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Solution: Deuterium oxide (D2O) is water in which the protons (1H) hav...

Question

Deuterium oxide (D2O) is water in which the protons (1H) have been replaced by their heavier isotope deuterium (2H). It is readily available and is used in a variety of mechanistic studies in organic chemistry and biochemistry. When D2O is added to an alcohol (ROH), deuterium replaces the proton of the hydroxyl group.

ROH + D2O ⇌ ROD + DOH  

The reaction takes place extremely rapidly, and if D2O is present in excess, all the alcohol is converted to ROD. This hydrogen–deuterium exchange can be catalyzed by either acids or bases. If D3O+ is the catalyst in acid solution and DO- the catalyst in base, write reasonable reaction mechanisms for the conversion of ROH to ROD under conditions of (a) acid catalysis and (b) base catalysis. 

 

Video Transcript

Alright everyone. So, here are our two reactions that we're going to draw, the acid catalyzed and base for our exchange here of our deuterium, so we're going to get ROH and our product should be ROD, alright? and it just talked about how deuterium is a heavier isotope of our hydrogen, so it should be treated very similar. Now, there's a difference between our acid catalyzed and our base catalyzed, it gave us that we are going to use our O, D, D, D plus and then OD minus as our base and our acid catalyzed, so what we're going to do is, how do you think each one is going to start out? what do you guys think? Well, one of the negative charge here, our base is going to deprotonate our acid is going to protonate, okay? So you're very familiar with how these reactants work whether it's acid or base catalyzed, so let's draw a mechanism for it, what's going to happen here is we're going to have our lone pairs is going to come in and it's going to grab one of our D's and kick the electrons over there, the difference in our base catalyzed this here, our O minus is going to be our base that's going to come and grab an H and kick the electrons onto our O, in both cases, let's draw our products from this reaction, what we're going to get into the top one is we're now going to have an R attached to an O, it may expect a negative charge, right? Note that's going to be for the bottom, for the bottom we're going to have that O, RO minus but for the top we're going to have is we're going to have an R, O, H, D, positive charge, and then of course this is going to be a D, so that's why I'm writing it like this and then we have our O, D and then D, okay? And now what did you get for the bottom? Well, for the bottom if we keep scrolling down we're going to get our O, D and then H, okay? And now what we need to do we want our product to be what? R, O, D, so what we're going to do in the top is we're now going to take this and we're gonna grab simply an H, okay? And kick the electrons back to the O and what about for the bottom? we're going to attack with this negative charge, and we're simply going to grab this D, kick the electrons onto our O. So, now what we get from that is we're going to get if we keep going in our reaction, same thing over here, and actually I'm going to just move this down a little bit, so we can see it better, and this, what do you think we're going to get in the top? what we're going to get what we want we have ROD and then we have left over, well, we have Plus our O, we have attached to D, D and then H with a positive charge and then for the bottom what we're going to get is again are OD but now we're going to regenerate our base, which is OH minus. So, as you see in both cases what we get is our final product, which is what we wanted, we're going to look up and what you see we get our OD, okay? So I hope this made sense and you're able to see the differences between our acid and base catalyzed mechanism for this, okay? So if you have any questions please, let me know.

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