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!