🎯 Objective

By the end of this guide, you’ll be able to rank molecules based on their S_N1 reactivity by evaluating:

• The stability of the resulting carbocation
• The ability of the leaving group

✋ Before we start

Make sure you’re comfortable with:

• Substitution reactions: Understanding how one group in a molecule can be replaced by another.
• Characteristics of S_N1 reactions: Knowing the mechanism and factors affecting the rate.

Let’s dive in!

Challenge 1: Ranking Molecules with the Same Leaving Group

Question:

Rank the following molecules in order of increasing S_N1 reaction rate (from slowest to fastest).

Step 1: Identify the carbon attached to the leaving group

Let’s start by figuring out the type of carbon the leaving group is attached to in each molecule. By identifying this, we can predict the type of carbocation that will form when the leaving group departs.

• Molecule I: Bromine is attached to a primary (1°) carbon.
• Molecule II: Bromine is attached to a secondary (2°) allylic carbon.
• Molecule III: Bromine is attached to a primary (1°) allylic carbon.
• Molecule IV: Bromine is attached to a vinylic carbon (part of a double bond).

Step 2: Evaluate carbocation stability

In an S_N1 reaction, the rate-determining step is the formation of a carbocation after the leaving group departs.

The more stable the carbocation, the faster the reaction.

Since we’ve already identified the type of carbon the leaving group is attached to, we can classify the carbocations accordingly. This step will help us better understand how carbocation stability influences the S_N1 reaction rate.

• Molecule I forms a 1° carbocation—not very stable.
• Molecule II forms a 2° allylic carbocation—stabilized by resonance and more stable.
• Molecule III forms a 1° allylic carbocation—stabilized by resonance but less stable than 2° allylic.
• Molecule IV would form a vinylic carbocation—highly unstable and unlikely to form.

Here’s the stability order for carbocations:

Quick note on vinylic carbocations

 

Vinylic carbocations have the positive charge on an sp² hybridized carbon of a double bond. They’re highly unstable because the electron deficiency is on a carbon that already has a high s-character, making it less able to stabilize a positive charge.

Step 3: Assess the leaving groups

Since all molecules have bromine as the leaving group, we can focus solely on carbocation stability.

Step 4: Rank the molecules

Since the leaving group is the same across the board, we focus on carbocation stability.

• Slowest (least reactive): Molecule IV (vinylic carbocation—very unstable)
• Next: Molecule I (primary carbocation—less stable)
• Then: Molecule III (primary allylic carbocation—resonance-stabilized)
• Fastest: Molecule II (secondary allylic carbocation—more stable due to resonance and being secondary)

Final Ranking:
IV < I < III < II

A quick note on nucleophiles 🧲

 

You might be curious about the role of nucleophiles. In S_N1 reactions, the nucleophile doesn’t affect the rate because the rate-determining step is the formation of the carbocation. Once the carbocation is formed, it reacts quickly with any available nucleophile. Therefore,

 

Rate = k [alkyl halide]

Challenge 2: Ranking Molecules with Different Leaving Groups

Question:

Which alkyl halide will undergo solvolysis in aqueous ethanol most rapidly?

Understanding solvolysis

 

Solvolysis is a type of S_N1 reaction where the solvent acts as the nucleophile. Here, aqueous ethanol is our solvent, which is polar protic, favoring S_N1 mechanisms.

Let’s dive in!

Step 1: Identify the carbon attached to the leaving group

Remember, by figuring out the type of carbon the leaving group is attached to in each molecule. By identifying this, we can predict the type of carbocation that will form when the leaving group departs.

• Molecule I: Bromine is attached to a 2° allylic carbon.
• Molecule II: Iodine is attached to a vinylic carbon.
• Molecule III: Chlorine is attached to a 2° carbon.
• Molecule IV: Bromine is attached to a 2° carbon.
• Molecule V: Chlorine is attached to a 3° allylic carbon.

Step 2: Evaluate carbocation stability

Let’s remember a key point: the more stable the carbocation, the faster the S_N1 reaction will go. This is our guiding principle as we look at each molecule.

• Molecule I forms 2° allylic carbocation, stabilized by resonance and more stable than 2° carbocation.
• Molecule II would form a vinylic carbocation, which is highly unstable and unlikely to form.
• Molecule III forms a 2° carbocation.
• Molecule IV forms a 2° carbocation.
• Molecule V forms a 3° allylic carbocation, stabilized by resonance and more stable than 2° allylic carbocation.

Here’s the stability order for carbocations:

Based on the stability of each carbocation, our ranking so far looks like this:

II < ? < ? < I < V

Now, what about Molecules III and IV?

 

Since they both form 2° carbocations, we’ll need another factor to decide which reacts faster. The answer lies in the leaving group! We’ll explore how leaving groups affect reactivity in the next step.

Step 3: Assess the leaving groups

Good leaving groups can better stabilize the negative charge after they leave. Here’s the general order of leaving group ability:

F-< Cl- < Br- < I- 

Since bromine is larger than chlorine, it can spread out (or “diffuse”) the negative charge more easily, making it a better leaving group. So, we’d expect Molecule IV (with bromine as the leaving group) to react faster than Molecule III (with chlorine).

With this information, our ranking looks like this:

II < III < IV < I < V

Step 4: Rank the molecules

We successfully combined both carbocation stability and leaving group reactivity for our final ranking:

II < III < IV < I < V

Why?

• Molecule II forms a vinylic carbocation, which is very unstable despite having a good leaving group (I⁻).
• Molecule III and Molecule IV both form 2° carbocations, but Molecule IV reacts faster since bromine is a better leaving group than chlorine.
• Molecule I forms a 2° allylic carbocation stabilized by resonance, so it reacts faster than Molecules III and IV.
• Molecule V forms a very stable 3° allylic carbocation due to resonance and hyperconjugation, making it the fastest to react.

Challenge 3: The ultimate challenge

Question:

Rank the following molecules in order of increasing reaction rate with methanol.

What’s happening here?

Methanol (CH₃OH) is a weak nucleophile and a polar protic solvent, which favors S_N1 reactions.
So, we’ll approach this as an S_N1 reaction ranking.

Step 1: Identify the carbon attached to the leaving group

Let’s start by figuring out the type of carbon each leaving group (bromine) is attached to:

• Molecule I: Bromine is bonded to a 1° allylic carbon.
• Molecule II: Bromine is bonded to a 1° carbon.
• Molecule III: Bromine is bonded to a 2° allylic carbon.
• Molecule IV: Bromine is bonded to a 2° allylic carbon.
• Molecule V: Bromine is bonded to a vinylic carbon.

Step 2: Evaluate carbocation stability

Next, let’s look at the carbocation each molecule would form in an S_N1 reaction:

• Molecule I forms 1° allylic carbocation, stabilized by resonance and more stable than 1° carbocation.
• Molecule II forms a 1° carbocation.
• Molecule III forms a 2° allylic carbocation.
• Molecule IV forms a 2° allylic carbocation.
• Molecule V would form a vinylic carbocation, which is highly unstable and unlikely to form.

Here’s the stability order for carbocations:

Based on the stability of each carbocation, our ranking so far looks like this:

V < II < I < ? < ?

Now, what about Molecules III and IV?

 

Both Molecules III and IV have bromine attached to 2° allylic carbons, and since they both have the same leaving group (Br), they’re similar. But to figure out which is more reactive, let’s consider the resonance forms of the carbocations they form.

Molecule III

As mentioned earlier, Molecule III forms a 2° allylic carbocation when the leaving group departs.

In this carbocation, the positive charge is spread out (delocalized) across both a 2° and a 3° carbon. This setup benefits from hyperconjugation provided by the tertiary (3°) carbon, which adds extra stability to the carbocation.

Molecule IV

Molecule IV also forms a 2° allylic carbocation when the leaving group departs.

In this carbocation, the positive charge is delocalized over two 2° carbons. While still stabilized by resonance, it doesn’t benefit from the additional stability of a tertiary position.

Therefore, Molecule III will react faster than Molecule IV due to the additional stability provided by the tertiary carbon.

Our revised ranking is now:

V < II < I < III < IV

Step 3: Assess the leaving groups

All molecules have bromine as the leaving group, so we don’t need to compare them here.

Step 4: Rank the molecules

Since all the molecules share the same leaving group, we can finalize the ranking based on carbocation stability alone:

V < II < I < III < IV

Great job! By understanding carbocation stability and the effects of resonance, you’ve successfully ranked the molecules by their expected S_N1 reactivity.

Keep practicing, and these concepts will soon become second nature!