Leaving Group Conversions - SOCl2 and PBr3

Since alcohols are terrible leaving groups we often want to convert them into Alkyl Halides and it turns out that there are two very common reagents that are used to turn alcohols into alkyl halides so let's go ahead and just look right into those so basically these reagents are SOCl2 and PVR3 and respectively they're going to turn alcohol into alkyl chlorides or alkyl bromides, alright? Now the only thing that's special about these is that these are always going to proceed through an SN2 mechanism, OK? So, what that means is that since its SN2 backside attack these are going to be limited to working on only primary and secondary alcohols, OK? So, if I have a tertiary alcohol I cannot use SOCl2 or PVR3 because I'm not going to get a backside attack remember the SN2 just means backside attack and remember that tertiaries suck at it because tertiaries don't have e a good backside so I needed to be either primary or secondary since this is a backside attack, what kind of stereo chemistry do think we could expect? We're always going to get inversion of configuration, OK? Remember that inversion of configuration and backside attack go hand in hand any time I say one I'm going to always be referring to the other so let's go ahead and analyze this mechanism, OK? I'm going to show you the general mechanism of SOCl2 but the same thing applies to PVR3, in fact I'm going to have you guys figure out the mechanism for PVR3 all on your own let's go and check out SOCL2 so overall the structure of SOCL2 or thionyl chloride is an S a double bond O and then two chlorines coming off the ends, OK? What's special about this molecule is that it actually has a very very electrophilic portion, OK? 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 those 4 on this thionyl chloride CL sulfur oxygen which 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? And if you drop dipoles it becomes painfully clear that you have dipoles pulling in all directions away from the sulfur, OK? So, what that means that the sulfur is going to have a very significant partial positive charge so my arrow is going to start from my oxygen and attack the sulfur, OK? 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, OK? it can either 4 or it can expand its octet to have six but it can't have five so the bond that we're going to break is the one the double bond leading up to the O so we're going to kick those electrons up to the O let's go ahead and draw the next step, so what I wind up getting is I wind up getting OH, OK? That H is still there from before but now that's attached to a sulfur an O negative, a chlorine and a chlorine, OK? So now in this next step......By the way we're forgetting a charge there should be 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 two really good leaving groups right there so what's going to happen is that this negative charge is going to go down and reform the double bond and kick out one of these chlorines, OK? So, what I'm going to get at this stage is I'm going to wind up getting a Leaving group that looks like this, O S double bond O CL with the H here and the positive charge, OK? 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 going to be the Cl negative that just got kicked out, this CL that we kicked out is now this C L and what's going to 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 the leaving group so what I wind up getting 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 because notice that my leaving group was on the wedge before so after I did the back-side attack I need to invert the stereo chemistry, I'm going to get that plus I'm going to get my leaving group which just looks like this, Cl on one side and OH 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 products or this guy over here especially that you know the stereo chemistry because it lets them know they really understand what's going, alright? So, I hope that mechanism wasn't too complicated I know it was a little weird how we kicked the electrons up to O and kicked out of chlorine but that's actually a really common pattern when it comes to double bonds to O and in Orgo 2 we're going to doing this mechanism all the time so it's actually really good that you're getting practice with it right away so that was the mechanism of Thionyl chloride, right? 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 guys know the PVR3 is going to react exactly the same way as the SOCL2 did in fact it's going to be a little bit easier, OK? And then NaN3 and it IÕm sorry the NaN3 is just a reaction that you should know from before that lets you integrate with this to try to figure out what's going on, OK? 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.