Kinetic vs Thermodynamic Control in Competing Reactions
In most of our reactions, we are simply concerned about the products of the
reaction. However, chemists study all aspects of reactions to include reaction rates
and energy changes.
Kinetics is the study of reaction rates. A reaction rate tells us how fast a
given reaction is occurring and includes time. For example, a reaction rate might
be expressed by how much of a specific reactant disappears in a given time. If the
concentration of the reactant R is in moles per liter, its concentration is [R], where
the square brackets [ ] mean concentration in moles per liter. The reaction rate can
be written in terms of the rate of disappearance of R, which is expressed as –
d[R]/dt and is read as the change in concentration of R with respect to the time t.
The minus sign means that the concentration of R is decreasing with time and
reflects the physical reality that R is being used up during the reaction. The rate
law of a reaction is an experimentally determined equation that has the form of rate
= k[R1]x[R2]y, where R1 is reactant 1 and R2 is reactant 2, k is the specific rate
constant and the exponents x and y are experimentally determined variables.
Thermodynamics is the study of energy changes that accompany a chemical
reaction or a physical change such as the melting of a solid. For example, the
enthalpy change (H) of a reaction tells us the change in heat content. The sign of
H tells us whether heat is added (+) to the reaction or evolved (-) by the reaction
(i.e., the reaction is endothermic or exothermic). In most cases, a reaction is
conducted in such a way that the more stable of two potential products is obtained
as the actual product. For example, the major product obtained when we dehydrate
2-butanol is trans-2-butene. trans-2-Butene is the most stable of the three potential
alkene products. Stability and internal (potential) energy are inversely proportional.
Thus, the lower the potential energy of a substance, the more stable it is. The Gibbs
free energy G is related to H by the well-known Gibbs equation G = H –
TS, where T is the absolute temperature and S is the entropy change of the
reaction. In many cases, S is very small, and the term TS becomes negligible
when it is compared to H. Therefore, in many cases G and H are very nearly
the same and either one of them can be used in those instances to describe energy
changes in a reaction. Therefore, it is largely the author’s choice to use H or G
to show changes in internal energy. G is more precise, but either term can
generally be used to make the same points. Going back to trans-2-butene. It is
more stable than cis-2-butene or 1-butene, the two minor products of the
dehydration of 2-butanol.
Lab 4
1
Let us consider a hypothetical reaction in which two reactants A and B
produce two products C and D. These products (C and D) form by two
significantly different reaction pathways and differ in their internal energy. Each
pathway proceeds through a single transition state, which is shown as a maximum
on a reaction profile or coordinate. The reaction is special, because the faster
forming product C is less stable than D and the slower forming product D is more
stable than C. This special case is shown in the reaction profile of Figure 1. Each
product is formed from A and B. Product C forms via the blue pathway and
product D via the red pathway.
C
Energy A + B
(G or H)
D
time
Figure 1. A reaction profile of a reaction that can go under either thermodynamic
or kinetic control.
In CHM 202, we learn that conjugated dienes can undergo either 1,4- or 1,2addition, depending on the temperature. The more stable 1,4-adduct forms at a
higher temperature than does the 1,2-adduct. Likewise, if two enolates can be
produced from a ketone, the more stable enolate forms at a higher temperature, the
less stable enolate forms at a lower temperature.
In this experiment, we are actually conducting two different reactions. An
aldehyde and a ketone compete with each other to form a product with a limited
Lab 4
2
amount of reagent. The aldehyde is 2-furfural, which is also called furfuraldehyde.
2-Furfural is a derivative of furan, a cyclic ether. The ketone is cyclohexanone, and
the reagent is semicarbazide in the form of its hydrochloride salt. Semicarbazide is
a derivative of hydrazine, H2NNH2. The structures below show several derivatives
of hydrazine that undergo condensation reactions with aldehydes and ketones.
Cyclohexanone and 2-furfural react similarly with all of these reagents.
O2N
H
H
H
H2NN H
H2NN
H2NN
hydrazine
phenylhydrazine
H O
NO2
2,4-dinitrophenylhydrazine
H2NN C NH2
semicarbazide
The Reaction
The reaction between an aldehyde or ketone and hydrazine or its derivatives
involves an addition followed by a dehydration so that the overall reaction is a
condensation reaction (i.e., one in which water is lost). Figure 1 shows the
generalized two-step reaction.

O
C


+

H  H
N N
H
R
addition
positive to
negative
O H
C
N
H N H
R
dehydration
minus H2O
analogous
to loss of
water from
a gem-diol
C
N
N H
R
Figure 1. General Reaction.
Mechanism
Figure 2 shows the mechanism of the reaction. In Step 1, the
Lab 4
3
O
C
H
H
N N
R
H
Step 1
H
O+ H
C
N
H N H
R
OC
N H
H N H
R
Step 4
H
H O
Step 2
fast
transfer
of H+
C
N
OH
C
N
H N H
R
H
H O+
H
Step 3
N H
R
Figure 2. Mechanism of Reaction.
hydrazine derivative adds to the carbonyl group. The non-bonded pair of electrons
on nitrogen forms a new sigma bond with the carbonyl carbon, and the  bond
breaks to leave a negative charge on oxygen. In the product of the first step,
nitrogen has four bonds and a positive formal charge. In Step 2, a proton rapidly
exchanges in the aqueous medium to form an aminoalcohol. In Step 3, the OH
group is protonated (i.e., the oxygen of the OH group abstracts a proton from H3O+
to make a good leaving group. In Step 4, the double bond between carbon and
nitrogen forms as the aqueous solvent abstracts a proton from the nitrogen and the
good leaving group (H2O) departs. Hydrazine and all of its derivatives follow this
general mechanism. The R group determines which compound is the reactant. For
example, if R = phenyl, the reactant is phenylhydrazine.
Simple Product Formation
Once we know the reaction involves two steps and we know the mechanism
of the reaction, we can look at the reactants and see that the double bond from
carbon to oxygen (i.e., the carbonyl double bond) becomes a double bond between
the carbonyl carbon and a hydrazine (or derivative) nitrogen atom.
Lab 4
4
C O
C NNHR
H2NNHR
+ H2O
Figure 3. Condensation reaction between a carbonyl group and a
hydrazine or semicarbazide.
The equation shown in Figure 3 is general and applies to the reaction of any
aldehyde or ketone with a hydrazine derivative, including semicarbazide.
Figure 4 shows the reaction of acetaldehyde with hydrazine,
phenylhydrazine, 2,4-dinitrophenylhydrazine and semicarbazide. All of the
reactions are condensation reactions of the type shown in Figure 3.
NNH2
CH3CH hydrazone
H2NNH2
NNHPh
CH3CH phenylhydrazone
H2NNH
O
CH3CH +
O2N
O2N
H2NNH
NO2
NNH
CH3CH
NO2
2,4-dinitrophenylhydrazone
O
H2NNHCNH2
O
NNHCNH2
CH3CH semicarbazone
Figure 4. Condensation reactions of acetaldehyde.
In each of the reactions, water is lost (condensation reaction) and a double
bond forms between the carbonyl carbon of acetaldehyde and the hydrazine
nitrogen atom. A reaction of an aldehyde or ketone with hydrazine produces a
hydrazone, with phenylhydrazine a phenylhydrazone, with 2,4dinitrophenylhydrazine a 2,4-dinitrophenylhydrazone, and with semicarbazide a
semicarbazone.
Lab 4
5
Competing Reactions
Equation 1 shows the reaction between 2-furfural and semicarbazide, and
Equation 2 shows the reaction between cyclohexanone and semicarbazide.
O
C O
H
O
H2NNHCNH2
O
O
C NNHCNH2
H
semicarbazone of
2-furfural
a yellow solid
2-furfural
Equation 1
O
O
H2NNHCNH2
O
NNHCNH2
semicarbazone of
cyclohexanone
a white solid
cyclohexanone
Equation 2
We can distinguish the products by their colors. The semicarbazone of 2furfural is yellow, and the semicarbazone of cyclohexanone is white. One
equivalent each of cyclohexanone and 2-furfural will be allowed to react with one
equivalent of semicarbazide at three different temperatures. In other words, two
different carbonyl groups will compete with each other to react with
semicarbazide. The product of thermodynamic control will be obtained at the
highest temperature, and the product of kinetic control will be obtained at the
lowest temperature. The color of the product reveals which semicarbazone forms.
The results can be explained by an energy profile curve, which shows two
reaction pathways—one for the product of thermodynamic control and one for the
product of kinetic control. The kinetic product has a lower energy of activation
(G‡) than does the product of thermodynamic control. The final energy of the
kinetic product is higher than the final energy of the thermodynamic product (i.e.,
the higher the energy, the less stable the product). So the kinetic product forms
Lab 4
6
faster (it has a lower energy barrier to overcome) but is less stable (it is at higher
energy) than the thermodynamic product. These results are only possible because
both reactions are reversible. We can cause the kinetic product to form by keeping
the temperature low enough so that the energy of activation for the thermodynamic
product is never exceeded.
Procedure
Reminder: Record all data directly in your notebook and turn in your data
sheet with your lab report!
Work in pairs as Student 1 and Student 2.
Student 1
1. Weigh exactly 3.0 g of semicarbazide hydrochloride on a pre-creased and tared
waxed paper and transfer the solid to a 150-mL Erlenmeyer flask, labeled #1.
2. Transfer exactly 6.0 g of K2HPO4, weighed on the same paper, to flask #1.
3. Add 75-mL distilled water to flask #1 and swirl the flask to dissolve the two
solids. Flask #1 now contains solution #1.
The K2HPO4 is added to buffer the solution (i.e., to control the pH).
Student 2
1. Measure 3.0-mL cyclohexanone in a 10-mL graduated cylinder and pour the
liquid into a 50-mL Erlenmeyer flask, labeled #2.
2. Without cleaning the graduated cylinder, measure 2.5-mL 2-furfural
(furfuraldehyde) and pour it into flask #2.
3. Without cleaning the graduated cylinder, transfer a total of 15-mL 95% ethanol
in two increments to flask #2. Flask #2 now contains solution #2.
Solution #1 contains the reagent semicarbazide as its hydrochloride and
solution #2 contains the ketone cyclohexanone and the aldehyde 2-furfural. We
shall allow reactions to occur at three different temperatures. The product of
kinetic control will predominate at the lowest temperature, and the product of
thermodynamic control will predominate at the highest temperature. The white
Lab 4
7
semicarbazone of cyclohexanone melts at 166 oC, and the pale yellow
semicarbazone of 2-furfural melts at 202 oC.
Low-temperature Experiment (Ice-Water Bath)
1. Pour 25-mL Solution #1 into a 50-mL Erlenmeyer flask and place the flask in an
ice-water bath to cool.
2. Pour 5-mL Solution #2 into a medium-size test tube and cool the test tube in the
ice-water bath.
3. After the two solutions have cooled to the ice-water temperature (0-2 oC), pour
the sample in the test tube into the sample of Solution #1.
4. After several minutes, a precipitate will form. Collect the precipitate on a
Büchner funnel.
5. Remove the filter paper containing the solid from the funnel and place it on a
paper towel. Label this solid “low temp.”
Room-temperature Experiment
1. Pour 25-mL Solution #1 and 5-mL Solution #2 into a 50-mL Erlenmeyer flask.
Swirl the flask to ensure mixing of the reactants, place the flask on your lab bench
and let it stand for approximately five minutes.
2. As soon as crystals form in the flask, cool the flask in the ice-water bath.
3. Collect these crystals as before, place them next to the low-temp crystals and
label them “room temp.”
High-temperature Experiment
1. Prepare a hot-water bath: fill a 400- or 500-mL beaker about three-fourths full of
water, place the beaker on a hot plate and heat the beaker at medium heat until the
temperature of the water is above 80 oC. Do not allow the water to boil but keep
the temperature above 80 oC.
2. Pour 25-mL Solution #1 into a 50-mL Erlenmeyer flask and place the flask in
the hot water bath.
Lab 4
8
3. Pour 5-mL Solution #2 into a medium test tube and place the test tube in the hot
water bath.
4. After 5 min., pour the 5-mL sample of Solution #2 into the 50-mL Erlenmeyer
flask.
5. Swirl the Erlenmeyer flask without removing it from the hot water.
6. Keep the Erlenmeyer flask in the hot water for about 15 min.
7. Remove the Erlenmeyer flask from the hot water and allow it to cool to room
temperature. Then, cool the Erlenmeyer flask in the ice bath and collect the crystals
as before, labeling this crop as “high temp.”
You now have three filter papers containing the solids obtained at the three
different temperatures. Carefully compare the colors of the three solids and
record them in your lab notebook. They vary in color from white (cyclohexanone
semicarbazone) to pale yellow (2-furfural semicarbazone). Use this information to
determine which one forms predominately at the low temperature, and which
forms predominately at the high temperature. You will not have calculations of
theoretical yield for this experiment. Instead, you will indicate which
semicarbazone is the product of kinetic control, and which is the product of
thermodynamic control.
Clean up: Scrape the solids from the filter papers into the non-halogenated waste
jar in the hood. Discard the filter papers in a beaker in the hood; do not throw the
filter papers into the non-halogenated waste jars. Flush all aqueous filtrates down
the drain with running water. Clean all glassware and replace it in its proper
storage location. Check the balance area. If chemicals and weighing papers are still
there, return them to their proper locations.
Lab 4
9
Kinetic vs Thermodynamic Control
Student last name___________________, First name___________________
Write equations for the following reactions.
1. Cyclohexanone and semicarbazide hydrochloride.
2. Acetone and 2,4-dinitrophenylhydrazine.
3. Propanal and phenylhydrazine.
4. Draw an energy-profile diagram with labels for the competitive reactions of
cyclohexanone and 2-furfural for semicarbazide.
5. Use your diagram from problem 4 and explain why you can cool the product of
thermodynamic control in an ice-water bath without altering the overall result.
Lab 4
10
6. A dehydration reaction and a condensation reaction both yield water as a
byproduct. Explain the essential difference between a dehydration reaction and a
condensation reaction.
7. What is the relationship between the color and the structure of the
semicarbazone of 2-furfural?
8. Draw the structures of pentanal and 2-hexanone and label them.
The bromination of butadiene produces the structures shown below.
Br
Br
Br
A
Br
B
9. Which structure, A or B, is the product of thermodynamic control? ____
10. Which compound A or B will give off more heat when burned in oxygen? ___
Lab 4
11