Acetal

In this page we're going to discuss another product that forms when carbonyls react with alcohols and that's called acetals. So guys, in general I'm just going to say a few facts and then we're going to move straight into the mechanism. The first fact I want you to know is that acetals once formed are actually stable in base so if you want to keep it an acetal for a long time, keep it in a neutral to basic solution however they're easily hydrolysed back to carbonyls using acid and that makes sense guys because remember that this is a reversible reaction, it's acid-catalysed, so it makes sense that if you use acid you're going to go back to the carbonyl and it's going to be in equilibrium. Now if you want to specifically make an acetal that not only has R R O R O R but it's actually cyclic meaning it forms a ring, then you're going to have to use a diol because the diol is going to have carbons in the middle that are going to link together and so for example the diol that I would need here would be a 1, 2 ethendiol because as you can see I've got my O, 1, 2, 3, 4, I've got my O, 2 carbons and an O and hat's what I have here. 1, 2, 3, 4. So I would need a diol to make my cyclic acetal. Now let's just go straight into the mechanism for acetals and what we'll find is that it actually the only way to get to an acetal is to use the acid-catalysed mechanism. The base-catalysed mechanism is fine if you want to get to a hemiacetal but it's not going to take it all way to a full acetal. To get to a full acetal, you're going to need to use the acid-catalysed mechanism.

So that being said let's actually start back from the beginning and just make it hemiacetal and then I'm going to show you guys how a hemiacetal can be made into an acetal. So at this point guys you guys should be pros. We know that the first step is going to be protonation, just you know I'm just reiterating here this is the same exact mechanism that we did for hemiacetals. So now I've got my resonant structure, I know that this positive on the O can resonate to the bottom so I could have a positive at the bottom as well. That shows how my carbon has been made hyperelectrophilic and the alcohol can do a nucleophilic addition. So I do my nucleophilic addition, I'm just going to put here a nucleophilic addition, and then, I'm sorry and then this was protonation, and then my last step was to do a deprotonation with my conjugate base. Now in this case my conjugate base was alcohol but whatever your conjugate base is just depends on which acid you use. Not a big deal. So then I would do this, this is my deprotonation step and we know that what we would get if drawn exactly the way we had it before was O H O R H H or how I like to draw it O H O R H H plus my catalytic acid. So we're done with the hemiacetal part, how do we turn this into an acetal? Guys, it's almost the same exact mechanism it's just going to happen again. So now that we have our hemiacetal, we have to figure out, we know that we're going to protonate again so we're basically going to go through a protonation, deprotonation step again but what do we protonate? Do we protonate the O H group or the O R group. How do I know which one? Well to be honest it could be either one. The O H could protonate or the O R could protonate but the direction of your reaction depends on which one you protonate. If you want to go towards the acetal, then that means you're trying to get rid of water. So then you should protonate the O H. If we're trying to go back to the original carbonyl, then you're trying to get rid of the O R so you can reform the carbonyl. So in this case I'm trying to go forward so I'm going to react with my O H. Does that make sense guys? Literally I'm choosing to use the O H because I'm trying to draw the forward reaction but if you were drawing the backwards reaction here then we would use the O R. This is going to give me a compound that looks like this so I'm going to have O H 2 positive, H, H, O R.

Now before in my other structure for hemiacetal I had a resonant structure. In this case I don't need one because the water is a great leaving group, it's just going to take off by itself making a full carbocation. So we could again say that this was protonation I'm sorry yeah protonation and then this would be you know that forming formation of leaving group and then I would get, I'm trying to use red here sorry, and then I would get nucleophilic attack, nucleophilic addition. So then I would get my nucleophilic attack of the O to the C and that's going to give me my tetrahedral carbon in the middle. At which point, what's going to be the last step? Deprotonation guys, you've got this. So then I deprotonate and I get my final compound that looks like O R O R H H or the way that I like to draw it, O R O R H H and you get your catalytic acid. Okay guys? So really the only, it's a lot of steps but it makes it easier the fact that you're doing the same thing twice. Also the only really tricky part is that middle step with the hemiacetal, figuring out which one to protonate. Once you figure out which one to protonate, moving forward isn't hard at all. So guys, I hope that made sense now you guys know how to do a full acetal mechanism just so you know this is one of the most highly tested mechanisms in all of organic chemistry so it's definitely one you want to commit to memory, it's definitely one that you could be asked to draw to some capacity or know about to some capacity on your exam. Alright, so that's it for this topic. Let's move on to the next one.