Organic Chemistry 51C. Lecture 17. More Structure, Stereochemistry, & Reactions of Carbohydrates.


>>Good morning.>>Good morning.>>So, today I’d like to
continue our discussion of the structure and
stereochemistry of sugars. We’re going to then talk
about reactions of sugars. And we’re going to come
back at the end to look, everything works out, how Emil
Fischer originally determined the structure and
stereochemistry of sugars in what really was a
tour de force work. And I’ll just give you a
teeny tiny taste of the flavor of the chemistry and
the logic involved. So, I mentioned Fischer. And last time we introduced
Fischer projections, this way of drawing sugars that
really pervades sugar chemistry. And it’s a very, very useful
tool for quickly thinking about structure and
stereochemistry. And so we’re going to use that
along with Haworth projections and other projections
on today’s talk. So I want to start by
just giving an overview of how we can look at some
stereochemistry of sugars. And I’ll take some very
simple, very small ones. We’ve talked about glucose. Glucose is an aldohexose. We’ve talked about — we’ve
talked about galactose. Galactose is another aldohexose. These are six-carbon sugars. At this point I just
want to start out to give you the
simplest thing where we can get all the
stereochemical relationships. And so I want us to look
at various tetroses. Tetroses are, of course,
four-carbon sugars. And specifically aldotetroses. [ Writing on Board ] Aldotetroses are sugars where
the 1 position is an aldehyde. We’ll later see keto sugars where you have not an
aldehyde but a ketone group. And all of these
are carbohydrates. All right, so give some flavor
of stereochemical relationships and maybe harken back to 51A, let me draw out the
four aldotetroses. Now, four I’m going to qualify because you could also
say it’s two depending on whether we’re
counting enantiomers. So let’s start with
the natural isomers. [ Writing on Board ] So this sugar, this
four-carbon sugar is D-threose. And remember, all of the
D-sugars, by definition, when we draw a Fischer
projection this way, all of the D-sugars have the
hydroxy group at the carbon next to the bottom position. The bottom stereocenter have that with the OH
pointing off to the right. Remember, this representation
has the backbone curving backwards, snaking backwards
in a series of zigzags. And so this means this OH is
pointing out of the board. This hydrogen is
pointing out of the board. This CH2OH is pointing back. And this remainder of the
chain is pointing back. And then in this
representation we continue to point the aldehyde
group further back. And so this hydrogen is pointing
out and this OH is pointing out. Now, what’s nice about the
Fischer projections is we can rapidly lay out all different
structures and compare them. It’s quicker and easier
and more intuitive than zigzag structures. And so I’m going to draw another
D-sugar, another D-aldotetrose. So it’s still going to be D, but
now instead of having this — so that means this hydroxy
group is still off to the right. But now instead of having this
stereocenter with the OH off on the left, we’re going to
have it off on the right. And this one here is
called D-erythrose. [ Writing on Board ] And of course, these
two molecules here are diastereomers. [ Writing on Board ] We’ve talked about
diastereomers last time. We talked about, for example,
glucose and galactose. Those are diastereomers that
differ at one stereocenter. We have a special name for
a stereoisomer that differs, a diastereomer that differs
at one of many stereocenters. We call that an epimer. And then we also talked
about stereochemistry at the carbonyl carbon
in the cyclic form, in the anomeric carbon where
you could have either the beta or the alpha stereochemistry. And we had a special
name for that, anomers. So we saw beta-D-glucose
and alpha-D-glucose. Those two diastereomers in
addition to being a type of epimer, you would also
refer to them as an anomer. Anyway, these are just generic
diastereomers, and epimers, I guess you would say. Now, imagine for a
moment that I flipped all of the stereocenters
in the molecule. I invert them, and
so we go like this. And again, I can very
rapidly represent this for our threose example. So now, for the threose, I’ve flipped this
stereocenter on the bottom. So that by, inherently, is
now going to be an L-sugar. And if flip the other
stereocenter like so, so now I’ve made the molecule
that’s the mirror image, or more specifically, the
non-superimposable mirror image. I have represented
the enantiomer here. And of course that means
this compound I’ve drawn in L-threose. Sometimes the L-sugars are
referred to as unnatural sugars and the D-sugars as
the natural sugars.>>Wait, shouldn’t it be L is
erythrose because the OH is on the left, like
with [inaudible].>>No. Because erythrose
— so okay. So good question. This is why we’re going
through this with some care. So here, if I now go ahead
and make this L-sugar, this is the enantiomer of — oops, that’s supposed
to be an aldehyde. So we’re going to make that CHO. So this is the enantiomer
of erythrose. And so this is L-erythrose. I saw another question.>>So the bottom
part that’s next to the CH2 is [inaudible]
that’s an S?>>So this is the S
stereocenter, right. Now, one thing that’s
interesting, and we’ll get to this when we talk
about the Fischer proof of stereochemistry
is until many years after the relative
stereochemistry, the diastereomers of sugars were
determined, people couldn’t tell which enantiomer was which. And so arbitrarily
it was chosen. Eventually it was
discovered what was real. But all of these relative
stereochemical relationships and the terms D and L apply
regardless of whether D is R, as we saw it was, or whether
D is S. Other questions at this point? So we would say that L-threose
is the enantiomer of D-threose. L-threose is a diastereomer
of L-erythrose. And L-threose is a
diastereomer of D-erythrose. So that gives us sort of a basic
introduction or re-introduction to stereochemistry and shows
us how we can use these Fischer projections very
nicely and quickly. So I want to take us
through a few other forms, a few other ideas in sugars and
then we’re going to come back to some reactions of sugars. So again, there are two
different diastereomers of erythrose — of aldotetroses. And then there are
four of aldopentoses and eight of aldohexoses. And we’re not going to learn
or think about every sugar, but I’ll give you a
few as an example here. So, I think in terms
of the aldopentoses — I’ll write aldopentoses,
five-carbon sugars. The one that’s got to be the
most important is ribose. And ribose happens to
have all of the OH groups on the same side in
the Fischer projection. So I will write D-ribose. Why is ribose important?>>It’s in RNA.>>It’s in RNA, deoxyribose
is DNA, deoxygenated at the 2 position. And so this is a tremendously,
tremendously important sugar. Now, we’ve learned to go
ahead and think about the — let me go ahead and do
this on this board here — we’ve learned to recognize
that sugars have a cyclic form. And generally, even though I’ve
been drawing them predominantly — or been drawing them
as the free aldehyde, remember we talked about how
the cyclic form, the hemiacetal and the aldehyde form
rapidly interconvert. We said glucose, for example, is only 0.003 percent
of the open form. Ditto for ribose. In water, ribose is
predominantly the closed form. And there are two
different structures, two different isomers
that exist. One of them is a
six-membered ring isomer. And this is — we’re
not going to focus. Remember I said this
squiggly lines means mixture of alpha and beta. [ Writing on Board ] Beta is up, alpha is down. We’re not going to focus on that
stereochemistry, that mixture of alpha and beta anomers
will exist in equilibrium. One with the OH up, the beta,
and the other with the OH down. That stereocenter isn’t fixed. It’s labile. But we will focus on the others. And if you just imagine
picking up the molecule and laying it down, you can see
in our Haworth projection here that our OH’s are
going to go down. And maybe the only one that’s
tricky in one’s mind eye to convert this Fischer
projection to the Haworth projection, maybe the only position
that’s tricky is that next to the bottom one. And so just imagine in
your mind’s eye we’re going to rotate, rotate and
rotate, like that. And how you say, okay, so
we bring this OH down here. The CH2 comes here. And that hydrogen comes here. And so if you can do that
rotation in your mind’s eye, you pick it up and
you say okay — actually, we’re going
to — I’m sorry. For this one — we’ll do that
next for the furanose form. For the pyranose
form we’re just fine because we just cyclized
on this oxygen. So we bring this oxygen
around to that hydroxy group. And so we have three
OH’s pointing down. So I was getting
ahead of myself. Okay, so here’s the
Haworth projection for the pyranose form. [ Writing on Board ] And in solution,
ribose exists in water; it exists as 76 percent of
the pyranose form as a mixture of alpha and beta anomers. But the other form that it
exists in is the furanose form. The furanose form is
the five-membered ring. It’s the one that when you
see DNA either in the chapter or in your biochemistry
courses, you think DNA, and RNA, I guess more specifically. So we’re going to draw
the pyranose form. So the pyranose form is
going to be cyclizing not onto this oxygen to make
a six-membered ring, but onto this oxygen to
make a five-membered ring. So now — and I guess
I’ll raise this up a little bit to give me room. So now we’ll draw the Haworth
projection of the pyranose form. So if we look at that, we’re
going to cyclize on this oxygen. Well, that’s still going
to — as we pick this up — see my finger is here,
they’re pointing out. I’m just picking that up. These two oxygens
are going down. Again, I’m going to leave our
stereochemistry unspecified at the anomeric position
because we’ll have a mixture of alpha and beta anomers. Now, as I said, to get over
there, we’re going to cyclize on this oxygen, so we’ll
just bring this oxygen down, rotate this CH2 up,
rotate that hydrogen over. We’re just rotating about a
bond as a mental operation. So when you do that,
that puts this CH2OH up and the hydrogen down. So that’s the pyranose form of
ribose — the furanose form. Boy, I can’t talk today. So let me — let me write
pyranose form and furanose form. And this is 24 percent
at equilibrium. And you should be proficient
enough, and of course this is at equilibrium in water. And you should be proficient
enough to do these operations in your head that
I’ve just done. Now, the best way to train
yourself is to begin by working with models, plastic models
are a great way to go. Dig them out from 51A. Computer models are
a great way to go. And I’ve given you those
tools to start with. A little bit of imagination. Remember, I gave you glucose,
just a little bit of imagination in rotating in [inaudible],
that structure I gave you for glucose will allow you
to see this relationship. Or, of course, you can just edit
it and put in the hydroxy groups in the right place
and delete it. But I’ve given you
that linear form which is representing
our Fischer projection. I’ve given you a cyclic form that represents the Haworth
projection of the pyranose form. So as you work through
these, you will become better and better at seeing the
relationship in your head. [ Erasing Board ] All right. All of the sugars that
I’ve draw thus far, all of the monosaccharide that I’ve drawn thus far are
aldo sugars, aldotetroses, aldopentose here, an aldohexose. Let me show you a keto sugar. So remember, keto sugars
are based on ketones. So we call them ketoses. And I’ll show you a ketohexose. And my philosophy has been to
teach you some important ones. We’re not going to
become sugar chemists. We’re not going to master
all eight diastereomers of the aldohexoses. And then all of the various
possible ketohexoses. But let me give you one of them
that you will know of by name. [ Writing on Board ] And again, I’m going to draw
it as a Fischer projection. And so this looks very much
like glucose except, remember, glucose had an aldehyde up on
top and had a hydroxy group off to the right at the
position number 2. Here we have a ketone group. We don’t have stereochemistry
at this position, but I have to draw the
carbonyl somewhere. So we have three stereocenters. This sugar is fructose, or
more specifically D-fructose. You’ve, of course, heard
of fructose, right? You’ve heard of it,
it’s the fruit sugar. It’s in fruit. You’ve heard of it, alas,
in high-fructose corn syrup, which is produced by
hydrolysis of starch. High-fructose corn
— starch from corn. Corn is very starchy. Starch is a big polymer
of glucose. And upon acid treatment,
fructose — glucose can isomerize
to fructose. Maybe I’ll give that to you
as a mechanism at some point. That’d be a good exam problem. Anyway, so coming
down to fructose — so okay, so fructose
is a keto isomer. And fructose exists
in a pyranose form and it’s 60 percent
in the pyranose form. And again, all of this is
at equilibrium in water. That would be cyclized
onto this hydroxy. This hydroxy going to form
an acetal with that carbonyl. And it’s 40 percent
in the furanose form. That would be with this hydroxy
cyclized into that carbonyl. Now remember, we’ve
had this theme that aldehydes are
more reactive. They’re less stable
than ketones. And we see this in the
sugars very nicely. So glucose, there’s very, very,
very little of the free aldehyde in solution, .003 percent. Fructose, there’s
still not a whole lot of the free ketone in solution. But there’s a good bit more. There’s about .25 percent
of the ketone in solution. I’ll say open form. By open form, of course,
I mean the ketone. All of this is in
aqueous solution. So I’ll write in water. And to me, in my mind’s eye, that kind of gives me this
feeling about stability. We saw this with
aldehydes and ketones. Acetone in water
is stable enough that you have very little
of the geminal diol. Acetaldehyde has
about 50 percent as the geminal diol in water. The aldehyde is much less stable because you don’t have two
electron-donating alkyl groups. And we see this here
as well that you have about 100 times more of the
keto form, of the open form of the carbonyl form in
fructose than you do in glucose. And so again, that
sort of says, yeah, aldehydes are a lot less stable,
a lot less happy than ketones. All right. This introduction to sugars, and indeed this whole
chapter is only going to be an introduction. This really sets us up to introduce some
chemistry of sugars. And the first chemistry
I think I want to introduce is the
hydrolysis of glycosides. [ Writing on Board ] We started, we looked
at lactose as a sugar. Your body, if it does
break down sugars, breaks it down into the
monosaccharide units. Some people end up not
being able to digest milk, meaning they don’t have
the enzymes to help break that glycosidic linkage
to break it down and then you have bacteria do
it for you in the wrong place in your body and give you gas. All right. So I don’t want us to
start with a big sugar, but I want us to
see another sugar. And so I’ll show you
another disaccharide here. So remember I mentioned
starch would be one that we can break down as well. But I’ll show you a simple one. And we’ll use one glucose
unit and one fructose unit. This is, of course, the
one with one glucose unit and one fructose
unit is sucrose. So another very,
very important sugar. And in many of the glycosides
you have an alpha linkage between the sugars. So starch has alpha linkages and cellulose has the
more stable beta linkage. So alpha means down. So here we have our
glucose unit. And now I’m going to
connect to another ring. And that ring is going
to be a fructose unit. And so we have a pyranose
ring, a five-membered ring. And I’m going to
go ahead and put in all my parts of
fructose here. So I have a CH2OH group. Here’s our anomeric position. I have an OH group off of
this position going down. Remember, people often
do a little broken line to show it’s in back. And then I have a CH2OH
group going up there. So that’s a glucose unit
linked by a glycosidic linkage. We call this oxygen
a glycosidic linkage. A glucose linked by a glycosidic
linkage to a fructose unit. Now acid catalyzes the breakdown
of glycosidic linkages. And so if we treat
sucrose with acid in water, I’ll write this as H3O plus. That would, of course, be a
strong acid dissolved in water. HCl is widely used,
although sulfuric acid in water could be used as well. Remember con. sulfuric acid dehydrates sugars. But in water, you
don’t have that. And if we go ahead and we treat
our sucrose with strong acid, for example boiling it up
in 6 molar aqueous HCl, which is popular as a good
way to hydrolyze a variety of different compounds, we
hydrolyze the glycosidic linkage and get our self one
glucose unit, like so and one fructose unit, like so. [ Writing on Board ] Oops, and I have dropped — oops
— I have in fact made a boo-boo in this structure here. And I want to correct
this right now. So, at this position,
we have a hydroxy group, at this position we
have a CH2OH group. So just to correlate
this with the structure on the other blackboard
starting from the top, our 1 position, our 2 position. That was the ketone. It’s now an acetal. Our 3 position, that’ the
hydroxy in the middle. Our 4 position. Our 5 position, that’s
that oxygen off on the bottom right next — you
know, this position over here. And the last one is our CH2OH. All right. So, let me finish
correcting this structure. And so here’s our
structure of fructose. Now, I’ve drawn this initially
as the beta anomer here. And I’ve drawn this
initially as one anomer. Eventually both of them
equilibrate to a mixture of alpha and beta anomers. [ Writing on Board ] [ Erasing Board ] All right. One of the things about these
molecules, one of the things about organic chemistry in
general is that molecules behave as their functional groups do. And if you look at the
molecule of sucrose, you will see embedded in the
structure two acetal linkages. So you will see an
acetal group over here. And an acetal group over here. And we know what acetals
do in aqueous acid. They hydrolyze. And we know the mechanism
for that. The molecule’s bigger, but
the chemistry is the same. And so the mechanism for
this is going to involve, and I’ll just sketch this out
briefly, the mechanism is going to involve protons going
onto oxygen, pushing apart to give an oxocarbenium ion,
a leaving group leaving. And then it’s going to
involve attack by water. So let me summarize that. Here’s our fructose molecule. And so we’re in aqueous
solution. And so we can imagine
protons go on, protons go off in
aqueous solution. And so the proton, all
the different oxygens, but since we’re breaking
the anomeric linkage, since we’re breaking
the glycosidic linkage, the point at which we’re set
up to break that linkage is when a proton goes onto the
glycosidic linkage oxygen. And so I’m going to walk you down that productive
pathway like so. In other words, oxygens — protons can go on and
off other oxygens, but this particular pathway
happens to lead us somewhere. And so I will show you that one. So we protonate this
oxygen like so. Now, that’s very good. Because that sets us up for
this oxygen to leave and for us to form an oxocarbenium ion. You got it in your head here? In other words, we push in with
that lone pair of electrons. It pushes out, pushes
electrons onto the oxygen. And now because our sucrose
molecule is so bloody big, I’m going to have to take us
across to the other blackboard. [ Erasing Board ] And so I guess — let me
continue here to get us to the other blackboard. I will show an equilibrium
arrow. And then we’ll just
swing over here. And now use your imagination, and we have a molecule
of glucose. [ Writing on Board ] Told you you’re going
to see a lot of glucose. And, we have the
oxocarbenium ion from fructose. And so I’ll draw that here. [ Writing on Board ] And now, to complete
our equation, the oxocarbenium
ion isn’t stable. And so to complete our
equation water can attack. And I’ll shorthand things. And we can lose a proton. And I will simply say here’s
our fructose structure. [ Writing on Board ]>>How do you do it from the
— on the opposite [inaudible].>>Thank you. Yes. Okay. And so, it has been very aptly
pointed out that you can do it by using this oxygen
to push out fructose. Now here’s why I did
it the way I did it. My way is a little more
right, a little more real. Just as the aldehyde is
less stable than the ketone, the oxocarbenium ion from
an aldehyde is less stable than oxocarbenium ion
from a ketone because of that electron donation
of the alkyl groups. And so I push to give the
better oxocarbenium ion, but at your stage
and at your learning, yes absolutely we
could go either way. [ Erasing Board ] All right. So that gives us kind
of an introduction. And we can see lots and
lots of other chemistry. For example, and I’m not going
to take you through the details of the mechanism
because I expect at this point you should
be able to go ahead and do this mechanistically. You’ve now had both
this little review of oxocarbenium ion
chemistry as well as all the acetal chemistry. But I want to show you
a fundamental reaction. If you take acid and a sugar — so here in the previous
example we saw a hydrolysis of a glycosidic linkage. Now I want to show
you a formation of a glycosidic linkage. [ Writing on Board ] Call it a formation
of a glycoside. If we take an acid and
use an alcohol as solvent, so you’ve got lots and lots
of alcohol, CH3OH, or ethanol. I’ll do this with methanol. And again, of course,
you can’t go to Kim’s stockroom
and buy the H plus. That means you do this with
HCl or H2SO4 in methanol. Now, by the exact same type of
chemistry, we can form a mixture of the beta and alpha
glycosides. [ Writing on Board ] So on the right — on the left
I have the beta methyl glycoside of glucose. On the right I have the alpha
methyl glycoside of glucose. And of course, both
of these form by protonating the
hydroxy group. You kick out to give the
oxocarbenium ion, like so. [ Writing on Board ] And then methanol can attack
from either the beta face to give the beta anomer,
attack and loss of a proton. Or from the alpha face
to give the alpha anomer. So, in your mind’s
eye, protons go on, protons go off all different
positions of the molecule. But if we protonate on this
hydroxy, we can push it out with a lone pair to
get the oxocarbenium ion. And now we’re in alcohol. We have a huge excess
of alcohol. And alcohol can attack
from one face or the other.>>So you’re sweeping up the
hydrogen, the [inaudible], the methanol [inaudible].>>The methanol is — oh,
attacking at the carbonyl.>>Oh, okay.>>So, I was only showing it
from one side or the other. So methanol attacks the carbon. We form a new carbon-oxygen
bond. At that point you have
protonated methanol. You lose a proton. We’re in this vast
sea of methanol. Protons go on, protons come off. Thoughts? Your mind’s
engaging with this, right? You’re figuring the
mechanism in your head. And this is good. This is good, because
ultimately, what you’re doing is
working through and wrestling with the motion of electrons
and the formation of bonds here.>>Can we reform the
glycosidic linkage with the same [inaudible].>>Can we reform the
glycosidic linkage with –>>With the same
mechanism, like, you know, H, like the OH gets like
protonated and gets dehydrated?>>Ah, in other words,
can we go back?>>Yes.>>Can I mix two molecules
of glucose, for example, or a molecule of
glucose and a molecule of fructose and go back? In theory — great question. Really important. And a contemporary super, super important problem
in sugar synthesis. Now, the problem with going
back is we’re always dealing with an equilibrium. We took out — we added water
in the hydrolysis direction. You want to take water out. And the problem is the things
that take water out also react with the other hydroxy group. Pure sulfuric acid is
great at taking water out. However, pure sulfuric
acid also ends up reacting and hydrolyzing the sugar. So yes, when people want to
reform glycosidic linkages, they do this by something
very special. It’s the same idea. Rather than having a hydroxy
group at the anomeric position, they make the hydroxy group
into a good leaving group, something that won’t go back. And then the reaction goes. They can use chloride
at that position. They can use various derivatives
of sulfur that often get used. It goes beyond the scope of this
course, but there’s a beautiful and rich sugar chemistry. And we’re just getting —
pardon my horrible pun — a tiny little taste
of sugars here. All right. So, I want to continue
with our taste of sugars talking
about some reactions. So in these two examples,
the hydrolysis of sucrose and the formation of
a methyl glycoside, we saw that all we’re getting
is the behavior of aldehydes, hemiacetals and acetals. Even the anomerization
was just acetal behavior. Sugars also have lots and
lots of alcohol groups. And we’ve learned lots
of chemistry of alcohols. Your textbook chose to
give you a few examples. And I’m going to take
you through those. There’s also some nice examples
in the homework problem. But there’s plenty more
rich chemistry of sugars. All right. Think way, way back to 51B. And you learned the
Williamson ether synthesis. You learned that if you took an
alcohol and a suitable base — and I’m deliberately being dicey
here because there are lots of bases and lots of conditions. And I’ll give you one
for sugars in a second. And an alkyl halide, I’ll call
it R prime X, suitable for SN2. Now what do I mean
by suitable for SN2? Well, methyl is great. Methyl iodide, methyl bromide
are great at SN2 reactions. Allyl, benzyl, very, very good. Allyl is double bonded. It’s a propenal halide. Benzyl is a benzene CH2 halide. Those are all really good. Ethyl, okay as well. By the time you get to
branch like isopropyl, not so good in the
Williamson ether synthesis, you go ahead and get an ether. Well, sugars can do
the same chemistry. And again, with a suitable base, some of the base chemistry’s
a little special for sugars. I’m going to give you an example
that actually is basically one of your textbook problems and
is very, very close to things that you’ve seen before in the
Williamson ether synthesis. Suffice it to say
there are many bases. So let me take our good friend
glucose here, and I’ll take it as a mixture of alpha
and beta anomers. And we’re going to treat it with
a strong base, sodium hydride. And I’ll put in parenthesis
excess. And I’ll give us benzyl bromide, or let’s say benzyl
chloride as a halide. And again, we’ll use
an excess of that. Normally, unless I
indicate it otherwise, even if I didn’t write
excess, I would mean that you use plenty of it. Basically, your sodium hydride
can pull off all the different hydroxy protons making
them into alkoxides. Sodium hydride is
a very strong base. And your benzyl halide is
a good alkylating agent in SN2 alkylation. So one can go ahead
and cleanly — and chemists abbreviate the
CH2 benzene group as Bn. So you can cleanly
SN2 alkylate all of the exposed hydroxy groups. And this is a beautiful example,
not only of the chemistry of sugars, but also of the fact that basically all the
stuff you learned in 51A and 51B still works even in
much bigger and more complicated and biological molecules. [ Erasing Board ] Now again, not trying
to get too diverse, not trying to show you too
many different reactions. I’ll keep to things that
are in your chapter. So another reaction
you learned, in fact, that we learned just
back in the quarter, earlier in the quarter was
that you can make an ester by reacting an acid chloride or an acid anhydride
with an alcohol. And so if I take some
alcohol ROH, and I take — I’ll write it out here, acetic
anhydride or acetyl chloride and we take a suitable
base like pyridine, that our alcohol reacts
to give an acetate ester. And sugars have lots
and lots of alcohols. And another reaction that
works very, very well is to make esters, particularly
acetyl esters, because acetic anhydride is
small and it’s highly reactive. And so again, I will take
our good friend glucose. You could do this with
galactose if you wanted. You could do this with mannose. You could do this with ribose. But if I take our good
friend glucose and I treat it with acetic anhydride, which
I’ll abbreviate as Ac2O. That’s just that structure
up there, and pyridine, which I will abbreviate pyr. Remember pyridine is the
heterocyclic base that’s less basic than triethylamine because
the lone pair of electrons is in an sp orbital and
held more tightly. It’s a very mild base
albeit somewhat toxic. Now we get the acetate. And we’ll abbreviate that OAc. Remember that’s just an
abbreviation for that. [ Writing on Board ] And so we can make the
acetate of glucose. [ Erasing Board ] All right. So, we’ve seen chemistry of
alcohols, we’ve seen chemistry of hemiacetals and acetals. Let’s focus in on some
chemistry of aldehydes. So the aldo sugars
have an aldo group. And if it’s not tied up,
if it’s in a hemiacetal, that aldehyde group
is available. And so, just to refresh, we learned that aldehydes
are very easy to reduce. When we were talking about
ketones and aldehydes, we learned, for example, that sodium borohydride NaBH4 is
a wonderful mild reducing agent. Now, one of the things
about sugars in solvents is, you know, like dissolves like. Generally sugars like to
dissolve in polar solvents, water, methanol,
things like that. They generally are not
so soluble in things like tetrahydrofuran,
which is much less polar. So methanol, water, great
solvents for sugars. So I’ll show you a reduction
of a generic aldehyde with methanol to RCH2OH. We add hydride anion and then
protonate the resulting alkoxide to give the alcohol. And so it shouldn’t surprise us that sugars have some
similar chemistry. And so I will take a sugar — I’ll get away from my
friend glucose here. We’ll come back to
him in a moment. So I’ll take a different sugar, one that you will have
unbeknownst to you, probably have heard of. [ Writing on Board ] Who’s chewed sugarless gum here? Okay. If you look on the
back of sugarless gum, you will often see that it
contains sorbitol and mannitol. Now this sugar is
called mannose, or more specifically D-mannose. [ Writing on Board ] Mannose is the same as
glucose except it’s epimeric at the 2 position. If we take D-mannose
and we reduced it with sodium borohydride, there are of course other
reducing agents including catalytic hydrogenation, which is probably how
this is done industrially. [ Writing on Board ] So if we reduce our D-mannose, we get D-mannitol,
M-A-N-N-I-T-O-L. Kind of makes sense as a name. Mannose goes to mannitol. Doesn’t make any sense
why the common name for the alcohol you get
when you reduce glucose is called sorbitol. But you will see that on
your gum there is a mixture of sorbitol and mannitol,
that is a mixture of epimers at this position. All right. So, all of this chemistry
we’re seeing here really is very familiar. And in the back of
your mind’s eye, in the back of your head
can be this little footnote. Glucose exists in equilibrium between the pyranose
form and the open form. Even though there’s very little
of the aldehyde form because of that equilibrium, the
aldehyde can form, be reduced. More aldehyde forms by Le Chatelier’s
principle, it’s reduced. And eventually all
of our glucose or all of our mannose is reduced
to sorbitol or mannitol. All right. So, other chemistry. And this will set
the stage for talking about sugar structure
determination. So, when we talked
about aldehydes, we learned that aldehydes
were easily oxidized by mild oxidizing agents. Even, for example, silver
oxide, Ag2O in aqueous ammonia, NH4OH or aqueous NH3,
however you like to write it, ends up oxidizing the
aldehyde to a carboxylic acid. And you can do this selectively
in the presence of alcohols. So, it shouldn’t surprise
us that you can go ahead and do this same chemistry
with sugar aldehydes. So for example, I’ll come back
to my friend glucose here. I just needed to
pick some examples for today’s class,
so I’ll take glucose. It works with all the sugars in which you have a
free aldehyde group. And so if I take glucose and I
expose it to these conditions, so I’ll write D-glucose
over here. And we expose it to
the same conditions. And just to save on writing,
I’ll write a ditto mark. Now we get the corresponding
carboxylic acid. [ Writing on Board ] And this has a much
more sensible nickname. This is called D-gluconic acid. [ Writing on Board ] Wouldn’t expect you to know all
of these names, but they’re kind of fun to roll of the tongue. All right. When one does this
reaction in the laboratory, and it’s often done as a
laboratory demonstration of this property, what happens
is the silver gets deposited on the flask, because you’re
going from silver oxide. It’s acting as an
oxidizing agent. You’re going to silver metal. It’s getting reduced. The silver metal gets
deposited on the flask. And it makes a beautiful
mirror of silver. The whole flask, in about
10 minutes of shaking, becomes a bright silver mirror. Well, nowadays there
are lots and lots of good ways to tell
sugars apart. But back in the 19th Century, people had only chemical
reactions and only properties. And so it was noted that
some sugars like glucose, reacted with silver
oxide, silver nitrate and ammonia make silver oxide,
reacted with silver nitrate and ammonia to give a
bright silver mirror. And others like sucrose,
under the same conditions — I’ll just put ditto
– go to no reaction. You don’t get a bright
silver mirror. And so it really,
really stands out. The sugars that reacted
were called reducing sugars. So glucose is a reducing sugar. Sucrose is not a reducing sugar because it doesn’t
react in this reaction. And this became a
common classification. If a sugar has what basically
is a free aldehyde group, in other words it’s a pyranose
or a furanose in equilibrium with an aldehyde,
then it reacts. If, on the other
hand, like sucrose, all of the aldehyde groups are
tied up by an acetal linkage, in sucrose it’s just
one aldehyde group, then it doesn’t react. And that would be called
not a reducing sugar. And so this test, this silver
mirror test got called the Tollen’s test after the
inventor, for reducing sugars. [ Writing on Board ] [ Erasing Board ] All right. So, one of the themes, and
we’ll see this for the rest of the today’s lecture. One of the themes of organic
chemistry historically has been what’s the stuff
of life made of? What are the molecules that
make up the food that you eat? What are the molecules that
make up your skin and your hair? And we now know the
starch and corn starch and wheat bread is
an oligosaccharide. We know that it’s made
up of glucose units. And we know that the proteins
in your hair and your skin and your arm are all
made up of amino acids. But, the end of the 19th
Century and the beginning of the 20th Century, this was
when people discovered that. And so the chemistry
we’re seeing now, and some of these questions like categorizing is a sugar
a reducing sugar or not, is part of that understanding. So I want to show you a
couple of more reactions and go somewhere with it. So, I’ll give you one more —
two more reactions of glucose. Now, aldehydes are very,
very, easy to oxidize. And so there are lots and
lots of oxidizing agents that can oxidize
a reducing sugar. And another one is bromine,
another one is copper, copper 2 or copper 1, I forget now, that’s the Fehling
test for sugars. But bromine is another
one from your textbook. And so I’ll just show you that. Bromine can also take
glucose to gluconic acid. So, just as we did
with the Tollen’s test. Now I want to show
you another one. And then we’re going
to go somewhere. So again, I’ll take
my friend glucose. [ Writing on Board ] And now we’re going to treat it with a different
oxidizing agent. This is a really neat reaction. We won’t talk about
how this occurs. We’re going to take this
with nitric acid in water. Your textbook doesn’t mention
heat, but usually you do this with boiling water
— in boiling water. So you do it at 100 degrees. Nitric acid is a
good oxidizing agent. And it’s not surprising
that the aldehyde group ends up getting oxidized
to the acid group. And it’s probably
not surprising, since we’ve seen lots
and lots of conditions that don’t touch alcohols
but do touch aldehydes. It’s probably not surprising that the remaining
alcohol groups in the molecule remain
intact, with one exception. And that is the primary
alcohol group also gets oxidized to the carboxylic acid. And so now we’ve
transformed glucose at the top end and
the bottom end. And again, things
have funny names. This compound here is
called D-glucaric acid. And again, I wouldn’t
expect you to know this. These names just are fun to say. What is fun — what is
interesting, of course, and what we want to pick out
in picking out the forest from the trees here is that it
is a diacid, that is a molecule with two carboxylic
acid groups in it. [ Erasing Board ] All right. So, I want to now take us back
to the beginning of the lecture. Now, at the beginning of
the lecture we started with the aldotetroses. And I drew out the structure
of the two D-aldotetroses. [ Writing on Board ] And I know people say
organic chemistry is about remembering things. Well, it’s not. It’s about thinking. We’ve seen mechanisms
of protons. But now I’m going to
ask you to do something. I’m going to ask you
to forget something. Forget which aldo
tetrose is which. [ Writing on Board ] And more, get into the way back
machine when we didn’t know which aldose is D-threose and
which aldose is D-erythrose. [ Writing on Board ] And get back into
the way back machine when we didn’t have
IR spectrometers and NMR spectrometers to look
at things like functional groups and coupling constants. People could burn molecules and measure how much
carbon dioxide was produced and how much water was produced. You could figure out the
empirical formula that way. And people knew the
colligative properties of matter and knew the freezing point
depression, and could figure out the molecular
weight that way. But, you couldn’t directly ask
questions of stereochemistry like you can now with an NMR
spectrometer telling you a cis alkene from a trans alkene. And come back to the
roots of organic chemistry where the question is what is
the stuff of life made from? And put yourself in the
laboratory of Emil Fischer, who won the Nobel Prize in
1902 for his work with sugars. His name permeates not only
the Fischer projection, but the Fischer esterification
and also a synthesis of proteins, of peptides
that interested him after he tackled sugars. He was going after all of life. He also got some of
the nucleic acid bases. And so now, we look at
what you could do then. And you go back to the
time of Louis Pasteur and tartrate and
optical activity. And Le Bel, and understanding
that carbon was tetrahedral, and understanding the
origins of optical activity. And only being able to measure
whether light is rotated, plane polarized light,
is rotated and how much it’s rotated. And melting points and formulas. And you ask, what are
the structures of all of the eight aldohexoses
before — I should say diastereomeric
aldo hexoses, because of course when we throw in
enantiomers there are 16 — – the four diastereomeric
aldo pentoses and two diastereomeric
aldotetroses? And we’ll just ask the simple
question for the aldotetroses. Fischer extended
this all the way up. So let’s look at
what you could do. All right, when you take one
aldotetrose, and you treat it with nitric acid and water and
heat at 100 degrees Celsius, you get an optically
active diacid. The diacid, and I’ll draw the
structure in just a second, is called tartaric acid. The diacid in which you oxidize
the upper and lower position. And when you take the other
aldotetrose, and you subject it to the same, you get a diacid that doesn’t rotate
plane polarized light, that’s optically inactive. [ Writing on Board ] All right. So let me go ahead and draw,
using Fischer projections, yeah?>>The other [inaudible], is
it in the same conditions?>>Oh, yeah. Ditto mark. Same conditions. All right. So, let me write the two
structures here of two diacids. [ Writing on Board ] Which of these molecules has
a plane of symmetry in it?>>The one on the right.>>The one on the right. It has a plane of symmetry here. [ Writing on Board ] Molecule that has
stereocenters but has a plane of symmetry is a meso compound. It’s optically inactive. A diacid with two carbons on
either end, a four-carbon diacid and two hydroxys in the middle
is called tartaric acid. The structure on the right,
then, is meso-tartaric acid. And the structure on
the left is optically — so I’ll write optically
inactive up here for the one on the right. The structure on the left
lacks a plane of symmetry. It’s optically active. I’ll write optically
active over here. And this particular molecule
is called D-tartaric acid. It’s found in grapes. It’s the structure
that Pasteur was able to resolve enantiomers from. Okay. So now we have a
little logic problem. Your textbook gives you
tons of logic problems. I’ve tried to cut them back
on the homework a little bit because I think they
go a little nuts on it. This is the history of sugar
chemistry they’re giving you. All right. So, that means if the one gives
an optically active product, the structure of that
aldotetrose had to be this. And so this, the one
that was called threose, then Fischer knew
was this structure. And the one that was
called erythrose — [ Writing on Board ] — Fischer knew was
this structure. [ Writing on Board ] And that’s pretty smart. All right. Now I will show you the real
genius bootstrapping of Fischer. And we’ll introduce
one last reaction. And that is the way that
Fischer could bootstrap himself up to all of the aldohexoses, starting with a very
small molecule. And so what I’ll
introduce finally is the Kiliani-Fischer synthesis. [ Writing on Board ] Your textbook introduces the
Wohl degradation, the opposite. And I’ll let you
read about this. But I want you to see the
forest from the trees. All right. So, how do you bootstrap
yourself up? How do you build up the sugars? And then hence, how do
you go ahead and puzzle out all the stereochemical
questions of the aldotetroses, the aldopentoses
and the aldohexoses? And ultimately, how do you
really answer what the stuff of life is made of? So Fischer started
with D-glyceraldehyde. [ Writing on Board ] And all the chemistry,
now, you have. In fact, you have all the
chemistry you need to get up all the way to the
aldohexoses, although you won’t. If you take an aldehyde and you
treat it with hydrogen cyanide, we already learned what happens. You form a cyanohydrin. [ Writing on Board ] And you do so in a
non-stereospecific fashion. In other words, you
get both diastereomers. So we formed a carbon-carbon
bond. We’ve introduced
a nitrile group. Now we’ve learned lots and lots of chemistry of the
nitrile group. We learned that the
nitrile group was in the carboxylic acid family. We learned that under the right
conditions you could with, for example, aqueous acid,
hydrolyze the nitrile group to the carboxylic acid group. We also learned various
conditions for reduction of nitriles. And one of those conditions
is catalytic hydrogenation. So imagine now that we
go ahead and we carry out catalytic hydrogenation
with hydrogen and palladium. Palladium’s a catalyst, but
you can attenuate the reducing ability of palladium by
poisoning it with a heavy metal. You learned this in the
reduction of alkynes with Lindlar catalyst. Palladium can be poisoned
by metals like lead and barium heavy metals
that reduce its ability to hydrogenate, that
attenuate it. And so if we have palladium on
barium sulfate and you do this in water, you end up
reducing the nitrile groups down from the nitriles slash
carboxylic acid oxidation state, that plus 3 oxidation state, to
the aldehyde oxidation state. Remember, carboxylic
acids are plus 3. It’s the highest oxidation
state you can go without getting to the carbon dioxide
oxidation state. So you reduce the triple bond
to a double bond and you get — and I’ll come to your
question mark here, what will inherently be
your implicit question mark in a second — and so you
bootstrap your way up one carbon to erythrose and threose —
or threose and erythrose, as I’ve written them
on this blackboard. [ Writing on Board ] And so Fischer could
start with one molecule. At the time, he didn’t know
the absolute stereochemistry, but he could start with a
molecule of D-glyceraldehyde and convert it to
erythrose and threose, which he could then tell apart by their optical rotation
properties upon oxidation. And he could bootstrap himself
up to the pentoses and so forth. So the one question here you
probably have is, wait a second. Where did that nitrogen go? Well, the primary
imines are not stable. And so in water, under
the conditions — your textbook shows acid. That’s fine, too. But honestly, in
water they hydrolyze. In water, the imine
hydrolyzes to the aldehyde. And so this sequence of
cyanohydrin formation followed by partial reduction
bootstraps you up a level. All right. Well, I think this is where
I would like to end up. We have now had all of the
basics of sugar chemistry. You’ll get some more
exposure through the homework. We’ll finish off our final week
talking about the chemistry of peptides and the chemistry
of organometallic reactions. And we’ll just get a little
taste of that chemistry from Chapter 28 and Chapter 26.
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