Leaving Group Conversions - SOCl2 and PBr3 - Online Tutor, Practice Problems & Exam Prep

Alcohols can be converted into alkyl halides using reagents like SOCl₂ and PBr₃, which operate via an SN₂ mechanism, suitable for primary and secondary alcohols only. This process involves a backside attack leading to inversion of configuration. The mechanism for SOCl₂ includes nucleophilic attack on sulfur, bond rearrangement, and the formation of a good leaving group, ultimately yielding an alkyl chloride. Understanding this mechanism is crucial for organic synthesis, particularly in reactions involving alcohols and halides.

Learning the mechanism of SOCl2.

Since alcohols are terrible leaving groups, we often want to convert them into alkyl halides. It turns out that there are 2 very common reagents that are used to turn alcohols into alkyl halides. So let's go ahead and just look right into those. Basically, these reagents are SOCl₂ and PBr₃. Respectively, they're going to turn alcohols into alkyl chlorides or alkyl bromides. The only thing that's special about these is that they always proceed through an S_N2 mechanism. What that means is that since it's S_N2 backside attack, these are limited to working on only primary and secondary alcohols. If I have a tertiary alcohol, I cannot use SOCl₂ or PBr₃ because I'm not going to get a backside attack. Remember, S_N2 just means backside attack, and remember that tertiaries are poor at it because tertiaries don't have a good backside, so it needs to be either primary or secondary. Since this is a backside attack, what kind of stereochemistry do you think we could expect? We're always going to get inversion of configuration. Remember, inversion of configuration and backside attack go hand in hand. Anytime I say one, I'm always referring to the other.

Let's go ahead and analyze this mechanism. I'm going to show you the general mechanism for SOCl₂, but the same thing applies to PBr₃. In fact, I'm going to have you guys figure out the mechanism for PBr₃ all on your own. Let's check out SOCl₂. Overall, the structure of SOCl₂ or thionyl chloride is an S, a double bond O, and then two chlorines coming off the end. What's special about this molecule is that it actually has a very electrophilic portion. Think of this oxygen here as a nucleophile. It has extra electrons that it wants to give away. If it were to attack one of the atoms on this thionyl chloride, Cl, sulfur, oxygen, which one would be the most susceptible towards nucleophilic attack? Or in other words, which one has the biggest partial positive charge that would be attracted to the negative charge of the oxygen? If you draw dipoles, it becomes painfully clear that you have dipoles pulling in all directions away from the sulfur. That means that the sulfur is going to have a very significant partial positive charge, so my arrow will start from my oxygen and attack the sulfur. Are we done with this mechanism or at least this part? No. Because if we make a bond, we have to break a bond because sulfur cannot have 5 bonds. It can either have 4 or expand its octet to have 6, but it can't have 5. The bond that we're going to break is the one the double bond leading up to the O. We're going to kick those electrons up to the O. Let's draw the next step. What I end up getting is I end up with O H. That H is still there from before, but now that's attached to a sulfur, an O negative, a chlorine, and a chlorine. By the way, we're forgetting a charge. There should be a positive charge right there because oxygen doesn't like having 3 bonds. The next step is that this oxygen does not like having that negative formal charge and it actually has two really good leaving groups right there. So, this negative charge is going to go down and reform the double bond and kick out one of these chlorines. What I'm going to get at this stage is I'm going to end up getting a leaving group that looks like this: O S double bond O Cl, with the H here and the positive charge.

It turns out that now this leaving group is excellent because that O has a direct positive charge, which means that after it leaves, it will be neutral, which is great. On top of that, the sulfur pulls electrons away. This just has an amazing leaving group. So what do you think could kick it out? What could attack the carbon and kick out this big leaving group? It's going to be the Cl negative that just got kicked out. This Cl that we kicked out is now this Cl. What's going to happen now, is that we get the backside attack. Now I'm going to get the Cl attacking that carbon and kicking out this entire thing as a leaving group. What I end up with is an alkyl halide. In this case, it's an alkyl chloride where my chlorine should be facing which direction? It needs to be on the dash. The reason is that my leaving group was on the wedge before. After I did a backside attack, I need to invert the stereochemistry. I'm going to get that plus my leaving group, which just looks like this: Cl on one side and O on the other. But guess what? Your chemistry professor does not really care about this one. I'm just going to put a little X. They don't care about that. What they really care about is your organic product over here. Especially that you know the stereochemistry because that lets them know that you really understand what's going on.

I hope that mechanism wasn't too complicated. I know it was a little weird how we kicked the electrons up to the O and kicked out a chlorine, but that's actually a really common pattern when it comes to double bonds to O. In Organic Chemistry II, we're going to be doing this mechanism all the time. So it's actually really good that you're getting practice with it right away. That was the mechanism of thionyl chloride. What I want you guys to do now is to use that knowledge to try to draw the mechanism for this entire reaction all on your own. I know that sounds really complicated, but just so you know, the PBr₃ is going to react exactly the same way as the SOCl₂ did. In fact, it's going to be a little bit easier. Okay. And then the NaN3 is just a reaction that you should know from before that I'd like you to integrate with this to try to figure out what's going on. So even if you don't get the second part, at least give the first step of this reaction your best shot and then I'll explain everything to you. So go for it.

Predict the mechanism of PBr3, and draw the final product.

All right. So let's draw the mechanism for the reaction of alcohol with phosphorus bromide. What we would get in this first step is that once again, I have very strong dipoles pulling away from my central atom, which in this case is phosphorus, so I do have a partial positive charge. And what I'm going to wind up getting is that the O attacks the phosphorus and kicks out a Br. Okay. So that actually all happens in one step here. So this one's actually even easier than the Cl₂ that we just did. So now what we're going to get is we're going to wind up getting the O still in the front attached to an H. It's the same H as before and that's going to be attached to P and then 2 Br's. So I'm just going to put them here. Br, Br. Am I missing anything? Well, I should draw the positive charge on the O, right? Because the O doesn't like having 3 bonds. And I should also draw the Br that left. So hopefully, I drew this, but for those of you that didn't, can you guess what happens now? Backside attack. Because now I have a good leaving group. So my bromine comes in and kicks out that whole big thing as a leaving group. And what I wind up getting is a bromine facing the other direction inversion of configuration backside attack plus my big complicated leaving group that's PBr₂ and an O on the other side. Once again, your professor does not really care about this one. They just care about your organic product. Okay?

Especially because, guess what, we have a second step to this reaction. Now we want to react this alkyl bromide with NaN₃. What's that going to be? Well, in order to figure out what this is going to be, we actually have to use the flowchart. Why is that? Well, because I have a secondary alkyl halide or an alkyl halide in general, a leaving group coupled with N₃^-, which is a nucleophile. Anytime you have a leaving group and a nucleophile together, you need to use the flow chart. Okay? So let's go ahead and figure this out. My first question on the flow chart is always is my negative or neutral? Well, NaN₃ looks neutral, but once you dissociate it, Na⁺ is a spectator ion, right? So it's negative. That means that on the flowchart, I'm going to go towards the left-hand side. So second step. Okay? Second step is, is this going to be one of my bulky bases? Is this one of the bulky bases that always favors elimination? No. Remember that I only had a select few bulky bases and N₃^- is not one of them. So I'm just going to say no. That means we keep going. So for 3, I ask what kind of alkyl halide do we have? Well, we have a secondary. Does a secondary alkyl halide always favor one type of reaction? No. Secondaries actually can go both ways. So let's go ahead and ask ourselves the last question. The last question is, is this a stronger base or a stronger nucleophile? And remember for this section, I just had you guys remember what were the strong bases. There were 5 of them. Okay? And then we also mentioned heat. There's no heat here and this is not one of my strong bases. So N₃^- is going to be a better nucleophile. Since it's a better nucleophile, it's going to favor SN2. K. So I know that seems like a long process, but now we know what the mechanism is. So it's going to be SN2, which means that I just have to take my N₃^- and do a backside attack on this guy. So now what's the final product going to look like? Well, I should replace the Br with the N₃. Correct? What's the stereochemistry going to be? The stereochemistry should be inverted again, which means that I should actually get my N₃ on the wedge again. Now I know you guys might not know how to draw N₃. You can just put N₃. That's fine. But

Here’s what students ask on this topic:

What is the mechanism of converting alcohols to alkyl halides using SOCl₂?

The mechanism of converting alcohols to alkyl halides using SOCl₂ involves several steps. First, the oxygen of the alcohol attacks the sulfur in SOCl₂, forming a bond and breaking the S=O bond. This results in a structure with a positive charge on the oxygen. Next, the negative charge on the oxygen reforms the double bond with sulfur, kicking out a chloride ion. This chloride ion then performs a backside attack on the carbon attached to the oxygen, displacing the leaving group and forming the alkyl chloride. This process results in inversion of configuration due to the SN2 mechanism.

Why are SOCl₂ and PBr₃ used for converting alcohols to alkyl halides?

SOCl₂ (thionyl chloride) and PBr₃ (phosphorus tribromide) are used for converting alcohols to alkyl halides because they facilitate the formation of good leaving groups and proceed via an SN2 mechanism. This mechanism is suitable for primary and secondary alcohols, leading to inversion of configuration. SOCl₂ and PBr₃ are effective because they create highly electrophilic intermediates that make the nucleophilic attack by halide ions more efficient, resulting in the formation of alkyl chlorides and alkyl bromides, respectively.

What are the limitations of using SOCl₂ and PBr₃ for alcohol conversion?

The main limitation of using SOCl₂ and PBr₃ for alcohol conversion is that they only work well with primary and secondary alcohols. This is because the reaction proceeds via an SN2 mechanism, which requires a backside attack. Tertiary alcohols are not suitable for this reaction because they are sterically hindered, preventing the backside attack necessary for the SN2 mechanism. Additionally, the reaction conditions may not be suitable for sensitive functional groups present in the molecule.

How does the stereochemistry change during the conversion of alcohols to alkyl halides using SOCl₂?

During the conversion of alcohols to alkyl halides using SOCl₂, the stereochemistry of the molecule undergoes inversion. This is because the reaction proceeds via an SN2 mechanism, which involves a backside attack by the nucleophile (Cl^-) on the carbon atom bonded to the leaving group. As a result, the configuration of the carbon center is inverted, meaning that if the original alcohol had a substituent on a wedge (coming out of the plane), the resulting alkyl halide will have the substituent on a dash (going into the plane).

Can tertiary alcohols be converted to alkyl halides using SOCl₂ or PBr₃?

No, tertiary alcohols cannot be effectively converted to alkyl halides using SOCl₂ or PBr₃. This is because the reaction mechanism for these reagents is SN2, which requires a backside attack. Tertiary alcohols are sterically hindered, making it difficult for the nucleophile to approach the carbon atom from the backside. As a result, these reagents are only suitable for primary and secondary alcohols, where the steric hindrance is minimal, allowing the SN2 mechanism to proceed efficiently.