SN2 Reaction - Online Tutor, Practice Problems & Exam Prep

The SN2 mechanism is a crucial organic chemistry reaction where a negatively charged nucleophile attacks an electrophilic carbon bonded to a leaving group, resulting in a one-step substitution. This backside attack minimizes steric hindrance, favoring less substituted alkyl halides. The reaction features a transition state with partial bonds, leading to inversion of configuration at chiral centers. The rate is bimolecular, depending on both nucleophile and electrophile concentrations, illustrating the importance of understanding nucleophilicity and leaving group ability in reaction dynamics.

Drawing the SN2 Mechanism

Alright, guys. So now we're going to jump into one of the most important mechanisms in all of organic chemistry, and it's a mechanism that you're never allowed to forget. So, what I teach you in the next 20 minutes, like, it's going to stick because you're going to need it for your Orgo, for your graduate exams, and even in graduate school. If you're planning on going to graduate school for anything pre-health, you're still going to need to know this reaction and that's called the SN₂ mechanism. So let's dive right into it. All right. So if I were to just give it a tagline and just say in one sentence what an SN₂ mechanism is, okay? What it is is that a negatively charged nucleophile reacts with an accessible leaving group. Now, leaving group, you should know what it is. Accessible, maybe you're a little confused, but we'll define it. To produce substitution, oh, you know what that is, in one step. Alright? So let's go ahead. I just want to get right into it. Let's just draw this mechanism out.

So I have this nucleophile that I'm just generally putting as NU^-. There are a lot of different nucleophiles out there. It doesn't really matter the identity right now. Now I have to figure out I'm reacting with an alkyl halide. What did I say alkyl halides were good at? Leaving. So what that means is I have to figure out what's the electrophilic part of this molecule because this is my electrophile. And what is it going to look like after it reacts? Okay. So, how do I find out which part is electrophilic? Does it have a positive charge already on it? No. So, I'm going to have to draw the dipole. What does the dipole look like? Well, remember that halogens pretty much always pull away from whatever they're attached to, so I would have my only major dipole pulling away from the carbon. So what I would have is a negative here, a partial positive there. Where is my nucleophile going to want to attack? It's going to want to attack the carbon. Okay? So the electrophilic part is not the X. It's the carbon. Okay? So I know I'm going to start off my arrow from my nucleophile, and I know I'm going to attack that carbon. But now, actually, we have a choice. Okay? Because what we have is a distinct set of sides.

So let's think about it this way. This is my carbon. And in the past, I haven't been very picky about exactly how I draw my arrows because I haven't been very picky. Okay? But if you think about it, there are actually 2 different sides to this carbon. Let's say that the X side, the one with the halogen, is called the front side. Okay? So the X has 3 lone pairs: 1, 2, 3. Okay? So that would be what I would consider the front side. So I'm just going to write here, that's the front. Okay? And the backside would be everything that's on the other side over here. And the backside, what it's going to have is just like a hydrogen and then some alkyl group. So this would be a methyl group and an ethyl group. Okay? Which of these two sides, front or back, do you think is going to be the easiest for my nucleophile to approach? Okay? And let's think about it this way. We know that it wants to hit carbon. So no matter what, it's going for the carbon. But what I'm asking is, is it going to try to go from the front side or the back side? It turns out that the front side is a really bad option. Why? Because the nucleophile, remember, it already has extra electrons. It's got extra electrons it's trying to get rid of. In order to go through the front side, it would need to pass through a bunch of electron clouds from the halogen. Do you think that's going to be very easy to do? It's actually going to be almost impossible. It doesn't happen. Those electrons are going to repel each other like crazy. So front side attack is actually impossible. It's never going to happen.

So what that means is that this is going to lead us to one of the most inappropriate phrases in all of science, and that is a backside attack. Alright? So as messed up as that sounds, alkyl halides are totally down with it. Backside attack is something that they're all about. And we're going to be doing this every day. Alright? For the rest of organic chemistry. So I hope you guys are cool with that. You've got to get used to it pretty quick. So, backside attack is the way to go because it's the way that's basically less hindered. It's going to be a lot easier for those electrons to pass through the backside where there are not as many electrons as the front. So now what we need to do is we have to draw the transition state of what this is going to look like.

Because let me just ask you this. Am I done with this mechanism? Do I need to draw any more arrows, or am I done? No. We should draw some more arrows. Why? Because remember that this is going to be a nucleophile and electrophile that does not have an empty orbital. Notice that there's no empty orbital here. This carbon already has 4 bonds. Okay? So this carbon already has 4 bonds. If I make a new one, that's 5. So if I'm making this bond, I'm going to have to break a bond, and you guys already know what I'm going to break. I'm going to break the halide off. I'm going to break the halogen off. So that means that I'm making a bond, and I'm breaking the bond at the same time. This is going to lead to something called a state. Okay? A transition state is just a high energy phase of the reaction that is very, very short-lived. Okay? What it means is that it never even really happens. What I'm trying to say is it cannot be isolated. Okay? It's a high-energy thing that it must happen because we know that it must go from one state to another. But if I tried to just isolate it in a test tube, I would never be to isolate transition states.

Understanding the properties of SN2.

Now, what I want to do is ensure you understand the mechanism we will use every time. It's crucial that you grasp everything about this mechanism. In addition, there are many facts we need to memorize because your professor will expect you to know a significant amount of conceptual information as well. You won't fully understand it unless you know all the concepts behind it. So, let’s start breaking down these different concepts, beginning with the nucleophile. You might already guess what this is, but do you think we need a strong nucleophile or a weak nucleophile to start off this reaction? When something is negatively charged, it is considered strong because this indicates its tendency to actively try to lose electrons. In the description, I mentioned that it's always a negatively charged nucleophile, which signifies it is strong. Thus, a weak nucleophile, one that is neutral or lacks a negative charge, would not be ideal for an SN2 reaction.

Then, the leaving group. Is it better if it's unsubstituted or highly substituted? "Substituted" refers to the presence of R groups. Unsubstituted means the alkyl halide, the leaving group, has fewer R groups, whereas highly substituted means it has many R groups. The mechanism is a backside attack, so think about whether it’s better to have minimal steric hindrance, like just small hydrogens, or bulky groups that take up a lot of space. Clearly, for backside attacks to be favored, you need ample space on the backside; otherwise, it's too cramped for the reaction to occur efficiently.

Next, let’s discuss the reaction coordinate. If I were to draw an energy diagram, what would represent the highest point—the transition state or an intermediate? As I explained earlier, we use a transition state because the reaction happens simultaneously; there's one molecule in between both sides. Is this a concerted mechanism or a two-step mechanism? Concerted means everything happens at once, aligning with the one-step process of SN2, where a bond forms and another breaks concurrently.

Now, let’s look at rate questions. Is the rate unimolecular or bimolecular? SN2 stands for bimolecular, indicating the rate depends on two species. If I increase the concentration of a certain reagent, it affects the rate of product formation. If you double the number of nucleophiles, it increases the chances of hitting the target—the electrophile or leaving group. Molecules engaging in these reactions respond to random motion; if you double the nucleophiles (arrows) and the electrophiles (targets), you quadruple the chance of a successful reaction, emphasizing the bimolecular nature of this rate.

Lastly, let’s tackle stereochemistry questions. If you have a chiral center and perform a backside attack, switching two groups' positions inverts the configuration. If an R configuration starts, it will change to S after the reaction, which is known as inversion of configuration. Retention would mean it stays the R configuration, but that’s not applicable here. Racemic means obtaining an equal mix of R and S enantiomers, but we achieve 100% of the other enantiomer due to the inversion. Always remember, a backside attack means inversion of configuration when discussing chirality.

We will now proceed with a practice question. Rank the following alkyl halides in terms of their reactivity towards an SN2 reaction. All leaving groups are identical in strength, so determine what else could influence their reactivity. Try to solve this, and I will discuss the answer in the next segment.

Predict the product of the following reaction.

• A.
• B.
• C.
• D.

Predict the product of the following reaction:

• A.
• B.
• C.
• D.

Here’s what students ask on this topic:

What is the SN2 reaction mechanism?

The SN2 (Substitution Nucleophilic Bimolecular) reaction mechanism involves a negatively charged nucleophile attacking an electrophilic carbon that is bonded to a leaving group. This attack occurs from the backside, leading to the simultaneous formation of a new bond and the breaking of the bond with the leaving group. The reaction proceeds through a single transition state, resulting in the inversion of configuration at the chiral center. The rate of the reaction is bimolecular, meaning it depends on the concentrations of both the nucleophile and the electrophile.

Why is backside attack favored in SN2 reactions?

Backside attack is favored in SN2 reactions because it minimizes steric hindrance. The nucleophile approaches the electrophilic carbon from the side opposite to the leaving group, avoiding the electron clouds of the leaving group. This approach is less hindered and allows for a more efficient overlap of orbitals, facilitating the simultaneous formation of the new bond and the breaking of the old bond. Frontside attack is hindered by the electron clouds of the leaving group, making it energetically unfavorable.

What factors affect the rate of an SN2 reaction?

The rate of an SN2 reaction is influenced by several factors: (1) Nucleophile strength: Strong, negatively charged nucleophiles increase the reaction rate. (2) Leaving group ability: Good leaving groups, which can stabilize the negative charge after departure, enhance the reaction rate. (3) Substrate structure: Less substituted alkyl halides (e.g., methyl or primary) are more reactive due to reduced steric hindrance. (4) Solvent: Polar aprotic solvents, which do not solvate the nucleophile strongly, increase the reaction rate by keeping the nucleophile more reactive.

What is the transition state in an SN2 reaction?

The transition state in an SN2 reaction is a high-energy, short-lived configuration where the nucleophile is partially bonded to the electrophilic carbon, and the leaving group is partially detached. This state is characterized by partial bonds and partial negative charges on the nucleophile and the leaving group. The carbon in the transition state has five bonds, making it highly unstable. The transition state cannot be isolated and exists only momentarily as the reaction proceeds from reactants to products.

What is inversion of configuration in SN2 reactions?

Inversion of configuration in SN2 reactions refers to the change in the spatial arrangement of atoms around the chiral center. When the nucleophile attacks from the backside, it pushes the existing groups to the opposite side, resulting in a mirror image configuration. For example, if the original configuration was R (rectus), the product will have an S (sinister) configuration. This inversion is a hallmark of the SN2 mechanism and is crucial for understanding the stereochemical outcomes of these reactions.