>> Good morning.
>> Good morning.
>> Well, I am really
happy today.
So, this is the first time that
I've used electronic homework
in one of the classes.
And I was so happy when I
looked this morning at 9 o'clock
and found virtually
everybody had engaged
with the homework assignments.
And so much of learning organic
chemistry is actively working
problems and actively
trying to understand things
that I was really,
really excited
to see people are
coming along with me
on this process of learning.
And it makes me very,
very happy.
So, today we're going
to continue our discussion
of Chapter 20.
And I'd like to introduce
organometallic reagents and talk
about their reactions
with carbonyl compounds
and their reactions in general.
And then we're going to talk
about reactions of members
of the carboxylic acid family
with hydride nucleophiles
and with organometallic
reagents.
And if there's time, we'll
conclude by talking about use
of these reagents in these
reactions in synthesis.
So, when we were
talking last time,
we talked about hydride reagents
with ketones and aldehydes.
And we said sort of generically,
if we have some ketone
or aldehyde, and I'll
write it generically,
just showing R groups
that could be alkyl,
that could be aromatic,
that could be hydrogen.
So generically a
ketone or aldehyde.
And we envision its reaction
with some hydride nucleophile.
Now remember, when I'm
writing something like this,
H minus in quotes, of
course there's no reagent
that is itself a
hydride nucleophile.
But these are things like
lithium aluminum hydride
and sodium borohydride that
serve as sources of hydride.
And then if we carry out a
workup with aqueous acid,
H3O plus or in some
cases, you can use water.
And again, I'll put this in
quotes because you can't go
and buy a bottle of H3O plus.
You could take sulfuric
acid and pour it into water
to make hydronium ion
and bisulfate ion.
You could take HCl in water and
make H3O plus and chloride ion.
And so you'd go to the stockroom
and ask for one of those.
When you do that, you get
a reduction of the ketone
or aldehyde and protonation
like so to give an alcohol.
Now, what I'd like to
do at this point is
to consider an analogous
reaction, again,
written at this point
as an abstraction.
So again, we'll take our
ketone or our aldehyde
and we'll imagine instead
of adding our hypothetical
hydride anion,
we're going to add something
that I'll say is R minus
with a lone pair of electrons
meaning a carbon-based anion.
We'll talk more about it,
not necessarily something
you can get as a free anion.
And certainly not
something that you could put
in a bottle on its own.
It would be part of an
organometallic reagent.
But again, right now
we're getting our view
from 30,000 feet.
And again, if you imagine
some type of aqueous workup
with H3O plus, now
instead of adding hydride
to our carbonyl compound,
we've added in an R group.
The use of primes
and double primes
and so forth is just
my way of representing
that they're various groups,
different or the same.
Anyway, but again, in the
view from 30,000 feet,
the profound thing about this
reaction as I've written it is
that we formed a
carbon-carbon bond.
[ Writing on Board ]
And so much of organic
chemistry,
both the practical aspects of
synthesizing useful molecules
and the real intellectual beauty
of the discipline is the fact
that we can create great
degrees of complexity,
valuable complexity, medicines,
analogs of natural products,
natural products that are
too scarce to get otherwise,
probes to probe reactions
and probe biology.
We can do all of this
through chemical synthesis
in which we take little
molecules that might be able
to be purchased and build
them up into big and complex
and useful molecules
that can cure cancer
or fight disease
or teach us things.
So now we come down to the
issue of what is our R minus?
What is our source of
our carbon nucleophile?
And the first really, really
valuable carbon nucleophiles
that were developed
were Grignard reagents,
developed by Victor Grignard.
He received the Nobel Prize
in 1912 in chemistry for this.
And the basic idea is that you
take some halide, a bromide,
a chloride or an iodide.
Fluoride is sort of the
oddball among halogens.
And if you go down
your periodic table
to astatine, it's not stable.
It's radioactive.
So organic chemists
would never work with it.
You can't really isolate it.
Anyway, if we take an alkyl
halide such as butyl bromide
and we treat it with magnesium
metal, if you've done one
of these reactions
in the laboratory,
you'll have seen your magnesium
metal is kind of bright
and shiny and lightweight.
It comes as turnings that have
been worked off a big block
of magnesium with a lathe.
And you'll put them with your
alkyl halide in a solvent.
The solvent will be an
ether solvent, either ether
or THF typically, although
sometimes other ethers can
be used.
THF is tetrahydrofuran.
It's a cyclic ether.
So it's an ether with
a five-membered ring.
And it has lone pairs.
It's kind of diethyl ether
with its ears pinned back.
And either of these work well.
The result is that you
get a Grignard reagent.
I started with butyl
bromide here
and so I'll write the Grignard
reagent from butyl bromide.
We call this butyl
magnesium bromide.
[ Writing on Board ]
And the ether solvent
coordinates to the magnesium.
The magnesium doesn't have
a complete octet here.
We have only four
electrons around it,
two from the alkyl group
and two from the bromide.
And so the ether
solvent, the diethyl ether
or tetrahydrofuran will
coordinate to the magnesium
and help give it the feeling
of having a complete octet.
Anyway, as I said,
the broader category
of this is a Grignard reagent.
And you can make these
from anything from bromo-
or chlorobenzene to alkyl
bromides to methyl iodide.
And so this typically --
typically one makes a
Grignard reagent as part
of a synthetic process.
They're actually
stable reagents.
You can put them in a bottle.
You can buy them.
But in the laboratory,
because they react with air
and because they
react with moisture,
you would typically go ahead
and immediately add
a carbonyl compound.
So let me show you a typical
synthetic carbon-carbon
bond-forming sequence that one
might do in the laboratory.
So we might take our butyl
bromide and treat it first
with magnesium and ether.
[ Writing on Board ]
You'll often see people
writing a slash in an equation.
Often that slash is a way
of saying the solvent
is below the slash.
Or for just writing a
simple line of equation,
we might put our
solvent below the arrow.
Then let's add a
carbonyl compound.
And for the purposes of
teaching, for the purposes
of this example,
I'll take acetone
as our carbonyl compound.
And finally, let's
do an aqueous workup.
And I'll give us -- I'll
choose an acid here.
Let's say aqueous HCl as
the acid I would choose.
And the product of this
reaction now is a carbon-carbon
bond-forming adduct,
or an adduct
with a new carbon-carbon bond.
If you wanted to
name the compound,
we've now formed this
six-carbon chain.
So it's 2-methyl-2-hexanol
as our product.
So this is very powerful because
now we've taken some small
compounds that you can buy
and we've made a
more complex molecule
that you might not
be able to buy.
Many molecules with
this type of structure,
these types of structures
like alcohols with long chains
on them are insect
pheromones, for example.
So some of these types of
synthetic products are used
to make traps for insects
like Japanese beetles,
where that will -- the pheromone
will lure the insect to the trap
and then they'll
fall in and die.
All right.
So I want to talk a little
bit about the properties
of Grignard reagents
and organometallic
reagents in general.
[ Erasing Board ]
So metals, of course, are
much more electropositive
than carbon.
So any bond between a metal
and carbon is either going
to be a polar covalent
bond or an ionic bond.
If you want an index value,
chemists like to keep
electronegativity values
in their head.
It's a useful way of
assessing the degree
of polarity of a bond.
Magnesium has an
electronegativity of 1.3.
Carbon, about 2.5 or 2.55.
So you can think of the bond
between magnesium and carbon
as a polar covalent bond.
In other words, you can think
of this as having a delta minus,
a partial negative
charge on carbon
and a partial positive
charge on magnesium.
Sometimes if you're writing
a mechanism and you want
to be quick about it,
you could say well,
I'll write a non-bonded
resonance structure even though
I know it's primarily
a covalent bond.
I could write an ionic
bond, resonance structure
so we could think of it as this.
Maybe I'll put this
in quotes here just
to remind us this is
sort of our thinking.
And so, if I'm thinking about
this from a mechanistic point
of view, one way we
could think about this --
I'll take our acetone and
our Grignard reagent written
generically -- one way we
could think about this is
that our organometallic
reagent serves as a nucleophile.
And of course, we draw a
curved arrow mechanism.
We help think about the
flow of electrons and bonds
by starting an arrow
at the lone pair --
at the available electrons,
either as a lone
pair or in a bond.
And then moving to the thing
that wants electrophiles,
in other words, from the
nucleophile to the electrophile.
As we did in thinking
about hydride mechanisms,
we can't go ahead and now
have five pairs of electrons
around this carbon atom.
So concurrently, as the
nucleophile moves in,
electrons move up onto oxygen.
And so we continue our
grammar, if you will,
of writing our curved arrow
mechanism now, writing a product
in which we have
our negative charge.
And if I want to be good,
organic chemists
are rarely good.
We love to throw away things
that aren't necessary
in our thinking.
But if I want to be good,
I'll draw the magnesium
counterion there.
We can also think of
this, just as we did
with lithium aluminum hydride,
maybe a little more correctly
or certainly a little more
sophisticatedly, we can think
of our covalent bond and
keep in the back of our mind
that it's a polar covalent bond
and write a curved
arrow mechanism
in which we simply go ahead
and now take the electrons
from that bond, move it
in like so, and again,
come out to the selfsame
product.
So these are good ways of
thinking, sound ways of thinking
about the mechanism
of the reaction.
[ Erasing Board ]
So, one thing to keep in
mind about Grignard reagents,
and indeed about most
organometallic reagents,
particularly ones in which
there's a large difference
in electronegativity between
the carbon and the metal is
that organometallic reagents
like Grignard reagents,
in general, act as
very strong bases.
[ Writing on Board ]
We can think about
basicity in terms of the pKa
of the conjugate anion.
So if you have butyl
magnesium bromide,
the conjugate anion
corresponds to an alkane.
In other words, where in the
conjugate acid you have a Csp3
carbon bound to a hydrogen.
The pKa for such a
hydrogen is about 50.
That's about at the very
end of the basicity scale.
That's about as basic as
you can get for a carbanion.
If you have an alkene, now
you have a Csp2 carbon bound
to a hydrogen.
The electrons are held
a little more tightly
in this type of structure.
The pKa now is about 44.
A way of thinking of this is
since the electrons are
a little more stabilized,
held a little more
tightly in an orbital
that has more s character,
right,
sp2 has 33 percent s character,
sp3 has 25 percent s character.
As you hold electrons
more tightly,
the CH bond is more
willing to give up H plus
and give you a carbanion.
So an alkene is a little bit
more acidic than an alkane.
These are all compounds
I call very weak acids.
And I usually put
quotes around the acid
because you would not get
any evidence of acidity
from the compounds, say,
dissolving it in water
and testing it with a pH meter.
And yet, in a Lewis acid-Lewis
base reaction, you can think
of an alkene, say,
as a proton donor
under certain circumstances.
Now, by the time you
get to an alkyne,
now you've got 50 percent s
character in your CH bond.
And now these are
reasonably acidic.
They're still very weak acids.
Your pKa is only 25.
That's really, really,
really weak, still.
It's not even like
water, which you think
about as an acid
forming hydroxide anions.
It's still a very weak acid.
Or alcohol forming an acid --
giving up a proton and
forming an alkoxide anion.
And yet, our pKa is about 25.
And I'll show you an implication
of that in just a moment
when we start to talk a
little more about alkynes.
But, the point of this comes
back to what I was saying before
about carbon -- about Grignard
reagents being very reactive
toward water.
A Grignard reagent acts
as a base with water.
And so if you expose
a Grignard reagent
like butyl magnesium bromide
to water, the water acts
like a Lewis -- acts like a
Bronsted acid and you get butane
and bromomagnesium
hydroxide, a mixed salt here.
So this is a Bronsted
acid-Bronsted base reaction
if you think about it.
We have water acting as an acid
on the left half
side of the equation.
We have our alkane as an acid on
the right side of the equation.
And because the pKa
difference is so humongous,
the pKa of water is 15.7, the
pKa of butane is about 50,
I don't even bother to
write an equilibrium arrow.
The equilibrium constant
is 10 to 34th, right?
It's 10 to the difference
in pKa's.
That reaction lies
so far to the right
that there's just no
component to the left on it.
And that's true whether it's
with water or an alcohol.
And so just by comparison,
imagine for a moment I wrote --
let me pick a particular
alcohol.
We'll pick ethanol
as an alcohol.
And so now we would get butane
and a ethoxymagnesium bromide.
Of course, this would be the
same with a carboxylic acid
or just about anything that
you would normally think
about as even mildly acidic.
[ Erasing Board ]
All right, well Grignard
reagents are one
of a broader family of
organometallic reagents.
Another member of the
family that reacts very,
very similarly are
organolithium reagents.
[ Writing on Board ]
And so if we wrote Grignard
reagents sort of generically
as RMgX, that would
be a generic way
of writing a Grignard reagent.
We can write organolithium
reagents generically as RLi.
Organolithium reagents are
formed by reacting alkyl
or aryl halides, again, we're
talking iodide, bromide,
chloride, with lithium metal.
If I write a balanced
equation, say, for bromobenzene
and lithium, it takes
two lithiums --
and that makes sense if
you think about it, right?
We're carrying out a two
electron process here,
and lithium has one electron.
So magnesium has two electrons.
And so a balanced
equation becomes
that we get phenyl lithium
plus lithium bromide.
And again, organic
chemists are awfully,
awfully bad about writing
products of reactions.
So I'll put the lithium
bromide in parentheses
because I might not write it.
So typically, if I were
writing this as, say,
a synthetic reaction,
and I imagine generating
an organolithium compound,
and say reacting it with
an organic compound,
I'll give you an example
of what I might write.
So I might take bromobenzene
or I might take chlorobenzene
if I wanted, treat it
with lithium metal.
Now, organolithium reagents
can be generated in ethers.
An ether can serve
to coordinate.
But they also form clusters.
And so they actually
can be generated
in other solvents
including hydrocarbons.
I'm just going to
skip the solvent here
because it's much less important
than in a Grignard reaction.
And let's say as my partner,
since I gave you a ketone
before, I'll take an aldehyde.
The aldehyde I've
chosen is pivaldehyde.
That's the trivial name, or 2,
2-dimethylpropanal
would be the IUPAC name.
And again, I'll imagine doing
some type of aqueous workup.
I'll just write H3O plus here
to indicate I haven't
specified the acid.
It could be aqueous HCl, it
could be aqueous sulfuric acid.
One that I personally
like to use
in my own laboratory is
aqueous ammonium chloride,
which is a very mild acid
and very good for workups
of reactions like this.
Anyway, after our workup,
the product now has a new
carbon-carbon bond like so.
And of course, because we've
generated a stereo center
in the molecule,
we've generated it
as a mixture of two enantiomers.
We've generated two different
enantiomers in equal amounts.
We've generated the racemate.
[ Erasing Board ]
All right.
In part because your textbook
mentions various different
organometallic reagents at this
point, I want to follow along.
And in part because
I want to remind you
of what I think are
really useful items.
And in part because I want to
tie into this concept of pKa,
I'd like to, at this point,
talk about acetylide ions.
[ Writing on Board ]
And I see I have a question.
>> Yeah, you sort of
erased the equation,
but like for the
second equation what
that lithium bromide is
attaching like an ethanol,
why doesn't the magnesium
bromide bond to the OH,
but instead bonds
to the [inaudible].
>> What's that?
Oh, the question was,
why doesn't the magnesium
bond to the OH.
It doesn't -- well,
there's no OH --
oh, you're saying
in that reaction.
Because the ethanol has
now given up its proton
to react with the butyl part.
So the proton has come off of
the ethanol onto the carbon
of the butane leaving an
ethoxide anion which combines
with the MgBr plus component.
>> Wouldn't the same
thing happen with water?
>> And it did.
So we had, in the case of
ethanol, we had EtOMgBr.
In the case of water,
we had HOMgBr.
So we had a very
analogous reaction.
Another question.
>> In that second equation,
are we supposed to
have two lithiums?
>> Ah, great question.
In that second equation, are we
supposed to have two lithiums?
This is very typical
of how organic chemists
will write a reaction.
So typically you would see
that one might not particularly
as you became more used
to writing reactions,
it might be implicit.
But indeed, you would
have two lithiums.
So you could easily envision
writing lithium parenthesis two
equivalents or two lithium.
And again, this is very
much part of the shorthand
of writing organic
reactions, particularly
when focused on synthesis.
Good questions.
>> So will organolithium
reagents not be created
who don't have two lithiums?
>> Ah, great question.
Would organolithium
reagents not be created
if you had only one lithium?
Well, now imagine, what would
happen would be you'd have one
mole of butyl lithium --
one mole of butyl bromide,
one mole of lithium, you'd get
half a mole of butyl lithium.
Now, here's where
the fun comes in.
If I let that sit at a low
temperature and used it quickly,
I would have a reaction of
one mole of butyl lithium --
of one-half mole
of butyl lithium.
But if I let it sit or tried to
put it in a bottle and I now had
that organolithium reagent
sitting for a long time
with more butyl bromide, I
would get E2 elimination or,
because the reagent
is strongly basic,
to give butene and butane.
Or I would end up
having SN2 displacement
to give octane or both.
Great question.
So yes, I would definitely
use two equivalents.
All right.
Well, at this point I want to
talk about acetylide anions
and sort of follow
along with our textbook,
but also because
thematically it fits in.
So as I said, acetylene,
alkynes in general,
are especially acidic.
While they're still very, very
weak acids, pKa of about 25,
they're strong enough
acids that very,
very strong bases can
pull off their proton.
So for example, if
I great an alkyne
with sodamide I get the
sodium acetylide anion.
[ Writing on Board ]
Now, sodium's a little more
electropositive than lithium.
Lithium has an electronegativity
of 1, sodium of .9.
By the point you get to
organosodium reagents,
they're pretty ionic
in the bond.
So I generally think
of these as ionic.
>> Sorry, is it NH2
or NH [inaudible]?
>> Ah, great.
NH2, NaNH2 is sodamide.
And if I'm going to balance
my equation, and as I said,
organic chemists
are usually very bad
about this, ammonia, thank you.
Thanks very much.
Ammonia is the other
byproduct of reaction.
And now we see ourselves
very much in the situation
of a Bronsted acid-Bronsted
base reaction.
So we have pKa of about
25 and pKa of about 38.
And so a difference of pKa means
that equation lies way, way,
way to the right, 10 to the
13th equilibrium constant
or thereabouts.
So basically I throw one
mole of sodium amide,
one mole of an alkyne and I get
essentially all acetylide anion.
And I'll write a
balanced equation,
or I'll write a synthetic
equation here in which I, say,
take propine, I treat
it with sodamide, NaNH2,
which you can make by dissolving
sodium metal in ammonia
with a little bit of iron and
it reacts to give sodium amide.
And for the heck of it,
again, I'm just trying
to give us a range of
carbonyl compounds.
For the heck of it,
I'll take 2-butanone.
And then, since I mentioned that
I like aqueous ammonium chloride
as a source of acid
for a workup,
I'll just demonstrate what I
would do in my own laboratory,
which is to use aqueous
ammonium chloride.
And the product of
this reaction is this.
It is the alkyne
with the alcohol.
This is, of course, racemic.
In other words, your textbook
points out very nicely
that you're going to
add your nucleophile
from both the front face and
the back face of the carbonyl.
And so we have one enantiomer
in which the OH is pointing back
and the alkyne is pointing out.
We have another enantiomer
in which it's added
from the back face.
And now the OH is pointing out
and the alkyne is pointing back.
And we will get equal
amounts of these,
both of the R and of the S.
Although this chapter
is focusing primarily
on carbonyl compounds, it's also
beginning to introduce ideas
of organic synthesis
and carbon-carbon bond
forming reactions.
And so your chapter reminds you
that you've already seen
certain carbon electrophiles.
So for example, you've
already seen epoxides.
So there are lots
and lots of types
of electrophiles you can
generate your compounds with,
that you can react
your acetylides
and other carbon
nucleophiles with.
And I'll just point out one
example here that brings
out a couple of additional
points.
So butyl lithium is
commercially available.
It's a common source of a highly
basic organometallic reagent
that can be used not only
as a nucleophile but also
as a very strong base.
And so butyl lithium
is a great reagent
for pulling off moderately
acidic protons,
for example protons from alkynes
and also protons from amines
like diisopropylamine,
which you'll see later.
Anyway, butyl lithium would
react with our alkyne.
Remember, our pKa of
the alkyne is about 25.
Our pKa of butane is about 50.
And so it would react again
in an acid-base reaction
to give now and an
alkynal lithium reagent.
And so we can use this as well
as a way of making anions.
And I'll just give you one
example, or this as well
as a way of making
carbon-carbon bonds.
So just to give you some
diversity in your chemicals,
in the molecules that you see,
I'll take, say, phenyl acetylene
and we could envision
treating it with butyl lithium.
You'll often see butyl
lithium written as NBuLi.
N means normal.
Normal is just a fancy way
of saying it's the regular,
it's the 1-butyl lithium rather
than say the lithium being
at the 2 position of butane,
which is sec-butyl lithium,
or being on a tertiary
butyl group,
which is called tert-butyl
lithium and is a stronger base.
Anyway, let's envision using the
common reagent N-butyl lithium,
treating our phenyl
acetylene with it.
That's going to give us our
lithiated phenyl acetylene,
our organolithium compound.
And just to demonstrate the
point of other reactivity,
we can picture the
reaction, say, with an epoxide
and then again an
aqueous workup, H3O plus.
And the product of this
reaction is an alcohol,
just as we've been
generating in all
of the reactions
I've shown thus far.
But what's interesting about
this is now the alcohol,
instead of being connected
directly to the carbon
that had been -- or
directly to the carbon
where the nucleophile
attacked, it's one over.
You can think of this as R
and the alkyne lithium
reacting with the epoxide.
And so we're going to
draw electrons flowing
from the carbon-lithium bond
into the carbon oxygen bond,
pushing electrons onto oxygen
and opening up our ring.
[ Writing on Board ]
[ Erasing Board ]
All right.
So what we've done at this point
is we've really overviewed a
basic, a fundamental
reaction of carbonyl groups,
and we've introduced these
compounds, these reagents
that make for very, very
strong nucleophiles,
hydride nucleophiles like
lithium aluminum hydride
in particular, to a
lesser extend less reactive
sodium borohydride.
And then various organometallic
reagents like Grignard reagents
and organolithium reagents.
At this point I want to sort
of broaden out our thinking
and start to talk not just
about ketones and aldehydes,
but more broadly
about the reactivity
of the carboxylic acid family.
[ Writing on Board ]
And to just put this
into context,
I mean carboxylic acids.
[ Writing on Board ]
Essentially all compounds
in which we have carbon
in the plus 3 oxidation state.
I will get to the several
questions in just a moment.
Esters being another
member of this family.
Acid chlorides.
So I'm going to paint
with a very broad brush.
And later on we're going to get
to a more specific understanding
of the reactivity of
this broad family.
But right now I'm going to
paint with a very broad brush.
I'll include in the
family acid anhydrides.
And maybe to wrap
up in main members
of the family I'll
talk about amides.
But, before we discuss
their reactions
with organometallic reagents,
I saw several questions.
I think there was one here.
One there?
Yes.
>> I was just wondering was
that [inaudible] whereas
with the lithiums -
the lithium-carbon bond the
electron attacks the epoxide.
Does it [inaudible] the
stereochemistry [inaudible]?
>> Ah, great question.
The question was when the
lithium attacks the epoxide,
does it attack from
the top or the bottom?
Does it occur with
inversion of stereochemistry.
And indeed it does.
Although there is no
stereochemistry at the center
that we formed here, you
could imagine, let's say,
having two different
substituents here
like a hydrogen and a deuterium
and we would get inversion
of stereochemistry.
Typically epoxides are
attacked by nucleophiles,
by basic nucleophiles at the
less sterically hindered carbon.
So for example, if I had the
epoxide with a methyl group,
called propylene oxide.
This one is trivially
called ethylene oxide.
If I had the epoxide with
a methyl group on one side,
propyl lithium -- the alkyne
would attack the carbon
that didn't have the
methyl group on it.
Does that make sense to you?
Another question?
I saw one young lady.
>> So it was the same question.
>> Same question?
And?
>> Why do we like to
use butyl stuff so much?
>> Why do we like
to use butyl stuff?
Great question.
So all of our hydrocarbons
ultimately come from petroleum.
And so one of the ways in
which butyl lithium is made is
by first cracking,
heating petroleum very hot
to get smaller fragments.
And one of the fragments that's
easily isolated is butene.
And the butene can then be
taken on to various types
of products including
butyl bromide for example.
So this is one of the
reasons that butyl is used.
You can buy propyl lithium,
but you can buy great big
bottles of butyl lithium.
Methyl is another common one.
>> So the chemistry it doesn't
-- it relies on fossil fuel?
>> Does the chemistry rely on
-- oh, I love that question.
Yeah, almost all of organic
chemistry ultimately goes back
to petroleum.
Some of the chemicals
that we use
as simpler building blocks
come from other sources
like carbohydrates
and, you know,
modern not ancient
plant sources.
But yeah, almost all of
organic chemistry comes back
to petroleum.
So the seats that you're
sitting in have a plastic
that maybe poly --
I don't know --
it's polypropylene or something,
and a covering that's a
synthetic fabric like nylon.
All of those have
come from petroleum.
And so I look at petroleum,
as much as I hate to pay $4.30
at the pump, when you
think about the amount
of stuff you're getting, you're
getting a gallon of gasoline.
This is too valuable to be
burning up because there are
so many things that
you can make from it.
You can't get a gallon
of anything.
You can't get 8 pounds or 7
pounds of anything for 4 bucks.
I can't get a gallon
of beer for 4 bucks.
And yet we go ahead
and we burn it.
So yes, petroleum is incredibly
valuable to organic chemists.
And one of my dear
colleague's favorite questions,
one that I won't be
asking you on an exam
because it's too open-ended and
far too ridiculously complex
for you at this point in your
sophistication, is to go ahead
and write a synthesis
of a steroid,
let's say cholesterol
or testosterone.
You've seen in your current
chapter some steroids starting
with petroleum.
So that would be one of
her favorite questions.
All right, but I
want to go now to --
I want to go now to
the reactions of esters
and of various members of
the carboxylic acid family.
And I thought your textbook's
presentation, particularly
in the reduction section, may
have been a little bit confusing
because there are a
lot of subtleties.
And when I think about a
subject, I like to think
about it in terms of
sort of the general rule
and then exceptions to it.
And there are a ton
of little exceptions.
And your textbook has picked
one and we're going to --
they're talking about
lithium aluminum hydride
with amines -- or with amides.
And we'll talk about that later.
But right now I want to paint
with a very broad brush a sort
of general reaction of
lithium aluminum hydride.
There are these strong
nucleophiles,
these potent nucleophiles,
hydride sources particularly
lithium aluminum hydride just
reduces the crap
out of everything.
Alkyl lithium reagents
like methyl lithium,
just add to everything
as much as they can.
Ditto for Grignard reagents
with certain exceptions.
So again, a very broad
brush view from 30,000 feet.
But we're going to take
a specific reaction.
We're going to take
methyl benzoate here
to exemplify our point.
And we're going to
imagine treating it
with lithium aluminum hydride.
I'm deliberately writing
this as a synthetic reaction.
We'll talk about what's
happening in a moment.
And then an aqueous
workup with --
I'll just write generically
H3O plus.
And you might do this
reaction, say, in THF.
And I'll be a good person and
write a balanced equation,
or at least -- actually, I
won't write a balanced equation,
but I will at least write
my two organic products
of the reaction.
Your organic chemist
is typically focused
on the big stuff.
But I'm going to write this
product, benzyl alcohol,
and the other organic product,
methyl alcohol, methanol.
And collectively, then,
these two constitute the
organic products of reaction.
And what this ends up
illustrating is a new property
that we haven't yet
seen called an
addition-elimination reaction.
[ Writing on Board ]
And we're going to see that this
addition-elimination reaction
goes through an intermediate
of benzaldehyde.
[ Writing on Board ]
And to a certain extent,
maybe with the exception
of amides here, if I took any of
the compounds that I'm erasing
and treated them with
lithium aluminum hydride,
you would get a similar
reaction of them.
All right.
So as I said, this
is a big mouthful.
And I like to break
big mouthfuls
into bite-sized pieces.
So let's think in sort of
broad mechanistic terms here.
All right.
I'm going to think about --
I'll write out our
benzoate component.
I'll write out our
methyl benzoate.
And I will at least for the
moment try to be a good person
and write a good mechanism
in which I write lone pairs
of electrons and try to
keep track of my charges.
And to keep things simple,
I'm going to write --
rather than writing out all
of lithium aluminum hydride,
I'm going to write hydride just
as this abstraction
of a hydride anion.
And the first thing that hydride
does is as a good nucleophile,
a potent nucleophile, we've
talked about the reactivity
in general of carbonyl groups.
The carbonyl group
is an electrophile.
Electrons flow from
the nucleophile
to the electrophile
and onto the oxygen.
[ Writing on Board ]
And I'll try to be a good person
and write all of my lone pairs,
all three lone pairs of
electrons around the oxygen.
And then two on this
other oxygen.
Now, this is not
a stable species.
This is not something
that you can isolate.
It's an intermediate.
And so I'm going to
remind us of the fact
that it's an intermediate
by drawing it in a bracket.
And we have a special name
for this intermediate.
Because we've gone
from a trigonal carbon,
a carbon with three
things around it,
to a tetrahedral carbon,
a carbon with four things
around it, I'm going to call
it a tetrahedral intermediate.
[ Writing on Board ]
And as I said, the tetrahedral
intermediate isn't stable.
The tetrahedral intermediate can
break down, electrons flow back
down from the oxygen,
they push out methoxide.
And now we get a new carbonyl.
At this point we've gotten
benzaldehyde, the intermediate
that I mentioned
and methoxide anion
with its three lone
pairs of electrons on it.
The reaction doesn't
stop at this point.
Esters are less electrophilic
than aldehydes.
In other words, aldehydes are
more electrophilic than esters.
We can write resonance
structures for esters
or a resonance structure,
in which the methoxy
group donates electrons
into the carbonyl and makes
it less electrophilic.
Aldehydes on the other hand
have less going for them.
And so we have more nucleophile,
and you can't just
stop at this point.
It's going to further
get reduced.
And so here's our aldehyde.
Here's our H minus again.
And again, I'll try to be a good
person and put it in quotes.
And electrons flow from our
hydride onto the oxygen,
now to give rise to
an alkoxide anion.
[ Writing on Board ]
And at this point,
that's what'll sit
around in your flask.
I haven't drawn the
counterions or anything.
Until you do a workup --
and I'll just write this
as workup meaning adding some
acid or adding some water.
I'll put this, again,
in our sort of 30,000
foot view of H plus.
And the product of this
reaction is benzyl alcohol.
And the other product
of the reaction is we'll
also protonate our methoxide.
So the other product of
our reaction is methanol.
And I'll just put that
in parenthesis here.
[ Erasing Board ]
All right.
So I want to show you
some generalities.
I want to show you
some analogy in this.
And so, at this point I'll write
essentially the same reaction
with just a slight
difference on it.
So before we could
have been thinking
about lithium aluminum hydride.
I said let's consider
lithium aluminum hydride.
At this point, let's
consider methyl lithium.
And so I'll take our same ester,
I'll take methyl benzoate,
but I'll treat it first
with methyl lithium.
And I'll write in
parenthesis two equivalents.
I'll try to remind us that
we're using a full amount of it.
And then secondly we'll treat
this with some aqueous acid.
And the product of this
reaction, now, instead of adding
in two hydrides, instead
of adding in two hydrogens,
we've added in two methyls.
It's essentially the
exact same thing.
And so now, we've gotten
an alcohol as a product
in which we've added
in two methyl groups.
The reaction's going to go
just like on the other one.
We went via benzaldehyde.
Here, the reaction's
going to go by way
of the ketone, called
acetophenone.
But just as in the case of
the other one, we can't stop
at the ketone with
one equivalent --
[ Writing on Board ]
-- of methyl lithium.
The ketone is more
reactive than the ester,
and so as it now
is sitting around,
it immediately reacts as well.
And so that's our sort of view
at 30,000 feet of the reaction
of these very, very strong
nucleophiles with members
of the carboxylic acid family.
As I said, there are some
exceptions, some differences.
But we can kind of catch
this general spirit of this
on the following equation.
[ Erasing Board ]
And so I'm going
to write this --
I don't always like the way
your textbook presents things,
particularly with a
lot of abstractions,
because from my way of
thinking, it's easier to start
at the concrete and then
work to the abstraction.
For a computer, I think
it's very good to like start
with an abstraction
and, you know,
it can spit out all
the examples.
But we've just looked
at two examples.
So now I'm going to
write the abstraction.
And I'll say many members of
the carboxylic acid family.
And I'm going to
write again this sort
of abstraction of
Nu minus dot, dot.
So some type of strongly basic
nucleophile that encompasses all
of the reagents that
we've been talking about.
So I'll say strongly basic
nucleophile that includes,
for example, lithium
aluminum hydride, RLi,
organolithium reagents, RMgX.
In other words, all of
these species have in common
that they have a bond between a
metal relatively electropositive
metal, and a highly
electronegative species, hydride
or -- I'm sorry a more
electronegative species, hydride
or lithium -- or carbon.
So basically, the generalities
of this type of reaction are
that in general, we get --
and I'll write parenthesis
excess here just
to avoid confusion.
When you have excess,
in general, you're going
to observe addition of two
equivalents of your nucleophile,
which I guess I've
written as Nu plus Z minus.
So that would sort of be
the biggest abstraction.
And I guess I'll try to be good
and keep my electrons in check.
So I'll write a balanced
lone pair.
All right.
So what we've looked at here --
what we've looked at here is an
addition-elimination reaction.
And we've seen this
general principle
that when you have something
like a methoxy group on a member
of the carboxylic acid family,
the reaction doesn't stop.
Things go on.
Now, to many students,
the first time they see an
addition-elimination,
they find it confusing.
And here's why they
find it confusing.
Here's our -- let's
say our ester.
And we've just added a
nucleophile to it, Nu minus,
to form our tetrahedral
intermediate.
[ Writing on Board ]
And that tetrahedral
intermediate isn't stable.
It breaks down.
It kicks out the OR group.
In other words, our electrons
flow like so to push out our
OR group that serves
as a leaving group.
[ Writing on Board ]
And the first time people
see this, having already been
through 51B and learned an
SN2 displace - was it 51B
or 51A you learned
SN2 displacement?
>> A.
>> A. 51A, and seen an
SN2 displacement reaction,
people start to think
about this and say, whoa!
What's going on here?
We don't see an OR group
like a methoxy group leave
in an SN2 displacement reaction.
But this is a little
bit different.
Here, a less basic
leaving group is okay.
See? In an SN2 displacement
reaction,
you've got to crowd five
things around carbon.
You don't have a
stable intermediate.
That leaving group was perfectly
happy attached to the carbon,
and yet something's coming
in and pushing it out.
And it's going over this
high energy barrier,
this transition state
to make that atom leave.
And so that atom that
leaves has to really,
really want the electrons.
In other words, a
good leaving group
in an SN2 displacement
reaction has to go ahead
and have a very stabilized
anion, or conversely,
because stability of
the anion means acidity
of the conjugate base,
conversely a strongly
acidic conjugate base.
So in this case, it's
a little different
because there's nothing bad
about the tetrahedral
intermediate.
We haven't had to
crowd anyone in here.
It's easy to get to this point.
But now it can happily
kick out the OR group.
And what it gets in return
is a carbon-oxygen pi bond,
which is very strong.
And so there's no problem
in getting your tetrahedral
intermediate together.
It's not pentavalent.
There's nothing bad about it.
But, it's very good to go
downhill and to breakdown
and kick out the leaving group
and get back your pi bond.
So, what I always like
to think about here is
in an SN2 displacement reaction,
you've got to have a
good leaving group.
But in the case of
this reaction,
an addition-elimination
reaction,
a less basic leaving
group is okay.
[ Writing on Board ]
And by comparison, I'd say
maybe less than 5 pKa is good
for a leaving group in
an SN2 displacement.
[ Writing on Board ]
Or maybe even an E2 elimination.
[ Writing on Board ]
Thoughts or questions
at this point?
>> Doesn't it work exactly like
a substitution [inaudible]?
Doesn't it work like a
substitution reaction
with the OR group [inaudible]?
>> Doesn't it work
like a substitution?
Indeed. This is a
substitution reaction.
So the very first step
of our methyl lithium plus
methyl benzoate reaction was
to substitute the methoxy group
for a methyl group
and get acetophenone.
And then, of course, we
couldn't stop at acetophenone
because it's even more
reactive than methyl benzoate,
so another equivalent
of methyl lithium adds.
But indeed, it is a
substitution reaction,
unlike an SN2 substitution
reaction,
here you can have a leaving
group that's a little bit less
acidic, like an alcohol
or an alkoxide.
[ Inaudible Audience Question ]
Ah, if you added a
water and acid --
>> Or an acid.
>> Guess I'm not
exactly following.
You mean?
>> Because there's
an [inaudible].
>> Oh, you mean in
the second step?
Would we -- yes.
So in the workup with acid,
we would protonate, indeed.
And one more question.
>> What does it mean
by the leaving group
have less than 5 pKa?
>> Okay. What do you
mean -- great question.
What do you mean by a leaving
group has less than 5 pKa?
In an SN2 displacement
reaction, chloride, bromide,
iodide are wonderful
leaving groups,
pKa of the conjugate acid,
respectively about negative 6,
whatever number you use,
about negative 6 for HCl,
for the conjugate acid about
negative 8 or thereabouts
for HBr, about negative
10 for HI.
And you can go a little bit
less good leaving group.
I can write an equation -- I can
give you an example of a case
where instead of having a
very strong acid, you know,
for the conjugate acid
in an SN2 displacement,
you could go slightly
from negative, you know,
negative 6 for chloride,
into the positive range.
I could give you an example
as low as 5 for the pKa.
Beyond that in an
SN2 displacement,
it pretty much isn't going to
occur unless, like in the case
of an epoxide, you
have ring strain.
pKa of the conjugate acid in an
epoxide is 17 for an alcohol,
but you've got that roughly
30 kilocalories per mole ring
strain associated with the
oxirane ring making it pop open.
So that's the rare case in an
SN2 displacement -like reaction
where you can actually have
what's essentially alkoxide
as leaving group, but
it's spring-loaded.
However, in the case of an
addition-elimination reaction,
absolutely, no problem
kicking out methoxy group
or ethoxy group, pKa of
the conjugate acid, 17.
Great question.
All right, at this point I want
to move on to the final point
that I want to make in
today's lecture and to bring us
to some ideas of synthesis
and showing you how powerful
organometallic reagents are
in carbon-carbon bond
forming reactions.
But also the process by
which organic chemists think
about using these reagents
and using reagents in general
to build up molecules.
And so I'm going to
give you a little bit
of a contrived example here.
But it's very much like the
type of thinking that people use
as they become more
sophisticated.
So, the example I'm
going to give us is
to synthesize a particular
compound.
I've chose it as an example
to illustrate a point,
but indeed something very
much like it could, say,
be a useful insect pheromone
that you might want to make
for a trap for, say,
Japanese beetles.
So we're going to try to develop
a synthesis of 4-ethyl-4-octanol
from compounds containing
four carbon atoms or fewer.
[ Writing on Board ]
There's nothing magical about
four carbon atoms, but many --
if you look
at commercially-available
compounds,
in general most of
them are small.
And you can buy a whole
range of different compounds.
And more big and more complex
compounds often are not
as available.
So you can buy many compounds
containing four carbon atoms
or three carbon atoms
or two or one.
Of course you can buy
plenty containing five,
but for purposes
of this example,
we're going to say
four or fewer.
And what we're going to do
with this is illustrate
the way chemists think,
the process of retrosynthetic
analysis.
[ Writing on Board ]
And we can call this the process
of thinking backwards about how
to synthesize something.
[ Writing on Board ]
And the reason retrosynthetic
analysis is so powerful,
the reason that the process
of thinking backwards is
so powerful is it's so easy
to get caught up in details.
When you try to look at things
from a forward point of view,
that it's very hard to see
how to get there from here,
from compounds containing
four carbons or fewer.
Oh well, we're using
organometallic reagents.
Do I use an organolithium,
do I use a Grignard?
Do I use an ester?
Do I use an acid chloride?
Do I use butyl lithium
as a base?
Do I use methyl lithium?
So retrosynthetic
analysis is kind
of like thinking your
way through a chess game.
We're going to start
with big pictures
and then work our
way to the details.
So let me show you
how I would think
about this particular example.
And you'll encounter on this
week's discussion problems
several examples very similar
to this that involve processes
of thinking backward at various
levels of sophistication.
All right.
So let me draw out my compound.
So it is -- let's see,
one, two, three, four,
five, six, seven, eight.
So here's my target molecule.
Now, when I first think about
this, I could look at this
and say well, okay, he
said four carbons or fewer.
And we just learned
about addition
of organometallic
reagents to esters.
So I could envision adding
-- taking some ester.
It would have to be a methyl
ester of propanoic acid
because that's four carbons.
I couldn't use any more.
And I could envision
forming this bond
by adding in butyl lithium.
And maybe again, for the
point of view of view
from 5,000 feet -- or 30,000
feet, I'm just going to think
in abstractions of some metal.
Whether it's a Grignard or
a lithium doesn't matter.
And again, I could think --
well, here we add in some metal.
And that kind of
catches the strategy.
But we've got a problem here.
And the problem is how
do we get selectivity?
[ Writing on Board ]
I just said you can't
really add one equivalent
of a Grignard reagent
or an organolithium
compound to an ester.
It won't stop at the ketone.
Was that your question?
>> Oh yeah.
No, that's fine.
>> All right.
So let me take this
same idea and see
if we can think backward
a little bit.
So, we know -- see, here I'm
trying to do it all at once.
And I kind of got
myself going here.
It's a lot better than saying
we're going to think forward.
But let me think a
little bit backwards.
We know that an alcohol
can be formed by a ketone
and an organometallic reagent.
So we could imagine -- so I'm
going to use this arrow here.
This arrow, this big open arrow,
means a retrosynthetic arrow.
It means a thinking
backwards arrow.
Organic chemists love
arrows, curved arrows,
equilibrium arrows,
resonance arrows.
This is a retrosynthetic arrow.
So I could envision going ahead
and going backward
to this ketone.
And I could envision --
let's see, did do that right?
Nope. I could envision
this ketone, 2-hexanone
and a butyl metal reagent.
And that would work.
I could add butyl lithium
or butyl magnesium
bromide to 2-hexanone.
And that would be okay
to make that alcohol.
But then I'd need a way
to make this ketone.
And oh, well okay.
We could make that ketone.
We've only learned reactions
that form carbon-carbon bonds
to make alcohols at this
point in this course.
So I could imagine making
that ketone by oxidation
of the corresponding
alcohol of 3-hexanol.
And again, right now I'm not
going to worry which reagent
to use, whether I use
chromium trioxide,
whether I use potassium
chromate, whatever.
But now I look and
I say, I'll wait.
And I can think backwards
again now to the point
of a propyl metal
reagent and propanal.
And at the strategic level,
we've now broken
this molecule apart.
We've used the process of
thinking backwards to figure
out how we can put
this molecule together.
And now having completed
our retrosynthetic analysis,
now we're ready to go forward
and worry from the strategy
to the tactics of, okay,
what reagent do we choose?
How do we do our details?
And so the last thing I'll do
in solving this hypothetical
problem is
to show you the synthesis
that I've worked out.
So I would start with propanal.
I've completed the requirement
with each of my three components
of four carbons or fewer.
I'd start with propanal,
it's commercially available.
It fits the requirement.
I'll add in, just for the fun
of it I'll use propyl
magnesium chloride.
I could use propyl lithium,
I could use propyl
magnesium bromide,
propyl magnesium iodide.
I'll carry out my
workup with aqueous HCl.
I could use aqueous
sulfuric acid.
I could use aqueous
ammonium chloride.
I could probably be lazy and
write H3O plus over the arrow.
But I'm going to
go to the stockroom
and get some chemicals.
And I'm going to ask
them for some HCl.
The product of that reaction
after workup is 3-hexanol.
I now am ready to
oxidize my 3-hexanol.
You learned lots of reactions
last quarter for oxidation.
They taught you potassium
dichromate in sulfuric acid
and water, sometimes
called Jones' reagent.
There are many, many
reagents based on chromium 6,
sodium dichromate,
chromium trioxide.
There are lots of
safer alternatives,
including alternatives
based on bleach.
But we're going to use
reactions that you know.
So we're going to use
potassium dichromate.
Then, at that point,
we have our 2-hexanone.
And question.
>> So I was just wondering,
could you also instead
of the [inaudible] use CrO3?
>> Absolutely.
We could use chromium
trioxide as an oxidizing agent.
Lots to choose from.
Nice in our retrosynthetic
analysis not having to worry
about getting that right yet.
And now, having the leisure of
going and choosing our reagents
and choosing our tactics.
Finally, to complete
our synthesis,
I'm going to take butyl lithium
and I'll do an aqueous workup.
And again, I'll use HCl.
I've written it below
the arrow here,
written it above
the arrow there.
Doesn't really matter.
A chemist is going to
read it the same way.
And lo and behold, I
have proposed a rational
and selective synthesis
of our target molecule
4-ethyl-4-octanol.
[ Writing on Board ]
And this art of recognizing
something in a molecule
and seeing where it comes
from is going to grow
and grow in the course.
Right now we've seen an
alcohol and we said oh,
I know how to make an alcohol.
I can make an alcohol by adding
in two carbon nucleophiles.
I can make alcohols by adding
in carbon nucleophiles.
Later on we're going to see
all sorts of other families
of carbonyl compounds.
All right.
I will see you on Thursday.
We'll start off with
a 10 minute quiz.
------------------------------1827f70bddb3--
