Objective

By the end of this guide, you should be able to rank molecules in order of their expected S_N1 reactivity by evaluating the stability of the resulting carbocation and the ability of the leaving group.

Let’s begin with a simple challenge.

Challenge 1: Ranking molecules with the same leaving group

Question:

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

Here’s how we’ll do it.

Step 1: Identify the carbon that is bonded to the leaving group.

First, let's find the carbon atom that's bonded to the leaving group. Is it a methyl, primary (1°), secondary (2°), tertiary (3°), allylic, benzylic, or vinylic carbon?

Molecule I

The leaving group, bromine, is bonded to a 1° carbon.

Molecule II

The leaving group, bromine, is bonded to a 2° allylic carbon.

Molecule III

The leaving group, bromine, is bonded to a 1° allylic carbon.

Molecule IV

The leaving group, bromine, is bonded to a vinylic carbon.

This step is essential because it helps us determine the type of carbocation that will form after the leaving group departs.

In an S_N1 reaction, the rate-determining step is the spontaneous loss of the leaving group, creating a carbocation.

The stability of this carbocation is key: the more stable it is, the faster the reaction will proceed.

Now, let’s draw the carbocation for each molecule to analyze their stability.

Molecule I

Forms a 1° carbocation.

Molecule II

Forms a 2° allylic carbocation.

Molecule III

Forms a 1° allylic carbocation.

Molecule IV

In molecule IV, the leaving group is attached to a vinylic carbon. This means that the carbon bonded to the leaving group is part of a double bond.

Now, here’s an important point to remember: vinylic carbocations are highly unstable. Because of this instability, it is very unlikely for a vinylic carbocation to form during an S_N1 reaction.

So, if you encounter a molecule with a leaving group on a vinylic carbon, you can confidently say that the S_N1 reaction will be extremely slow or might not happen at all.

Step 2: Assess the leaving group

The best leaving groups are those that can stabilize the negative charge after departure. Generally, they are the conjugate bases of strong acids.

Recall that in the S_N1 mechanism, the rate-determining step is the loss of the leaving group to form a carbocation. Hence, a good leaving group means a faster reaction.

All molecules in this example here have the same leaving group, bromine. This is not a determining factor.

Step 3: Consider the solvent

Polar protic solvents such as water and alcohols stabilize the transition state and the carbocation intermediate, thus favoring S_N1 reactions.

However, since no solvent information is provided in this question, we won't consider this factor for now.

Step 4: Combine carbocation stability and the leaving group's ability to determine the expected S_N1 reactivity

Now, let's combine the knowledge of carbocation stability and the leaving group's ability.

Since all the molecules have the same leaving group, carbocation stability is the determining factor here.

Here's the stability order for carbocations:

Using this order, we can rank the molecules by their increasing relative rate for the S_N1 reaction (from slowest to fastest reacting).

Molecule IV - Bromine on a vinylic carbon

Vinylic carbocations are highly unstable because the positive charge is on a carbon that is part of a double bond. This makes it very difficult for the carbocation to form, resulting in the slowest reaction rate.

Molecule I - Bromine on a 1° carbon

Primary carbocations are more stable than vinylic carbocations but are still quite unstable. This makes Molecule I react slowly, but faster than Molecule IV.

Molecule III - Bromine on a 1° allylic carbon

Allylic carbocations are stabilized by resonance, which means the positive charge can be delocalized over multiple atoms. Even though this is a primary carbocation, the allylic position provides extra stability, making Molecule III more reactive than Molecule I.

Molecule II - Bromine on a 2° allylic carbocation

Secondary carbocations are more stable than primary carbocations due to additional alkyl group stabilization. Additionally, being allylic means this carbocation benefits from resonance stabilization. This makes Molecule II the most reactive among the given molecules.

Final ranking from slowest to fastest S_N1 reactivity:

IV < I < III < II

Challenge 2: Ranking molecules with different leaving groups

Alright, the first challenge was just the warm-up! Now, let’s dive into a more complex question, similar to what you might see on an actual organic chemistry exam.

Ready? Let’s go!

Sometimes, you might encounter a question that doesn't specify the type of reaction you're dealing with.

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

Step 1: Identify the carbon that is bonded to the leaving group.

First, let's find the carbon atom that's bonded to the leaving group. Is it a methyl, primary (1°), secondary (2°), tertiary (3°), allylic, or vinylic carbon?

Molecule I

The leaving group, bromine, is bonded to a 2° allylic carbon.

Molecule II

The leaving group, iodine, is bonded to a vinylic carbon.

Molecule III

The leaving group, chlorine, is bonded to a 2° carbon.

Molecule IV

The leaving group, bromine, is bonded to a 2° carbon.

Molecule V

The leaving group, chlorine, is bonded to a 3° allylic carbon.

Step 2: Assess the leaving group

Now, let’s evaluate the leaving groups. Remember, the best leaving groups are the most stable anions, typically the conjugate bases of strong acids.

Let's look at the leaving groups in our molecules:

• Molecule I - Bromine on a 2° allylic carbon

• Molecule II - Iodine on a vinylic carbon

• Molecule III - Chlorine on a 2° carbon

• Molecule IV - Bromine on a 2° carbon

• Molecule V - Chlorine on a 3° allylic carbon

When comparing Molecules III and IV, we see that both have their leaving groups bonded to 2° carbons. However, bromine is a better leaving group than chlorine, so Molecule IV will react faster than Molecule III.

Step 3: Consider the solvent

In this question, the solvent provided is aqueous ethanol.

Since we are comparing the relative rates of each molecule in the same solvent, the solvent is not a determining factor for this specific comparison.

By focusing on the carbocation stability and the ability of the leaving group, we can accurately rank the reactivity of the molecules without worrying about the solvent in this case. This keeps things simple and allows us to concentrate on the key factors that affect the reaction rate.

Keep this in mind as you practice, and you'll become more confident in predicting reaction rates!

Step 4: Combine carbocation stability and the leaving group's ability to determine the expected S_N1 reactivity

Now, let's combine what we know about carbocation stability and the ability of the leaving group.

Here's the stability order for carbocations:

Based on the stability order for carbocations and the ability of the leaving group, we can rank these molecules from slowest to fastest S_N1 reactivity as follows:

Molecule II - Iodine on a vinylic carbon

Vinylic carbocations are highly unstable because the positive charge is on a carbon that is part of a double bond. This makes it very difficult for the carbocation to form, resulting in the slowest reaction rate.

Molecule III - Chlorine on a 2° carbon

Secondary carbocations are more stable than primary but less stable than tertiary. Chlorine is a decent leaving group but not as good as bromine. Therefore, this molecule will have a slower reaction rate compared to those with bromine as the leaving group.

Molecule IV - Bromine on a 2° carbon

Like Molecule III, this forms a secondary carbocation. However, bromine is a better leaving group than chlorine, making the reaction faster than Molecule III.

Molecule I - Bromine on a 2° allylic carbon

Allylic carbocations are stabilized by resonance, making them more stable than regular secondary carbocations. With bromine as a good leaving group, this molecule will react faster than Molecules III and IV.

Molecule V - Chlorine on a 3° allylic carbon

Tertiary allylic carbocations are highly stable due to both hyperconjugation and resonance effects. Even though chlorine is not as good a leaving group as bromine, the stability of the tertiary allylic carbocation makes this the fastest reacting molecule.

Final ranking from slowest to fastest S_N1 reactivity:

II < III < IV < I < V

Great job! By understanding carbocation stability and the abilities of different leaving groups, you've successfully ranked the molecules by their expected S_N1 reactivity.

Keep practicing, and these concepts will soon become second nature! Ready for the next challenge? Let's go!

Challenge 3

Now, let’s tackle another challenging question.

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

Step 1: Identify the carbon that is bonded to the leaving group.

First, let's find the carbon atom that's bonded to the leaving group. Is it a methyl, primary (1°), secondary (2°), tertiary (3°), allylic, or vinylic carbon?

Molecule I

The leaving group, bromine, is bonded to a 1° allylic carbon.

Molecule II

The leaving group, bromine, is bonded to a 1° carbon.

Molecule III

The leaving group, bromine, is bonded to a 2° allylic carbon.

Molecule IV

The leaving group, bromine, is bonded to a 2° allylic carbon.

Molecule V

The leaving group, bromine is bonded to a vinylic carbon.

Step 2: Assess the leaving group

All molecules in this example here have the same leaving group, bromine. This is not a determining factor.

Step 3: Consider the solvent

In this question, the solvent provided is methanol.

Since we are comparing the relative rates of each molecule in the same solvent, the solvent is not a determining factor for this specific comparison.

Step 4: Combine carbocation stability and the leaving group's ability to determine the expected S_N1 reactivity

Since all molecules have the leaving group bromine, the deciding factor here is the stability of the carbocation formed.

Here's the stability order for carbocations:

Based on the stability order for carbocations, we can rank these molecules from slowest to fastest S_N1 reactivity as follows:

V (vinylic) < II (1°) < I (1° allylic) < ? < ?

Now, what about Molecules III and IV?

Both Molecules III and IV have leaving groups bonded to 2° allylic carbons. To determine which one reacts the fastest, we need to consider the resonance forms of the carbocations they form.

Molecule III

• Molecule III forms a 2° allylic carbocation.

• The positive charge is delocalized over a 2° and a 3° carbon. This carbocation benefits from hyperconjugation provided by the tertiary carbon.

Molecule IV

• Molecule IV also forms a 2° allylic carbocation.

• In Molecule IV’s 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.

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!