Leaving Group Conversions - SOCl2 and PBr3 - Video Tutorials & Practice Problems
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Don’t you wish there were a method to convert alcohols into alkyl halides without all these complications? Ah, looks like we’ve got just the reagents for you!
1
concept
Learning the mechanism of SOCl2.
Video duration:
7m
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since alcohol's air terrible leaving groups, we often want to convert them into alcohol Allied's And it turns out that there are two very common re agents that are used to turn alcohol's into alcohol slides. So let's go ahead and just look right into those. So basically these re agents are S O C. L two and PBR three and respectively, they're gonna turn alcohol's into alcohol chlorides or alcohol bromides. Alright, Now, the only thing that's special about thes is that these air always gonna proceed through an S n two mechanism. Okay, so what that means is that since it's s and two backside attack thes, we're gonna be limited toe working on Onley primary and secondary alcohols. Okay, So if I have a tertiary alcohol, I cannot use S o c 02 or PBR three because I'm not going to get a backside attack, Remember? That s and two just means backside attack. And remember that tertiary These suck at it because tertiary is don't have a good backside. So I needed to be either primary or secondary. Since this is a backside attack, what kind of stereo chemistry do you think we could expect we're always going to get inversion of configuration. Okay. Remember that inversion of configuration and backside attack go hand in hand. Anytime I say one, I'm gonna always be referring to the other. So let's go ahead and analyzed this mechanism. Okay? I'm gonna show you the general mechanism for S L C L two. But the same thing applies to PBR three. In fact, I'm gonna have you guys figure out the mechanism for PBR three all in your own. Let's go ahead and check out S L C L, too. So, overall, the structure of S L C L two or three Ionel chloride is, um, and s adult bond. Oh, and then two coins coming off the end. Okay. What's special about this molecule is that it actually has a very, very Electra Filic portion. Okay, think of this oxygen here as a nuclear file. It has extra electrons that it wants to give away. If it were thio, attack one of the atoms on those four on the spinal Cloyd c l sulfur oxygen, which would be the most susceptible towards nuclear filic attack. Or, in other words, which one has the biggest partial positive charge that would be attracted to the negative charge of the oxygen. And if you draw die polls, it becomes painfully clear that you have die polls pulling in all directions away from the sulfur. Okay, so what that means is that the sulfur is gonna have a very significant partial positive charge. So my arrow is going to start from my oxygen and attack sulfur. Okay, so can I just leave it like that? 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 five bonds. Okay, it can either of four, or it can expand. The talk tests have six, but it can't have five. So the bond that we're gonna break is the one the Dole bond leading up to the Oh, So we're gonna kick those electrons up to the oh, let's go ahead and draw the next step. So what I went up getting is I would end up getting oh, h k. That h is still there from before, But now that's attached to a sulfur. An own negative, a chlorine and a chlorine. Okay, so now in this next step. Oh, by the way, we're forgetting a charge. There should be a part, a positive charge right there, because oxygen doesn't like having three bonds. Alright, so now what's the next step? Well, the next step is that this oxygen does not like having that negative formal charge, and it actually has to really good leaving groups right there. So what's gonna happen is that this negative charge is going to go down and reform the double bond and kick out one of these chlorine. Okay, so what I'm gonna get this stage is I'm gonna wind up getting a leaving group that looks like this. Oh s double bond O C L with the h here and the positive charge. Okay, so it turns out that now this leaving group is awesome 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. So this just has an awesome There's an amazing leaving group. So what do you think could kick it out? What do you think could attack the carbon and kick out this big leaving group, and it's gonna be the c l negative. That just got kicked out. This seal that we kicked out is now this cl And what's gonna wind up happening is that now we get the backside attack. Now I'm going to get the cl attacking that carbon and kicking out this entire thing as leaving group. So what I wind up getting is alcohol. Hey, Light. In this case, it's an alcohol chloride where my chlorine should be facing which direction it needs to be on the dash. The reason is because I notice that my leaving group was on the web before. So after I did a backside attack, I need toe invert the stereo chemistry. I'm gonna get that. Plus, I'm going to get my leaving group, which just looks like this seal on one side and ohh on the other. But guess what? Your chemistry professor does not really care about this one. I'm just gonna put a little X. They don't care about that. What they really care about is your organic product or this guy over here. Especially that you know, the stereo chemistry, because that lets them know that you really understand what's going on? All right. So I hope that mechanism wasn't too complicated. And it was a little weird how we kicked the electrons up to the O and kicked out of chlorine. But that's actually really common pattern when it comes to double bonds toe Oh, and in or go to we're gonna be doing this mechanism all the time, So it's actually really good. They're getting practice with it right away. So that was the mechanism of final 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 in your own. I know that sounds really complicated, but just you guys know the PBR three is going to react exactly the same way as the S L C L two did. In fact, it's gonna be a little bit easier. Okay? And then the any anything in it I'm sorry. The n A and three is just a reaction that you should know from before that. I'd like to you to integrate with this, to try to figure out what's going on. Okay, So even if you don't get the second part at least give the first the first step of this reaction your best shot, and then I'll explain everything to you, so go for it.
These reagents proceed through SN2 mechanisms, so they will only work on 1° and 2° alcohols. That said, we expect inversion of configuration from this conversion.
2
example
Predict the mechanism of PBr3, and draw the final product.
Video duration:
6m
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All right, so let's draw the mechanism for the reaction of alcohol with the phosphorus bromide. What we would get in this first step is that once again, I have very strong die pools pulling away from my central atom, which in this case, is phosphorus. So I do have, ah, partial positive charge. And what I'm gonna wind up getting is that the O attacks the phosphorus and kicks out of BR in case. So that actually all happens in one step here. So this is actually even easier than the S l c l two that we just did. So now what we're gonna get is we're gonna wind up getting the oh still in the front attached to an H. It's the same ages before, and that's gonna be attached to pee and then to BRS. So I'm just gonna put them here. VR pr um, I missing anything. Well, I should draw the positive charge on the oh, right, because the oh doesn't like having three bonds and I should also draw the br that left. Okay, so, hopefully as 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 bro mean comes in and kicks out. That whole big thing is a leaving group, and what I wind up getting is a bro. Mean, facing the other direction. Inversion of configuration, backside attack. Plus my big, complicated leaving group. That's PBR, too, And an O H on the other side. Okay, 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 steps of this reaction. Now I want to react this alcohol bromide with n A and three. What's that gonna be? Well, in order to figure out what this is gonna be, we actually have to use the flow chart. Why is that? Well, because I have a secondary alcohol. Hey, lighter and alcohol in general. Ah, leaving group coupled with and three negative, which is a nuclear file. Any time you have a leaving group and a nuclear fall together, you need to use the flow chart. Okay, so let's go ahead and figure this out. My first question in the flow chart is always is my nuclear fault. negative or neutral, will any and three looks neutral. But once you associate it and a positive is a spectator ion. Right, So it's negative. So that means that on the flow tracking going to go towards the left hand side. Okay, so second step. Okay, second step is is this gonna be one of my bulky basis? Is this one of the bulky basis that always favors the elimination? No. Remember that I only had a select few big, bulky bases. And And three negative is not one of them. So I'm just gonna say no. That means we keep going. So for three, I ask what kind of alcohol Hail I do We have. Well, we have a secondary. Okay, Does secondary occupy highlight? Always favor one type of reaction? No. Okay. 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 nuclear file and member for this section? I just had you guys remember what were the strong bases. There were five of them. Okay. And then we also mentioned heat. There's no heat here, and this is not one of my strong bases. So and three negative is gonna be a better nuclear file, since it's a better nuclear file it's gonna favor s into Okay, so I know that seems like a long process, but now we know what the mechanism is. So it's gonna be SN two, which means that I just have to take my end three negative and do a backside attack on this guy. So now what's the final product gonna look like? Well, I should replace the BR with the M three. Correct. What's the stereo chemistry gonna be? The stereo chemistry should be inverted again, which means that I should actually get my n three on the wedge again. Okay, Now, I know you guys might not know how to draw. And three you could just put in three. That's fine. But notice Here's the biggest thing. I started off on a wedge and they ended off on a wedge. Okay. How is that possible? Well, because whenever you do to s and two reactions back to back, we wind up. Getting is retention of configuration. And guess what? This is a big deal to your professors. Your professors really want you to be able to use a re agent that's either going to produce retention of configuration or inversion of configuration, depending on what you're starting with. Okay, so in this case, if I want retention of configuration, I could do SL seal tour PBR three to get one s and two. And then I could use a nuclear file to produce my second s and two and that would give me a substitution while keeping retention of configuration. Okay, so I hope that makes sense, Yes. So far, um, I will highlight that more in a little bit, but I want us to be aware of the fact that stare chemistry is a big deal in this topic. In fact, stare chemistry might be the most important part of the entire topic, because that's gonna be what determines if you get the question right or wrong. So let's go ahead and move on.
A few notes regarding the above video:
The flowchart being referred to is "The Big Daddy Flowchart" which can be found here – https://www.clutchprep.com/organic-chemistry/sn1-sn2-e1-e2-chart
The list of bases we should look out for are:oxides (OR-), alkynides (RC≡C-), NH2-, H-, and ∆ (heat)
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