Concept: Concept: Mutorotation and Optical Activity9m
Now I want to dive deeper into the process by which an alpha anomer can become a beta anomer and so forth. So, guys pyranose and furanose rings are constantly hydrolyzing back and forth between their cyclic and straight chain forms. Now, you guys know what that mechanism looks like, you know what the Rings look like, and I'm here to tell you that it's in constant equilibrium, it's not just like one ring is formed then it stays like that, it's a constant equilibrium of hydrolysis, cyclization decyclization, etc. okay? Well, the specific process by which an anomeric position would convert between the alpha and beta forms is called mutorotation, it's basically the backwards reaction of cyclization but it's so that you can achieve the other anomer, okay? Now, here in this example what I've done is I'm showing you how, let's say you start off with a beta d-glucopyranose and you expose it to either acid or to base, just a little bit of acid or just a little bit of base, this ring is going to start, is going to begin to hydrolyze and it's going to turn to the straight chain and then it's going to eventually mutorotate forming the other anomer. Now, what I did in this drawing is I didn't draw just the Alpha anomer, I actually drew this squiggly line, what does that squiggly line mean? it means that you lose the Alpha and Beta anomer information, because you're going to get some mixture of both. Now, it's important to note, this is not going to be a racemate, remember what racemate means? it means that you're going to get 50 of 50 of each, I'm not saying that, that's not what the squiggly line means here, it just means you're going to get a combination. So, here instead of putting beta d-glucopyranose I just put d-glucopyranose because you're going to get a mixture of the two, mutarotation is going to make it so that you're just going to have some kind of mixture of the two. So, guys it turns out that in your textbook the explanation of mutarotation is very, very closely tied with this idea of optical activity. So, what I want to do is instead of having you read that and be confused I just want to explain it. So, you can understand why they've always talked about optical activity when they talk about mutorotation and what is that optical activity provided the proof for mutorotation, which I'm going to show you now. So, first of all it's important to note that anomers are always going to differ in optical activity, okay? You're always, like an alpha anomer is always going to rotate light different than a beta anomer, okay? for example, the Alpha anomer of a D-glucopyranose rotates light at a positive 112 rotation in a polarimeter, okay? Whereas a beta D-glucopyranose rotates light at positive around positive 19, okay? Guys these are super random numbers that you do not need to memorize, okay? But this just shows you the completely different optical activity of the Alpha versus the beta. Now, if you weren't paying attention and maybe you just kind of like thought very quickly you might think, you might wonder why are they not just opposite rotations, because remember that back when we talk to optical activity back a long time ago, we talked about how if one enantiomer shows the positive rotation, the opposite enantiomer should then show the negative rotation of that same rotation of plane polarized light, in this case they're both positive and they're very different for each other, why is that?
Well, guys remember, that enantiomers are not enantiomers of each other, right? They're not enantiomers, like a bunch of the chiral centers stays the same, what they actually are is remember they're diastereomers, okay? So, that's why they have unrelated activities you can't predict that one is going to be positive and one is going to be negative, they just have random numbers, here I've given you the random numbers for glucopyranose but other types of rings you would need to actually use a polarimeter to get that number, okay? So, it's not something that you're to memorize. Now, why is this important? Well, because back in the day, when they were first discovering kind of this relationship between optical activity and chirality it was noticed that, it was observed that D-glucopyranose always equilibrated to a positive 52.5 rotation, which is weird, because that's not the rotation of either one of these, right? 112 is one, the other one is 18.7 and somehow it would always rest, if you let it sit around long enough, no matter which one you started, this whether you started with 100% of the A or whether you started off with 100% of the B, they would always end up at positive 52.5. Now this is strange because this is not even the middle point, this is not even the halfway point of rotation between the two optical activities. So, what I actually did is I made a little graph for you guys can really understand this, so notice that the beta had a rotation of positive 18, right? And the Alpha had a rotation of positive 112, right? So, that means at the very beginning, let's say that we started off with 100% of the beta anomer? Well, that would give us 18.7 optical activity, but what was noticed is that after several days of it be sitting in an aqueous solution, the optical activity would get more and more and more positive until it reached this magical number of 52.5. Now why, why is that? and then the same thing happened with this guy, with the Alpha, the outflow would get lower and lower and until it got to 52.5? Well, guys the reason is because we learned. Remember, that we learned that specifically for a beta glucopyranose the beta prevails at a 64% equilibrium and the Alpha prevails at a 36% equilibrium, meaning that if you were just start with one of them and allow it to mutorotate you're eventually going to reach an equilibrium or 64% of the pyranose is in the beta form and 36% is in the alpha form and that is why at the end the rotation that you end up getting is 52.5 because that happens to be right at the 64% mark of the difference between the two, what this shows is that 64% of it is the beta and only 36% of it is the Alpha, if it had been 50/50, let's say it equilibrated exactly 50/50 then we would expect a rotation right in the middle of these two, which is positive 65, but we don't see positive 65, what we see is that it's a little bit more than 60, it's a little bit less than 65, what it must mean that the one that's on that side is the majority because it's the one with the lower optical activity and when you actually do the numbers it shows that 64% is equal to the beta, okay? So, anyway, this is proof that needle rotation was occurring, you know, chemists back in the day, they realized, oh, this must mean that if I have an alpha somehow it's going to equilibrate and become a beta. So, then they started figuring out. Well, what is a potential mechanism for this to take place, how could an alpha potentially become a beta? Well, that's going to be the next video, so hopefully this makes sense so far in terms of optical activity proving that mutarotation exists and in the next video I want to go into the mechanism.
Concept: Concept: Mechanism4m
Alright guys. So, while mutorotation is possible in both acid and base I'm only going to show you the active catalyzed mechanism and that's because it turns out that the base catalyzed mechanism has a bunch of cross reactions that are possible that complicate products. So, usually, this is just shown in acid, okay? So, let's go ahead and get started, the very first step is protonation as with any acid catalyzed mechanism. So, I'm just going to redraw my hydronium so that I can have one accessible proton and the protonated O will be the O in the ring because remember, we need to break this ring somehow, okay? So, what this is going to do is it's going to give me a positively charged O and at this point I can use the electrons from my anomeric oxygen, remember, that's the one we're trying to flip up and down to make a double bond, okay? So, two electrons from my O can come down and make a double bond and then I can break this bond in the ring to form an alcohol. So, at this point what's going to happen is that the chirality of this carbon is now lost because it turned into a double bond, which means it's trigonal planar.
Now, it could, it's right in between, it could be up or it could be down, okay? Also this O is going to have a positive charge because it still has an H, cool? Awesome. So, now guys, we're just going to start doing the reverse and get rid of that double bond. So, this O can now reform the ring kicking electrons back up, so this is going to do is it's going to give me once again a positive O and now I'm going to get the other OH. Now, we're going to show that now it;s shifted to the up instead of down and that's because this OH could attack from either the top or the bottom, so it could make the OH go up or it can make it go down, in this case since you started from the down position, I'm trying to rotate it mutorotate it to the up position. So, that's why I throw it up here and then finally guys all we have to do is deprotonate. So, then deprotonates we can regenerate that catalytic acid and we're going to go ahead and achieve a final product, which is now a beta D-glucopyranase, okay? And I'm sorry, pyranose, and this is the process that turns, this is the mutorotation process that turns in output to a beta and this is what's constantly taking place and eventually they would reach an equilibrium based on the stabilities of each, we know that for glucose it happens to be 64% of the beta but if this was any other sugar the number would be different and the anomer might even be different, maybe the alpha would be favored with another sugar, okay? Awesome guys, great job, let's move on.
Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of α-D-galactopyranose is +150.7°, and that of the anomer is +52.8°.When either of the pure anomers dissolves in water, the specific rotation gradually changes to +80.2°. Determine the percentages of the two anomers present at equilibrium.
What is the specific rotation of an equilibrium mixture of fructose? (Hint : The specific rotation of an equilibrium mixture of glucose is +52.7.)
Draw the full arrow-pushing mechanism for the following transformation.
Consider the two sugars below. On the left is ß-(D)-glucose. What is the best description of the sugar on the right?
Mutarotation (inversion of the anomeric center) is a process that happens spontaneously when a sugar is put into aqueous solution. Which of these reactions is mutarotation?
Which compound will not be able to undergo mutarotation?
The interconversion of α- and β- diastereomers of a carbohydrate is called mutarotation. Which statement is true?
1) Only enzymes can perform mutarotation – in solution these structures do not interconvert.
2) Mutarotation requires an acid catalyst (H+) to proceed.
3) Mutarotation proceeds through an acyclic (no ring) intermediate.
4) The reaction converts a D–sugar into an L–sugar.