Good Leaving Groups - Online Tutor, Practice Problems & Exam Prep

Substitution reactions rely on leaving groups, which facilitate bond breaking to allow new bonds to form. A good leaving group is typically a stable conjugate base, influenced by factors like electronegativity and atomic size. The best leaving groups are found further down and to the right on the periodic table, with iodine being a prime example. Understanding the element effect, along with inductive and resonance effects, is crucial for predicting leaving group stability and reactivity in organic synthesis.

How to use the factors affecting acidity to predict leaving group ability.

As I previously mentioned, substitution reactions would not be possible without leaving groups. Why? Because remember that the electrophiles do not have an empty orbital. So what that means is, in order to make a bond, I'm always going to have to break a bond. And I can't break a bond unless something's going to leave. Okay? So what that means is that it's going to be important in this chapter that we know exactly how to recognize when something is a good leaving group and when something is a bad leaving group. Okay? So let's go ahead and get started and learn more about leaving groups.

As I said, leaving groups are what break a bond to make it reactive. And in general, we can use the same rules that we learned from acids and bases to predict when something's going to make a good leaving group. We just use the rules that would make something a good conjugate base. Because what a conjugate base is, it's something that's stable after it gains electrons. Okay? And that's exactly what a leaving group is as well. After accepting an electron pair, is it going to be stable or not?

So, how do we determine if a leaving group is good or bad? We use the same exact factors that affect acidity that we used for the acid-base chapter. So remember that we had a few factors that affected how stable conjugates were after they left. The one that we're going to use most often is actually just the element effect. Because it turns out there aren't that many good leaving groups and we can usually use the element effect to describe all of them. Alright?

So I know it's been a long time, but what were the trends that we used for the element effect? We used electronegativity. We said, as something goes to the right towards fluorine, it's going to be more electronegative, so it's going to like to have a negative charge more. It will be more stable. But also we use the size trend. We said that as a conjugate gets bigger, the easier it is for it to accept more electrons because it doesn't care. I use the word it's a very squishy atom. It's got tons of electrons everywhere. So if it gains two extra electrons, it's not really going to care. Ideally, the best leaving group would be the one that's furthest to the right and furthest down. So, that would be around down here around iodine. Now, you can't get much lower than iodine because then you start getting into these weird transition metals that are radioactive and probably give you cancer. But iodine is a good place to end in terms of the trends.

This doesn't explain all leaving groups because remember we have other types of factors affecting acidity as well. Remember that there was the inductive effect, there was the resonance effect; all that stuff still applies. But, in general, the element effect does a really good job explaining the really common leaving groups. So what I want you guys to do is go through this exercise. I've given you pairs of molecules. All you have to do is compare the two different pairs and see which one would have the better leaving group. Go ahead and just try to use the factors affecting acidity to figure that out and then I'll explain each one to you. So go for it.

Predict the best leaving group

Alright. So if you were paying attention, this first one was pretty easy. We know that basically the leaving group is probably going to be the oxygen or the nitrogen. Why? Because I do have CH₃s present. But just letting you know, carbon is never going to be a good leaving group. Why? Because think about it, carbon on the periodic table is pretty far to the left. So what that means is it's going to be a terrible leaving group. It does not like to have a negative charge. So never think of carbon or hydrogen. Hydrogen is even worse. Hydrogen is all the way over there. They do not like to have electrons. So, please never make those a leaving group. But what I do have is I have an oxygen and I have a nitrogen. What effect could I use to figure out which one's the better leaving group? The element effect. I can just say, okay, which one's more electronegative? Which one's bigger? They're both actually roughly the same size because they're on the same row, but one is more electronegative and that's O. So what that means is that the right answer for a is that the O is going to be a better leaving group than the NH₂. Alright? So that's the end of that question. Go ahead and try to use that same logic for b. Go for it.

Predict the best leaving group

Alright. So b was actually even easier than a. Why? Because for b, I don't have any good leaving groups. All I have is CH_3 on both sides. So this would once again be terrible if it picked up electrons. It would have to be a c^-. That just doesn't really happen. Okay? Then, whereas the other compound has a chlorine. Okay? And obviously, if chlorine picks up electrons, that's going to be a lot more stable. Why? Because of the element effect. Once again, chlorine let me erase this little chlorine is here and then carbon is here. So chlorine has an advantage both with its much more electronegative, and it's also bigger. So it's just all around a much, much better leaving group than carbon could ever dream of being. Alright? Magnitudes of order are better. Alright? So let's go on and see.

Predict the best leaving group

I'm making these too easy for you. Alright? So for this one, once again, we could use the element effect, and hopefully you picked iodine. I mean, I gave you that one right at the beginning. I told you that iodine is pretty much one of the best leaving groups I could ever have. They both in terms of electronegativity, actually fluorine is going to be more electronegative than iodine. Okay? But the size effect is going to take over, and iodine is just going to be so much bigger than fluorine that it's going to be a better leaving group than the fluorine will because the iodine will tolerate those electrons better. Okay? Cool. So that's it for c. Last one d, and then we'll be done with this page.

Predict the best leaving group

So I had to throw one tricky one in there. I hope this one didn't freak you out too much. Basically, this one was a challenge because you may have noticed you actually could not use the element effects for this one. Because the element that's leaving is the same for both of them. They're both oxygen. Okay? So what effect could I use? And it turns out that this is going to be one example. So sorry if you didn't think about that, but let's say that the O leaves, let's say that this weird thing leaves. Okay? The O would just look like this: O^-. Trying to avoid my head. Perfect. Okay? This other one would look like O^-, but that's attached to S=O=O-CH₃. Okay? Does any of these have an advantage? Well, the O^- is what we call localized. Localized is a really bad thing. That means that the charge can only be in one spot. It would just be right here. That's not very stable. Whereas, this negative charge could resonate on this other compound. What I could do is I could take these electrons and move them here, so I would get a negative charge at the top and then I could also get a negative charge at the bottom. So this one is what we called remember the word was delocalized. When you can resonate, that's delocalized. That means that the electron, the negative charge, can spread everywhere. So it turns out that because of the crazy amount of resonance that this molecule can have, this is going to be a much better leaving group than O^-. In fact, this is actually going to be one of our good leaving groups. Later on, when we talk about leaving groups in more depth, this is a group called a sulfonate ester. A sulfonate ester, Like I said, I'm going to go more in detail later. But just so you know, even though O^- is kind of not that great of a leaving group, because it has so much resonance, it becomes an amazing leaving group. Are you guys cool with that? So I'm just trying to show you how really all the factors affecting acidity are fair game, but this is one where you use resonance to figure that out. Alright. So now, let's simplify things. I've given you the general rule, but let's simplify things because you have a lot of mechanisms to learn. Okay? And a lot of stuff that you have to memorize. So I'm just going to simplify everything we just talked about and summarize it in one line. Are you guys cool with that? This is the line. K? Due to their high electronegativity, alkyl halides will be the primary leaving group for this chapter. Alright? Didn't that simplify things? So what I'm trying to say here is I'm really trying to take your entire thought process out of it. What I'm saying is this, now you know how to find out if something is a good leaving group or a bad leaving group. If it's better or worse. But I'm just going to simplify things and tell you this. Alkyl halides 90% of the time are going to be your leaving group. So if you see an alkyl halide, just imagine that is my leaving group. If you see something else, ignore it for now because we're going to be dealing with alkyl halides basically for this entire chapter. At the very end, then I'll go ahead and add a few other leaving groups. But for right now, let's just focus on alkyl halides because that's going to simplify things a lot. Cool? Alright. So let's move on.

Here’s what students ask on this topic:

What makes a good leaving group in organic chemistry?

A good leaving group in organic chemistry is typically a stable conjugate base. This stability is influenced by factors such as electronegativity and atomic size. A more electronegative atom can better stabilize the negative charge after leaving, making it a good leaving group. Additionally, larger atoms can distribute the negative charge over a larger volume, further stabilizing it. The best leaving groups are often found further down and to the right on the periodic table, with iodine (I^-) being a prime example. Other factors like inductive and resonance effects can also play a role in determining the stability of the leaving group.

Why is iodine considered a good leaving group?

Iodine (I^-) is considered a good leaving group because it is large and highly polarizable, which allows it to stabilize the negative charge effectively after leaving. Its position on the periodic table—further down and to the right—means it has a lower charge density and can distribute the negative charge over a larger volume. This makes it more stable as a free ion. Additionally, iodine's high electronegativity helps it to stabilize the negative charge, making it an excellent leaving group in substitution reactions.

How does electronegativity affect the stability of a leaving group?

Electronegativity affects the stability of a leaving group by determining how well the atom can stabilize the negative charge after it leaves. More electronegative atoms are better at stabilizing negative charges because they have a stronger tendency to attract electrons. This makes them more stable as free ions. For example, fluorine (F^-) is highly electronegative and can stabilize the negative charge well, although its small size makes it less ideal compared to larger halides like iodine (I^-).

What is the element effect in determining good leaving groups?

The element effect refers to the influence of an atom's position on the periodic table on its ability to act as a good leaving group. Atoms that are further to the right and down the periodic table tend to be better leaving groups. This is because they are more electronegative and larger, which allows them to stabilize the negative charge more effectively after leaving. For example, iodine (I^-) is a better leaving group than chlorine (Cl^-) because it is larger and can better distribute the negative charge.

How do inductive and resonance effects influence leaving group ability?

Inductive and resonance effects influence leaving group ability by stabilizing the negative charge on the leaving group. Inductive effects involve the electron-withdrawing or electron-donating properties of atoms or groups attached to the leaving group. Electron-withdrawing groups can stabilize the negative charge through the inductive effect, making the leaving group more stable. Resonance effects involve the delocalization of electrons across multiple atoms, which can also stabilize the negative charge. For example, a leaving group that can participate in resonance will be more stable and thus a better leaving group.