SN1 and SN2 Practice Problems with Detailed Answers

SN1 vs SN2 Practice Problems That Actually Teach You Something

Most textbooks throw 50 problems at you and hope you figure it out. That's garbage. This guide walks through real mechanism problems with answers that explain why the reaction goes one way or the other.

If you don't know the basic difference between SN1 and SN2 yet, stop here. Come back when you've memorized the mechanism types. This is for practice and reinforcement.

Quick Reference: SN1 vs SN2

Here's the comparison table most students need but textbooks bury in chapter 7.

Problem 1: Predicting the Mechanism

Question: 2-bromo-2-methylpropane reacts with methanol. What mechanism? Why?

Answer: SN1. Full stop.

Here's the reasoning:

• The substrate is tertiary — the carbon bearing the leaving group has three methyl groups attached
• Steric hindrance makes backside attack (SN2) impossible
• Methanol is a weak nucleophile anyway
• The reaction forms a stable tertiary carbocation

The mechanism goes: leaving group leaves → carbocation forms → nucleophile attacks from either face → racemic product.

Problem 2: Secondary Substrate with a Strong Nucleophile

Question: 2-bromopropane reacts with hydroxide ion (OH⁻) in methanol. SN1 or SN2?

Answer: SN2. Usually.

Secondary substrates are the gray zone. You have to look at the other factors:

• Hydroxide is a strong nucleophile — that pushes toward SN2
• Methanol is polar protic — that pushes toward SN1
• The substrate is secondary — neutral territory

When you see a strong nucleophile like OH⁻, CN⁻, or RS⁻ attacking a secondary halide, SN2 wins most of the time. The strong nucleophile overwhelms the substrate's mediocre SN1 potential.

Problem 3: Drawing the SN2 Mechanism

Question: Draw the product when (S)-2-bromobutane reacts with iodide ion (I⁻) in acetone.

Answer: The product is (R)-2-iodobutane. Here's why the mechanism matters:

• Iodide attacks from the backside — opposite the C-Br bond
• The leaving group (Br⁻) leaves as the bond forms
• Stereochemical inversion occurs at the reaction center
• (S) starting material gives (R) product

This is the classic "umbrella flip" — the three substituents invert like an umbrella in the wind.

Problem 4: Identifying When Rearrangements Happen

Question: 2-bromo-2-methylbutane is treated with water. Predict the major product and explain any rearrangements.

Answer: SN1 with rearrangement.

The carbocation that forms is secondary — but it can rearrange to a more stable tertiary carbocation via methyl shift.

After rearrangement, water attacks the tertiary carbocation. The major product is 2-methyl-2-butanol, not the unrearranged 2-methyl-1-butanol you'd get if no rearrangement occurred.

SN2 reactions never show rearrangements. If your mechanism requires a shift, it's SN1.

Problem 5: Solvent Effects in Action

Question: Why does DMF (dimethylformamide) favor SN2 reactions?

Answer: DMF is polar aprotic.

Polar protic solvents (water, alcohols) stabilize the nucleophile through hydrogen bonding. This makes the nucleophile less reactive. Polar aprotic solvents don't do this — the nucleophile stays "hot" and attacks aggressively.

Common polar aprotic solvents:

• DMF
• DMSO
• Acetone
• Acetonitrile

Common polar protic solvents:

• Water
• Methanol
• Ethanol
• Acetic acid

Problem 6: Predicting Major Products

Question: Predict the major product when 1-bromo-1-methylcyclohexane reacts with ethanol.

Answer: 1-ethoxy-1-methylcyclohexane via SN1.

The substrate is tertiary (on the ring, with a methyl and two ring carbons attached). Ethanol is a weak nucleophile. Polar protic conditions. SN1 dominates completely.

The product forms as a racemic mixture at the reaction center because the carbocation is planar — the nucleophile attacks from either face with equal probability.

Getting Started: How to Solve Any SN1/SN2 Problem

Follow this checklist every time:

Step 1: Identify the substrate

Count the carbons attached to the carbon with the leaving group. Methyl = 0, primary = 1, secondary = 2, tertiary = 3. This is your starting point.

Step 2: Evaluate the nucleophile

Strong nucleophiles (OH⁻, CN⁻, RS⁻, halides) favor SN2. Weak nucleophiles (H₂O, ROH, NH₃) don't have enough oomph for SN2 unless the substrate is really reactive.

Step 3: Check the solvent

Polar aprotic = SN2. Polar protic = either, but pushes toward SN1 when combined with weak nucleophiles.

Step 4: Apply the rules

• Tertiary substrate + weak nucleophile = SN1
• Primary/methyl substrate + strong nucleophile = SN2
• Secondary substrate = it depends. Look at nucleophile strength first, then solvent.

Step 5: Draw the mechanism if asked

For SN2: show the single step with the nucleophile approaching from the back, leaving group departing simultaneously.

For SN1: show the carbocation intermediate clearly, then the nucleophile attacking from either side.

Common Mistakes to Avoid

• Forgetting stereochemistry — SN2 inverts, SN1 racemizes
• Ignoring solvent — same substrate can react differently in different solvents
• Assuming secondary always means SN2 — weak nucleophile changes everything
• Missing rearrangement possibilities — carbocations can shift before the nucleophile attacks

Work through 20-30 problems using this checklist and you'll have it cold. There's no secret — just pattern recognition built through repetition.