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🎯 By the end of this guide,

 

You’ll learn how to confidently rank alkyl halides and nucleophiles by their reactivity in S_N2 reactions. By understanding key factors like steric hindrance, leaving group strength, and nucleophilic strength, you’ll be ready to spot trends and solve S_N2 problems with ease!

 

✋ 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.

 

 

Challenge 3: Ranking nucleophiles by relative strength

 

Question:

What is the order of rates, from slowest to fastest, for the reactions of the four nucleophiles with ethyl iodide?

 

 

Now, let’s move on to the nucleophile—the molecule that’s doing the attacking. In S_N2 reactions, a stronger nucleophile speeds up the reaction.

 

Remember: A nucleophile is an electron donor. Atoms that are willing to share their electrons make strong nucleophiles.

 

To rank these nucleophiles, we need to consider the factors that affect nucleophilic strength. We’ll break these down as we work through this question step by step.

 

Sound good? Let’s get started!

 

Step 1: Identify the atoms that can donate electrons

 

The first thing to figure out is which atoms in each molecule have lone pairs that can be donated. These are the nucleophilic atoms.

 

Note:

Lone pairs on heteroatoms like nitrogen, oxygen, and sulfur are often omitted in chemical structures. Be sure to recognize that these atoms almost always have lone pairs, and it’s important to know how many each one has in a given structure.

 

 

 

Here’s the breakdown:

 

• Choice A: Oxygen
• Choice B: Oxygen
• Choice C: Nitrogen
• Choice D: Sulfur

 

So, we’ve identified the key players. Let’s move on to how their charge affects their nucleophilic strength.

 

Step 2: Consider charge

 

Key rule: Negatively charged nucleophiles are stronger than neutral ones. Why? Because they have more electron density, which makes them more eager to share electrons and attack an electrophile.

 

Quick Check!

Which choices are negatively charged?

 

 

Answer:

• Choice A: Neutral (Oxygen)
• Choice B: Negatively charged (Oxygen)
• Choice C: Neutral (Nitrogen)
• Choice D: Negatively charged (Sulfur)

 

Since neutral nucleophiles are weaker than negatively charged ones, we know: A and C (neutral) will react slower than B and D (negatively charged).

 

But we still need to figure out:

1. Which is slower between A and C?
2. Which is faster between B and D?

 

Let’s tackle these one at a time.

 

Step 3: Compare different atoms

 

Now, let’s figure out which of the neutral nucleophiles (A and C) is the slowest, and which of the charged nucleophiles (B and D) is the fastest.

 

A vs. C: Oxygen vs. Nitrogen

 

 

Now we’re comparing A (oxygen) and C (nitrogen). Both are neutral nucleophiles, but they’re in the same row of the periodic table.

 

 

When comparing atoms in the same row:

 

Key rule: Nucleophilic strength decreases as electronegativity increases. More electronegative atoms hold onto their electrons tightly and are less willing to share them.

 

• Oxygen is more electronegative than nitrogen, so it’s less willing to donate its lone pairs.
• Therefore, A is a weaker nucleophile than C.

 

Ranking so far:
A < C < ? < ?

 

B vs. D: Oxygen vs. Sulfur

 

 

Next, let’s compare B (oxygen) and D (sulfur). Both are negatively charged, but they’re in the same column of the periodic table.

 

 

When comparing atoms in the same column:

 

Key rule: Nucleophilic strength increases as atomic size increases. Larger atoms have valence electrons that are farther from the nucleus, making them more reactive.

 

• Sulfur is larger than oxygen, so it’s a stronger nucleophile.
• Therefore, D reacts faster than B.

 

Final Ranking:
A < C < B < D

 

Looking ahead 👀

 

You might wonder: What happens when the nucleophilic atoms are the same? In that case, we consider resonance stabilization, induction, and orbitals which we’ll explore in the next challenge. 😊

 

 

Challenge 4: Ranking anions by nucleophilic strength

 

Question:

Rank the anions according to nucleophilic strength from weakest to strongest

 

 

 

 

Here’s how we’ll do it!

 

Step 1: Identify the atoms that can donate electrons

 

The first thing to figure out is which atoms in each molecule have lone pairs that can be donated. These are the nucleophilic atoms.

 

 

Here’s what we’ve got:

 

Choice A: Oxygen

Choice B: Oxygen

Choice C: Nitrogen

Choice D: Oxygen

 

Now that we’ve identified the key players, let’s see how their charge affects their nucleophilic strength.

 

Step 2: Consider charge

 

Recall that negatively charged species are stronger nucleophiles because they have extra electron density to share.

 

In this challenge, all the nucleophilic atoms are negatively charged, so we’ll need to move on to the next step to differentiate them.

 

Step 3: Compare different atoms

 

We’re comparing nitrogen (Choice C) and oxygen (Choices A, B, and D).

 

 

Since these atoms are in the same row of the periodic table, electronegativity plays a key role.

 

Quick check!

What’s the key rule for comparing atoms in the same row of the periodic table?

 

When comparing atoms in the same row of the periodic table, nucleophilic strength decreases as electronegativity increases. This is because more electronegative atoms hold onto their electrons tightly and are less willing to share them.

 

 

Nitrogen is less electronegative than oxygen, making it more willing to share its lone pairs and the strongest nucleophile.

 

Ranking So Far:
Weakest to strongest: ? < ? < ? < C

 

Now, let’s figure out how to rank the oxygen-containing anions (Choices A, B, and D).

 

Step 4: Compare the same atoms

 

To differentiate between the oxygen anions, we consider Resonance, Induction, and Orbitals in that order. These factors help determine how stable the negative charge is on the oxygen atom. The more stable the charge, the weaker the nucleophile.

 

These factors will help us determine how stable the negative charge is on the oxygen atom. The molecule with the most stable negative charge is the least nucleophilic anion.

 

Resonance

 

Is the negative charge delocalized over multiple atoms?

 

• Choices A and B: The negative charge is localized on the oxygen atoms

 

 

• Choice D: The negative charge is delocalized over two oxygen atoms, which makes it more stable and less nucleophilic.

 

 

Result: Choice D is the most stable anion and, therefore, the weakest nucleophile.

 

Ranking So Far:
Weakest to strongest: D < ? < ? < C

 

Now between A and B which is the weaker nucleophile? Let’s consider the next factor, induction, to find out.

 

Induction

 

Are there electronegative atoms nearby that stabilize the charge?

 

 

• Choice A: No nearby electronegative atoms to stabilize the charge.
• Choice B: Three fluorine atoms nearby stabilize the negative charge via induction, making it more stable (and weaker) than Choice A.

 

Result: Choice B is weaker than Choice A.

 

Final Ranking:
Weakest to strongest: D < B < A < C

 

 

You did it! 🎉

 

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