed guide on how to predict whether a reaction follows S_N1, S_N2, E1, or E2. Check it out here!

For quick reference, here’s a table summarizing how to determine the reaction mechanism:

Let’s apply the table to predict the reaction mechanisms!

Step 1: Classify the alkyl halide (substrate)

First, we identify the type of alkyl halide present. In this case, the leaving group (Br) is attached to a secondary carbon, meaning our substrate is secondary.

Step 2: Determine the nature of the nucleophile/base

Pathway 1: The reagent CH₃OH is a weak nucleophile and a weak base, which favors the S_N1/E1 pathway.

Pathway 2: The reagent CH₃O^– is both a strong nucleophile and a strong base. This means the reaction follows an S_N2/E2 pathway, with E2 being the major product and S_N2 the minor product.

Since we’re only focusing on substitution products (S_N1 and S_N2) in this guide, we won’t discuss elimination (E1/E2) here.

Now that we’ve determined the reaction pathways, let’s move on to predicting the S_N2 product!

Predicting the products of an S_N2 reaction

Step 1: Identify a good leaving group

The first step in predicting an S_N2 reaction is checking if the leaving group is good. If the leaving group is poor, the reaction won’t proceed.

In this case, Br (bromine) is a good leaving group, meaning we can expect this S_N2 reaction to occur.

However, for nucleophilic substitution to happen, the leaving group must be attached to an sp³ hybridized carbon.

Vinylic and aryl halides (where the leaving group is on an sp² hybridized carbon) do not undergo S_N2 reactions.

Step 2: Replace the leaving group with the nucleophile and invert the configuration

Now that we’ve confirmed the leaving group is good, let’s predict the product.

We previously identified CH₃O^– as the nucleophile in this reaction. CH₃O^– attacks the electrophilic carbon, forcing the Br to leave.

Net result: Br will be replaced with OCH₃.

Because this is an S_N2 reaction, the nucleophile attacks from the opposite side of the leaving group (backside attack). If the electrophilic carbon is chiral, this results in inversion of configuration at that center.

And that’s it! 🎉 You’ve successfully predicted the product of this SN2 reaction.

Predicting the products of an S_N1 reaction

Step 1: Identify a good leaving group

Just like in S_N2 reactions, S_N1 reactions require a good leaving group—if the leaving group is poor, the reaction won’t proceed.

In this case, Br (bromine) is a good leaving group, so we can expect the S_N1 reaction to proceed.

Again, the leaving group must be attached to an sp³ hybridized carbon.

Vinylic and aryl halides (where the leaving group is on an sp² hybridized carbon) do not undergo S_N1 reactions.

Step 2: Draw the carbocation intermediate

A key feature of S_N1 reactions is the formation of a carbocation intermediate after the leaving group departs.

Drawing the carbocation intermediate is crucial—you’ll see why in the next step!

Step 3: Watch out for rearrangements and resonance

Carbocations are unstable and will rearrange if a more stable carbocation can form. The three common types of carbocation rearrangements are:

• Hydride shifts
• Methyl shifts
• Ring expansions

A skill you must develop is recognizing when a carbocation rearrangement will occur. We cover this in more depth in another guide, but here’s the key takeaway:

 

If a rearrangement can lead to a more stable carbocation (e.g., secondary → tertiary), it will occur.

In this case, the carbocation can rearrange via a hydride shift to form a tertiary carbocation.

In the next challenge, we’ll look at how resonance affects where the nucleophile attacks.

Step 4: Draw products with retention and inversion of stereochemistry

Now that we have our carbocation intermediates, it’s time to predict the final products.

Because the carbocation is planar (flat), the nucleophile can attack from either side. This results in both inversion (>50%) and retention (<50%) of configuration.

Net result: Br will be replaced with OCH_3.

Because there are two possible carbocation intermediates—one secondary and one tertiary—we expect a mixture of products.

The major products come from the more stable (tertiary) carbocation.

The minor products come from the less stable (secondary) carbocation.

And there we have it! We’ve successfully predicted the products of both S_N1 and S_N2 reactions. 🎉