Cyclic Structure of Glucose and Fructose Video Lecture - Class 12

Leah here from leah4sci.com/MCAT, and in this
video, I'm going to show you how to convert
a sugar in a Fischer Projection to a Haworth
Projection and a Chair Confirmation.
You can find my entire series on Fischer Projection
alongwith my practice quiz and cheat sheet
by visiting my website at leah4sci.com/Fischer.
Knowing how to convert between Fischer, Haworth
and Chairs is a critical skill for the MCAT
not just the drawing but actually recognizing
how the structure should look when you convert
it so that you don't have to waste any time
during the actual drawing on your exam.
But to learn how to recognize it, you actually
have to take the time and learn how to draw
it, learn the patterns to look out for and
then you can use the shortcut to save time.
Let's start with the Fischer Projection for
D-glucose.
Recognize that D-glucose is ad aldohexose
which means it has an aldehyde at the end,
it has 6 carbons and the ose means sugar.
The trick for recognizing glucose especially
D is that you simply alternate right, left,
right, left and then D means that it goes
on the right.
So we have an OH on the right, an OH on the
left, an OH on the right and then an OH on
the right for the D, L is on the left and
then everything is reverse for L. I'm purposely
keeping the colors different for the right
and the left because that will be the key
behind the trick, say I want you to be able
to differentiate the groups.
And the final carbon is just a CH2OH.
Carbons 1 and 6 are 8 chiral so they're not
written out in the Fischer Projection form,
but carbons 2 through 5 are and that's why
we have it drawn out this way.
Since aldehyde is higher priority, we start
counting from the top.
And without going through the complete reaction
I want you to simply recognize that to convert
between the Fischer or linear form to the
Haworth or cyclic form, the OH on carbon 5
is the one that attacks.
So the OH will attack the Carbonyl carbon,
kick off the bond between carbon and oxygen
making that an O which gets protonated to
an OH.
And the OH that forms can be facing forward
or rear and that's because the aldehyde has
an Sp2 hybridized carbon and if it's from
one side, you get the Alpha, if it gets attack
from the other side, you get the beta but
the reaction is still the same and for the
trick it doesn't matter which one you want
to use.
Let's look at it from the side to make sure
that you understand but keep in mind, you
can't just rotate a Fischer Projection ninety
degrees, you're going to get the enantiomer
but for the sake of the trick, it doesn't
matter because the trick will still have the
molecule fall into place.
I'd turn everything 90 degrees so I can see
what's going on, but then I'll show you the
trick that lets you avoid this turn.
We have an aldehyde on the right, that's carbon
number 1, we have the CH2OH on the left, that's
carbon number 6 and now let's look at carbons
2 to 5 and the trick is we're going to take
all the purple carbons and drop them right
down.
Whatever's on the right is going to be down.
So we have an OH on carbon 2, H on 3, OH on
4, OH on 5 and whatever did not wind up on
the right side will be lift up, meaning the
left side stays up.
Carbon 2 gets a Hydrogen, 3 OH, 4 Hydrogen
and 5 Hydrogen.
Now look at the attack again.
The OH on carbon 5 reaches around, the entire
molecules swings around so that the OH attacks
the the Carbon.
The carbonyl now has just a single bond between
the carbon and Oxygen making that another
OH.
And now let's see how to draw that ring or
the Haworth Projection.
We'll start the structure by looking at the
bonds between carbons 2 and 3.
So we'll draw a line to represent the bonds
between 2 and 3, and then we'll wrap the entire
molecule around so I want you to imagine that
this attack here happens where the molecule
swings around towards the back so the 2 and
3 are the two forward most carbons and everything
else is into the page meaning it wraps around
the back of the molecule.
Attached to carbon 2 on the right we have
carbon number 1, Carbon number 1 got attacked
by the Oxygen from carbon number 5 so let's
swap that out for a red Oxygen and show that
over here.
On the other side we have Carbon number 3
is attached to carbon 4, 4 is attached to
5, 5 is attached to the Oxygen which did the
attack.
Make it bigger so we can see where all the
substituents go and add them in.
For carbons 2 and 3 nothing's changed, we
have an OH down on 2, Hydrogen down on 3,
Hydrogen up on 2 and OH up on 3.
Carbon is 4 is also the same but it's slightly
off to the side.
We have an OH going down and a Hydrogen going
up.
Carbon 1 can be up or down depending on Alpha
or Beta.
I like to think of them as follows; when I
draw an Alpha, it goes down, it has a tail
that goes down and kinda looks like a fish
which is down on a sea.
And then Beta has a really big head.
The head goes up kind of like a bird that
flies high so if we put the OH up, it'll be
a beta glucose, if we put the OH down it'll
be an Alpha or in this case an Alpha D-glucose.
For the sake of keeping it generic, we'll
put the OH out to the side meaning it's not
up or down and I'll just put in a Hydrogen
and I'm drawing this quickly a line to show
it could be up or down but for this molecule,
we don't care.
Carbon 5 is the tricky one.
When I learned this in Biochem, we were taught
that the molecules twist, so you have to look
at the chirality this way in that way, and
honestly it's too tedious to follow . So what
I want you to think of is this; everything
that we saw so far stay the same.
We use the trick to drop the right down and
the rest of left up.
Right down, left up.
For Carbon 6 which is attached to carbon 5,
we're going to put 6 opposite where the OH
was.
The OH is on the right, meaning it would be
dropped right down.
And that means if CH2OH is opposite, that
will be left up and that is CH2OH.
And I want you to think of it this way rather
than memorizing.
With Beta D-glucose OH and the CH2OH are on
the same side, Alpha D-glu goes CH2OH and
OH are opposite.
That Mnemonic works but if you're given a
molecule that's a little different or for
example not a glucose it gets confusing.
But if you can look at the Fischer Projection
and recognize that the CH2OH which is carbon
6 sitting on 5 is going to be opposite of
where that OH was because it's twisted on
the attack, you'll know if the OH would have
been dropped right down the CH2OH stays up
and that means default, the Hydrogen goes
down.
This was a long and tedious method and on
the MCAT you don't have time for this.
You don't wanna draw it, you don't wanna waste
that kind of time.
So now let's use the same trick without re-writing
it sideways, now let's look at how to go straight
from the Fischer to the Haworth.
So here we have D-glucose again and we're
going to draw a 6-membered ring with an Oxygen
up right.
We then number so that carbons 2 and 3 are
two forward most carbons that you see.
One is towards the right and then just continue
around the circle and then let's add the substituents.
For carbons 2,3, and 4 we drop it right down.
2 gets an OH down, 3 gets an H, 4 gets an
OH.
For the other side, the green substituents
are left up.
Hydrogen up on 2, OH up on 3, Hydrogen up
on 4.
And carbon 6 goes opposite where the OH on
Carbon 5 would have gone.
The Oxygen on carbon 5 to be attacked that's
in the ring.
If the OH would have been dropped right down,
the CH2OH gets left up.
And that means we have a Hydrogen going down.
Last but not the least see if a molecule asks
for Alpha or Beta and then put your OH on
carbon 1 accordingly with the Hydrogen opposite.
If OH is up for Beta, Hydrogen is down, if
OH is down for Alpha, Hydrogen is up.
So that's the Haworth.
Now let's see how to turn this into a chair
conformation.
The first thing you have to is be very comfortable
with chairs.
I have detailed tutorial for this on my website,
the link is on the description.
But we'll do a quick overview here.
We want to start with a basic chair skeleton.
But because it's a sugar, we want an Oxygen
here.
I like to have the substituents go down from
the Oxygen.
That means here we have a parallel line, this
side is the opening.
Alright, that's the opening right there, so
we have the carbon going down and on this
side you have the carbon going up.
The numbering works the same way, so you start
at the Oxygen and then clockwise, we have
1,2,3,4,5.
6 will be our substituent and then just like
with the Fischer to the Haworth, we're going
to take the right side from 2 to 4 and drop
it right down.
But here you have to be careful because in
a chair confirmation you have your axial and
equatorial substituents.
So let's the skeleton lines first so we know
where we have axials up and down and where
we have equatorials up and down.
And again this is covered in detail in my
tutorial the links in the description.
So now we know where the pieces are going
to go, let's fill them in.
Carbons 2, 3, and 4 gets drop right down so
the purple substituents on the right will
go down.
2 gets an OH down which winds down equatorial,
3 gets a Hydrogen down winds up axial, 4 gets
an OH down winds at equatorial.
On the same carbons we have the green substituents
that get left up so Hydrogen axial up on 2,
OH equatorial up on 3, Hydrogen axial up on
4 and then carbon 6 is opposite of OH on 5.
The OH would have been dropped right down
so the CH2OH gets left up and look at that,
it winds up equatorial which means we have
a Hydrogen down.
Now for the sake of this molecule let's show
Beta E-glucose and the reason I'm showing
this is this confirmation helps to understand
why glucose is so stable in that cyclic form.
In this molecule every single OH group including
the CH2OH wind up equatorial making it very
very stable and the tiny Hydrogen atoms are
axial so you don't have many unfavorable diaxial
interactions going on.
Since these are 6-numbered rings, they're
called pyranose and this represents a 6-membered
ring sugar where it's not 6 carbons but instead
it's heterocycle because you have an Oxygen
on the ring.
But what happens if you start with the molecule
like the Fructose which gives you 5 numbered
ring?
D-fructose is a ketohexose that's because
it's a ketone on a hex 6, ose sugar.
Carbons 1 and 6 are both CH2OH and the carbonyl
occurs at carbon 2 which is not chiral but
we still show it on a Fischer Projection so
that we can have the rest of the molecule
flow the same way as we did with D-glucose
which is an Isomer of the Fructose.
On D-Fructose, we only have 3-chiral carbons.
Carbons 3, 4 and 5 and they follow the same
pattern as Glucose.
That's another reason I draw this on the right
because it reminds me of the pattern.
OH to the right, OH to the left, OH to the
right and then D, it means the last one is
OH to the right.
Fill in the Hydrogens for the rest and that's
D-Fructose.
The trick for a 5-membered pyranose is the
same as a 6-membered pyranose except we draw
a 5 membered ring instead of 6.
So we have 5 numbered ring with the Oxygen
top and the numbering is also different.
For this molecule we have 3 and 4, rather
than 2 and 3 as the carbons that we look at.
And the reaction here once again is the Oxygen
on carbon 5 swinging around to attack the
carbonyl on carbon 2.
So the anomeric carbon here is carbon 2 and
that's the one that can go up and down.
Carbon 5 could be attacked so it's still attached
to the Oxygen that attacked and Carbon 6 will
come off at carbon 5.
The only carbon that actually stay the same
are 3 and 4.
So 3 and 4 will be dropped down, that means
we have a right, down, hydrogen on 3, a right
down OH on 4.
The other side gets left up so we have the
OH up on 3, Hydrogen up on 4.
Now what about the rest of the molecule?
Just like with the pyranose, the CH2OH on
Carbon 5 will be opposite where the OH was.
If the OH would have been dropped right down,
the CH2OH will be left up, and that means
we have the Hydrogen going down and Carbon
2 will vary depending on if we get out for
a beta depending on the direction of the attack.So
in this case we'll just show a squiggly line
but it's not just the OH that we're showing,
we're also looking at the CH2OH coming off
of that carbon.
So one of the groups will be a CH2OH and the
other group will be an OH where the OH is
what tells you Alpha or Beta, where Alpha
is down and beta is up and the CH2OH simply
goes in the opposite direction.
That conclude the series on Fischer Projections.
If you feel confident with the material make
sure you try the practice quiz which you can
find on my website along with this entire
series and cheat sheet leah4sci.com/Fischer
and don't forget to subscribe and to share
this series with all your premed friends so
that they to can learn this trick and save
some time on their MCAT.