 Okay, let's do this problem, this virubillary or oxidation problem, or these two problems. So in it you're taking a ketone or an aldehyde and you're reacting them with a peroxy acid. And what that's going to do is we're going to take this oxygen, remember the mechanism, and put it in between one of the two groups that is alpha to the carbonyl. So if you remember your migratory aptitude, right, it's going to be H first, then tertiary, secondary, then phenyl, then primary, then methyl. So you have to remember that, of course, if not then you won't be able to do these problems. But anyway, so now all you've got to do is figure out what the carbon is, alpha or the lowercase hydrogen is alpha to the carbonyl group. So in this case we're distinguishing between, distinguishing between a primary and a secondary. So if we go over here and we look, secondary's got the greater migratory aptitude. So the product will be carbonyl with an oxygen there, and just this portion after the oxygen. So you can see you just stick the oxygen in between there. And same process with this one, you've got to remember this, H always comes first, it's the little guy, so it's going to go in between there. So taking benzaldehyde and making benzoic acid, of course, H has a greater migratory, H has a greater migratory aptitude than pH. And of course on the other one, we were looking at the difference between secondary and primary. So if you want to, of course, remember the mechanism too, that'll help you out. Any questions? Okay.