t-Butyl Ether Protecting Groups

So the first type of protecting group that you need to know and probably one of the more common ones is a tertbutyl ether protecting group. Now what this does is it adds an ether to the oxygen making it unreactive because if you guys remember or if you guys just – we've learned about functional groups in the past, alcohols participate in a lot more reactions than ethers do. What that means is that if I can turn my alcohol into an ether, it's going to be protected as long as it is an ether.
Now the reaction that we usually use for this is an acid-catalyzed alkoxylation. Just so you know, an acid-catalyzed alkoxylation is a lot like an acid-catalyzed hydration except that we're using an alcohol as our solvent. In this case, the alcohol actually comes from my molecule.
So let's go ahead and draw out this mechanism. We're going to react with a molecule called isobutylene which is just this four-membered hydrocarbon with a double bond. And what we're going to wind up getting is an ether. Let's figure out how.
In our first step, we're going to protonate our double bond through a normal addition mechanism. What this is going to give me is a Markovnikov carbocation. Remember that Markovnikov states that your carbocation goes in the more stable position. After I've done that, given the electrons to the O, what happens next. Well, not it's time for my alcohol to step in. My alcohol is actually going to wind up attacking that carbocation.
What I'm going to make is something that looks like this, where now I have a tertbutyl group on one side, the ring structure on the other. I still have one H and a positive charge. Now how do you think we could get rid of that positive charge? Smart. What we could do is we could use the conjugate of my original acid. So I'm going to go ahead and use the conjugate of my sulfuric acid. I'm going to deprotonate. And lo and behold look what I've got. I now have an ether instead of an alcohol.
Now, why do you think this might be helpful, having it look like that? Well, because it turns out that this ether that I'm looking at right here, is completely unreactive to strong bases like alkynides. Remember that I said an alkynide would react with an alcohol, it won't react with an ether. So now that means if I were to introduce my alkynide to this molecule after the ether's in place, guess where it's going to react? Not with the ether. The ether's protected now. This is my protecting group. That's my protecting group.
So now what's going to happen is the only thing that it can possibly react with is my alkyl halide through an SN2 reaction. So that's the advantage of protecting groups. They allow us to react with just the thing we want and to ignore the thing that we don't want it to react with.
Now you might be wondering, “Well, Johnny, what does the final product look like?” Well, what we would do at this point is that we could – after this reaction is over, we could remove the protecting group. Why is that? Because we said this reaction has to be easily reversible, right? So what that means is that see how this is drawn with a forward-looking arrow? Well actually, it would be truly in equilibrium. It wouldn't be just a forwards arrow.
So for example, here were I drew a forwards arrow here, that should really be, technically, it should be in equilibrium because we know that it's going to go forwards now, but we can make it go backwards later.
So after we do this step, how do we get it back to the original alcohol? Well, if adding our protecting group was step one, and if adding our alkynide was step two, then we have a third step. The third step is just to add mild acid, so I could just say H2SO4 and water. And what that's going to do is that's going to deprotect. Whenever you protect, you always have to deprotect.
What does deprotect mean? It just means that I'm going to take that ether completely off. Now I'm not going to show you the whole mechanism to deprotect, but you can imagine it's just the reverse mechanism of everything we've drawn to protect it. So what that means is that I would actually protonate the O first, then it would leave and then it would get protonated. The tertbutyl group would leave and then it would get protonated and eliminated.
I hope that makes sense guys. For the purposes of your test, you will need to know when you have to use a protecting group and when you don't. In terms of synthesis, your professor could ask you, how do I make this final product. And just using that one reagent wouldn't be enough. You would need to use – first you'd need to protect. Second, you could use your alkynide. Third, you would have to deprotect using acid and water.
I hope that made sense, guys. Let me know if you have any questions. If not, let's go ahead and move to the next topic.