Aldol Condensation Video Lecture - NEET

Voiceover: Aldol condensations
are extremely important
reactions in organic chemistry.
Let's look at, first, an "aldol addition,"
where if we take something
like acetaldehyde,
and that's some sodium hydroxide,
we form this aldol product.
So the term, "aldol" comes from
the fact that we have an "aldehyde" here,
and then, the "ol," because
we have an alcohol as well.
Let's look at the mechanism
for the formation of this:
So, if we start with our acetaldehydes,
let me go ahead and draw it out here,
and we look for our alpha carbon,
it's the one right next to our carbonyl,
so here is our alpha carbon,
there are three alpha
protons on that alpha carbon;
I'm just gonna show one here,
and then we're gonna have sodium hydroxide
come along, and act as a base.
So we have the hydroxide anion,
so negative one formal
charge on the oxygen;
it's gonna take this proton,
leaving these electrons
behind, on our alpha carbon
so let's go ahead and show the
carbanion that would result:
So we have our carbonyl,
and then we have our
hydrogen on the right side;
on the left side, we have
a lone pair of electrons
on this carbon, so we have a carbanion.
So the electrons in magenta, here,
moved off, onto this carbon,
we can draw a resonance structure,
these electrons move into here,
these electrons go off onto our oxygen,
and we would form the oxyanion.
So we have a double bond here,
and then our oxygen, up here,
would have three lone pairs of electrons,
so, negative one formal charge,
and then a hydrogen.
So the electrons, in magenta,
moved in here, to form our double bond,
and so we know, that the negative charge
is best supported on an oxygen,
so the oxyanion is probably the best way
to show an enolate anion,
however, for the aldol condensations,
I'm always gonna use the carbanion,
as our enolate anion;
it's actually a little bit easier
for students to look at
the mechanism that way.
So, we're gonna start with our carbanion,
so I'm just gonna redraw it down here,
so here's our carbanion,
and once you form your enolate anion,
your enolate anion can
act as a nucleophile,
'cause this carbon here
is negatively charged.
And, because we use something
like sodium hydroxide,
sodium hydroxide is not basic enough
to completely form enolate anions,
so only a small amount of
your enolate anion is formed,
and so this can act as a nucleophile,
and attack another
molecule of acetaldehydes,
so I'm gonna go ahead
a draw another molecule
of acetaldehyde,
and I'm gonna draw it,
in a particular way:
I'm going to leave off the hydrogen,
and I'm going to put
it right next to this;
it just helps me, when I'm
doing an aldol reaction here.
Alright, so if I know the carbonyl,
of my other molecule of
acetaldehyde, is polarized;
the oxygen is partially negative,
and the carbonyl carbon
here is partially positive.
So the carbonyl carbon is electrophilic:
it wants electrons,
and it's going to get electrons
from our nucleophilic enolate anion.
So, these electrons here are
going to attack our carbon,
push these electrons
off, on to our oxygen,
so let's go ahead, and
show the result of that:
So, on the left side, we
would have our oxygen,
with three lone pairs of electrons,
and then we form a
carbon-carbon bond here,
so that's extremely important,
and then we also have,
over here on the right, our aldehyde.
So let's show the movement of electrons:
So these electrons, right
here, this is our carbanion,
attack our carbonyl carbon,
form this bond, a carbon-carbon
bond forming reaction,
to form an alkoxide intermediate,
and so we can protonate
our alkoxide with water.
So water is right here,
so we could show a lone pair of electrons
picking up this proton,
and we can go ahead and
draw our aldol product now,
so we will draw our aldol
right here, like that.
So the formation of a carbon-carbon bond
makes an aldol addition,
and then next an aldol condensation,
a very important reaction
in organic chemistry,
so, there's our aldol product.
Alright, let's look at
an aldol condensation,
which is kind of a continuation
of what we were just talking about,
and you can form different products,
depending on reaction conditions.
So, for example, here we're starting off
with acetaldehyde again, and
sodium hydroxide, and water,
we're adding heat,
so we're changing the reaction conditions,
and so, we're gonna get
a different product,
and so we call this an aldol condensation.
So you can see, the product
that we get is an "enal,"
so we get an enal,
we get a double-bond here, for the "en,"
and then we have an "aldehyde,"
so you also hear it referred to
as an alpha-beta unsaturated,
because this carbon, right
here, is the alpha carbon,
and this carbon is the beta carbon;
it's unsaturated, because we
formed a double bond here,
so this is an aldol condensation.
And to form this enal product
we're going to start with the aldol
that we just made in
the previous reaction.
So acetaldehyde, with sodium hydroxide,
can form an aldol, as we just saw,
so let's go ahead and
draw in the aldol product:
So the same mechanism
that we just discussed,
and I should point out, that we are doing
a base-catalyzed aldol condensation here,
so we have a hydrogen here,
and then our aldol had an oxygen,
and then a hydrogen right here,
so there's our aldol.
And if we look at our aldol,
the carbon right next to the carbonyl
is our alpha carbon,
and there's still two acidic
protons on that alpha carbon,
so I'm just gonna draw one in there,
and then we can think
about sodium hydroxide
acting as a base, and forming
another enalate anion,
so I'm gonna have sodium
hydroxide come along,
so, OH minus,
it's gonna function as a base:
it's gonna take this proton,
leave these electrons
behind, on this carbon,
so let's go ahead and draw that.
So we would have our oxygen right here,
with two lone pairs of electrons,
we would have our carbon now,
with a lone pair of
electrons making a carbanion,
and then we have our carbon
EEL over here, like that.
Alright, so let's once
again show those electrons,
these electrons right in here,
forming our carbanion.
And then, to form our enal product,
we can just think about,
let me go ahead and draw an arrow here,
we can just think about these electrons
moving in here, to form a double bond,
and kicking these electrons off
onto the oxygen, to form
hydroxide as a leaving group,
so we can say that hydroxide leaves
and then we form our double bond,
so I can go ahead and show that:
Let's say these electrons right here,
were the magenta ones
to form our enal product.
And so, once again,
I'm using the carbanion
form of our resonance structure,
instead of the oxyanion,
so again, it's probably more appropriate
to use the oxyanion,
but it's just easier,
from a student perspective
to use a carbanion,
to go ahead, and form your
enal product that way.
So, in terms of another product,
this is the major product
that we would form;
this is the trans product,
so if you think about this double bond,
these hydrogens are on opposite sides,
so this is the trans product,
so that's the major one.
You would form a small amount
of the cis product, as well,
so let me go ahead and draw that in,
but the cis product would
be the minor product,
because it has more steric hindrance
than the trans product,
so, this is the major product,
and I'm sure the one that your professor
wants you to put on an exam question,
so the trans product has
decreased steric hindrance.
Alright, let's do one more
of these aldol condensations:
So here, I've drawn out their reaction,
and so, the question would be,
"Write a plausible mechanism
"for the formation of this enone?"
So we have an "enone" this time,
we're starting with a ketone,
and so, we have a double
bond, and then a ketone,
so an enone product.
So we're gonna use sodium hydroxide,
and we're going to heat things up here,
so let's think about what would happen:
so we have our ketone, and
we have two alpha carbons,
which are symmetrical,
so it doesn't really matter
which one we take a proton from,
so I'm just gonna say, take
a proton from this one,
so we take a proton from here,
to form our enolate anion,
and, once again, I'm
gonna form the carbanion,
just to make things simpler.
So we have our carbanion now,
and then we have another molecule,
cyclohexanone's gonna come along,
so let me go ahead, and draw that in,
and we know we have a
polarized carbonyl situation:
partial negative, partial positive.
So, our nucleophilic enolate anion
is going to attack right here,
so we can show these electrons
attacking right here,
pushing those electrons
off, onto your oxygen,
so I'm just going a little bit fast
through this mechanism here,
and so we have, now, a bond that formed,
and now this oxygen right here
is going to have three lone
pairs of electrons around it,
negative one formal charge,
and I can go ahead,
and draw in the rest
of my ring, like that,
so let's follow some of
those electrons here:
so, I'll make these
electrons in blue, this time,
forming the all-important
carbon-carbon bond right here,
and then, these electrons
came off, onto our oxygen,
forming our alkoxide.
So we can think about water
protonating our alkoxide intermediate:
so lone pair takes that proton,
leaves those electrons behind,
and, hopefully we have enough room
to draw the aldol,
so, over here, we would
have our aldol intermediate,
we have our OH, we have
our ring, like that.
Okay, thinking about what happens next,
the carbon next to the carbon
EEL is the alpha carbon,
so that's our alpha carbon,
and so there's still a
proton on that alpha carbon.
I don't really have a lot of room here,
but let me see if I can draw
in an alpha proton, like that.
And so, we could have hydroxide,
once again, functioning as a base:
so picking up this proton,
and then it leaving
these electrons behind,
on that carbon in red,
so let's get some more room here.
So, now what we would have is our ring,
we have our carbonyl there,
and then we have a lone pair
of electrons on this carbon,
a negative one formal charge;
over here, we have our OH, like that,
and then we have our ring.
Okay, so in the next step,
once again, I'm not drawing an oxyanion,
so I'm not gonna draw the oxyanion form
of this enolate anion, here,
but we can show these
electrons moving into here,
and pushing these electrons
off, onto the oxygen,
and so let's go ahead,
and draw our product,
once we do all of that:
So we would have our ring,
we would have our carbonyl here,
and then, we would form a double bond,
and then, now we can see
that we have our product,
so let's follow those electrons:
So, electrons in blue
here, these electrons,
move in here to form our double bond,
and then, let's use red
to show these electrons
in here, coming off,
and so we get hydroxide
as our leaving group,
and we form our conjugated
enone as our product.
And so, this is one way to
represent the mechanism,
and, once again, the reaction conditions
will determine what
kind of product you get:
our enone, here, is conjugated,
we have an alternating
single-double bond situation,
and so, that favors this product, here,
so that shifts the equilibrium,
once again, to make more of your enone.
So, practice doing a lot of
these aldol condensations.