So let's go through the enolate mechanism first. And actually, my first question for you is, can you think of any reasons why specifically the C2 position of a monosaccharide is susceptible to epimerization and base? What we're talking about is this position right here. Can you think of any other principles in organic chemistry that would make that position specifically susceptible to base? And guys, we do know of a principle that would apply here and that would be the principle of alpha carbons. Remember that alpha carbons, more than usual, are uniquely acidic due to their proximity to a carbonyl. Remember that the alpha carbon is always going to have hydrogens that are much more easy to deprotonate than other hydrogens. While these other hydrogens have pKas of I don't know, you know, around 50, this guy here is going to have a pKa of around 20. Specifically, when I say this guy, I'm talking about this hydrogen right here because that's the only Alpha Hydrogen. So in the presence of base, it's not going to be difficult for my, for my O to remove that hydrogen and form and in, takes away an H, I make a double bond, and I kick electrons up to the O. So what that's going to do is it's going to form an enolate that now looks like this, double bond here, oops, no, double bond here and negative charge here. Notice that now I've lost the stereochemistry of this O. This O is now trigonal on a trigonal planar carbon, it could either be on the right or the left because it's now double-bonded. Now guys, there may be some of you that are saying Johnny that's not how I usually draw enolates, usually how I draw an enolate would just be with a double bond H and with a negative here. Is that also possible? Totally. That's the same thing. These are actually resonance structures of each other. So it's totally fine if you draw it that way. The only reason I drew it this way is because that's the way your book tends to draw it and that would be like the major contributor of the enolate. So your book tends to draw it like that but if you in principle this is the same exact reaction if you just draw the negative on this carbon because once again what's going to happen is that O is going to be able to flip back and forth because now it doesn't have stereochemistry.
Okay? Does that make sense? Awesome. So now guys, what happens? Well, now we would just protonate again. Or not protonate again, we would now protonate. So then in the next step, you have water, okay, and what you're going to do is just reform the double bond. So this negative charge is going to come down, make a double bond, and then this double bond is going to grab the H. And now it's going to become the aldehyde once again. But notice that because we lost the stereochemistry the stereochemical information on that C2 carbon, now it's possible for the O to basically racemize and go to the other side. And by the way, guys, this is not unique to sugars. One thing that we've learned in enolate chemistry is that enolates always racemize the alpha position. So essentially, nothing new is happening here. All we're doing is we are racemizing the alpha position of the sugar which happens to be the C2 position. So this totally falls in line with all the principles that we learned about enolate from enolate chemistry from your alpha carbon chapter.
Awesome guys. So that was the enolate mechanism, and next, I want to show you guys the indiol mechanism which is a competing mechanism that accomplishes the same thing.