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This content explains the cyclic structure of glucose, contrasting it with its linear form and detailing the reasons why the cyclic structure is more accurate. It also introduces other carbohydrates like fructose, disaccharides (sucrose, maltose, lactose), and polysaccharides (starch, cellulose), explaining their structures and linkages.
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In today's video, we will look
at the cyclic structure of glucose in our biomolecule chapter.
That is, have you seen all these
furanoses, pyrones? Let's look at that structure. I will tell you how to write it easily. Not only that, but there are also starch, cellulose, lactose, maltose. Look, we will see about that in our video. When we say this cyclic structure of glucose, in the last class video, we
saw this linear structure. Did we see the pistol structure? Is
that correct? If you
ask, it is
not possible to have a linear structure. It seems to have a cyclic structure.
We see some limitations in that. So
why is there a
possibility of having a cyclic structure?
We are
going to look at the structure first of glucose
explain most of its property but the
following reactions and facts could be could
not be explained by the structure
What are they saying on what
basis it is
not possible for it to be a linear structure
despite having the aldehyde
group the glucose does not give skips
test and it does not from the hydrogen sulphide addition product
with sodium
hydrogen sulphide that is to say you have
seen the skip test so
when we study our aldehyde group we
will do our skip test to check whether there is aldehyde or not so we will do our skip test skips test that is what we are saying is
that solution which we are actually saying is aqueous rosaniline hydrochloride we
will call it our skips reagent we
will call it aqueous rosaniline hydrochloride come to it and pass sulfur dioxide
then it If there was a pink color, the
pink color disappeared.
If you look at the Scripps reagent, if you
pass the sulfur dioxide present in the pink color, the pink color disappears. If
you add Skips agent to the compound,
then it will reappear. If that is the case with the
Skips agent, then
according to our Scripps agent, we can say that it is an aldehyde group. If that is the
case, then it
it
will not respond to us. If that is the case, then what does it mean? Is
that aldehyde free? If you
go and look at the Fischer structure, then it seems
like the aldehyde is free. But if that is the case, then it should have
responded to the Scripps agent, that is,
what should have been the Scripps agent, and the
pink color should have appeared again.
But I don't know how it happened. It doesn't form the
the
hydrogen sulfide addition product with
nhso3. What is it? Sir. So,
we have seen this, that is, when we
read about this sodium hydrogen sulfide,
we have read about this reaction. There were
some people who said that aldehydes or ketones were present in the aldehyde or ketone. So, you add
sodium hydrogen sulfide to it, that is, we add
sodium bisulfate. So,
we have a bisulfate addition product. If you
look at the o that can be there, if
you look at one hydrogen in another bond, if you look at the so3 na, it
changes. But if
you heard that something like that happened,
we don't know that anything like that happened. If you
heard that, as they say, if there
was a linear structure,
that reaction would not have happened. So what does it
mean that the aldehyde is not free for you, it has
no chance of having a linear structure. Next point is the
penta state of glucose does not
react with hydroxylamine indicating the absence of free C group.
group.
That is, you saw it in the hydroxylamine test. So if there was a free C group, it would
definitely have reacted. That is,
pentaacetate must have reacted.
We have already seen this. Look, is there a
video of our last class here? They must have
given it to you in your textbook.
Look, they must have given you the hydroxyl amide test in your textbook. Look, look
here.
Here, pentaacetate reacted.
So, how did the five carbon atoms behave in
that acetic
anhydride? If
you look at ch3co, then it would have reacted instead of oh, would
n't it? Then
instead of 5oh, you would have created pentastate. Okay,
there would have been a free C
group. If there was a free C group,
then it would have definitely
come to us in the hydroxyl
hydroxyl
amine test and produced cyanohydrin. If there was a hydroxyl amine test, it would have definitely produced cyanohydrin.
But that is, it would have come to
us as noh, right? If you look here, we
call it oxygen. Look, this must have
come to us because we must have had a free C
group because
when you had pentastate, ch3co would have come and
reacted instead of oh. So there would have
definitely been a free cho group because
in the middle place, you would have had a ch3co bond instead of oh.
Five carbons. At
that time, would there have been an aldehyde group in the residue or
not? If so, I
I
did a hydroxyl amine test. When I did that,
I definitely had an oxygen
product. If
you asked me if something like that was formed,
I have no evidence for that. So
what does that make us suspicious about? If so, would there definitely have been a
free C group?
That makes us suspicious. Next, glucose is bound to exist in two
different
crystal forms, which are called alpha and beta. They say that glucose has two crystal forms,
alpha one and beta one.
Have you seen alpha glucose? The melting
point is 419 Kelvin. It is obtained by
crystallization from a concentrated
solution of glucose at 303
Kelvin. Have you seen the beta form?
Melting at 423 Kelvin, it is not a
crystalline form, so at
371 Kelvin,
we get it in a saturated aqueous solution. Okay, so
we see that there are two forms of glucose. This behavior could
not be explained by the open
chain structure of glucose. It was proposed that
one of the OH groups may add to
the CH and form a cyclic hemistal
structure. So, on that basis, it is
unlikely that this is a linear structure like Pister says. What have the first carbon atom and the second carbon atom done here? They have created a
link for you.
What have the hydrogen in the OH on the second carbon atom done? So the hydrogen in the
O on the first carbon atom has converted to OH. They have converted to
hydroxyl. So, they are
finding that. Look,
you know, if you look at the pistol structure,
then you should have just
one CHO group. On the
first carbon
atom, there is a second, third, fourth, fourth, and sixth
carbon atom.
So, when they
look at this testing, they
see that there is no free CH, that is, a free aldehyde group. But I know that they look at it that way. If they look at the
pistol structure, when they learn, they see that there is a free
aldehyde group.
What is the reason for that? So, is the aldehyde group that can be here actually
actually
free or not? What happened to
this oxygen?
A hydrogen from here has gone to create a bond and
created a link.
An oxygen from the second carbon atom has come and
created a link. This is how it should
have been. We call it alpha beta and say
two structures. We call it alpha
beta and alpha d glucose. We call it
next beta d
glucose. You see alpha d glucose. So in this
first carbon atom, there is
a hydrogen on the left side, which is on the right side. If you
look at beta d+ glucose, there is a hydrogen on the left side, which is on the
right side, which is on the left side. So the remaining
parts will be the same.
This will come and
establish euequilibrium. Sir, when will this
euequilibrium be established? Is it
actually linear or cyclic?
I don't understand what you
are saying. If you ask me, this is a
glucose molecule that can actually be a cyclic structure,
but in an aqueous solution,
when equilibrium is established, you will find water. You
dissolve a sugar molecule, you
dissolve glucose, so until
then it remains in a cyclic structure,
but what happens when you dissolve it in water?
You know, it converts
from that cyclic structure to a linear structure.
When it converts to a linear structure, like this, alpha-D
glucose and beta-D glucose are
two glucose structures. They
can establish equilibrium.
Okay, so we call this plus glucose, this
plus glucose is glucose,
but this is alpha-D and this is beta-D.
Okay, so how is this? Sir, is it easy for
me to write? I don't understand. When I
look at this, you say that another
cyclic structure is going to come after this.
Tell me this first, is it easy? If you
asked me that, there is a way to do that too. If you
pay attention, you can write it easily.
Okay, it's not that difficult. Be
patient. If you
take notes, you can write it easily.
So, look here. First, write a
six-carbon atom.
1 2 3 4 5 6.
Bring an aldehyde group to the first carbon atom. Let's
write the first five. The last carbon atom.
Bring a primary alcohol to the sixth carbon atom. Okay, is there a third
carbon atom?
First, second, third. You saw that. So, just bring an
oh on the left side.
Bring hydrogen on the right side.
Bring this first. After this,
everyone should have only an oh on the right side.
After this,
everyone should see the right. There should be only oh on the side,
only hydrogen on the left side, okay, you should
remember this, there should be an aldehyde on the first carbon atom, there should be
a primary alcohol on the last
carbon atom, only on the third
carbon atom, you should have oh on the left side, hydrogen on the right side, and so on, everyone should have oh only on the right side. Now I will convert the Haworth structure from Fischer to Sapposa. Haworth means cyclic structure. He is the one who says that the Haworth structure is like that. How is the cyclic structure linked? Sir, if you have seen that, is there this second carbon atom or not?
1 2 3 4 5. Look, is there this hydrogen on this second
carbon atom or
not? What is he
doing? If so, there is this aldehyde.
Look, this If
you have asked how aldehyde actually looks,
I am writing because I really want to understand. If
you have seen this aldehyde, is there a
c double bond or hydrogen or
not? What does this hydrogen do? They are also
making a panda form near this o. Okay, so what are they doing with
the center of the carbon atom that can be here? If so,
they are also creating a link for this oxygen. If you have
drawn this and seen how it looks, sir, you have brought
six carbon atoms as usual.
1 2 3 4 5 6 6 You have brought
an alcohol primary alcohol in the carbon atom.
Now in the third carbon
atom, you have brought an oh on the left side and
brought a hydrogen on the right side. First, let it be as it is.
What did I say here? What does this hydrogen that can be in this
second carbon atom become?
Oh convert Didn't I
say it was okay? So there's
a hydrogen here, these are just like that.
Okay, these are
all as usual for you. Oh, okay, okay,
you brought this as if it was enough.
This second carbon atom
brought this and
what did you do here? Only one oxygen is
gone. Now
bring it from here and link it here.
Okay, this second carbon atom
is there, isn't it? This second carbon
atom is first, second, third, fourth, fifth.
Okay, this sixth carbon is
linked to you like this.
First, you
learned to write this easily. So, from this, it's easy to
write a cyclic structure from this. You just
have to bend it like that. Okay, that is,
there is a purone or a pyronose. So, what do you do with this pyron
structure and the purone structure?
Bend this structure like that. You brought it,
so you
can write an easy cyclic structure.
You will understand this easily when you see the diagram.
Okay, if you see the structure here,
you will understand. Look, did you
notice this? Now that's what
I wrote here, I said alpha di
beta. If you look at this alpha,
you will have an oh on the right side, beta,
you will have an
oh on the left side. Remember only that first carbon atom.
atom.
Now, pay attention to this cyclic
structure. This is alpha
beta. We say alpha di
glucose. Beta di
glucose. That is,
alpha di plus glucose pyranose. We say beta di plus glucose pyranose.
Why is this called
pyranose? If so, we
call this pyranose. In general,
these are similar to this pyran structure. But
you don't have to imagine all the double bonds.
Just look at this, it looks
like a benzene ring. There is
only one
oxygen atom at the top. That's how it's written.
Why is it written as a cross? It's
easier for you to make notes on this bond. That's why it's written as a cross. If
you draw it straight like this,
you don't have this hooh bond. You
You
can write it correctly on each carbon. That's why you
write it as a cross. Do you understand now how they always came up with this structure?
structure?
Do you have this straight chain? What should happen to this structure? All this should come to
you at the bottom. All this should come to the top.
Remember well.
This is what is on the right side. Everything should go to the bottom.
All this should come to the top. So
imagine it like this. Look,
everything on the right side has come to the bottom.
Look, oh oh, they've
come to the bottom. Look here, oh oh, it's at the bottom.
So, you've got hooh, look, look, there's
always this
ch2oh primary alcohol here,
right? They always have to bring the sixth carbon atom
above the pip. If they bring
it as ch2oh, look, the same way, beta-d-glucopyranose.
beta-d-glucopyranose.
Everything is the same,
only this first carbon atom.
What did I say? Oh, I
said it would be flipped, didn't I? Look, here, there's
oh on the left side,
what happens when you say oh, you have it on the top? If you
see that,
is it on the top? Look, oh, it's on the top.
Look, what do you do? Bring
this pyran structure. Bring the
sixth carbon atom, the
primary alcohol that can be on the sixth carbon atom, to
the top.
Okay, very simple. First, you
learn to write linear structures, that
is, learn to write pistol structures.
After that, we will come up
with a Howarth structure and
call it our Howarth structure. This is how we have a
ring structure.
Look here, that's what we
have been told. The cycle structure of
glucose is more correctly represented by the havth structure as given below. So,
how is fructose?
So, fructose is also similar. It has six carbon atoms.
But on the
second carbon atom, there is a keto
group. Look here, look here too. It has six carbon atoms. As
usual,
what did you do? Write six carbon
atoms. On the last carbon atom,
bring a primary alcohol.
Here, on the first carbon atom, bring a
Bring the primary alcohol to the
first and last carbon atoms.
Bring the primary alcohol to the
second carbon atom. Give a keto
group to the second carbon atom. What will happen to the second
second
carbon atom as usual? What will happen to the
second carbon atom? You will have a link to the second carbon atom.
Look here. Look here. If you
saw this, in glucose, that is, in alpha-
D glucose, you saw that. If you saw that, from the first
carbon atom to the
second carbon atom, you will have a linkage. But if you saw that in fructose, then what will happen to the
second carbon atom? You will have a linkage from the second carbon atom to the second carbon atom.
How is this easy to
write? How do you say this? Now, let's
follow the method I told you. So,
look here. Now,
how is fructose present? So the writing is the same
1 2 3 4 5 6 6
carbon atom, bring a primary alcohol
here, what did you do to the first carbon
atom, bring a primary
alcohol, bring ch2oh, bring a
a
double bond o to the second carbon atom, bring an oh on the third carbon atom, as usual, bring an oh on the opposite side, bring a
hydrogen on the right side,
like everyone else,
oh comes on the right side, okay,
draw this first, because
after writing like this, it will be easy for you to
write the structure,
fructose or glucose, right? So when it is like this,
where is the
linkage going to happen now? So
what is this hydrogen going to do with this oxygen that is on the second carbon atom, is it going to create a bond?
bond?
Right, in the same way, now
you will get the linkage from there.
Look here, it's a verification,
you don't need to look, you understand the same thing.
Look, what's happening at the
second carbon atom? You're
linking at the
second carbon atom. So where's the oxygen that was here? Where's the
hydrogen? You said there's an oh on the
second carbon atom. Where did this hydrogen go? So they went to the
o on the second carbon atom and created a hydrogen bond.
Okay, so
where's this hydrogen going? They're going to go to the
oxygen that was there on the second carbon atom and
create an oh. Okay, so
1 2 3 3 4 5 6 6 You've brought an oh on the carbon atom. Okay, you've
brought a primary alcohol on the first carbon atom.
Primary alcohol. Primary
alcohol. Okay, so you saw this.
So here's the oh.
Now what's the second carbon atom from here? It's
going to happen, so here you have brought
hydrogen here, as usual,
hydrogen here,
what happens from hydrogen here, so the link is created, do you understand, and
from here
you have a link created, now can I
write this like this, why is this
first, okay, this first, second, third, fourth, fifth,
6, right, sir, can I write this like this,
no, the textbook
has given it like this, look, it's bent like this, can I write this like this, so I have
drawn it like this to understand this, and
you still want to
write it like it is in the textbook, so
so
what did you do with this first carbonate,
bring it to the left side? So look how it looks like, look
here, there is a
ch2 oh, so let's say
this is the first carbon atom, so this is the second carbon atom, okay, so here you have an
oh, the third, fourth, fifth,
sixth carbon atoms have an oh, look
here, so here
you have an
oh in the first, second, third, opposite, there will be hydrogen,
so these are oh, these are
hydrogen, here hydrogen, so from here they are creating an
oxygen link for you,
okay, and from here they are creating an
oxygen link for you,
right, I'll write it once, so they will
come to you like this, okay,
sir, it's bent like this, is everything carbon
here, we have to
imagine it like that, because
organics, let's
say that everything in this corner is carbon,
or not, there is nothing like that,
just that They bend it like that to make it clear that there is no
carbon in it. Okay, now
compare it.
How is this ch2oh in the textbook? First, second, third, fourth,
fifth. This
second, We
call it puran, there is no double bond here, nothing else comes
here, just a
light cross that can be drawn here,
so how do you bend this to that
glucose, similarly, everything that is on the right side, even if it comes to the bottom, should be on the
left side or on the top,
similarly, if you
have seen all these primary alcohols, they
are mostly on the top of the ring structure, look here, look here,
all primary alcohols are always on the top,
this is the same, only when you come to beta d,
you will get a flip, you have
seen only this second carbon atom, then it will flip,
now how do we say that the oh that is on the first carbon atom in our glucose
comes and
gets flipped to both the left side and the right side, didn't we say
α d + glucose beta dd +
glucose, didn't we?
If you have seen this, there will be oh on the right side,
this is the left Did you see the same fructose with oh on the side? Then we
call it puronese, right? If you
see the same beta d- fructo puronese with ch2oh on the left side and oh on the right
side, then on the right side, ch2 primary alcohol, there
is an oh on the left side. Okay, so do this
comparison and study a little.
Generally, you should first learn to draw a
linear structure. To write a linear structure, that is, you should learn to write a pistol structure. Then you should learn to
link it. That is, you should learn to link it. Then
after that, you should practice writing this ring structure. Only by
writing a little and
practicing, you will be able to understand it easily. Okay,
look at this. Why is this? Sir, if you put a minus
sign, then this is a levorotatory compound, so
say minus, if you saw that, you would have put a plus sign. They
said alpha digluco pyronose, right? What is the
reason for that? If so, texto rotatory
compound, this is a levorotatory compound,
this capital D is
like that. As
usual, you have given us glyceraldehyde by comparing it to you. As you saw in our last class video, so in the glycerol structure, you have given us this name by comparing that lower asymmetric carbon. The capital D is like that. Is that right? We
call this fructofuranose. Next, let's
look at disaccharides.
You have seen disaccharides. So, it is formed by the
linkage of two monosaccharides. We have
already seen in our last class
video how sir
that linkage is there, then
there is a linkage through oxygen, that is, we will
say glycosidic linkage, so when that oxygen
linkage comes, a water
molecule is eliminated, okay, if you
saw here, sucrose comes and we
call it a disaccharide, you
hydrolyze this or you use
an enzyme called sucrase and you make it react,
then you hydrolyze it,
so if you saw here, you
get two monosaccharides called d+ glucose and d- fructose,
okay, both of them are c6h12o6, but
this one is d+ glucose and this one is d- fructose, okay,
then fructose and glucose are
two monosaccharides There
is a linkage,
if you look at that linkage, we call it a glycosidic linkage.
Here it is. Look, this is alpha-glucose. This is beta-fructose.
beta-fructose.
Beta-fructose. This is the primary
alcohol. The sixth carbon atom is at the bottom.
Oh is at the top. Okay, okay. You should
pay attention to this. You saw in the linkage.
So, a water molecule here would have been eliminated. When it
comes to the oh molecule, both of them would have been eliminated. And an
oxygen molecule would have created a linkage. We
call this our glycosidic linkage. Okay, what is the
link? C1C2 linkage. A glycosidic linkage is between
carbon one in glucose and
carbon two in fructose.
What do we call it? So, sucrose is a disaccharide.
disaccharide.
Sucrose is a text.
Rotary But After Hydrolysis
Generally, sucrose comes and is a text
rotary molecule. If you
look at the molecules obtained after hydrolysis, glucose
and fructose, i.e., alpha-d
glucose and beta-d fructose, are
they not dextrorotatory?
Glucose and glucose are liver-
rotated fructose.
Since the liver rotation of
fructose is
-924°, i.e., beta-fructose is
not present, if you look at its optical rotation, it
is at -924°. It is
more than its text rotation of glucose, which is +525°.
If you look at its anti-clockwise rotation, it is at 924°.
The mixture is liver rotatory,
thus hydrolysis of sucrose brings about
a change in the sign of rotation from dextrotory plus to
levo - and the product is named as inverse sugar. We say that they produce two monosaccharides that can have two rotations, called plus and minus. What is the name for them? We can say that invert sugar. Okay, in general, when this person is in his personal system, sucrose is a dextrorotator, but
after hydrolysis, what is available is
glucose and fructose, which can have dextro and levo. Next,
Next,
maltose. If you have seen maltose, you can say that it is
a best example of a disaccharide. If you
have seen this person, both of them are
alpha-glucose. There is a
glycosidic linkage between them.
Here you have a c1c2
linkage, that is, a linkage between
carbon one in alpha-glucose and
carbon 2 in beta-fructose.
But you saw in maltose, so the
linkage between carbon one in alpha D glucose and carbon 4 in alpha D glucose, that is, when there is an
oh here and there is an oh
in this carbon, a
water molecule is
emitted. So, when the oxygen
linkage comes, we
say it is a glycosidic linkage. Okay, compare it.
If you see c1c2 here, c1c4.
But here there is fructose. This is the only
alpha D glucose. Do
you understand? So, if you
compare it, we call this maltose.
Next, lactose. Lactose is
a disaccharide. If you
see here, we call it beta D galactose and
beta D glucose.
Do you understand? We call this our beta D
galactose. Beta D We
call it glucose.
If you see this, there is a linkage between carbon one and
carbon four, that is, carbon four in beta-D glucose and carbon one
in beta-D galactose.
Do you understand?
Compare this glucose and galactose
and study it. Look here. The second carbon atom is the
same as usual. There
is no big difference in appearance. Look here. Look at the third
and fourth. So, pay attention to the third and fourth. Pay attention. All of this is like this.
Compare it with alpha and beta. Okay, compare it and study it. If you
say that it is the same and write down the same structure, you
will not get marks.
So, we call this beta-D galactose and we
call this beta-D glucose. This is also glycosidic. They
are in linkage. Next polysaccharides
are polysaccharides. You
know, if a number of monosaccharides are
linked together and a number of
linkages are formed, we
call them polysaccharides. Let's look at a
best example of polysaccharides.
Then we can
say starch. Starch is a polysaccharide. If you
look at this starch, there are two
types of polymer compounds. Then there are
two people who say amylose and amylopectin. Amylose is
like a chain molecule.
Glucose is like a chain molecule. If you look at amylopectin, there
are chains.
Look at this, we call it our amylose. If you look at amylose, then it is
formed in alpha linkage.
Everyone knows that alpha diglucose is the first one.
Our amylose is Let's see
starch is the main storage polysaccharides of
plants it is most important diet
resource of human beings high content of
starch is found in serials fruits tubers
and some vegetables it is a polymer of
alpha glucose mostly alpha glucose it
consists of two components amylose and amylopectin amylose is water soluble compound but
amylopectin if you look at it
water insoluble ok both are
starch water
soluble also water insoluble
amylopectin if you look at it
water insoluble which constant is
suppose 15 to 20% of starch amylose mostly
15 to 20% only
chemically amylose is long a
chain with 200 to 1000
alpha glucose molecules
mostly 200 to 1000 monomers
mono Unit molecules, monosaccharide
molecules, that is, all the
alpha-D glucose
molecules that are present in amylose are linked,
so we call it mostly amylose, and they
all have c1 c4 linkages. They all have glycosidic
glycosidic linkages.
Look here, they all have c1 c4
linkages. There are no bridges here,
but if you look at the same amylopectin, there
are bridges. Amylopectin is
insoluble in water and
constant about 80 to 85% of starch.
In general, all starches, including corn starch, have
this much
more. That is, if you
look at amylopectin than amylose, it is more in corn starch. That is
why most of our
corn flour gets fried in some places or when
something is fried, it is a
coating on top. Why do we use it when cooking at home? If so, it
has a good strong polymerization, so you can coat it with some spices, if so,
it will be well
cooked, so it will be like French bread, in
general, the corn flour is
high in that, then we
call it corn flour, right? If so, it is a favorite because it has
a little more amylopectin in it, so it is
insoluble in water, sir, you call it insoluble, I think that kind of
soluble is not like that,
you dissolve it in water, if you wait for
a while,
that starch will come down and
deposit, then it should not be
called soluble, so
okay, so soluble, it will
be completely combined with water, no matter
how long you
leave it at rest, it will not deposit, so
amylopectin Because of the presence of amylose, if
you dissolve starch in water and
then leave it without heating, it
will go back and set.
So you cannot say that it is soluble.
Only amylose is soluble.
So amylopectin is
85% of starch in water. It is a branched
chain polymer of alpha d glucose in which in which chain is formed c1c4
glycosic linkage where as branching occurs by c1c6 glycosidic linkage.
Look here. This is also alpha d glucose. It will be in the
c1c4 linkage. If you look in the middle, the c1c6
linkage will come. That is, you have brought two. You have brought the
chain mallikula c1c4
c1c4 and you have brought the same c1c4.
Similarly, you can buy another c1c4. Correct,
c1 which is on top is also at the bottom.
What did you do by bringing the c6 that may be there and
creating a bond? When you
create it like that,
eliminate one water. There is
an oh that may be there and an
oh that may be there. When the water is
eliminated, an o linkage has come.
Look, this is what we call a glycosidic linkage.
Okay, there would have been a ch2oh here,
similarly, in c4, there would have been an
oh in c6. Okay, there would have been an
oh from here. From c one,
so when two ohs come, you
release a water molecule, h2o.
Then there will be only one o.
That is what
creates the linkage. So we call this our France.
Okay, writing is easy. It is very
critical. If you learn to write a structure and a
ring structure, it is enough.
Alpha diglucose, beta, and the same goes for
fructose. After that,
everything you write will be easy.
Okay, just be
careful with galactose, learn to write next.
Cellulose is the same. The number
of monosaccharide linkages is what creates this.
So cellulose is exclusive to
plants and it is the most abundant
organic substance in plants kingdom. It is the predominant constant of
cell wall of
plant cells. Cells are a straight chain.
chain.
Most of the cells are straight chain. We
call it cell wall. Have
you seen that cell wall? It is made of cellulose.
You must have
known this when you studied biology. If you
were a biology student or
even if you studied up to 10th, you must
have heard all this. So then it is a
predominant concept. Cells are plant cells.
Cellulose is a straight
chain polysaccharide compound only.
Beta D glucose units If you
look at this, all cellulose
has beta D glucose. If you look at this,
you saw amylose and amylose in pectin, then alpha D glucose
was there. Pay close
attention. If you look at this,
you have
alpha D glucose in both alpha D glucose and amylose and amylopectin. But if you
look at cellulose, all of them
have beta D
glucose. If you
ask who has a sir linkage in beta D glucose, then mostly c1 c4
linkages are created in this too.
So differentiate beta D glucose and
write it correctly. Okay, there is
nothing wrong with writing this straight. There is nothing
wrong with writing it crassly like this. Let's
say that everything is a glycosidic link, beta linkage. Glycogen
Glycogen
You know about glycogen. Generally, plants
store a lot of carbohydrates.
So where do they store them? If
you have seen all the plants that convert starch into energy, if
you have seen all the plants and trees,
when they get excess carbohydrates, they
store them as starch.
When they need energy, they
break down the starch. What do they do? They
take carbohydrates. But
animals are the same. So, animals,
animals,
other than plants and trees, if you
have seen all the animals, do they
store starch when they get excess carbohydrates?
No, they convert it into glycogen.
That glycogen is there. We
are in some kind of fasting. It is
time for us not to eat, so it is a little delayed.
That's how energy is. The body needs it, so we have
stored that glycogen, have
n't we? Because
what we
have done is that we have stored glycogen, we have
broken down that glycogen and taken
glucose, what we have done,
we have taken energy back. So
there is this difference between animals and plants, that's why they say here,
look, glycogen is the carbohydrate or
store in the animal body as a glycogen, it is also known as animal starch
because its structure is similar to amylopectin and is rather more highly branched. It is the same as glycogen, right? Because amylopectin is in starch, it is found
only in plants, it is found only in
plants and trees. The
same amylopectin structure is
found in glycogen, but because it is
found in our animals, we call it our
glycogen, almost like this animal.
Starch can be called okay. It is present
in the liver, muscles and
brain. It is everywhere. When
the body
needs glucose,
what do we do with it by using enzymes? We break down that glycogen and
take carbohydrates and convert them into glucose and use them.
This is mostly glycogen. Glycogen is also found in
yeast and fungi. This is the
next importance of
carbohydrates. If you have seen this,
you will know it. It is very easy. If you have read it, you will understand.
understand.
Mostly, what are all these carbohydrates and
where are they used? If you
have seen it in Ayurveda, we use it as whites. That is, they use it in medicine. Is it in
honey or not? We also use it directly from that. So, we are
talking about starch and glycogen in plants and animals. All this is already
seen. You read it, it's not
difficult at all.
We have to look at the two altopentoses.
Next, we have to look at the triposes
and two deoxy triposes. We
call them nucleic acids, right? DNA RNA.
So the starting point for that is here.
Look at the two
altopentoses that are triposes and two
deoxy triposes. They are present in the
nucleic acid. Carbohydrates are bound in the biosystem
in combination with
many proteins and lipids. In DNA,
wherever you have them, there are carbohydrates.
carbohydrates.
Okay, we will talk about that next. Is that a
starting point for that? In
this importance of
carbohydrates, you have
given it in the last three lines. Okay,
if you look at the next index, glucose is the first.
141 Glucose or
sucrose are soluble in water but
cyclohexane or benzene are insoluble in
water. Explain
cyclohexane. That's why you have a ring
structure. Okay, glucose and sucrose, sir.
Why are they water soluble? It's very
simple. Oh, are there any molecules?
Hydroxyl molecules. Oh, if
there are, then hydrogen bonding will definitely create
water molecules. So,
hydrogen bonding will create water molecules. So, hydrogen bonding
will create water molecules. And
all molecules that
have water molecules can easily form water molecules. You
can create hydrogen bonding and make it easily soluble, so
glucose and sucrose
become soluble quickly, but benzene is
not like that, so you have a restructuring,
okay, so you don't
have any oxygen, so you have very, very little chance of
creating hydrogen bonds, so that's what
you have in your insole.
Next, what are the expected products of
hydrolysis of lactose? Do you remember when we hydrolyze lactose,
what is this product? Look here, look,
two products come out, beta-D-
galactose and beta-D-
glucose, and you get two different
products from lactose.
That's what they gave us at that place in Costin. They
gave us the linkage, but you
hydrolyzed lactose. Do that, then we should definitely get two monosaccharides, beta-
DD galactose and beta-DD glucose,
glucose,
right? Next, how do you explain the absence
of an aldehyde group in the
pentaacetate of D glucose? If the
group is free, there is
no chance for it to react there.
How do you prove that the pentaacetate of D glucose is free? If so, the
hydroxylamine test was the only oxygen producer for me.
If so, I would definitely
say that there must be a free aldehyde group in it. So, in the hydroxyl limit
test, that's why
it came up and said, "The aldehyde group is
not there." Let's come and tell you something else.
So it is a ring structure, there is
no possibility of it being a linear structure. Let us tell you that.
Okay, some people will be very confused.
Sir, whether it has an aldehyde group or
not, I will ask you to
separate it by saying it is a reducing sugar. Then we
look at the tolerance reagent test, we look at the falling falling
solution test. If you
see that, they say that reducing sugar is like that. So on what
basis do they
say this? We don't understand. If
you ask that,
I will upload a separate video for you. Next time, I will tell you
how to solve it. Okay, so
I will stop here. In the
next class, we will talk about this
reducing sugar. It is a common
question that can be asked in most competitive exams. Okay, come and
see it, and then we will go to our proteins.
Okay, this video is mine. I'm stopping, let's
see in the next video. If you want to continue receiving videos like this, do
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