- 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.