Carbohydrates – cyclic structures and anomers | Chemical processes | MCAT | Khan Academy

February 19, 2020 0 By Jose Scott

– Alrightie, so we’ve been speaking so far about carbohydrates as
chains of carbon atoms. And these are chains of
carbon atoms that feature an aldehyde or a ketone functional group. And that falls into this general kinda one-to-two-to-one ratio of
carbon, hydrogen, and oxygen. And of course I’ll keep
using glucose as an example. Now I’ve also used the
term polyhydroxylated to refer to the numerous hydroxyl groups that are in these carbohydrates. And really I bring all
of this verbiage back up to hopefully spark your
ability to see that carbohydrates have all the
makings of an internal, or intramolecular, I
guess, reaction between the carbonyl carbon here, and
one of the hydroxyl groups. Because essentially what we
have is carbonyl and alcohol chemical reactions kinda
just waiting to happen. What happens when an alcohol nucleophile attacks an aldehyde or a ketone? Well, if there’s an excess of alcohol, we end up with a product that is either an acetal, or a ketal. But what happens if there’s
only one nucleophilic attack by an alcohol, if we
just have one alcohol? And that’s gonna be the
case in the ring-closing, intramolecular reaction
we have going on here. Well in that case, we end up with a hemiacetal, or a hemiketal. And really that
terminology is just kind of a review of acetal and
ketal chemical reactions that would fall under, I
guess if you’re looking in an organic chemistry book,
aldehyde or ketone reactions, probably in the carbonyl section. So let’s show how this
process is happening in the context of our glucose, over here. First, I’m gonna highlight
the particular hydroxyl oxygen that’s gonna act as the nucleophile. So, we’ll make that pink. And after being deprotonated,
so after losing this proton, this oxygen is gonna have an extra set of electrons right here. And those electrons are gonna
target that carbonyl carbon. So I’ll draw the carbonyl carbon in green. And remember that the carbonyl carbon has a partial positive charge on it. It has a partial positive charge, because a lot of the electron
density in this double bond is being hogged by this oxygen, so the oxygen has a
partially negative charge, and the carbonyl carbon
is partially positive. And that makes it a perfect
target for the nucleophile that’s been created in
the deprotonation process of this oxygen. And so after the oxygen’s
electrons attack this carbonyl carbon, what’s
gonna happen is the electrons from this double bond are kinda gonna kick back up to the oxygen up here, and eventually they’re gonna
attract another proton, and will form another hydroxyl group out of some of the
electrons from that bond. Now you might be asking, and it’s a perfectly valid question, why it’s this particular oxygen, the one I highlighted, that’s
acting as the nucleophile. And you’re gonna see as soon
as I get the product drawn, that we’ve formed a six-member ring, so it really has to do
with product stability. And if you remember, the basis for the formation of the
ring in the first place, was the increased stability
over the straight carbon chain. So it makes sense that we’re gonna form the most stable ring that we can. Now when we end of with a
six-membered carbohydrate ring, such as the case with glucose here, we call the product a pyranose. The -ose again, as the suffix for sugar. And the pyra- part, to
indicate that this ring is a sugar with six carbons. And then if the carbohydrate ring is a five-carbon ring,
we call it a furanose, which is a bit easier for me to remember, because furanose and five
both start with the letter F. So that’s kinda the memory jogger for me. And maybe a good example for that would be ribose, with
its five-carbon chain, but I’ll kinda stop there, because almost every
ring-forming carbohydrate that I can think of with
biological implications, at least, forms either a five-
or a six-membered ring. So pyranoses and furanoses. So just by convention, you
can see that I have placed the O in this corner up here, and that places the
formerly carbonyl carbon down here right below it. And it’s actually no
longer the carbonyl carbon, but it’s still significant
because it’s the only carbon here that is bonded to two oxygen atoms, the highlighted oxygen and it’s bonded to another hydroxyl group, as well. So I’ll keep a distinguishing color. And we also distinguish its name now as the anomeric carbon. So that’s the anomeric carbon. And then we can go ahead and fill in the rest of the
substituents in the diagram. So, a hydroxyl group and
another, and another. And we call this diagram
a Haworth diagram. So, Haworth diagram. And the Haworth diagram
doesn’t show us the actual configuration of the ring, because in reality, six-membered rings are gonna show up in a
more-stable chair shape. But it is beneficial in telling us which substituents are
above or below the ring. So to keep this convention
straight in my mind, I remember the phrase,
“Downright uplefting.” So, “Downright uplefting.” Kind of a play on, I guess, the phrase, “That’s downright uplifting.” But, “Downright uplefting,” because as I fill in the substituents, those on the right side
of the Fisher diagram will point down, and
those on the left side of the Fischer diagram are gonna point up. So we can actually see that that one’s up, and we’ll make sure that
this numbers off right. This one’s up as well, And maybe we’ll start numbering with one, two, three, four, five, six. And we can do that over here. This would be one, two,
three, four, five, six. So our three carbon in the Haworth diagram is pointed up, and our three carbon on the Fischer diagram has its
substituent on the left. So down, right, up, left. And as we get to the last carbon group, which kinda forms this tail down here, I remember that if it’s a D-sugar, that group is gonna point up
in the Haworth projection. So this is a D-sugar. And you can see in the Haworth projection that this last carbon points up as well. And really this is gonna be the case for a lot of sugar chemistry
that you deal with, because again, we’re
entomatically programmed to digest D-sugars, so we often end up with this last group pointing up. Now the last thing I want to show you is the chair confirmation,
so the chair confirmation. And that’s because this
is the kind of diagram that’s gonna give us
a sense for the actual configuration of D-glucose. But it really does just follow suit with the Haworth projection as far as the substituents being
above or below the ring. So let me just kinda keep
filling in the substituents here. I’ll number them off, again,
just so you can kind of see that there’s some consistency here. So we’ve got one, two,
three, four, five, six. And again this three-carbon right here is the only one with the
hydroxyl group pointing up. And I guess I better change
the color of our one-carbon, to keep that consistent, as well. Now I didn’t indicate the position of the anomeric carbon’s hydroxyl
group yet, because I think it makes more sense to
show it in this diagram. Remember that the original
nucleophilic attack by the oxygen way back over here, that could’ve created
two different products: one with an R configuration
about the anomeric carbon, and the other with a S configuration. So that last hydroxyl
group can actually be in two different positions. One one hand, the hydroxyl
group would be cis to the last carbon in
the equatorial position. So it’d be cis to this
last carbon over here. And it’s in a equatorial position. And we call this the beta anomer. Then on the other hand,
I guess, it can be trans to that last carbon group,
which would place it in the axial position down here. So I guess it could be down
here in the axial position, and we call it the alpha anomer when the hydroxyl group is in the axial. And I kinda remember
that a little bit easier: alpha for the axial position
of that substituent. And I guess I’ve also
heard that if fishies are down in the sea and birds
are up in there air. So if that helps you keep them straight, you might be able to use that also. Now you gotta remember
that what caused this ring to close in the first place, was some amount of acid or base. And the amount of acid and base in water is actually kinda capable of doing that. Because that’s what facilitated this ring-closing process in the first place. And in water, the ring can actually open and close spontaneously. And when it opens up, the
C1 and C2 bond right here can actually rotate,
and when it closes again you can form either the
alpha or the beta product. So this thing is constantly
opening and closing to form the two different products. And we call that process, where it opens, and rotates, and closes again, mutarotation. So this thing is mutarotating
in the water at all times. So mutarotation. And the outcome is that we end up with both configurations,
the beta and the alpha, in the equilibrium concentration. So for glucose, that’s gonna
be about 36 percent alpha, and about 64 percent beta. And the reason that the alpha
configuration is less favored in equilibrium for glucose, is because the transpositioning
of the hydroxyl group creates some stearic hindrance. But this is pretty individualized
for different sugars. So, I guess the most
general rule, I suppose, that you could apply to all cyclic sugars, would be to say that the
beta anomer, again anomer, that the beta anomer is the
one with the anomeric oxygen, in the cis position with
respect to the last carbon.