Organic Chemistry / Leaving Group Conversions - SOCl2 and PBr3
  • Solutions 18
Sign up for free to keep watching this solution

Solution: Write a chemical equation for the reaction of 1-butanol with...

Question

Write a chemical equation for the reaction of 1-butanol with each of the following: 

(d) Phosphorus tribromide

Video Transcript

Hey everyone, let's take a look at this reaction, we have our 1-butanol reacting with a PBr three. Now, in this case we have a primary alcohol, so we know that a primary alcohol is not going to be a leaving group, so we're going to have to do something to that in order to react here and get our final product. Now, the cool thing about primary alcohols is whenever reacting with PBr3 or what also would work is at SOCl2, this is going to be a leading group conversion and it's going result in an inversion of configuration. So, say if our alcohol was on a wedge in the reactant we would get as a final product is what you think in alcohol on a dash, okay? So this is going to result in an inversion of configuration, here we don't need to worry about that but just as a quick note, so we know that we're going to be doing a leaving group conversion with PBr and SOCl2, so you can probably predict what our final product is going to look like, what it's going to be is we're going to get an alkyl bromide 3,1 from 1-butanol, right? this is 1-butanol, we're going to get a final product a bromine here, in place of bit and this is going to be our 1-bromobutane, let's see a little bit on how that works. Now, we said that this would also work with SOCl2, so with SOCL2 what we get is we get a, when do you guys think? 1-chlorobutane, okay? Now, a little bit on how the mechanism works is what's going to happen first is we can redraw PBr3 and you can expect that there's going to be an electrophilic portion because we have all these dipole moments, right? to the electrophilic bromine, so we're going to have a partial positive on this phosphorus atom, so what's going to happen first is we're going to use this alcohol as our nucleophile, we're going to come in and attack and what do you think we're going to do? we're going to make a bond and break one, so we're going to kick off one of our bromine atoms, what we expect to get an intermediate it's going to be something that looks like this, we have on an oxygen connected to a hydrogen, it's also connected to a phosphorus that's connected to two bromines. Now, this is actually going to be a good leaving group and what's going to happen is that bromine that left, right? there going to be around and it's going to proceed to it SN2 mechanism and kick off this group, right? So that is why we get an inversion of configuration. So, our primary alcohol is going to be converted to a primary alkyl halide here, which we expect and we get as our final product, okay? So this is a little bit on how the mechanism works, that's going to work the same way with SOCl2, because if we redraw that out we know it's going to be a double bond to oxygen and have two chlorine groups. So, again we're going to have those dipole moments going on in all these directions and then we have a positive, pressure positive sulfur, okay? So the same thing would happened there, okay? But in both cases is just a way to convert an alcohol it could be primary or secondary to an alkyl halide through an SN2 mechanism, let's not forget that, and we know that SN2 mechanisms are going to result in an inversion of configuration, okay? So I hope that makes sense and let me know if you have that questions.