University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Reaction Efficiency
Chemical Concepts
Mass, molecular weight, the mole, balanced equations, stoichiometry, density.
Green Concepts
Reaction efficiency – atom economy and other metrics; safer products; paradigm shifts –
asking new types of questions.
Introduction
In Green Chemistry, our emphasis is on changing the thought processes about chemistry –
thinking about the implications of what we do as chemists instead of just the immediate
outcomes. We teach students to ask questions like the following.
Are we designing and using safe chemicals?
What is “safe” with respect to human health and the environment?
Are we making our chemical products wisely, avoiding dangerous chemicals in their
manufacture and avoiding the generation of hazardous waste?
By the very nature of these thought processes, we address many things that are important to
both teachers and their students. Most importantly, we address chemistry from a “real life”
perspective, and, by worrying about these real life issues, we explore chemical experiments
that can be carried out without the need for specialized equipment or laboratories.
How do we determine how “good” a chemical reaction is? This question can form the basis
for both valuable experience with the fundamentals of chemical calculations and essential
thought processes regarding green/sustainable chemistry. In this lesson, we will look at
several ways to assess reaction efficiency – the quality of a reaction in terms of its production
of a desired product. Through this analysis, new and more exciting ways to introduce old
(boring?) chemical concepts (e.g., mass, molecular weight, the mole, balanced equations,
stoichiometry, density) may be revealed.
In this lesson, we will work through a set of exercises. We will look at a seemingly simple
chemical reaction in order to develop our thinking about reaction efficiency, guided by
worksheets. This sort of work would be ideal for adaptation to the “peer led” small group
approach to problem solving that is being widely adopted throughout the United States, and
elsewhere, and the following description identifies places in the lesson where it would be an
excellent idea to stop talking, arrange for students to break into smaller groups, and discuss
the questions being raised, then report back to the full group.
The following discussion provides references to a worksheet that is provided as the last few
pages of this packet.
Copyright Kenneth M. Doxsee, University of Oregon
Page 69 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Here is an experimental procedure for a simple inorganic reaction, reported in a standard US
laboratory textbook used by college undergraduate students, but not atypical of experiments
that might be performed by high school students as well. (Note: I do not recommend that
students actually perform this experiment. We will simply analyze it using pencil and paper.)
(NH3)2PtI2 + Ag2SO4 + 2 KCl → (NH3)2PtCl2 + 2 AgI + K2SO4
Experimental Procedure
Prepare a solution of 63 mg (0.202 mmol) of silver sulfate in 10 mL of water
in a 25-mL beaker containing a magnetic stirring bar. Add 100 mg (0.207
mmol) of the cis-diiodo derivative, in small portions, to this Ag+ solution.
Heat the suspension, with stirring, on a sand bath (70-80 ºC) for 10-12 min.
Filter the mixture to separate the precipitate of AgI.
Isolation of Product
Concentrate the filtrate to a volume of about 2.0 mL. Treat this solution with
330 mg (4.43 mmol, a large excess) of KCl. Heat the mixture on a sand bath at
70-80 ºC for 2-3 min. Bright yellow crystals of cis-diammine
dichloroplatinum(II) should precipitate out. The heating is continued for an
additional 5-8 min. Cool the mixture to ) ºC in an ice-water bath. Filter the
product using a Hirsch funnel. Wash the crystals with 500 mL of ethanol
followed by 1 mL of ether and dry them under suction in air. Determine the
percentage yield.
Let‟s say a student has run this reaction and has isolated 50 mg of the desired product. Since
we started with 100 mg and ended up with 50 mg, a simple attempt to quantify the efficiency
of the reaction might be simply to compare the two amounts, calculating a “mass yield” (part
1 on the cisplatin worksheet):
(50 mg/100 mg)*100% = 50%.
For a student just beginning studies of chemistry, this would represent a good – and perhaps
the only way to attempt to quantify how “good” the reaction was. As such, it can serve as a
good starting point for discussion, guided by questions from the teacher or, even better,
questions and comments from the students.
Is this a good measure of reaction efficiency?
If the reaction went perfectly, with every molecule of the starting compound converted
into a molecule of the desired product, would this measure give a perfect “score” for
the reaction (=100%)?
Does a molecule of the product weigh the same as a molecule of the starting material?
Copyright Kenneth M. Doxsee, University of Oregon
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University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
The end result of these discussions, of course, should be the recognition that we need to
consider both the amount of product obtained and its molecular weight relative to that of the
starting material. This leads to the concept of a “theoretical yield” – the maximum amount of
product that could be obtained given the amount of starting material used. As appropriate, this
could serve as the point of introduction of basic chemical concepts such as molecular weight
and the mole, but realistically it may be better to have students already acquainted with these
sorts of concepts, as otherwise this lesson may be slowed too much by the need to discuss all
these other issues.
1 molecule diiodide = 482.960 amu
1 molecule dichloride = 300.057 amu
… so 100 mg of diiodide could theoretically provide 100 mg * (300.057/482.960) = 62.1 mg
of the dichloride. Suddenly our actual yield of 50 mg looks a lot better than we thought, since
it is (50 mg/62.1 mg)*100% = 80.5% of the theoretically possible amount of product.
Part 2 on the cisplatin worksheet:
“corrected mass yield” = 50 mg / [100 mg * (MWproduct / MWstarting material)]
= mass yield * (MWstarting material / MWproduct)
= 50% * (482.960/300.057) = 80.5%
This, now, again represents an opportunity for student discussion about the merits of the
calculation and any issues that it may present. Perhaps the most obvious thing that has been
missed is consideration of the stoichiometry of the reaction and the amounts of all the
reactants used, not just the amount of the starting material forming the focus of our
transformation. In short, we need to consider two other key chemical concepts – we need to
ensure that we are looking at a balanced chemical equation, and we need to examine the
amounts of each reagent used to determine which is serving as the limiting reagent.
The full chemical reaction we are considering involves the treatment of the diiodide with
silver (I) sulfate and potassium chloride. Students can be presented with the opportunity to
write a balanced equation, showing all starting materials and products and the relative
amounts of each needed. Small group discussion is always a good possibility, and can help
weaker students get up to speed. The end result (part 3 on the worksheet):
(NH3)2PtI2 + Ag2SO4 + 2 KCl → (NH3)2PtCl2 + K2SO4 + 2 AgI
Now, to determine the limiting reagent, we need to convert the quantities used (in grams) to
the number of moles used. Setting up a simple table (as in the worksheet) is a nice, organized
way to do this.
Copyright Kenneth M. Doxsee, University of Oregon
Page 71 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
Compound
(NH3)2PtI2
Ag2SO4
KCl
(NH3)2PtCl2
molecular wt
482.960
311.794
74.551
300.057
milligrams
100
63
330
2010
millimoles
0.207
0.202
4.43
Examining the calculations, we see that we used a large excess of KCl (only two equivalents
were required by the reaction stoichiometry), but that we actually should have used a bit more
silver sulfate. The silver sulfate thus serves as the limiting reagent, and our theoretical yield of
cisplatin was actually only 0.202 millimoles, or 60.6 mg, and our actual yield then is (50
mg/60.6 mg)*100% = 82.5% of the theoretically possible amount of product (worksheet parts
4 and 5).
This is typically where we end with such discussions – the chemical yield of a reaction,
measured by comparing how much of the desired product was obtained to how much could
theoretically have been obtained given the reaction stoichiometry and the amounts of
reactants used. But does this number really capture the “quality” of the chemical reaction?
Does 82.5% of theoretical yield represent a reasonably efficient reaction, perhaps worthy of a
grade of „B‟?
Professor Barry Trost, a chemist at Stanford University, felt that reliance on „yield‟ as a
measure of reaction efficiency represented an inappropriate measure, suggesting that we are
better at carrying out chemical transformations than we in fact are. He developed a concept
called “atom economy,” looking at a chemical reaction from the perspective of how many
input atoms are incorporated in the desired product, vs. how many are discarded as waste.
Let‟s set up another table, this time considering this issue (worksheet part 6):
Compound
(NH3)2Pt I2
Ag2SO4
2 KCl
Totals
atoms used in
pdt
(NH3)2Pt
Cl2
(NH3)2PtCl2
wt used
229.151
70.906
300.057
atoms
discarded
I2
Ag2SO4
K2
I2 K2 Ag2SO4
wt discarded
253.809
311.794
78.197
643.800
Looking at the reaction this way, it doesn‟t look like such an efficient process any more, does
it? Less than one third of the total mass of the reactants ends up as our desired product; the
rest is converted to waste.
Atom Economy = 300.057/(300.057 + 643.800)*100% = 31.8%
How is this as a measure of the efficiency of our reaction? [Another good point for discussion,
questions, etc.] It iss getting better, but it ignores at least two really important things, both
arising because this calculation is based only on the balanced chemical equation, without
consideration of the actual experiment.
Copyright Kenneth M. Doxsee, University of Oregon
Page 72 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
First, it does not take into account the actual quantities of reagents we used. For
example, we used a large excess of KCl, presumably in an effort to increase the yield
of the reaction. In so doing, all the excess KCl becomes waste.
Second, it does not take into account the amount of product we actually obtained.
Let‟s inject this experimental reality into the calculation:
Experimental atom economy (worksheet part 7) = (theoretical yield/mass of all
reactants)*100% = [60.6 mg/(100 mg + 63 mg + 330 mg)]*100% = 12.3%.
And if we look at how much we actually obtained rather than the theoretical amount:
Actual atom economy (worksheet part 8) = (actual yield/mass of all reactants)*100% = [50
mg/(100 mg + 63 mg + 330 mg)]*100% = 10.1%.
What does this number – 10.1% - really mean? It means that in carrying out this reaction, we
converted 89.9% of all of our starting materials into waste! Put another way (worksheet part
9), this number means that for every gram of desired product (cisplatin) that we produce, we
produce (89.9%/10.1%)*1g = 8.9 g of waste. We paid for those starting materials! And we
will pay to dispose of the waste! So we pay twice for the inefficiency of our reaction, and get
that much less desired compound that could be sold.
This is beginning to look like a respectable measure of our reaction efficiency. I find it very
interesting that this is what a student would calculate if you neglected to tell them anything
about structure, stoichiometry, balancing equations, moles, etc.! However, we have still
neglected to consider one additional important factor – the reaction is not carried out as
written in the balanced equation – we also use a solvent, and when we isolate the product by
filtration, we use more solvent. (And if we were to recrystallize our product, still more
solvent.)
In many cases, solvent usage is the single largest source of waste in chemical production. Our
solvent usage, should we actually follow the written procedure exactly as written, is 10 mL of
water for the reaction, and 0.5 mL of ethanol and 1 mL of diethyl ether in the isolation step.
How much waste is generated from this solvent use (worksheet part 10)?
10 mL H2O * 1g/mL = 10 g H2O
0.5 mL CH3CH2OH * 0.794 g/mL = 0.397 g CH3CH2OH
1.0 mL (CH3CH2)2O * 0.715 g/mL = 0.715 g (CH3CH2)2O
Total solvent waste = 11.112 g
Let‟s now, as the last stage of our analysis, fold this information into our calculation, and
simply look at how much desired product is obtained from how much chemical input
(worksheet part 11).
Copyright Kenneth M. Doxsee, University of Oregon
Page 73 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
[50 mg/(100 mg + 63 mg + 330 mg + 11,112 mg)]*100% = (50/11,605)*100% = 0.43%
Les than half a percent of all of our chemical inputs ended up as a desired product!! Put
another way, known as the “environmental factor”, a measure of how much waste a reaction
produces relative to the amount of desired product (worksheet part 12):
E factor = waste/product = (11,605 mg – 50 mg)/50 mg = 231
We‟ve come a long way from our initial observation of an 82.5% yield, haven‟t we?
A final, related measure that may be catching on, courtesy of Green Chemistry Institute‟s
Pharmaceutical Sciences Roundtable:
Process Mass Intensity Metric (worksheet part 13) = quantity of raw materials input/quantity
of bulk active pharmaceutical ingredient (API) out, where „process‟ = all steps from
commonly available materials, raw materials = everything used, including water, and „bulk
API out‟ is the final salt form, dried to specifications. In our simple case, this is = (11,605)/50
= 232, and in general one might imagine it will be very similar to the E factor.
Before we move on, let‟s spend just a moment more thinking about this experiment and what
it teaches.
On the positive side, it gives experience with balancing equations, calculating mole ratios, and
classical inorganic coordination chemistry, as well as introducing the concept of
chemotherapy. It provides laboratory experience with running reactions at elevated
temperatures and isolation by filtration.
On the negative side, it presents a reaction that requires elevated temperatures (energy
consumption), and that generates large amounts of waste relative to the amount of desired
product (232 grams of waste per gram of product!) Finally, the product is an active
anticancer drug! It has dramatic biochemical activity at low dosage. Why are we having
beginning students risk potential exposure to a compound with proven biological
activity???
This discussion of the experiment represents some of the key concepts in green chemistry – in
short, we learn to ask questions at every stage of an experiment, synthesis, or process.
Why are we making this compound?
What are its properties – particularly in terms of health and environmental impact?
Can we accomplish the same goals by making a different and safer compound?
Can we make the compound from safer starting materials, or using safer solvents, or
using less energy input (either heating or cooling)?
Can we reduce the amount of waste that is generated, or eliminate it entirely?
Given time to think about the 12 principles and our discussions today, you would probably
conclude that we have illustrated principles 1, 2, 3, 7, 8, and 9.
Copyright Kenneth M. Doxsee, University of Oregon
Page 74 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Cisplatin worksheet
(NH3)2PtI2 + Ag2SO4 + KCl → (NH3)2PtCl2 + K2SO4 + AgI
compound
(NH3)2PtI2
Ag2SO4
KCl
(NH3)2PtCl2
1.
molecular weight
482.960
311.794
74.551
300.057
mass (milligrams)
100
63.0
330
50.0*
*
Experimental result
millimoles
0.207
0.202
4.43
“Mass yield”
= (mass of product/mass of starting material) * 100%
= (_______ mg/_______ mg) * 100 % = _______ %
2.
“Corrected mass yield”
= mass yield * (MW of starting material/MW of product)
= _______ % * (__________ g-mol-1/__________ g-mol-1) = _______ %
3.
Balanced equation
__ (NH3)2PtI2 + __ Ag2SO4 + __ KCl → __ (NH3)2PtCl2 + __ K2SO4 + __ AgI
4.
Theoretical yield
= (millimoles limiting reagent) * MW of product
= _______ millimoles * __________ mg-mmol-1 = _______ mg
5.
Actual yield
= (mass of product/theoretical yield) * 100 %
= (_______ mg/_______ mg) * 100 % = _______ %
6.
Atom economy
compound
(NH3)2Pt I2
Ag2SO4
2 KCl
Totals
atoms used in pdt
(NH3)2Pt
Cl2
(NH3)2PtCl2
formula wt used
229.151
70.906
300.057
Copyright Kenneth M. Doxsee, University of Oregon
atoms discarded
I2
Ag2SO4
K2
I2 K2 Ag2SO4
formula wt discarded
253.809
311.794
78.197
643.800
Page 75 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Atom economy
= [formula weight of product/(formula weight of product + formula weight discarded)]
* 100 %
= (__________/__________ + __________) * 100 % = _______ %
7.
Experimental atom economy
= (theoretical yield/mass of all reactants) * 100 %
= [_______ mg/(_______ mg + _______ mg + _______ mg)] * 100 % = _____ %
8.
Actual atom economy
= (actual yield/theoretical yield) * experimental atom economy
= (actual yield/mass of all reactants) * 100 %
= [_______mg/(_______ mg + _______ mg + _______ mg)] * 100 % = _____ %
9.
Waste generated for every gram of desired product
= [(100 % - actual atom economy)/actual atom economy] * 1 g
= [(100 % - _______ %)/_______ %] * 1 g = _______ g
10.
Solvent usage in this reaction:
Water: 10 mL * density = 10 mL * 1.00 g-mL-1 = _______ g = _______ mg
Ethanol:
0.5 mL * density = 0.5 mL * 0.794 g-mL-1 = _______ g = _______ mg
Ether: 1.0 mL * density = 1.0 mL * 0.715 g-mL-1 = _______ g = _______ mg
Total solvent usage = _______ mg + _______ mg + _______ mg = _______ mg
11.
Yield of product based on all chemical inputs
= [mass of product/(mass of all reagents + mass of all solvents)] * 100 %
= [_____ mg/(_______ mg reagents + _______ mg solvents)] * 100 % = _____ %
Copyright Kenneth M. Doxsee, University of Oregon
Page 76 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
12.
2010
E-factor (Environmental factor)
= mass of waste/mass of product
= (mass of all reagents and solvents – mass of product)/mass of product
= (_______ mg - _______ mg)/_______ mg = _______
13.
Process mass intensity
= mass of raw materials input/mass of product output
= _______ mg/_______ mg = _______
Copyright Kenneth M. Doxsee, University of Oregon
Page 77 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
B. Aniline oxidation
An example of an organic reaction – note that organic chemists are notorious for focusing
only on the reagents and products of interest, ignoring other inputs and byproducts and not
always even balancing equations.
Oxidation of aniline is a classical method for the production of hydroquinone:
NH 2 + MnO2
H2SO4
Fe
O
O
HO
OH
The balanced equations:
2 PhNH2 + 4 MnO2 + 5 H2SO4  2 quinone + (NH4)2SO4 + 4 MnSO4 + 4 H2O
2 quinone + 2 Fe + 2 H2O  2 hydroquinone + 2 FeO
And the overall equation:
C6H5NH2 + 2 MnO2 + 2.5 H2SO4 + Fe 
C6H4(OH)2 + FeO + 0.5 (NH4)2SO4 + 2 MnSO4 + H2O
I don‟t have full experimental details for this reaction, so let‟s just use it for practice in the
calculation of atom economy.
Calculation of the atom economy for this one may seem more complicated, but as before, all
we need to do is balance the equation, then look at how many input atoms are used and how
many are discarded. (Refer to the worksheet on the following pages.)
Imagine how this one would look upon full analysis, given that the best it could be is 19.4%,
and certainly there are large amounts of solvents involved in the process.
Experimental details are required to calculate the other measures of reaction efficiency –
Experimental Atom Economy, Actual atom economy, Waste generated for every gram of
desired product, Solvent and other chemical usage, Yield of product based on all chemical
inputs, E-factor, and Process Mass Intensity. You may wish to have students try to find such
experimental information through the Internet. I have provided on the following page two
abbreviated reports of the reaction that include most (but not all) of the information needed.
Having students discuss and identify what essential information is not provided, and having
them make educated guesses, would be valuable exercises. (The first procedure does not
report how much steam is needed, and the second procedure omits the type and amount of
base used to adjust the pH of the reaction mixture to 5-6. Neither procedure says how much
iron was needed, and the second procedure does not provide the final yield of hydroquinone.)
Copyright Kenneth M. Doxsee, University of Oregon
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University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Abstract (Quinones. Gibbs, C. F. (The B. F. Goodrich Co.). (1944), US Patent 2,343,768;
Chemical Abstracts 38:22768).
A mixture of MnO2 (11.6 g) and 50% H2SO4 (36 g) is placed in a suitable vessel and
maintained at 50-60° over a water bath. A mixture of MnO2 (58.8 g), H2SO4 (50 g) and H2O
(40 g) is added over a period of 1 hour concurrently with a mixture of aniline (24 g), 50%
H2SO4 (140 g) and H2O (165 g). The pressure is maintained at 40 mm and a constant stream
of water vapor is passed into the bottom of the vessel. The quinone collects in a water-cooled
receiver immediately after the start of the addition of aniline sulfate; the yield is 73% or
higher. This process makes it possible to carry out the oxidation at a higher temperature
without the formation of large amounts of by-products. It is also speedier.
Abstract (Quinone production from aniline and its reduction to hydroquinone. (Instytut
Przemyslu Organicznego). Brit. Patent 1,151,673 (1969), 3 pp.; Chemical Abstracts
71:61012).
In the preparation of quinone (I) by the oxidation of Ph-NH2 (II), unreacted MnO2 is reduced
to water-soluble manganous salts as an alternative to steam distillation to remove the I
formed. Thus, 400 L water, 115 kg concentrated H2SO4 and 15 kg MnO2 (85% pure) is
stirred, cooled to 5° and 30 kg II added over 3 hours at 5-8°. Simultaneously, an additional 60
kg MnO2 is added in 15 kg-portions at 30-minute intervals, then the mixture stirred 7 hours at
5-10°; 3.13 kg H2O2 (calculated as 100% H2O2) is added over 30 minutes and the mixture
neutralized to pH 5-6, then reduced with Fe filings at 65-75° until one half the I present was
reduced to hydroquinone (III). Cooling to 10° and filtration yields a precipitate containing 30
kg quinhydrone which is then further reduced to III. NaHSO3 may be used in place of H2O2.
Copyright Kenneth M. Doxsee, University of Oregon
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University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Synthesis of Hydroquinone
Worksheet
NH 2 + MnO2
H2SO4
Fe
O
O
HO
OH
Balanced equations
C6H5NH2 + 2 MnO2 + 2.5 H2SO4  C6H4O2 + 0.5 (NH4)2SO4 + 2 MnSO4 + 2 H2O
C6H4O2 + Fe + H2O  C6H4(OH)2 + FeO
Overall equation
C6H5NH2 + 2 MnO2 + 2.5 H2SO4 + Fe 
C6H4(OH)2 + FeO + 0.5 (NH4)2SO4 + 2 MnSO4 + H2O
compound
C6H5NH2
2 MnO2
2.5 H2SO4
Fe
Totals
atoms used in pdt
C6H4
O2
H2
C6H6O2
wt used
76.098
31.999
2.016
110.112
atoms discarded
NH3
Mn2O2
1.5 H2 + 2.5 SO4
Fe
NH3Mn2O2Fe + 2.5 SO4 + 1.5 H2
wt discarded
17.030
141.875
243.169
55.847
457.921
Atom economy
= (formula weight of product/formula weight of all reactants) * 100 %
= (__________/__________) * 100 % = _______ %
Experimental details are required to calculate other measures of reaction efficiency, as in the
cis-platin section. Consider having students try to find such information through the Internet.
On the previous page are short reports of the reaction that include most (but not all) of the
information needed. Ask your students to discuss and identify what essential information is
not provided, and having them make educated guesses. (The first procedure does not report
how much steam is needed, and the second procedure omits the type and amount of base used
to adjust the pH of the reaction mixture. Neither procedure says how much iron was needed,
and the second procedure does not provide the final yield of hydroquinone.)
Copyright Kenneth M. Doxsee, University of Oregon
Page 80 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Solventless Aldol Condensation
Chemical Concepts
Carbonyl chemistry; the aldol reaction; phase behavior; melting points of solids and mixtures;
recrystallization.
Green Concepts
Solventless reactions; atom economy. (Consider Green Principles 1, 3, 4, 5, 7, 8, and 9.)
Introduction
The aldol condensation represents a powerful general method for the construction of carboncarbon bonds, one of the central themes of synthetic organic chemistry. In the base-catalyzed
aldol condensation reaction, deprotonation alpha (adjacent) to a carbonyl group affords a
resonance-stabilized anion called an enolate, which then carries out nucleophilic attack at the
carbonyl group of another molecule of the reactant. (Analogous acid-catalyzed reactions are
also well-known.) The product, a beta-hydroxy carbonyl compound, often undergoes facile
elimination of water (dehydration), affording an alpha, beta-unsaturated carbonyl compound
as the final product.
H
H
H
H
R
- H+
-
H
R
O
H
R
R
R
O
H
R
O-
O
O-
R
+ H+
R
O
R
H
- H2 O
H
R
OH O
O
enolate
Mechanism of the base-catalyzed aldol condensation
Aldol condensation reactions between two different carbonyl compounds can lead to complex
product mixtures, due to the possibility of enolate formation from either reactant and to the
possibility of competing “homo” coupling rather than the desired “cross” coupling.
R'
R
H
H
H
R
O
R
H
+
R'
OH O
H
crossed aldol products
H
homo aldo products
OH O
H
+ R'
O
R
R'
H
R
OH O
+
R'
OH O
“Crossed” aldol condensation can afford complex mixtures
If, however, only one of the carbonyl compounds has alpha hydrogens available for
deprotonation and enolate formation, the “crossed” aldol reaction can provide synthetically
Copyright Kenneth M. Doxsee, University of Oregon
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University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
useful yields of products. Thus, for example, benzaldehyde cannot be converted to an enolate,
yet reacts readily with enolates of other carbonyl compounds, including acetone.
O
OH O
O
H
+
H 3C
O
CH3
CH3
CH3
A successful “crossed” aldol condensation
Homo coupling of acetone (or other ketones) is generally not a problem in such reactions, as
the aldol condensation of ketones is generally not a very efficient reaction. (More specifically,
each step of the aldol condensation is reversible under the reaction conditions – at least until
the dehydration step – and thus equilibrium is established. With aldehydes, equilibrium favors
the aldol product, but with ketones, primarily for steric reasons, very little aldol condensation
product is present at equilibrium.)
In this experiment, you will explore the aldol condensation reaction of 3,4dimethoxybenzaldehyde and 1-indanone.
O
H3CO
O
H
+
NaOH
H3CO
3,4-dimethoxybenzaldehyde
O
OCH3
1-indanone
OCH3
In contrast to typical experimental procedures for aldol condensation reactions, this reaction
will be carried out without solvent. Ongoing research is revealing a number of reactions that
proceed nicely in the absence of solvent, representing the best possible solution to choice of a
benign solvent. Although these reactions are frequently referred to as “solid-state” reactions,
it has been noted [1] that in many cases, mixture of the solid reactants results in melting, so
that the reactions actually occur in the liquid, albeit solvent-free state. This melting
phenomenon is interesting and actually represents one of the key points of this experiment.
You have learned that impurities lead to lower melting points. Here, you will experience this
in a vivid way – as you mix the two solid reactants, they will melt. In addition to providing a
memorable demonstration of the impact of impurities on melting points and illustrating the
possibility of carrying out organic reactions in the absence of solvents, this experiment
highlights another key green concept – the design of efficient, atom-economical reactions.
The aldol condensation, if effected without dehydration, has an atom economy of 100% and
requires only a catalytic amount of acid or base, and even with dehydration, the atom
economy remains quite high.
1.
G. Rothenberg, A. P. Downie, C. L. Raston, and J. L. Scott, “Understanding Solid/Solid Organic
Reactions," J. Am. Chem. Soc. 2001, 123, 8701-8708.
Copyright Kenneth M. Doxsee, University of Oregon
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University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
Laboratory
SAFETY PRECAUTIONS: Use care to avoid contact with solid sodium
hydroxide or the reaction mixture.
Reaction
1. Place 0.25 g of 3,4-dimethoxybenzaldehyde and 0.20 g of 1-indanone in a test tube.
Using a metal spatula, scrape and crush the two solids together until they become a
brown oil. Use care to avoid breaking the test tube.
2. Add 0.05 g of finely ground (using a mortar and pestle) solid NaOH to the reaction
mixture and continue scraping until the mixture becomes solid.
Workup and purification
4. Allow the mixture to stand for 15 minutes, then add about 2 mL of 10% aqueous HCl
solution. Scrape well in order to dislodge the product from the walls of the test tube.
Check the pH of the solution to make sure it is acidic.
5. Isolate the solid product by vacuum filtration, continuing to pull air through the solid
to facilitate drying. Determine the mass of the crude product.
6. Recrystallize the product from 90% ethanol/10% water, using the hot solvent first to
rinse any remaining product from the test tube. You should not require more than 20
mL of solvent to effect this recrystallization.
Characterization
7. Determine the mass and melting point of the recrystallized product. (A typical melting
point range is 178 – 181 °C.)
Questions
1. Describe the physical properties (color and state) of your crude product. Report the
mass and percent of theoretical yield of the crude product.
2. Report the color and melting point range of your recrystallized product. Report the
mass and percent of theoretical yield of the recrystallized product.
3. Calculate the atom economy for the reaction.
4. Perform an economic analysis for the preparation of your product.
Research Questions
1. What is a eutectic mixture? A eutectic liquid? Are there practical applications of
eutectic mixtures? If so, give an example, including an explanation of why the
properties of a eutectic mixture are helpful to the particular application.
Copyright Kenneth M. Doxsee, University of Oregon
Page 83 of 84
University of Oregon – Thailand Distance Learning Program – Green Chemistry
2010
2. The introduction to this experiment comments on the difference between a true “solidstate” reaction and a reaction involving two solids that in fact proceeds through
formation of an intermediate liquid state. Find other reported examples of “solid-state
reactions,” “solventless reactions,” and related concepts and attempt to analyze them
with regards to whether they have been described correctly. (For example, a “solidstate reaction” between two low-melting solids might well be expected to proceed
through formation of a liquid, as did our experiment.)
References
This experiment was adapted from the primary literature and has been presented in Green
Organic Chemistry: Strategies, Tools, and Laboratory Experiments, Kenneth M. Doxsee and
James E. Hutchison, Brooks-Cole, 2003. Any number of solventless aldol reactions are
possible, and it may be attractive to allow students some latitude in choosing their reactants,
taking care to avoid unexpectedly hazardous reagents or products. The reactants reported here
were chosen deliberately to highlight the melting point depression phenomenon; other pairs of
reagents may or may not visibly melt upon mixing.
Copyright Kenneth M. Doxsee, University of Oregon
Page 84 of 84