Ranking molecules by SN2 reactivity

✋ Before we start

Make sure you’re comfortable with:

• Substitution reactions
• Characteristics of S_N2 reactions

Challenge 1: Ranking alkyl halides with the same leaving group

Question:

Rank these chlorides in order of S_N2 reactivity from slowest to fastest

Step 1: Identify the carbon bonded to the leaving group

The first thing we want to do is figure out the type of carbon bonded to the leaving group (in this case, a chlorine atom). This is the carbon where the nucleophile will attack.

S_N2 reactions are sensitive to steric hindrance. The more crowded the carbon, the harder it is for the nucleophile to attack, and the slower the reaction.

• Methyl halides are the most reactive because they have the least crowded reactive carbon.
• Tertiary halides are unreactive because their reactive carbon is the most crowded.

Here’s what we’ve got:

• Choice A: Chlorine is attached to a secondary carbon
• Choice B: Chlorine is attached to a primary carbon
• Choice C: Chlorine is attached to a vinylic carbon
• Choice D: Chlorine is attached to a primary allylic carbon

Remember, S_N2 reactivity decreases as steric hindrance increases. This means primary alkyl halides react faster than secondary alkyl halides.

Vinylic carbons do not react in S_N2 at all—the pi electrons in the double bond repel the nucleophile and prevent the attack.

Step 2: Assess the leaving group

In this challenge, all the molecules have chlorine as the leaving group. That means we only need to focus on steric hindrance for ranking.

Let’s rank the choices based on what we’ve analyzed:

• Choice C: Vinylic chloride → No S_N2 reaction.
• Choice A: Secondary chloride → Slower because it has more steric hindrance than the primary chlorides (B and D).
• Choice B: Primary chloride → Faster, less steric hindrance compared to A.
• Choice D: Primary allylic chloride → Fastest. The allylic position provides additional stabilization, making it more reactive than B.

Final ranking:

C < A < B < D

Quick Check!

Why do you think the allylic position in Choice D makes it more reactive than Choice B?

Challenge 2: Ranking alkyl halides with different leaving groups

Question:

Ranking the following alkyl halides in order of increasing S_N2 reactivity

We’ll follow the same steps as before.

Step 1: Identify the carbon bonded to the leaving group

Let’s start by figuring out the type of carbon the leaving group is attached to in each molecule. This is where the nucleophile will attack, so understanding this step is key!

Here’s what we know about each choice:

• Choice A: Chlorine is attached to a primary allylic carbon
• Choice B: Iodine is attached to a vinylic carbon
• Choice C: Bromine is attached to a primary allylic carbon
• Choice D: Chlorine is attached to a primary carbon

Key rule:
Steric hindrance is always our first consideration. When two molecules have similar steric hindrance (like Choices A and C), we look at the leaving group reactivity to decide which reacts faster.

Even though iodine is the best leaving group (we’ll explain why in Step 2), Choice B reacts the slowest because vinylic halides do not undergo S_N2 reactions.

Based on steric hindrance, here’s what we know so far:

• Choice B: Slowest → Vinylic carbon prevents S_N2.
• Choice D: Faster than B → Primary carbon, minimal steric hindrance.
• Choices A and C: Tie → Both are primary allylic carbons.

Ranking so far:

B < D < ? < ?

Now, to break the tie between Choices A and C, we need to evaluate the leaving groups. Let’s move on to Step 2!

Step 2: Assess the leaving group

For S_N2 reactions, the leaving group needs to stabilize the negative charge once it departs.

The most commonly used leaving groups are halides. When comparing leaving groups, halides follow this trend for S_N2 reactivity:

• Fluorine is the worst—it is the smallest of the halides and has a high charge density.
• Iodine is the best leaving group because it’s large and spreads out the negative charge (lower charge density).

Quick Check!

Look at the chart of leaving group reactivities. Which molecule do you think will react faster: Choice A (chlorine as the leaving group) or Choice C (bromine as the leaving group)?

Breaking the tie:

• Choice A: Chlorine as the leaving group.
• Choice C: Bromine as the leaving group → Better than chlorine.

So, Choice C reacts faster than Choice A because bromine is a better leaving group.

Final ranking:

B < D < A < C

Quick Check!
Can you explain why bromine’s size makes it a better leaving group than chlorine? Leave your thoughts in the comments section.

You did it! 🎉

Awesome work making it through this guide! You’ve learned how to confidently rank alkyl halides by their reactivity in S_N2 reactions. You also gained a deeper understanding of how steric hindrance and leaving group strength affect reaction rates.

Got questions or thoughts? Share them below, and I’ll be happy to help! Organic chemistry can be tricky, but with practice and patience, you’ll master these concepts in no time. Keep up the great work, and see you in the next challenge! 😊