>> Good morning.
>> Good morning.
>> So I'm excited today,
we begin the second half
of the course today
which, of course,
means yes today's lecture
won't be on the mid-term.
So okay, the usual question
but no, seriously I'm excited
because now we get
on to the topic
that I hinted back
way in lecture 2.
I said that basically the bulk
of this course is covering
carbonyl chemistry.
And even beyond lectures,
the 6 weeks of the class,
we continue with the theme
of carbonyls in the very end
of the class where we talk about
carbohydrates and about peptides
and proteins but we've
talking for the past,
essentially 5 weeks,
about the electrophilicity
of the carbonyl group.
And now we're getting
into this second half
of carbonyl compounds, the
fact they're nucleophilic
at the alpha carbon under
the right conditions.
So back in our first week,
I drew a diagram
sort of like this.
I said okay, here's some
generic carbonyl compound.
Not necessarily a ketone
or ester or aldehyde.
We could mean an ester.
And the general gist of it is
that the carbonyl
compound is electrophilic.
It can react with
the nucleophiles.
And now we come to
this second half
and so I'll say now we
bring in this notion
that the alpha carbon under
the right conditions can react
as a nucleophile.
In other words, it can
react with electrophiles.
[ Background sounds ]
And we're going to be
exploring this idea extensively
over the next 4 lectures.
We're going to be spanning
this idea from Chapter 23
in your Smith textbook
to Chapter 24
and discovering the richness
of the chemistry of enols
and enolates which
are the intermediates
by which we have this reaction.
Okay. So we talked before
about various aspects
of the carbonyl compound
structure.
And we said -- we talked
about the reactivity
of a carbonyl compound.
We talked a little bit
about in water equilibrium
with a hydrate species and
we said that, for example,
in acetone there's very
little of the hydrate.
And typically it's not a
big factor in the behavior.
In other words, if I took an NMR
[phonetic] spectrum of acetone
in water, I wouldn't
see the hydrate
because it's just a minuscule
amount of the compound.
Well, there's another
miniscule amount
that can be present that's
very important in reactivity
and that is the enol tautomer
so I'll write an equilibrium
where we have and let's just
take this as acetone right now.
We'll be going through
with various examples
but let's just pretend you
have acetone or some ketone
or aldehyde let's say.
And so we have this equilibrium
where we have another species
and now this species is
obviously very different.
In other words, you have
the carbonyl compound
and then you have a species that
has a double bond and an alcohol
and that's why it's
called an enol
because en means a double
bond and ol means alcohol.
And so we refer to this
as the keto tautomer
and this as the enol tautomer.
So tautomer means an isomer
where you've moved
hydrogen around.
And so you notice, of course,
if this is acetone here,
we have 3 hydrogens attached
to this carbon and then
in this particular
enol of acetone,
we would have 2 hydrogens
attached and one here.
So that it's the same formula.
It's the same molecular
formula and
yet a different arrangement
of atoms.
Tautomers are a type of
constitutional isomer.
Remember we've learned
about stereoisomers
and then we've learned about
constitutional isomers.
[ Background sounds ]
And by constitutional isomer,
I mean we have a
different connectivity.
Stereoisomers have
the same connectivity.
Now as I said the enol
tautomer is an equilibrium
with the keto tautomer and
that equilibrium lies way,
way to the left.
I'll tell you just how far
to the left in just a moment
but I wanted to introduce
our players.
Now, if we imagine
pulling off the hydrogen
of the enol tautomer.
In other words, deprotonating,
removing with that hydrogen
with a base, we get what we call
an enolate anion [phonetic].
[ Background sounds ]
And I'll be sort of
semi generic here.
In other words we
could be acetone.
We could be something else
but I want to show you
that there are 2 substituents
attached to the carbon
of the double bond, to what
we call the alpha carbon.
And the enolate anion
can be thought
of as having 2 different
resonance structures.
So we can envision 1
resonance structure
where we have our negative
charge on the oxygen
and a second resonance structure
where simply move
these electrons down
and move the electrons
onto the alpha carbon.
Remember we use a double headed
arrow not a reaction arrow,
not an equilibrium arrow
but a double header arrow
to indicate resonance
structures.
[ Background sounds ]
And I will very explicitly
draw my lone pairs
and draw my negative
charges like so and just
to be a good person, I'm
going to put this in brackets
to remind us -- to
help remind us
that these are 2 pictures
of the same thing.
It's not 1.
It's not the other.
Unlike the keto and
enol tautomer,
they are not 2 distinct species
that you can identify
distinctly.
They are 1 in the
same at the same time.
And so I will write
resonance structures here
to remind us of this.
[ Background sounds ]
And I will also remind us
that since this is the anion
of an enol, we call
it an enolate anion.
[ Background sounds ]
All right and these 2 sets
of variance on the carbonyl,
the enol tautomer and the
enolate anion are going
to be occupying our thinking
for the next couple of weeks,
for the next 4 lectures as
we explore the reactivity
of carbonyl compounds and
specifically their ability
to react as nucleophiles
at the alpha carbon.
[ Background sounds ]
All right.
In terms of understanding
reactivity
and this reactivity
is nucleophiles,
I think the enolate is a
very obvious place to start.
So if I draw the structure
of an enolate like so.
And again, I'm keeping
it kind of generic.
You could think of
this as acetone enolate
or you could think of as some
other enolate but in fact,
this is pretty much
universal behavior of enolate.
And we imagine some
electrophile.
Now again in the spirit
of keeping things generic,
I'm going to write it as E plus.
That doesn't necessarily
mean we have a positively
charged species.
You've seen positively
charged electrophiles.
You've seen carbocations
in the past.
Protons or H3O plus, of course
are cationic electrophiles
but by the end of
today's lecture,
you'll also be seeing
methyl iodide
and SN2 type alkylating agents
acting as electrophiles.
So here in very cartoony,
very loose fashion,
I'll write E plus
in quotation marks.
And the reactivity
at the alpha carbon
as a nucleophile can be thought
of as bringing our electrons
down from the oxygen,
bringing our electrons
over from the double bond
over to the atom bearing
the positive charge.
I guess technically if I want
to do this right my arrow
should end up at the atom
and again this is sort of
an abstraction of an atom.
And so if I want to
complete my drawing,
that constitutes our reaction.
Now if you're not
comfortable with moving all
of these curved arrows around,
realize that this is
just the representation
that this first part
of moving the electrons
and moving these electrons
is just the embodiment
of how we get from the
resonance structure on the left
to the resonance
structure on the right.
In other words, when I'm
interconverting these 2
resonance structures,
in my mind's eye,
I'm bringing these
electrons down
and bringing these electrons
onto the alpha carbon
but I don't necessarily
need to draw that out
because by this point in
my course and by this point
in your thinking, you
should be comfortable enough
with that operation
in your mind.
Okay. So that constitutes
the reactivity of enolates
and I just want to take a moment
before I take questions to talk
about the reactivity of enols.
Again, overviewed in this
very, very generic fashion.
So in the case of an enol, this
minor species that's present
in equilibria with ketones and
aldehydes and to a lesser extent
with other carbonyl compounds,
we can imagine here's our enol.
And again I'll have my
generic electrophile, E plus.
And we can imagine electrons
flowing down from the oxygen
into the carbon oxygen
bond giving rise
to a carbon oxygen
double bond concurrently
with electrons flowing from
the carbon carbon double bond
to the electrophile.
And if we continue with our
grammar of electron pushing,
our grammar of the curved arrow,
now that's going to leave us
with a protonated carbonyl and
a bond to our electrophile.
And now we can imagine
losing a proton.
Again, I'm talking about very
much in abstractions here.
So I'm not talking about
a proton just popping off
but a proton being taken off
by base and so at this point,
we can imagine going
ahead and getting to --
and I'll write that proton
in quotes because again,
one would envision
something taking it off.
So we can envision getting
to the carbonyl compound that's
reacted at the alpha position.
Question?
>> Yeah. Can a negatively
charged carbon also grab the
electrophile from one
of the same product?
>> Can the negatively
charged carbon,
absolutely it is
one in the same.
Whether I write this reaction
as a flow of electrons like such
or whether I write
the reaction as --
I guess again technically
to be correct
since it's a resonance
structure, I'm going to write it
in the exact same geometry.
Whoops, the exact same geometry
because the 2 resonance
structures share a geometry.
Okay. If I want to think
about it like this.
This is the self same thing.
It is the exact same
way of writing this
and so these are not
really different.
They're just different ways of
representing the same thing.
And as one moves in
their understanding
of organic chemistry, you
go from sort of needing
to explicitly write that
latter one out to recognizing
that the one image embodied
in the former which happens
to be the major resonance
structure,
the one image embodied in the
former ends up being enough
for us to think our way through.
Other questions?
[ Background sounds ]
All right.
Let's move on to some specifics.
So this provides an overview.
This provides a very generic
introduction to the chemistry
of enols and enolates.
And now let's talk a little bit
about some specifics
about the prevalence.
[ Background sounds ]
Of all the carbonyl
compounds with the exception
of dicarbonyls which
we're talking
about very soon in
a lecture or so.
With all the carbonyl
compounds, aldehydes
and ketones are the most enolic
with aldehydes being a little
bit more enolic than ketones.
Your textbook just gives
sort of a generic answer
that there's less than
1 percent enol present
for a typical carbonyl compound
but it's a lot less
than 1 percent.
There's very little enol.
So acetone is an equilibrium.
If you spray acetone
on your glassware,
you're spraying mostly acetone
but you're spraying a
little bit of acetone enol.
It's an equilibrium and the
position of that equilibrium,
k is equal to the
equilibrium constant is equal
to 1 point 5 times
10 to the negative 7.
In other words there's less
than 1 part in a million
of acetone enol present
at equilibrium.
Cyclohexanone, in general
more substitution makes
for a more stable double bond.
We learned this in the
chemistry of alkynes.
That, in general, when
you put alpha groups
on a double bond
it's more stable.
So cyclohexanone exists in
equilibrium with the enol
and it's a little bit
more enolic than acetone.
The equilibrium constant is
5 times 10 to the negative 5.
Again just a hair
more than 1 part
in a million whereas before
I guess we had a hair less
than 1 part in a million.
As I said in general, aldehydes
are a little bit more enolic
than ketones.
That's not necessarily
surprising.
Remember, the aldehyde carbonyl
group is pretty darn unhappy
as a carbonyl group.
It's only got 1 alpha
group donating in.
We saw this when we talked
about the chemistry of hydrates
and we said that acid aldehyde
exists about 50 percent,
about half hydrate and
half as the aldehyde form
in water whereas acetone is
vastly predominantly the ketone
form not the geminal diol.
And so similarly, we see
that acid aldehyde is a little
bit more enolic than say acetone
which would probably
be the best comparison.
So in acid aldehyde, you
again have an equilibrium.
Again, that equilibrium
lies way, way to the left
but now you have just a hair
more of the acid aldehyde enol.
Equilibrium is at 2 times
10 to the negative 5
for your equilibrium constant.
So compared to acetone,
a couple of orders more,
orders of magnitude more.
[ Background sounds ]
In spite of being a minor
equilibrium component,
the reactivity of enols is very
important in the reactivity
of carbonyl compounds.
And that makes sense.
Even if you have a
little bit of something,
when it reacts more
gets generated
and so a reaction can proceed
and proceed and proceed.
Being small doesn't
mean you're unimportant.
And so we can take a look
at how enols and ketones,
keto and enol tautomers
interconvert and look
at the mechanisms of formation.
Another question?
>> Yeah. So does formaldehyde
has more like enol form
because it's unstable?
>> Who wants to answer
this question?
Someone. Enol form
of formaldehyde.
>> I mean sorry --
>> Can no alpha carbon
on formaldehyde.
>> Sorry not the formaldehyde.
[Inaudible].
>> Acid aldehyde.
So --
>> Does acid aldehyde have
more like amino counterparts
at its acetone counterparts
because acetyl is more stable?
>> That's a very good
way of thinking about it.
Whenever you have
an equilibrium,
you're comparing the energy of
a ground state and a product.
And so the bigger the
difference in energy,
the bigger the equilibrium
constant though bigger whether
it's positive or
negative, you know depending
on whether it's greater than
1 or less than 1 depending
on how the energy
difference goes.
So anything that either lowers
the energy of the reactant
or raises the energy
of the product makes
that equilibrium
constant greater
in magnitude whether it's,
you know, in one direction
or the other direction.
So when we talked about acetone
versus cyclohexanone
they're both ketones
but we're lowering
the energy of the enol
by making the double
bond more stable.
So here's the acetone enol,
here's the cyclohexanone enol.
In the case of acid
aldehyde versus acetone,
now you can think of it as
we're raising the energy
of the reactant because acid
aldehyde is higher in energy.
It's less stabilized relative
and it's always a question
of reference frame but that's
a good way to think about it.
So later on we're going to
see things that provide a lot
of stabilization of the enol
specifically conjugation.
All right.
So that's talking about
reactants and about products,
about keto and enol tautomers.
Let's now look at how we get
from the keto tautomer
to the enol tautomer.
And this reaction can
be casually catalyzed
by either acid or base.
I want to start off by talking
about acid catalyzed
enol formation.
And I'm going to write out
the mechanism very slowly
and carefully, very meticulously
because I think this
really is important
to think your way through.
We've already seen on the quiz
how important mechanism is
and clarity and expression
and thinking on mechanism.
The mechanism on the
quiz was kind of long
and complicated involving
many steps.
The mechanism of acid catalyzed
enol formation is relatively
simple and easy.
If you have an acid, hydronium
ion for example, protons go on
and off everything with
lone pairs of electrons.
The carbonyl of acetone or
the carbonyl of a ketone
or the carbonyl of an
ester is weakly Lewis.
Weakly Bronsted basic and so
you'll have an equilibrium
where you can protonate
the carbonyl.
[ Background sounds ]
Or if we're looking at
hydronium ion as catalyst,
hydronium ion has been
consumed in the first step
of this reaction and now
we're going to recreate it
in the second step
of the reaction.
I'm going to go again draw
our protonated acetone
and now I will explicitly
draw in my alpha hydrogens
because we're going to use them.
Very often when you're
writing a mechanism or thinking
about a mechanism, you
will write different parts
of a molecule at one time.
So here's our water.
Here's our alpha proton and
now water is weakly basic,
very weakly basic.
Water can pull off
the alpha proton.
We just push our electrons up
on to the oxygen and again,
we have an equilibrium.
Whoops.
[ Background sounds ]
And I will be very, very
explicit and try to draw
in my -- all of my
relevant hydrogens here
to help keep us on track.
[ Background sounds ]
I will try to remember all of my
lone pairs and all of my charges
so I'm not a bad
person and here we go.
That's an example of how we go
from the keto tautomer
to the enol tautomer.
[ Background sounds ]
Thoughts or questions
at this point?
[ Background sounds ]
>> Is water equal to
the [inaudible] or --
>> Is water able to
pull off the hydrogen?
So you mentioned the term
addition elimination.
Now we've seen addition
elimination in acid chlorides.
That's where a nucleophile
adds to a double bond.
We kick up a lone
pair of electrons.
It comes back and
pushes something out.
Nucleophile like an amine
adds to an acid chloride.
We kick electrons
up onto the oxygen.
They kick back down.
Kick out the chloride.
So this reaction mechanistically
is an acid base reaction.
And the only thing that's
maybe surprising or confusing
to you is the fact
that we're continuing
to push our electrons
all the way up.
Now technically, technically
I could go ahead and stop --
[ Background sounds ]
-- at this point.
But, of course, if I
stop at this point,
I recognize in my mind's
eye well this is a really,
really funny resonance
structure.
A really unimportant
bad resonance structure.
[ Background sounds ]
-- of that.
And so we just keep pushing.
And of course, protons come on
and off all different
acidic positions
in acid base equilibria.
And so the proton -- whoops
I forgot my positive charge.
So much for being a
good person over here.
So the protons can come on and
off the oxygen if then we head
in the reverse pathway
and guess what?
Most of the time your
equilibrium partitions back.
In other words, when you have
an equilibrium like this --
[ Background sounds ]
So here's our energy.
Here's our reaction coordinate.
When you have a reaction
like this where we're going
from reactant from our
acetone plus H3O plus
to our protonated acetone and
then to the enol and technically
if I'm writing this out
correctly, this is acetone,
protonated acetone plus H2O.
What's happening
in our mechanism is first we're
going uphill to an intermediate
and then we're coming
down to our product,
to our enol that's
higher in energy.
So when you're at this
point, at the intermediate,
you can partition back.
You can partition forward.
It goes both ways.
Most of the time it actually
ends up going back down
and ultimately, of course,
it's the difference in energy
between our product
and our reactant
that determines the
position of the equilibrium.
There was another question --
>> Yeah. Can you explain why the
water can take off an H proton,
the bottom left structure
and not --
and it doesn't take the
H from the [inaudible]?
>> It does indeed take
the H from the oxygen
and the mechanistic reverse
of this step is taking
the H from there.
Most of the time it
takes the H from there.
Some of the time we
go forward and it goes
in the other direction
or should I say,
well it's not technically the
partitioning, it's not going
to be most of the time.
It's going to be half
and half because most
of the time we're not
forming this intermediate.
So technically can be a little
more of one or a little bit
of the other but the point
is that heads us back.
Sometimes we pull off this
proton and it heads forward.
This proton is acidic.
Other question?
>> Can the water also
perform a nucleophilic attack
on the electrophilic
[inaudible] carbon?
>> Can the water also
perform a nucleophilic attack?
Great question.
This is what's so profound.
This is why people can find
organic chemistry confusing
because yes it can
and yes it does.
And we saw that.
We saw -- we talked about acid
catalyzed hydrate formation,
acid catalyzed geminal-diol
formation.
So in your acetone, in water
with a little bit of acid,
quite honestly even
without a little bit of acid
because water has hydronium on
it 10 to the negative 7 molar,
in your acetone and water, there
is this gamish, this mixture
of acetone, the main component
and then a little
bit of the hydrate.
And even less of the enol.
So absolutely and so
keeping this in one's mind,
oh yeah this is the pathway
we're thinking about now
because this leads to the
good stuff we're going to see
over the next 4 lectures is
often confusing to initial --
to beginning students.
Great question.
Really important
because this is the part
of understanding
all [inaudible].
And you look at this mechanism
here, this manifestation
of catalysis and
realize what this means
for the uncatalyzed mechanisms.
It means that in the uncatalyzed
mechanism you're not actually
changing the difference
in energy.
I'm trying to draw this curve
for the uncatalyzed mechanism
between acetone and acetone
enol at the exact same level.
You're not actually changing
the difference in energy
between acetone and
acetone enol.
What we're doing is providing
a low energy pathway that's not
accessible without acid.
In other words, there is
hypothetically a one step
mechanism that converts
acetone to acetone enol
but the energy barrier
is much higher.
So when any acid is present
and that includes that 10
to the negative 7th molar
H3O plus hydronium ion
in pure water, acid can
catalyze this reaction.
[ Background sounds ]
Now, the other thing and
I've been beating on this,
beating on this like
a dead horse.
And the other thing that's
important is reactions proceed
through the -- reverse
reaction proceeds
through the exact
same mechanism.
When cyanohydrins form by
adding cyanide to a carbonyl
to give an oxy anion and then
protonating the oxy anion
to get the alcohol, we learn
that cyanohydrins break
down under the same conditions
by deprotonating the oxygen
to give an oxy anion
and then kicking
out cyano -- kicking cyanide.
When enols form by first
protonating the carbonyl
to give a protonated
carbonyl compound
and then deprotonating the
alpha carbon to give the enol
that means that the reverse
pathway proceeds backwards
by the same mechanism.
And so we can write and
you should be able to write
in your sleep, the
reverse mechanism.
We simply go ahead and
start with the enol --
[ Background sounds ]
-- and hydronium ion, protons --
the proton goes to
the alpha carbon,
electrons flow down
from the oxygen.
[ Background sounds ]
And now in the second step,
that's the step in going
from our intermediate
water back to our acetone.
Now water just acts a base
and pulls off the
protons on the carbonyl.
[ Background sounds ]
Electrons flow from the lone
pair on water to the protons,
electrons flow from the
bond back onto the oxygen.
[ Background sounds ]
Thoughts or questions?
[ Background sounds ]
All right.
Let's see where this gets us
in terms of some reactivity.
And then we'll get to
enolates which get really fun.
[ Background sounds ]
All right.
I'm going to take probably
the simplest reaction
that I can think of and
that reaction is going
to be deuteration.
So let's take cyclohexanone
as a variation here.
And we'll envision
dissolving it in D2O.
I'll put parentheses as solvent.
D2O is just heavy water.
It's just deuterium oxide.
It's just the isotopomer
of water with deuterium.
Every glass of water you drink
contains zillions of molecules
of deuterium or zillions
of deuterium atoms.
One out of 7,000 hydrogen
atoms is deuterium
in natural abundance meaning in
your proteins, in your lipids,
in your carbohydrates.
There is a minuscle
amount of deuterium.
And one can by electrolysis
or distillation concentration
the deuterium to get pure D2O.
If we treat are cyclohexanone
with D2O
and a little bit
of catalytic acid.
I'm getting to get away from
H3O plus and D3O for a second
to remind us one actually
has to go into the laboratory
and get a real compound.
I've written DCL.
DCL is just hydrochloric
acid made with deuterium.
So it's a strong acid.
It disassociates in D2O.
What will happen is
you will replace all
of your alpha protons
with deuterium.
And this is kind of cool.
You can see this for yourself.
If you ask your lab TA to
give you some cyclohexanone
and some D2O and a
little bit of DCL
and you dissolved
your cyclohexanone
in D2O you would be able
to take an NMR spectrum
and you would see the alpha
protons, the hydrogens next
to the carbonyl at about 2
and a half parts per million.
And as the reaction
proceeded, the peak in the NMR
from those hydrogens,
the peak at 2
and a half parts per
million would disappear
as those hydrogens were
replaced by deuterium.
All right.
So how does this occur?
Well it's the exact
same mechanism
that we saw before just
repeated over and over again.
I'll write this in
abbreviated fashion.
We can envision that in D2O
we protonate the D2O with DCL.
Remember, when you
pour HCL into water,
you have H3O plus and CL minus.
The D3O plus is just like
H3O plus in its reactivity.
You can put a deuterium onto the
carbonyl to give a protonated
or in this case,
deuterated cation.
And now in the second
step of this reaction,
if I'm balancing my
equation, in my second
of my other product is D2O.
In the second step
of the equation just
as we have pulled the
protons off the alpha position
of acetone with H2O
for protonated acetone.
Now the D2O can pull off our
alpha protons and so again,
we have an equilibrium.
Now we take ourselves
to the enol plus,
if I want to balance my
equation, D2O H plus.
[ Background sounds ]
All right.
I think I will continue
the sideboard here.
[ Background sounds ]
All right.
So one thing I should
point out is we have lots
and lots of D2O present.
Water is 55 molar.
Deuterium is practically
the same as water.
D2O is going to be
essentially 55 molar.
Protons, as I've been
saying again and again,
go on and off every
protonated species.
In other words, when we have
D2O H plus and lots and lots
of water, so I will write plus
XS D2O, you have an equilibrium
and of course mass action
drives this equilibrium.
And so you'll get a little bit
of H2O, HOD plus D3O plus --
in other words, basically
you've got tons and tons of D2O.
Your water just spreads out.
Your protons just spread out
to make little bits of HOD.
If I do a reaction
with 55 molar D2O
and I have a 1 molar solution or
a 10th molar solution of ketone
as my protons go in, I
get this little bit of HOD
in this mass of D2O
and D3O plus.
All right.
Let's continue our
mechanism here.
So we have our enol with
the deuterium on it.
And we have lots and lots
of D3O, D2O and a little bit
of D3O plus we can
protonate our enol.
I'm just writing the same
mechanism as I wrote before
without the curved arrows
and without filling in all
of my lone pairs of electrons.
And now we can have
further proton --
further deuteron transfer.
[ Background sounds ]
And you can envision this
mechanism just continuing.
We form the enol.
We protonate the
enol with deuterium.
We form the enol some more.
We protonate with deuterium and
eventually we've washed out all
of those enolizable
alpha positions
and now we have the fully
deuterated molecule.
[ Background sounds ]
And organic chemists
are terribly bad
at balancing equations
but if I want
to write a balanced
equation, I would write
that ketone plus 4 D2O
goes with catalytic DCL
in D2O solvent goes to the fully
deuterated ketone plus 4 HOD.
And lots and lots of D2O.
[ Background sounds ]
So, one of the other properties
of carbonyl compounds is
that their alpha
proton is acidic.
So, for example, in acetone
the alpha proton has a pKa
of about 19 and that's
interesting
because if you think about it.
If you think about say a
regular methyl compound
or a regular alkyl
compound, the pKa is about 50.
In other words the pKa
of an alkane is while nominally
acidic, very weakly acidic,
it is very, very,
very weakly acidic.
It's only acidic in the
sense that you can think
of its conjugate base of a
very, very, very strong base.
And yet by the time you
get over to acetone,
putting that carbonyl
there, shifts by 31 orders
of magnitude the
acidity of that proton.
It shifts the equilibrium
massively because when I think
of the conjugate base
of acetone I don't --
[ Background sounds ]
-- just think of this.
I think about that special
resonance stabilization
that we get as the enolate.
And you can see how
picturing these two resonance
structures together.
The structure on the left
is really very inadequate
to explain this stability.
The structure on the right
explains it beautifully.
We have a negative
charge on the oxygen.
We have oxygen is
electronegative.
We know alkoxide has or
alcohols are reasonably happy
to lose a proton.
They're weak acids but they're
not very, very, very weak acids.
pKa of an alcohol is 16 or 17.
The pKa of an alkane is 50.
That oxygen does a heck
of a lot and with acetone,
with pKa of 19, we're
getting a lot of stabilization
from this resonance structure.
In other words, this is the
major resonance structure.
This is the one that's important
in determining the reactivity
and this one is the --
or the stability and this one is
the minor resonance structure.
[ Background sounds ]
Now more generally, the carbonyl
group provides stabilization.
And your textbook gives
a number of compounds.
And I'll give you these
compounds and then I want
to give you a generalization
that I keep in my own head.
So your textbook gives you
acid aldehyde with a pKa of 17.
And it gives you ethyl acetate
with a pKa of 25 for the methyl.
And acetonitrile with a pKa
of 25 and dimethylacetamide
with a pKa of 30 for
the alpha proton.
And really the numbers that I
keep in my head, these are a lot
of details but the
numbers that I keep
in my head are as follows.
I keep in my head two
numbers for all of this.
That the alpha protons
for the ketone
and aldehyde family
are about pKa 20.
[ Background sounds ]
Twenty's good enough.
Yeah, I'll keep in the back
of my head aldehydes are
a little bit more acidic.
I'll keep in my mind that 20 is
typical for say cyclohexanone.
That acetone is 19 but
those details are small.
Those are unimportant.
The other big picture while
aldehydes and ketones are
about pKa 20, in general the
carboxylic acid family, esters,
nitriles, even amides,
even carboxylic acids
under the right conditions you
first remove the acidic proton
so it's a carboxylate and then
you remove a second proton.
In general, a good number
to keep in mind is about 25.
Yeah, there are some variations
but it's not as super important
as keeping these
pictures in mind.
And one of the pictures from
this is in general ketones
and acid -- and aldehydes
are less acidic than water.
In other words, hydroxide will
take off only a small amount
of the protons, only an
equilibrium concentration.
Ethoxide will pull off only
an equilibrium concentration
but stronger bases, those can
pull off protons quantitatively.
[ Background sounds ]
Since the alpha protons
of ketones
and aldehydes are acidic,
it shouldn't surprise you
that base can also
catalyze enol formation.
[ Background sounds ]
So, let's come back to
our acetone molecule
as our sort of archetypal
ketone.
And now let's envision
hydroxide as a base.
I could do with this
alkoxide too but some sort
of moderately strong base
not super, super strong.
We'll talk about super, super
strong in a moment but something
like hydroxide or an
amine like triethylamine.
We can pull off the
alpha proton.
The alpha proton
is weakly acidic.
Remember pKa of 20 is the number
you want to keep in your head.
So we have an equilibrium
here with the enolate.
[ Background sounds ]
But that equilibrium lies
slightly away from the enolate
and so we can protonate.
In other words you generate
a little bit enolate
but that you don't generate
the enolate quantitatively.
And so we can protonate if
we protonate back on carbon,
we're heading back
in the same position.
We're headed back to the
start of the reaction.
If you protonate on
oxygen on the other hand,
now we've proceeded on the enol.
[ Background sounds ]
And just as in our acid
catalyzed enol formation,
hydroxide here although
it's being consumed
in the first step, is being
recreated in the second step
and so it is not being taken
up or destroyed or created.
And the reaction overall
is acting as a catalyst.
Similarly, we're not
changing the position
of the keto enol equilibrium.
We're only creating a
lower energy pathway
to allow the ketone and
enol forms to interconvert.
>> Would there be more enolate
in your solutions that enol
since the solutions
is basic and the --
like the water can
be like [inaudible]?
>> Okay. Good question.
If you're working your
equilibria, technically yes.
In this particular example,
that is an astute question.
In this particular example
with these particular pKas
technically you're right.
There will be more
enolate than enol present.
If you want to come an example
where now you have
very little enolate
and the equilibrium still goes,
substitute in this reaction
as base, triethylamine
because now --
remember we said
that for acetone,
we're going to have 1 point
5 times 10 to the negative 7
as our equilibrium constant.
So there's not a lot of
enolate, not a lot of enol.
And we're having our
pKa for acetone of 19
which means you actually do
have, you know, 1 in a 1,000
or 1 in 10,000 enolate,
1 in a 1,000.
So if you substitute
triethylamine, pKa about 10
or 11 for the conjugate acid,
you'll still catalyze this
reaction but now you'll very,
very little of the enolate.
>> Can you [inaudible]?
>> Do you use a Dean-Stark
apparatus?
No need here.
We're not actually trying
to change the position.
And okay, could one?
You're -- now you're
asking a smart question.
Can one drive this reaction?
And the answer is you can't
drive it by removing water
because the reaction is
unimolecular overall.
There's no net change
in the molecule.
It's an isomerization.
So the way to drive the reaction
to the enolate is to go ahead
and to use a very strong base.
And unfortunately there's
very little one can do to try
to generate the enol
as an isolated species although
there's an Israeli chemist
who actually is focused
on stable enol.
It's because another thing
about organic chemists,
in addition to being very
bad at balancing equations is
if someone says it
can't be done.
Enols are unstable.
You can't make a stable enol.
You can bet you an organic
chemist will go out there
and say how do I
make a stable enol?
We'll see some stable enols
that are very special enols
but for a regular one.
All right.
So where is all of this leading.
Let's go and do some
enolate chemistry.
So just like with DCL and
D2O, if I take cyclohexanone
with catalytic sodium
deuteroxide just the deuterium
analog of sodium hydroxide in
D2O as solvent and I let it sit.
Again, you can do this
for real in an NMR tube.
Johnny or Kim could go
back to my laboratory
and do this right now and show
you those alpha protons would
wash out and be replaced
by a deuterium.
Similarly, if I imagine for
a moment that I had something
that had [inaudible]
at the alpha position.
Let's imagine for a moment
that I have this ketone
where I have a methyl group.
And let's say that I have
just the S enantiomer.
And I treat this ketone with,
let's say, sodium hydroxide
since we're talking
enolate chemistry.
Sodium hydroxide in water.
Then what'll happen will be
I'll lose my enantiomer purity.
I will get a mixture
of the two enantiomers.
I will get the racemic.
I will get an equal
mixture of the R
and the S. What's happening
here in this latter example,
we're forming little bits
of the enolate repeatedly.
The enolate is flat.
The enolate loses
its [inaudible].
[ Background sounds ]
Doing here.
[ Background sounds ]
Whoops.
[ Background sounds ]
I will say that we go via this.
In other words, you can
picture in your mind's eye,
the base pulls off
the alpha proton.
We get the enolate.
The enolate is flat.
We have no polarity
[phonetic] to the enolate.
If a proton comes from the back,
we get the original enantiomer.
If it comes from the front,
we get the new enantiomer.
And so our compound
racemizes an aqueous base.
[ Background sounds ]
All right.
So, in addition to being
bad at balancing equations,
in addition to wanting
to do what can't be done,
organic chemists
are control freaks.
We want to control the molecules
and make them do stuff.
And often we want to do stuff
to make stuff that's
useful like new medicines.
So whereas weak base doesn't
make a lot of enolate,
a strong base does
and we can use that.
If you have something like
cyclohexanone in a base.
I'll write it as B minus,
let's say, a very strong base.
In other words pKa
of the conjugate acid,
much greater than 20.
Then we have an equilibrium
that lies way, way to the right.
Often so far to the right
that we can just
ignore the left arrow
to give us the enolate anion.
And our protonated acid.
[ Background sounds ]
And the base that's
extremely valuable
to organic chemists is
diisopropylamide anion
or more specifically the lithium
salt, lithium diisopropylamide.
For those of us who are control
freaks, for those of us who want
to make molecules do stuff,
this base is really good
because it's a strong base.
pKa of the conjugate S, the pKa
of diisopropylamine is about 40.
So I'll write pKa iPr2NH
is a nice shorthand
and I'll say tilde
approximately 40.
Remember we're comparing.
I said if you want to keep
one number in your head,
keep 20 for the pKa
of cyclohexanone --
for the pKa of a ketone,
that's actually the pKa
of cyclohexanone to
within a unit but again,
just keep that number
in your mind.
The other thing that's
really important
about the diisopropylamide
anion is it's big.
Those isopropyl groups and
there are 2 of them, are bulky.
They get in the way.
In other words, lithium
diisopropylamide is a good base
but not a good nucleophile
and again, I'm going to come
down this confusion in
organic chemistry students
when you're starting out.
Oh my God, there's all
this stuff to know.
How do I know that this compound
like cyanide is acting
a nucleophile?
How do I know
that diisopropylamide
is acting as a base?
Well you start to
learn these patterns.
The very bulky ones are
better at pulling off protons.
The little bit ones and cyanide
is just a little bullet is
better as a nucleophile.
So the reaction here lies
so far to the right --
[ Background sounds ]
-- that our lithium
enolate, I won't even bother
to write the reverse arrow.
Our lithium enolate
is essentially forms
stoichometrically and of
course, the other component
in this reaction is
diisopropylamine.
So we have an acid
base reaction.
Stronger acid plus
stronger base reacts
to give weaker acid
plus weaker base.
How do you know where
that equilibrium lies?
You just look at the
pKa of the acids.
Okay, we've got the weak
acid pKa 40 on the left.
We have a strong
acid on the right.
And remember, the other number
I like to keep in mind is
about 10 orders of magnitude.
In other words, if you've
got things within a pKa range
of about 10 units, I'd say
yeah you've got a little bit
of an equilibrium but if you
go to 20 units apart and it's
like bam, that reaction goes
essentially all the way.
[ Background sounds ]
Often when organic chemists are
writing out synthetic reactions,
we'll write in a
little bit of shorthand.
For example, I will say
LDA, it's widely used
in tetrahydrofuran as a solvent.
In order to minimize its
reaction as a nucleophile
and minimize equilibration
processes,
often you'll do these reactions
at negative 78 degrees Celsius.
That's the temperature
of a dry ice bath
which makes it particularly
popular.
Remember early on, I said that E
plus is sort of an abstraction.
I said you'd see methyl iodide.
So you can envision
a reaction like this.
In the first step, we
add cyclohexanone to LDA
at negative 78 degrees.
In the next step, we
drip in methyl iodide
and allow the reaction
to warm up.
Maybe perform an aqueous workup
on it with a little bit of acid
and we've methylated
our cyclohexanone.
What's happening here, same
thing we saw with protons.
We've generating our
enolate stoichometrically.
Here's our methyl iodide
whether it's a proton,
whether it's methyl iodide.
We have a good electrophile.
Electrons from the
alpha-carbon to the methyl,
to the electrophilic
methyl carbon
and we do an SN2 displacement.
[ Background sounds ]
To give the methylated
cyclohexanone and iodide anion.
[ Background sounds ]
Most of the time organic
chemists don't go ahead
and buy lithium
diisopropylamide.
We typically make it
by acid base chemistry.
n-Butyllithium is
widely available.
It's the most common
alkyllithium, widely used.
You can mix it with
diisopropylamine, NTHF,
if you like before you
add your cyclohexanone.
And you pull off a proton,
you get butane plus
lithium diisopropylamide.
And again, I'm now getting into
writing things in shorthand
so I'm not writing as many
lone pairs of electrons here.
And this, too, is an
acid base reaction.
n-Butyllithium is very,
very strongly basic.
The pKa of an alkane
is about 50.
pKa of diisopropylamine
we already said is 40.
That equilibrium lies
way, way to the right.
You might ask why don't we
use butyllithium as a base?
It comes back to what I
was saying about sterics.
If I use butyllithium, the main
reaction, the only reaction
that you would detect, that
you would write on say an exam
if you want to get
down to the brass tacts
of being an O chem
student would be addition
of the butyllithium
to the carbonyl.
It's not sterically
hindered enough to not add.
We go to lithium
diisopropylamide
with those big isopropyl groups
and now it's much more basic
than nucleophilic
and butyllithium toward
a carbonyl is much more
nucleophilic than it is basic.
[ Background sounds ]
All right.
Let's have some fun now with
variations among enolates.
So let's imagine we have
our methyl cyclohexanone
and we treat it with base.
Now we can get two
different enolates.
We've got two different
types of alpha protons.
We've got alpha protons
that are opposite to methyl
and alpha protons that
are next to the methyl.
In other words, we
can get this enolate
or we can get this enolate,
the less substituted enolate
or the more substituted enolate.
Now, in general, we know
that more substituted
double bonds are more
thermodynamically stable.
The more substituted
enolate is what we call the
thermodynamic enolate.
[ Background sounds ]
It's lower in energy.
Among an equilibrium between
the two enolates, it would form.
But we also know
that size matters.
We have a bulky side
of the molecule.
We have a non-bulky
side of the molecule.
The base, particularly a
sterically hindered base needs
to get in to pull off a proton.
It's going to grab the
easier one if it's bulky.
So the enolate then is less
substituted is often referred
to as the kinetic enolate.
The one way that forms first.
The one that is easier to form.
So we can go ahead
and think of it as --
[ Background sounds ]
These protons are more
accessible to base.
[ Background sounds ]
-- but this proton is
technically a little bit
more acidic.
[ Background sounds ]
So I will write here.
[ Background sounds ]
Just for comparison.
More substituted enolate,
more thermodynamically stable.
[ Background sounds ]
So let's take a moment
to play with this notion
because as I said,
organic chemists love
to control reactions, love
to control reactivity.
[ Background sounds ]
And it's important because
when you're making molecules
for medicines or for drugs, you
need the molecule that you want.
And I'll take a couple of
examples from your textbook here
because they're good examples.
If we take our methyl
cyclohexanone and we treat it
with LDA and THF at negative 78,
you may see it written
just as LDA.
You may see it written
as LDA THF.
You may see it written
as LDA THF negative 78.
And we treat it with
methyl iodide.
We deprotonate to form the more
-- to form the kinetic enolate
and we end up with a
2,6-Dimethylcyclohexanone.
So I'll write via
the kinetic enolate.
Borrowing an example from your
textbook which if you want
to ask questions about after
class I will tell you some
of the subtleties of.
But if you treat with a weaker
base that can equilibrate --
and your textbook uses
sodium methoxide in ethanol.
And basically the way
I will write this is
to mix sodium methoxide
methyl iodide in ethanol.
Then because we're
under conditions
where we're thermodynamically
equilibrating our enolates,
your predominant product,
your main product is going
to be the
2,2-Dimethylcyclohexanone via
the thermodynamic enolate.
And so this is a taste
of the type of control
that organic chemists can get.
And I'll write this
as major product.
Technically here, if I do
this reaction with LDA,
this is the major product.
In other words, I'm
going to get 99 parts
of this and 1 part this.
Technically this is
the major product.
I'll probably get about 80 parts
of this and 20 parts of that.
[ Background sounds ]
I know there's one question but
I want to wrap up with a couple
of last points about
stereochemistry
and stereoisomers at this point
because I think it's pretty
and I think it's cool.
And I think it's intellectually
deep and I think that a lot
of this is what organic
chemistry is all about.
So technically I'm
lying in that structure.
Or maybe to be more precise,
I'm telling you half a story.
And the other half of the story
is when I draw this structure.
[ Background sounds ]
I'm living in flat lands.
Each of those carbons
with the methyl group is a
stereogenic center.
And so we have different types
of species that we can get.
We have the species, the
stereoisomer to be more specific
in which the 2 methyls are
on the same side of the ring.
We have the cis stereoisomer.
We have the stereoisomer in
which the 2 methyl groups are
on the opposite side of the
ring, the trans stereoisomer.
And so when we form our enolate
and now I will specify my
methyl stereochemistry.
If our methyl group adds
from the back we get
the trans stereoisomer.
If our methyl group
adds from the front,
we get the cis stereoisomer.
[ Background sounds ]
And there's one more level of
subtlety and I'll show you that
and then that's going to
wrap up today's lecture.
So imagine for a moment
I take this compound.
And almost invariably
it's going to be racemic
if I haven't specified
otherwise.
And so I treat with LDA and
THF at negative 78 degrees
and then I treat with methyl
iodide just as I've done.
And you'll get a mixture
of the cis trans compound
but of course, the
trans compound is going
to be the racemic.
In other words, we generate our
enolate and the methyl group
if it adds opposite the
methyl group depending
on whether this methyl
group is out or back,
we're going to get
the racemic compound.
In the case of the
cis, the cis compound,
if it adds from the same face,
the cis compound
is a mesocompound.
So in other words whether I
draw it like this or I draw it
like this, this is just
the self same thing.
All right well that gives
us a taste of the richness
of stereochemistry as
well as a beginning
on this notion of control.
And you can see even though
we've talked as this point
about control of the
regiochemistry of addition,
we've only seen that there
is a still depth and richness
to the stereochemistry
of addition
and that will go largely
beyond the scope of this class.
Thank you.
[ Silence ]
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