We begin our lecture on carbohydrates today.
We will have two lectures on carbohydrates.
We will see how important they are in the
process as we go on especially when we do
bioenergetics where we will be studying metabolism
of these carbohydrates.
The first thing that we want to know is what
are carbohydrates?
Carbohydrates are actually aldehyde or ketone
compounds with multiple hydroxyl groups.
They have a lot of functions in our body.
The first one which is the most important
is as energy stores in the body as fuel and
metabolic intermediates that give us our basic
energetic, the energy we need actually to
get along.
The next important thing is that they form
the part of the structural framework of RNA
and DNA, the essential molecules for life.
When we consider the central dogma of biology,
it was to go from DNA to RNA to protein.
These carbohydrates actually form the basic
structural framework of RNA and DNA and when
we do the structures of nucleic acids and
their components, we will see how this actually
comes into the picture.
They also form the structural elements in
the cell walls of bacteria and plants and
in the exoskeleton of arthropods.
This is what forms the chit in which you would
have probably heard about the exoskeleton
of these arthropods like cockroaches and it’s
that hard shell that is formed.
That is also a polysaccharide which is basically
a carbohydrate.
They are also linked to many proteins and
lipids and when we study lipids and the fluid
mosaic membrane of lipids, we found out there
were proteins which we called glycoprotein
that adds sugar attached to the lipid moieties
that was part of the fluid mosaic membrane
that allowed the transfer of material from
the inside to the outside of the cell.
They also play a key role in cell-cell recognition
processes.
So these are the major function of these carbohydrates
that are basically aldehyde or ketone compound
that are multiple hydroxyl groups.
Basically these polyhydroxy alcohols, as you
probably might all know they have the general
formula (CH2O)n and n is the number of carbon
atoms that are present.
For example; glycerol which was something
we also used in the lipids can be considered
to be the parent compound even though itself
it is not a carbohydrate.
What can happen to this glycerol is it can
be oxidized to form an aldehyde or a ketone
and these actually form the basis for all
carbohydrates.
If we go look at structure of the glycerol,
this is what glycerol is CH2OH CHOH CH2OH.
When we study the fatty acids and lipids,
we found that these were linked to fatty acids
forming an ester and then we had a phosphate
linkage and then some extra linkages.
That actually gave us the polar head groups
and the two hydrophobic tails of our lipid
molecule.
When we come to carbohydrates again, we see
that glycerol again here is the parent compound.
What is happening here is there is oxidization
to form glyceraldehydes.
We can also form CH2OHC=OCH2OH which is called
dihydroxyacetone.
We know that acetone is CH3COCH3.
This is dihydroxyacetone.
We have again the parent compound being glycerol
and from this we can get these two aldehyde
and ketone components that actually form the
basis for all carbohydrates.We have carbohydrates
that contain an aldehyde like glyceraldehyde
that are called aldoses.
Glycerolaldehyde is an aldotriose because
it has three carbon atoms and aldose with
three carbon atoms.
We have ketoses like the one I just drew,
dihydroxyacetone where the carbohydrate contains
a ketone.
It is a ketotriose.
It is a ketose with three carbon atoms.
This is just nomenclature that you would look
at.
So you have an aldose.
In this case glyceraldehyde is an aldotriose.
We have ketoses.
Dihydroxyacetone is it ketotriose.
These are the structures that we just looked
at.
We have our glyceraldehyde and aldotriose.
We have dihydroxyacetone, a ketotriose.
If we look at the carbon atoms here, this
carbon atom the one that in centre here is
an asymmetric carbon atom because it has four
different groups attached to it.
We can form therefore isomers of glyceraldehydes
if the OH is on the other side.
If the OH is on the left, then we form an
isomer.
That is not true in the case of the ketotriose.
We do not have a chiral center here because
we have identical groups; there is no carbon
atom that has four different groups attached
to it.When we look at aldoses and ketoses,
the numbers of isomers are different depending
on the number of carbon atoms present.
If we have three carbon atoms like in this
aldotriose, there is a possibility of two
isomers being formed because this OH can flip
from the right to the left.
Whereas, for the ketotriose, this is the only
possible isomer.When we have an aldotetrose,
we have four carbon atoms and four isomers
possible.
Similarly when we have a ketotetrose ,which
is a ketone with four carbon atoms, in that
case you could have a possibility of two isomers
depending on the number of chiral carbon atoms
that you would have and where you could place
the OH in 3D structure.
If we look at glyceraldehyde once more, we
have these four different groups attached
to it.
It means there are two such stereoisomers
of glyceraldehyde possible.
One is D- glyceradehyde and the other is L-glyceradehyde.
In L-glyceradehyde the OH is on the left D-
glyceradehyde the OH group is on the right.
So this is the stereochemistry that is followed
by convention for all carbohydrates.
By convention the sugars are written in such
a manner that the most oxidized carbon, that
is the aldehyde or the ketone is written to
the top of the structure.
The chiral centre that is farthest from the
oxidized group is the one that decides whether
D or L, not any other OH group.
We have the sugars written with the most oxidized
carbon i.e. the aldehyde or the ketone at
the top.
The chiral center that is farthest from the
oxidized carbon will determine whether it
is D or L. If the hydroxyl group is to the
left, it is the L configuration, if it is
to the right it is the D configuration.
In general, we mostly use D isomers in biological
components.
We have some other aldoses that we can consider
here.
This is three carbon atoms.
We have D-glyceraldehyde.
When we look at four carbons, we have D-erythrose,
the D because the most oxidized carbon is
at the top.
The carbon atom that is chiral that is farthest
away from the oxidized carbon has the OH to
the right which means this is D-erythrose.
This is also D-threose because the OH attached
to the chiral carbon farthest away from the
oxidized carbon is on the right i.e. your
nomenclature.
In D-ketoses we have the three carbons, you
have dihydroxyacetone.
When we are looking at four carbons again,
we have carbohydric called erythrulose where
we have ketotetrose .It has four carbon atoms.
It is a ketose.
We have now the oxidized carbon as for the
top is possible.
We have the chiral carbon with the OH to the
right.
So it is a D-erythrulose.
The other types of D-aldoses, where we look
at 6 carbons that are going to consider, we
have allose, altrose, glucose, mannose, gulose,
idose, galactose and talose.
What we need to know is it what is important
for most biological carbohydrates is glucose,
mannose and galactose.
We see all these are actually D structures
that have been drawn here.
The most oxidized carbon is at the top.
The farthest chiral centre away from the oxidized
carbon has the OH on the right, so this is
the D glucose.
The glucose one is the one that we are going
to be most interested in and we look at the
structure in much more detail as we go along.
These are other sugars that these structures
are important now for all of the sugars that
we are going to do later.
Glucose is an aldohexose why because it has
aldehyde group and it has six carbon atoms.
So it is an aldohexose.
Fructose is a ketohexose.
It has six carbon atoms and it has the ketone
group to it.
Ribose which forms the basis of the ribonucleate
acid, the basic sugar in ribonucleic acid
is an aldopentose and the sugar that forms
the basis for all DNA molecules is 2 deoxyribose
where the numbering begins from always the
most oxidized carbon atom.
So this is number 1, this is number two.
The deoxy means it has lost this.
It is now CH2 so this is deoxyribose.
What we have for ribose that is going to form
the basis for ribonucleic acid is the ribose
sugar which is an aldopentose for deoxyribo
nucleic acid; the basis is two deoxyribose
which is missing and OH.
We will see how important this is in a lot
of structures.
For example, when we did the mechanism of
ribonuclease, we found out that the ribonuclease
because it acts on this two prime OH could
not act on DNA and cleave DNA.
It could only cleave ribonucleic acid, if
you remember when we did enzyme mechanism.
So these are the small bio molecules that
are extremely important or probably the most
important ones where we have glucose, fructose,
ribose, and 2 deoxyribose.
These are some other important monosaccharides.
D-glyceraldehyde is simplest of sugars.
D-glucose is the most important one in the
diet.
We will see how this is also stored and how
it is even broken down in the body.
Then we have D-fructose which happens to be
the sweetest of all sugars.
We have the D-galactose that forms of part
of milk sugar.
And as I said, D-ribose is used in RNA.
And deoxyribose is used in DNA and all of
these you see are D-enantiomers.
When we looked at proteins, we found that
most of them were L-amino acids.
Now this is what we mean by something called
epimers.
First of all, let us look at the numbering
of glucose.
If we look at the numbering of glucose, we
find that it is basically an aldohexose.
The aldohexose has an aldehyde group on the
top.
This is the way it is written and the rest
of the molecules follow.
The chiral carbon that is number 5 here which
is the farthest away from the oxidized carbon
will determine whether it is D or L. This
is on the right so it is D-glucose.
We have now two other aldohexoses is here;
D-mannose and D-galactose.
What we have here is, if we look at the structures
carefully, in this case the change in the
orientation of the OH is on carbon number
2.
The rest is the same.
We have the OH on the left in mannose and
on the right in glucose.
However for carbon number 3, 4 and 5, the
OH directions are exactly the same.
The only change that you see is in carbon
number 2 between D-glucose and D-mannose.
What you say is when you have the difference
in the stereochemistry at only one carbon
atom; it is referred to as what is called
an epimer.
So you would call this an epimer at C2 meaning
that at carbon number 2 this would be an epimer
of D-glucose at carbon number 2.
Since we all have to remember the structure
of the D-glucose and if it what to be mentioned
that, you know the structure of the D-glucose
and you know that you have D-mannose which
is an epimer of D-glucose at carbon number
two, you would know the stereochemistry carbon
number 2 is the only one that is different.
Instead of having OH on the right you would
write the OH on the left.
If I say now that there is another epimer
of glucose that is an epimer at C4 that is
called D-galactose.
Now since we know the structure of glucose,
the only stereochemistry of the carbon that
is going to change the stereochemistry is
going to be at carbon number 4.
In carbon number 4, instead of having the
OH on the right I am going to have the OH
on the left.
So that would be the epimer at C4 which is
D-galactose.
All we would have to remember actually is
the structure of glucose and then know that
D-galactose is an epimer of glucose at C4
and D-mannose is an epimer of glucose at C2.
Let’s look 
at other structures of glucose.
We have what is called hemiacetal formation.
This is some thing that you probably studied
before also but nevertheless we go through
it because it is going to be important in
the understanding of what goes on later on
in terms of its breakage and cleavage.
We have an aldehyde.
An aldehyde reacts with an alcohol.
When it reacts with this alcohol, it can form
what is called a hemiacetal.
What is this hemiacetal?
It has an OH and an OR2.
What is attached here is the rest of the chain.
We have the aldehyde group and an alcohol
group here.
What happens here is this OR2, the lone pair
on the oxygen is going to attack this carbon.
So 
what happens here is this goes and takes up
this proton.
So it forms you see this proton is marked
in red which is actually coming from the alcohol
and the OR2 is now linked to the carbon to
form what is called a hemiacetal.
We can also form an acetal.
What is going to happen to acetal?
We are going to have again this attack on
the carbon here and what is going to be released
now.
The OH is going to take up this H and form
H2O and an acetal.
This is extremely important in the cyclization
of glucose which will see in a moment.
We have two such forms of glucose called an
and a form.
Basically what is happening is, in the glucose
structure, we have not only an aldehyde group
but also an alcohol group.What does it is
actually look like?
We not only have an aldehyde group.
We can write this actually in a different
fashion.
Our aldehyde is 
sitting here and our alcohol is here.
So what can actually happen?
I can have cyclization.
We write this in a manner that is going to
be like this actually it’s not the one that
actually react this alcohol.
We have OH CH2OH COH, this would be apparent
when I show you the slide also but you should
draw it yourself once.
We have an H also sticking here.
We just number the carbon.
Let us be careful in numbering.
We have this carbon number 1, 2, 3, 4, 5 and
6.
Which is the alcohol that is actually forming
the hemiacetal?
It is this alcohol with this aldehyde because
this CH2OH is up 3 numbers 6.
So the numbering is 1, 2,3,4,5 and 6.
If we have the same reaction going on, what
is going to happen is, the oxygen is going
to come here.
What is this going to do?
It is going to take up this proton.
What have 
we 
actually done, if we go back to the hemiacetal
formation, we have the aldehyde with the alcohol
forming an hemiacetal.
If we look at the this is, what is happening
we have the cyclization where is this OH attached
to five that is doing what participating with
the aldehyde so the hydroxyl oxygen and the
aldehyde of carbon number one is going to
cyclize and we will get a six membered ring.
One of the member is going to be the oxygen,
a six membered ring with the CH2OH being the
sixth carbon.
We can also five membered rings because we
can also have the cyclization of D-fructose.
We will look at what those ring look like
in a moment so six membered rings resemble
pyran, they are called pyranosides.
So glucose is also called glucopyranoside.
Furanoside are five membered rings with oxygen
because they resemble furan.We usually have
them drawn as Haworth projection as shown
before.
We have six membered rings and five membered
rings that resemble pyran and furan so they
are also called pyranosides and furanosides.
This is what we have.
You should try and work out the structure
for the case of furanose and see what it look
likes.
You should get a five membered ring in that
case because we have the C=O.
We have a CH2OH attached to it.
If we look at the -D glucopyranose, we will
see what the is in a moment.
but for now we have to know that we have a
six membered ring and the oxygen resembling
the structure of a pyran, so we have glucopyranose.
The five membered ring in case of fructose
which is the ketohexose is forming a fructofuranose.
There are somethings that we can notice here
especially at carbon atom number one.
In the first diagram, we have the OH that
is trans to the CH2OH.
In the second diagram, we have the OH cis
to the CH2OH which is the sixth carbon.
What do we have, what is called an anomer
and anomer.
You form anomers of glucose only when you
have the closed ring structure.
In this case the OH on carbon atom number
one that we have to take care of, if this
OH is trans to the CH2OH.
It is called an anomer.
If it is cis to the CH2OH, it is known as
a anomer.
The same goes with fructofuranose, the CH2OH
if it cis to other CH2OH.
It is the , if it is trans then it is the
form.
This is what the anomers are.
The different stereoisomers and are called
anomers.
We have the aldehyde or ketone carbon which
will give us or decide whether we have the
anomeric in the position or in the position.
You have to remember that this is only possible
when there is cyclization.
So we have the OH down compare to CH2OH, that’s
trans.
If the OH group is up compared to CH2OH, it
is basically a cis conformation when it is
and when we have the .We will see how this
is going to help us in all the glycosidic
linkages that we will talk about later.
We have glucose here, we have the cyclization.
What is the cyclization leading us to, when
we have the OH trans to the CH2OH, we have
D glucose, when we have the OH cis to the
CH2OH, it is beta D glucose.
When we have the ketose sugar, in this case
they shown the OH trans to the CH2OH and the
OH cis to the CH2OH.
We have six membered rings for R glucose which
means that we can have either a boat conformation
or a chair conformation.
This is something you already know with six
membered ring and what do we have is we can
have axial bonds or equilateral bonds depending
on the disposition.
This shape is because this carbon atom is
sp three hybridized.
So it has a tetrahedral disposition conformation
to it which means that you can have an axial
path and an equatorial path.
We also know the chair conformation is favored
over the boat conformations because in this
case we have steric crowding.
Of these two major conformations that are
available, this would be the structure of
glucose.
We have the six membered rings which is our
pyranose ring because we have the oxygen sitting
here.
We have the OH and we have the OH in this
case it is in the equatorial form here it
is in the axial form.
We have the stable chair conformations because
it is obviously going to reduce the steric
clash between the ring substitutents, This
is something you know already.
We have here the axial and the equatorial
axis.
There are two possible chair conformations
where we have the oxygen.
What we can have here, this is D glycopyranose.
We have to look at the name carefully pyranose
means it has a six membered ring with oxygen
unit.
We have the OH trans to the CH2OH.
It is . Why is it D?
It is D because the last chiral carbon atom
away from the most oxidize carbon atom is
having the OH on the right.
It is D glucopyranose.
Another thing we have to consider is in aqueous
solution, the different anomers can actually
interconvert between themselves.
This inter conversion actually gives us a
transient formation of the open chain form
because you realize that once it goes through
the reaction there is going to be certain
intermediate that is going to have the alpha
chain in the open conformation.
When it has it in the open conformation, when
it closes again the OH could either be up
or down i.e. cis or trans to the CH2OH that’s
on the sixth carbon atom.
What can happen is in aqueous solution, when
we have the different anomers that can actually
interconvert due to the transient formation
of the open chain form.
this process is known as mutarotation.
The ratio of the two forms is actually determined
by the relative stability and in the case
of D-glucose; the anomer is more stable because
it results in the C1 hydroxyl group in the
equatorial position.
Between the boat and chair, we know that the
chair conformation is more stable.
When we are looking at the chair conformation
if you want to look at which of the chair
conformations are stable.
We have to look at the chair conformation
which has the most bulkiest substituent at
the equatorial positions.That is exactly in
the case of the anomer whether hydroxyl group
is in the equatorial position and that will
make it the stable form for D-glucose.
This is what we have in looking at the structure
of glucose.
We have aldehyde group and the OH that is
attached to the carbon number 5.
We have the formation of the basically have
our cyclization.
In this cyclization, it is possible that the
OH either goes trans to the CH2OH or it goes
cis to the CH2OH.
We usually refer to this as a pyranoses.
It is based on pyran and then we also know
that when we look at the chair conformation
of this, it would be more convenient to have
the OH group at an equatorial position so
that there will not be any bulky groups at
axial position would be unfavorable.
Let us look at the reactions of the glucose
and other monosaccharides.
There are certain reactions of glucose that
are important for the metabolic breakdown
of glucose.
There is oxidation and reduction that can
occur.
Oxidation and reduction required for the complete
metabolic breakdown of glucose, this is extremely
important and another important about glucose
is storage.
It is not stored as glucose because glucose
breaks down.
There is a metabolic breakdown of glucose
so when it is stored it is stored as a glycogen.
We will see in the next class what the different
types of methodologies are or the different
types of linkages are and how we can actually
store them.
We have our different reactions one is oxidation
reduction.
What is the oxidation reduction?
We will see later on when we do bioenergetics
and consider metabolism of carbohydrates how
this is important there.
We have esterification where we have the production
of phosphate esters.
We have amino derivatives where we have the
use to produce structural component and glycoprotein.
Glycoproteins are those that have glucose
linked to the protein.
We will see how they are actually linked to
the protein.
Then the most important thing that we will
be considering for disaccharide formation
of polysaccharide formation is glycosidic
linkages.
When we consider glycoside formation, we will
understand what these linkages actually are
in their formation.
For example; when we consider that we have
an form we know that we have a form.Now if
we just look at the structure of glucose in
different forms, consider the beta form.
Why is it the form because the OH is cis to
the CH2?
So it is .When we talk of these linkages,
we have to first of all remember the numbering
of the carbon.
What is the number of this carbon?
It is number 1, 2, 3, 4, 5 and 6.
What happened here?
We had hemi acetal formation because we have
the aldehyde and the OH of C5 link together.
When we have glycosidic linkages, we will
see in the next class, we are forming now
from one unit we are going to link in another
unit.
in the linkage of the other unit, we are going
to first of all look at whether this carbon
is or , then we are going to determine what
it is linked into or which two carbon atoms
are linked.
If you link one and four and the first carbon
atom that you are linking is in the beta form,
this forms what is called beta one four linkage.
If I were to write another form of this, I
have to number them again.
I have 1, 2, 3 and 4.
If I have a link between the carbon atom at
1 and the carbon atom at 4 in the other monosaccharide,
this is one monosaccharide and this is the
other one, if I link these two together, at
1 and 4, I would have a 1,4 glycosidic linkage.
Depending on whether the first monosaccharide
being linked to the second one was in the
or position, I would have the numbering as
(1, 4).
We will look at this in bit more detail when
we do the next lecture.
What are the reactions of these glucose?
These are the specific reactions that can
actually go on.
We look how we can actually have glycosidic
formation?
What is this glycosidic formation?
If we go back to the slides here, we have
the linkage of the monosaccharides in this
case what we looked at was the formation of
a possible disaccharide.
If we keep on linking in this fashion, if
we have a series of say 1, 4 linkages, then
obviously I am going to form what is the polysaccharide.
We will look at the different linkage as I
said in the next class.
Now one of the reactions that is extremely
important when we consider aldehydes is their
oxidation to carboxylic acids.
In their oxidation, the sugar that has a free
aldehyde is usually referred to as reducing
sugar.
Sugar with free aldehyde for example; like
glucose which is the sugar that has the free
aldehyde.
So, all aldoses are sugars with free aldehyde
groups.All these aldoses are reducing sugar
and what happens is actually, when you have
the suffix ose, it is referred to as sugar
whether it is a ketose or aldoses.
It has the ose and aldose is where you would
have the sugar in the aldehyde form a ketose
is where would have it in the ketone form.
The name basically is made by changing the
ose ending i.e. ose the end of the ose to
onate.
So if you have glucose it would become gluconate.
The test for reducing sugar is something called
Fehling’s reaction.
What happens here is you have a reduction
of Cu(II) to Cu(I).The basic reaction for
reducing sugar for aldehydes is Fehling’s
Reaction.
It is a very common reaction that is used
in the determination of whether you have glucose.
For example; it can be quantitative even when
you have say, pathological test or biochemical
test.
In this case we are looking at glucose.
It works for any reducing sugar.
So reducing sugar would be one that would
have an aldehyde to it.
What would be an aldehyde with a sugar and
aldose in this case?
We are looking at glucose which is an aldohexose.
That is basically all the nomenclature that
we would know for a monosaccharide what is
basically happening.
Fehling’s reaction that are going from Cu(II)
to Cu(I).What happens is you have RCHO + 2
Cu2+ + 5OH– ? Is basically an alkaline medium
where this goes to RCOO– so what we have
formed gluconate.
We have Cu2O as well.
Then what happens here is this actually gives
you a red coloration to your experiment.
What we have in this case here, in D glucose,
now you recognize this is beta D glucose.
So any structure given to it should be recognizable
in terms of where the OH is, where this CH2OH
is and whether it is fructose or a furanose
or a pyranose.
We have 
a linear and a cyclized form.
So we have D glucose in the linear form forming
the gluconate.
The 
monosaccharides are the simplest carbohydrates.
We found out that they are all formed from
glyceraldehydes.
The basic form is glycerol, and then from
glycerol, we form glyceroldehyde.
We know that we have two forms of glyceroldehyde;
D-glyceroldehyde and L-glyceroldehyde.
It is the disposition of the OH at the chiral
carbon that determines whether it is D or
L. If it is on the left it is L glyceroldehyde
if it is on the right it is D-glyceroldehyde.
Then we have the aldehydes or ketones that
would have two or more hydroxyl groups attached
to it.
Basically that would be the carbohydrates
and the empirical formula is (CH2O)n, where
n is the number of carbon atoms present.
We have aldoses and ketoses depending on whether
we have the oxidized carbon in the aldehyde
form or in the ketone form.
The open chain forms of glucose and fructose
cyclize into rings.
We looked at that also where the aldehyde
can react with an alcohol to form an intramolecular
hemiacetal.
We looked at reaction, where it is possible
for the aldehyde and the alcohol to form a
hemiacetal.
In this case, we have the aldehyde and alcohol
both in the same molecule.
Since they are both in the same molecule we
form an intramolecular hemiacetal and then
a ketone can react with an alcohol to form
an intermolecular hemiketal.
When we form these structures, we form either
a pyranose depending on whether we have a
six membered ring or we form a furanose if
we have a five membered ring.
For the fructose case we will have a furanose
and for the glucose case we have a pyranose.
Then we also looked at how the pyranose ring
can actually adopt either a chair conformation
or a boat conformation and the furanose actually
adapts what is called an envelope conformation.
What is this envelope conformation?
I am sure all of you are aware of this.
We have the furanose it looks like this and
when we have the pyranose it looks like this
right so this would be a pyranose and this
would be a furanose.
What happens in this case is it is these parts
that are level and in this case we have either
the chair or boat conformation but in this
case we have what is known as an envelope
conformation what is the envelope conformation.
The oxygen is slightly lifted up from the
rest of the carbon atoms like flap of an envelope.
This would adapt an envelope conformation.
This would adapt a chair or boat conformation.
What did we find out?
We know that this is going to adapt a chair
conformation why because the boat conformation
would give it some steric hindrances.
Then we also found out that the conformation
would be more stable for glucose.
Then we looked at certain reaction that could
go on different reaction at glucose can actually
have.
