Chemistry 242 Exp. 4
Solventless Aldol Condensation
We talked some about “green chemistry” last term. This lab is another “green”
procedure in that it uses no reaction solvent even though the reactants are all solids. This
saves expenses in terms of both the initial purchase of the solvents and their disposal. In
addition, there is no need to separate the product from the reaction solvent, because there
isn’t any. We still use solvent for recrystallization, though.
The placement of this lab presents a bit of a quandary. It could come towards the
beginning of the organic chemistry lab (start of 241) because it illustrates the depression
of mixture melting points: mixing two solids together and having them become a liquid
solution (because their mixture melting point was depressed to below room temperature).
Or, we could wait until the third term when we study carbonyl compounds.
Alternatively, we could take a middle route, covering this lab after not only melting
points, but after mechanisms, too, so we can understand what is happening in the
reaction. I chose this third route, because it is kind of short and may allow some catch up
time.
This will serve as a good reminder of mixture melting point behavior and we have
recently studied enol mechanisms, so this is a good time to have this lab. I don’t expect
you to be able to reproduce this mechanism yet (not until chemistry 243), but I do expect
you to follow along with it. It has a couple of concepts that are relative to a lot of organic
chemistry reactions we have studied so far: nucleophile/electrophile attractions and
acid-base chemistry.
Alpha Hydrogens to a carbonyl are acidic because the product conjugate base is
resonance stabilized. “Alpha” is just a way of counting from a functional group (or other
designated point in the molecule). Alpha is one carbon away, beta is two, gamma is
three, and so forth. The conjugate base product of an acid/base reaction with these acidic
hydrogens is called an enolate ion. The “en” from alkene and the “ol” from alcohol.
This part is similar to the enols we have been studying, but with a negative charge on a
carbon adjacent to the carbonyl, hence the “ate ion” on the end of the term. In the first
step of this reaction, the initial reaction with base removes a proton from the alpha carbon
creating a deprotonated enol, hence the name enolate ion.
Step 1. Formation of the enolate ion
OH
O
H
C
H2
O
O
CH2
CH2
resonance for enolate ion
In the next step, the enolate acts as a nucleophile, adding to the carbonyl carbon
of another molecule. If the carbonyl the enolate adds to is a different molecule, it is
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called a “crossed aldol” reaction. If the enolate adds to a molecule of itself that has not
been deprotonated yet, it is called a “self aldol” or “homo aldol” reaction. A
deprotonated enolate ion will not add to another enolate ion. The negative charges
repulse each other too much.
Step 2. Addition of the enolate to a carbonyl carbon. In this example, if both R’s =
-CH3, then this is a self aldol. If either or both R groups are anything other than methyl
groups, this is a crossed aldol.
O
R
O
O
O
CH2
R
R
R
This addition product is treated with acid to give a β-hydroxyketone (hydroxy is
the substituent name for an alcohol – see Bruice, p. 791).
Step 3.
O
H
O
OH2
O
OH
R
R
R
R
The β-hydroxyketone can then undergo a condensation reaction, eliminating
water. This condensation reaction is not reversible and is driven by the conjugation of
the ketone and alkene pi bonds. It goes even more readily if the R groups are also
conjugated with the emerging alkene, such as benzene rings would be.
Step 4.
O
H
OH
R
OH2
O
R
R
R
H
HOH
2
To avoid all the different possible products, usually only one of the two carbonyl
compounds has alpha hydrogens. Otherwise, the number of different crossed aldols and
self aldols makes isolation of the desired product difficult and also lowers the yield of the
desired product. See page 874 in Bruice for more on aldol reactions.
The reactants in this experiment will be 3,4 dimethoxybenzaldehyde and 1indanone.
O
O
O
+
NaOH
OMe
H
OMe
OMe
OMe
3,4-dimethoxybenzaldehyde
product
Me = methyl
mp = 180 C
Safety precautions: Although green chemistry reduces safety precautions, it is
difficult to eliminate them all together. Solid sodium hydroxide is a very strong base.
Skin contact should be avoided. NaOH is also very hygroscopic (readily picks up
moisture from the air. Close the NaOH container lid IMMEDIATELY after removing
your solid product and BEFORE doing anything else. This is VERY IMPORTANT.
This is a synthesis, and the procedure for synthesis prelabs will be followed again.
You need a balanced reaction showing stereochemistry and a table of reactants, like the
one below:
reagent
mw
(g/mole)
166
mL
used
-
density
132
-
-
NaOH
40.0
-
-
product
280.0
-
-
3,4 dimethoxybenzaldehyde
1-indanone
Limiting reagent
g used
moles
used
-
-
mole ratio
theor
actual
-
________________________________________________
Topics to review: aldol reactions (preview, actually), recrystallization, melting pts
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Reaction Instructions:
1. Place 0.25 g of 3,4-dimethoxybenzaldehyde and 0.20 g of 1-indanone in a test
tube. Use a stirring rod or metal spatula to scrape and crush the two solids
together until they become a brown oily mixture. Remember that your strength
and enthusiasm may exceed the beaker’s limits: don’t push out the bottom of the
beaker. Do not continue until step two until all the solid is gone. This may take
as long as 20 minutes, but if not done, you may have to start all over again.
2. Add 0.05 g of finely ground (using mortar and pestle) solid NaOH to the reaction
mixture and continue scraping until the mixture becomes solid. This much NaOH
is less than one pellet; you can easily share with fellow students to avoid waste.
Isolation:
3. Allow the mixture to stand for 15 minutes, then add about 2 mL of 10% HCl
solution in water. Scrape your product off the walls of the beaker to expose it to
the HCl. Check the solution pH with 1 cm or smaller pieces of paper to be sure
that it is acidic (less than 7). Remember to record exactly how much HCl is used
as you would with anything that you add to your reaction. The whole point of
science is to reproduce your results. If you do not know what and how much you
added of each ingredient, you cannot possibly expect to reproduce your results
with accuracy.
4. Isolate the solid with vacuum filtration. Let the solid stand in the funnel with the
water drying to pull air through in order to dry the product a little.
5. Recrystallize the product with a 90% ethanol/10% water mixture, first using some
hot solvent to rinse any remaining product out of the beaker. Remember that
recrystallizations use already hot solvent. Add only barely enough hot solvent to
get your product dissolved or your yield will be greatly reduced. You should not
need more than 20 mL of solvent. After dissolving, let cool to room temperature,
then place in an ice bath.
6. Filter product crystals and spread them out on your watch glass with a scrap of
paper (weighing paper does nicely for this) so that the crystals can dry for a week
before doing step 7.
Characterization.
7. Determine the mass of your product and its melting point
Post lab analysis:
Analyze purity and yield of your product. In other words, what was your percent
yield? Do you think that is good or bad? If your percent yield is poor, what do you think
happened that reduced your yield? What do you think you could do next time to improve
your yield? Did the color and mp of your product match the literature mp and color?
Was it even close?
Post Lab Questions:
1. In this reaction, there is only one carbon with acidic alpha Hydrogens. Draw
the reactant with it and circle the carbon with the acidic Hydrogens.
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2. What are the products of these crossed aldol condensation reactions (only
one possible answer)?
O
O
+
(H3C)3C
C(CH3)3
O
O
+
O
O
+
3. In other reactions, both reactants may have acidic Hydrogens, leading to two
different products. What are the two products of these crossed aldol
condensation reactions?
O
O
+
O
O
+
O
O
+
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