Carbonyl Condensation Reactions.

Carbonyl Compounds:

R

O

H

2

C

!

+

!

-

R'

Earlier, we talked about how the carbonyl carbon was such a great electrophile. Later,

we talked about how we can make the carbons alphato the carbonyl group into great

nucleophiles (enolates). Does anyone see the problem here?

A very common problem with aldehyde enolates in particular is that they tend to react

with themselves - to “dimerize.” This can be a good thing! This self-reaction process is called

an Aldol Reaction. Why Aldol? Because one aldehyde reacts with one enol. The first step in

this reaction can be summed up as follows:

O

H

KOH

Ethanol

O

H

HO

KOH

O

H

O

H

O

H

O

H

2

O

Remember, of course, that this reaction is reversible (a retro-aldol):

O

H

O

H

HO

O

H

O

+

The “Forward” reaction, i.e. formation of the aldol product, is initially only favorable for

non-sterically hindered aldehydes. Ketones undergo the retro reaction quite easily. You should

also note that any carbonyl compound with alpha hydrogens can undergo the aldol reaction (i.e.

esters, acids, etc.). As a note to make your life easier, every time you see an O-C-C-C-O

arrangement in a molecule, you should be thinking aldol... (you’ll understand this better when I

explain it in class).

The next question to ask is, how do I tell whether a molecule is going to undergo an

aldol-like reaction, or an enolate-like reaction? There are a couple of clues:

1) STRONG base. If bases like LDA are being used, and a decent electrophile is

present, then you’re likely looking at an enolate type reaction.

2) GOOD electrophiles. If there’s another strong electrophile around (A halogen (Br

2

, for

example), or something like iodomethane (CH

3

I), you’re looking at an

enolate type of reaction.

3) Catalytic amounts of a weak base. If there are no other electrophiles added,

then the use of a “weaker” base, such as sodium ethoxide (NaOEt) or hydroxide

(HO

-

) is usually a clue that an aldol-type reaction will be taking place.

The Aldol Condensation.

This is what I would call a “money reaction.” If you get the details of this one down,

you’re golden.

First off, what do you think of when you hear the word “condensation”? Yup, water. So

it makes sense that the aldol condensation is an aldol reaction that gives off water. And that’s

exactly what it does – the initial aldol product is dehydrated to give a conjugated carbonyl

compound:

O

H

HO

H

HO

O

H

HO

O

H

As you see, the ene-al is simply the aldol product minus the elements of water - a condensation

reaction. This dehydration reaction is your fourth method for forming double bonds - although it

only works under very specific conditions (i.e. it only makes conjugated carbonyl compounds -

although this is still VERY useful!).

So, one more time . . . the aldol condensation takes a carbonyl compound (usually an

aldehyde, ketone or ester) and joins two of them together to form a conjugated carbonyl

compound. Here’s another example (with the carbons from one of the pair marked in bold, to

show where everything is going...):

O

NaOEt
EtOH

O

O

+

Generally, with aldol condensations you are combining two of the same molecule to form

a dimer. However, in a few cases it is possible to combine two different molecules to form the

aldol product. There are two MAJOR restrictions: 1) One of the pair must not have any

enolizable (i.e. alpha) protons (for example, aryl aldehydes & ketones like benzaldehyde or

benzophenone) OR 2) One of the pair must form an enol MUCH more easily than the other

(i.e. a 1,3-dicarbonyl compound versus a normal ketone). Some examples:

O

H

+

O

OEt

NaOEt

EtOH

H

O

OEt

Gaaa! It's that *&%*! Ethyl
Cinnamate again!!!

(There must be a million ways to make
this stuff...and you'll see 'em all!)

No Enolizable Protons here...

O

O

Protons here much more easily enolized...

+

O

t-BuOK

t-BuOH

O

O

But what about the enolates of other carbonyl compounds, such as esters? Do they

undergo these “self-condensation” reactions as well? Of course they do! However, the thing to

note is that esters have essentially a “leaving group” attached to the carbonyl carbon. Thus the

products of the condensation reaction are slightly different. In essence, the product of an aldol-

type reaction between two esters (this reaction is called a Claisen condensation) is almost always

a 1,3-dicarbonyl compound (remember these?). In its simplest form:

O

OEt

NaOEt

O

OEt

O

EtO

O

OEt

O

OEt

O

OEt

O

H

3

O

+

OEt

O

OEt

O

H

H

EtO

O

OEt

O

(note - why do we use NaOEt, and not NaOH?).

Again, this is a fairly general reaction – just about any ester can undergo this type of self-

condensation. The “driving force” of this reaction is not dehydration, however – it is the

formation of the stable enolate anion (i.e. the one between the two carbonyl groups) that makes

the final step irreversible. Is it possible to do “mixed” Claisen reactions? Of course, provided

the same rules are followed as for the mixed aldol. One component must either have no

enolizable protons, or must be enolized VERY rapidly. Here’s an example (in addition to what’s

in your book):

N

OEt

O

Ethyl Picolinate

O

Acetone

2

+

2 NaH

O

O

O

N

N

Mechanism Practice!!! Draw the mechanism for the reaction above, as well as for:

O

O

O

N

N

O

O

NH

4

N

O

H

N

N

(Hint: ammonium acetate is basically a mixture of ammonia (NH

3

) and acetic acid (CH

3

COOH))

The nature of the electrophile is also worth considering. What if we have a conjugated

carbonyl compound? Recall that we then add a stabilized nucleophile to the alkene portion - a

1,4 addition. Malonates and keto-esters are particularly good at this type of reaction, called a

Michael addition:

O

+ H

2

C(COOEt)

2

NaOEt

O

CH(COOEt)

2

NaOEt

HC(COOEt)

2

O

CH(COOEt)

2

If a large excess of the conjugated compound is used, it is of course possible to get a double-

Michael (dialkylated) product:

O

OEt

O

+ excess

O

NaOEt

O

OEt

O

O

O

The Stork enamine reaction is just another way of performing Michael-type additions

with less-activated ketones and aldehydes. You should familiarize yourself with this reaction

from the text, but don’t put too much emphasis on it...

You now know how aldehydes, ketones and esters can react to give addition or

condensation products, both in conjugated and non-conjugated systems, as well as a very limited

way of doing “mixed” condensations. Another way to get a reaction between two different kinds

of carbonyl groups is if they are both part of the same molecule. This is called intramolecular

condensation, and is an excellent way to make 5- and 6-membered rings:

NaOH

EtOH

NaOH

O

O

H

O

O

H

O

H

H

O

OH

H

H

H

OH

O

O

H

O

H

H

As stated above, this reaction generally only works to make 5 and 6 membered rings -

you generally won’t form 3,4,7 or 8 membered rings in this way. Also note that you could have

nearly any substituent hanging off either of the carbonyl groups, or off of the carbons in between

the carbonyl groups - this reaction is VERY general. Also note - you can do this reaction with

estersas well (in which case, it is called the Dieckmann reaction). Take a look at the synthesis

below - you need to be able to recognize what types of reactions are taking place at each step,

here:

C C C

O

MeO

O

MeO

O

OMe

O

OMe

+

O

MeO

O

OMe

1) NaH

MeOOC

MeOOC

COOMe

COOMe

COOMe

COOMe

2) H

3

O

+

H

2

/ Pd

MeOOC

MeOOC

COOMe

COOMe

COOMe

COOMe

3 NaH

3 BrCH

2

COOMe

MeOOC

MeOOC

COOMe

COOMe

COOMe

COOMe

COOMe

MeOOC

MeOOC

H

H

H

H

3

O

+

Heat

HOOC

COOH
COOH

COOH

HOOC

HOOC

H

3

O

+

MeOH

MeOOC

COOMe

COOMe

MeOOC

MeOOC

COOMe

1) NaOMe

O

COOMe

O

MeOOC

O

MeOOC

H

3

O

+

Heat

O

O

O

2) H

3

O

+

We can now put all of the knowledge we’ve gained in the last chapter together to look at

the world-famous Robinson Annulation. This is a classic organic reaction which works

splendidly to bring two different molecules together to form a ring, or to form several rings from

a single molecule. It involves both Michael additions and Aldol condensations. Let’s watch one

in action:

CH

3

O

H

+

O

KOH

O

CH

3

O

CH

3

O

OH

CH

3

O

Okay, that’s the overall picture. Let’s look at a second example step-by-step, complete

with mechanistic commentary:

OCH

3

O

+

O

KOH

OH

H H

OCH

3

O

O

OCH

3

O

O

OCH

3

O

O

O

H

H

The first step is the enolization of one of the ketones. The trick is - - - which ketone?

Well, you have to do a little thinking here. Which enolate anion will be the most stable? In this

case, the protons on the carbon next to the aromatic ring are the most acidic - the resulting

double bond is in conjugation with the benzene ring, which helps. We then do a Michael

addition to the conjugated ketone, and the resulting enolate tautaumerizes to the ketone. What

happens next? An aldol condensation:

OCH

3

O

O

OH

H H

OCH

3

O

O

OCH

3

O

O

O

H

H

OH

H

OCH

3

HO

O

OCH

3

O

Fairly standard aldol condensation chemistry. The “dangling” ketone is enolized, and the

resulting enolate attacks the ring-ketone to form the initial aldol product. Following a proton

transfer (to protonate the alcohol, and deprotonate to form a new enolate), hydroxide is lost to

form the new conjugated ketone, which in the case of the Robinson, is in a ring.

That about wraps it up for this chapter. As you have seen, it is all basically one reaction,

just applied to a variety of molecules. Rules to remember - look for the most easily enolized

carbonyl group, attack the most electrophilic center, then spit out the smallest molecule possible.

That’s it!