Fischer to Haworth and Chair for Glucose and Fructose (Vid 5 of 5)

Fischer to Haworth and Chair for Glucose and Fructose (Vid 5 of 5)

November 1, 2019 100 By William Morgan


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.