E x p e r r i i m e n t t : : K i i n e t t i i c v s s .
.
T h e r r m o d y n a m i i c C o n t t r r o l l o f f O r r g a n i i c R e a c t t i i o n s
During your study of reactions this year you have examined many mechanisms. You have used these mechanisms to predict which one of two possible organic products would be formed exclusively or as the major product. You have also used them to predict which product would form faster. Usually, the product formed in greater quantity is also the product formed faster.
This is because, in the rate determining step, one of the two products is more stable than the other. The transition state for this step resembles the products, so the more stable product has a more stable transition state leading to it, which in turn lowers the energy of activation. The now familiar set of potential energy curves below shows this clearly.
Potential
Energy less stable TS with higher E a
Progress of Reaction less stable product more stable product
The more stable product is called the thermodynamically controlled product. The product formed by way of the lower energy of activation pathway forms faster and is referred to as the kinetically controlled product. In the case discussed above, the same case you have seen over and over this year, the kinetically controlled product (the one formed faster) and the thermodynamically controlled product (the more stable one) are the same. The more stable product is formed faster.
But there are exceptions to this common scenario. One of them is the formation of 1, 2- and
1, 4-addition products from conjugated dienes. You have probably studied this reaction already, and it would be a good idea to review it before doing this experiment. Another is the formation of semicarbazones, reactions that you will be doing in this experiment. In these exceptions, the product that forms faster (the kinetically controlled product) is the less stable product, while the more stable product (the thermodynamically controlled product) has a higher energy of activation leading to it and is formed more slowly. In the reactions to be studied in this experiment, you will allow an aldehyde (2-furaldehyde) and a ketone (cyclohexanone) to react

with semicarbazide hydrochloride to form semicarbazone products called 2-furaldehyde semicarbazone (FS) and cyclohexanone semicarbazone (CS) respectively.
O O
H
+
C
O + NH
2
NH C NH
2
NNH C NH
2
O O
H H
2-furaldehyde
(F) semicarbazide
(S)
2-furaldehyde semicarbazone
(FS)
+ H
2
O
O
O
NH
2
NH C NH
2
H
+
O
NNH C NH
2
+ H
2
O cyclohexanone
C semicarbazide
(S)
One of these reactions is kinetically controlled. The product of this reaction is the one that forms faster. The other reaction is thermodynamically controlled. The product of this reaction is the more stable product. For this pair of reactions, the thermodynamically controlled product is not the same as the kinetically controlled product. One of the products, either FS or CS, will be the kinetically controlled product. The other product will be the thermodynamically controlled product. Let’s look at a generalized reaction in which A and B react with semicarbazide to form
S and BS.
Potential
Energy B + S
A + S
Progress of Reaction
A + S → AS
B + S → BS
AS (kinetic control)
BS (thermodynamic control) cyclohexanone semicarbazone
(CS)

As these curves are drawn, AS is the product of kinetic control. It has the lower energy of activation for formation, but it is less stable. It forms faster. BS is the product of thermodynamic control. It is the more stable product, but it is harder to form because the energy of activation for its formation is higher.
At all temperatures, AS, the kinetically controlled product, forms faster. BS, the thermodynamically controlled product, forms more slowly. However, a careful look at the AS energy curve shows that the activation energy for the reverse reaction of AS forming A and S is not much larger than the activation energy for the forward reaction, the formation of AS. At many temperatures, the A, S, AS system is a very reversible reaction. On the other hand, looking at the BS energy curve shows that the reverse reaction has a very high energy of activation, one that markedly slows down the reverse reaction of BS to form B and S. Thus the BS, once it forms, reverses itself very little. Any equilibrium between BS, B, and S lies far to the BS side.
In summary, at all temperatures, AS more easily reverses back to A and S than BS does. While
AS forms faster, it is less likely to “stay formed.” BS, on the other hand, forms more slowly but is more likely to “stay formed.”
In this laboratory period, you will be doing several experiments. In parts 1 and 2, you will be preparing the pure semicarbazones of cyclohexanone and 2-furaldehyde. In parts 3a, and b, equimolar amounts of cyclohexanone and 2-furaldehyde will be “competing” with each other to react with a limited amount of semicarbazide. In other words, there is not enough semicarbazide to react with all of the cyclohexanone and 2-furaldehyde present. Which semicarbazone forms in greater quantity at different temperatures will tell us which one is the kinetically controlled product and which one is the thermodynamically controlled product. In 3a, the competitive reaction occurs at 0 o
C. Here, both reactions will be slow, but the kinetically controlled product will form faster and stay formed more, because the low temperature will slow down the reverse reaction. Because of the low temperature, the higher activation energy needed to form the thermodynamically controlled product will greatly reduce the formation of this product.
The competitive reaction in part 3b occurs at a temperature between 80-85 o
C. Here, the temperature is high enough to form the thermodynamically controlled product while both the forward and reverse reactions of the kinetically controlled product occur faster too.
Part 4 consists of two parts. In 4a, cyclohexanone semicarbazone is mixed with 2-furaldehyde and the mixture is allowed to sit together at 80-85 o
C for a few minutes. If cyclohexanone semicarbazone is the kinetically controlled product, you would expect it to reverse itself, forming cyclohexanone and free semicarbazide. The free semicarbazone could then react with 2furaldehyde to form the less reversible thermodynamically controlled product, 2-furaldehyde semicarbazone. On the other hand, if cyclohexanone semicarbazone is the thermodynamically controlled product, you would expect it not to reverse itself, forming no semicarbazide to react with the 2-furaldehyde. You would be left with the unreacted cyclohexanone semicarbazone.
From the melting point of the product of part 4a, you will be able to determine whether cyclohexanone semicarbazone is the kinetically or thermodynamically controlled product.
Similarly, by repeating this procedure in part 4b with 2-furaldehyde semicarbazone and cyclohexanone, you will get supporting evidence for your conclusions. From parts 3a, b, and c alone, you should be able to determine which semicarbazone is kinetically controlled and which

one is thermodynamically controlled. Similarly, your results from 4a and b alone should confirm your conclusions from part 3. Together, your results from parts 3 and 4 should enable you to conclude with confidence which semicarbazone is kinetically controlled and which is thermodynamically controlled.
In all parts of 3 and 4, you will determine the composition of the products obtained from the melting points of the product mixtures, so you will need to know something about how mixtures behave when they melt.
Theoretically, a pure substance melts at a constant temperature. The melting point of pure cyclohexanone semicarbazone is 166 o
C. This means that if heat is applied to a pure sample, it will begin melting at 166ºC and the temperature will remain at 166ºC until the last bit of solid has been melted. For pure 2-furaldehyde semicarbazone, which has a melting point of 202 o
C, the same is true. [

Note: In practice, when taking a melting point of a pure substance, the melting point technique we use makes it nearly impossible for the start and finish of the melting process to be at the same temperature. If the sample is heated slowly at the melting point, approximately 2ºC or less per minute, then the melting point range of the pure sample will be
1ºC or less. This 1ºC melting point range is indicative of a pure substance.]
However, when solid X is contaminated with an impurity that is soluble in molten X, the melting point of the impure sample is lowered and the melting point range, the range of temperatures between the appearance of the first drop of liquid and the disappearance of the last bit of solid, is broadened. For example, when a sample of pure 2-furaldehyde semicarbazone is contaminated with a little bit of cyclohexanone semicarbazone, the melting point will drop below 202ºC and the melting point range will broaden. Depending on the amount of contaminant present, the melting point range might be 192-196ºC. Similarly, when a sample of cyclohexanone semicarbazone is contaminated with some 2-furaldehyde semicarbazone, the melting point will drop below 166ºC, even though the “contaminant” is the higher melting2-furaldehyde semicarbazone.
Thus, a sample of cyclohexanone semicarbazone (166ºC) with a small amount of 2-furaldehyde semicarbazone (202ºC) may have a melting point range of 157-163ºC. The more impurity that is present, the more the melting point is lowered and the broader the melting point range becomes.
For mixtures containing two components, melting point diagrams like the one shown below can be drawn. Consider two compounds, X (mp 125ºC) and Y (mp 175ºC). If a series of mixtures of
X and Y of known composition are made, and the melting points of the mixtures are determined, a plot of melting temperature vs. composition might look like this.

175ºC
Melting point of Y
125ºC
Melting point of X
Temperature
Eutectic Point
100% X
(0% Y)
Percentage of X (by mass) 0% X
(100% Y)
As you can see from this melting point diagram, as more and more Y is added to X and the percentage of X decreases, the melting point of the mixture decreases. Similarly, as more and more X is added to Y, and the percentage of Y decreases, the melting point of the mixture decreases. The point where the two melting point lines meet is called the eutectic point.
(Eutectic means “easy melting” in Greek). The mixture of X and Y that has the composition shown at this point is called a eutectic mixture. The eutectic mixture has the lowest melting point that a mixture of two substances, in this case X and Y, can have. Unlike all other mixtures of X and Y, the eutectic mixture has a constant melting point, much like that of a pure substance.
It begins and finishes melting at the same temperature.
In your pre-lab assignment, you will be making a melting point diagram like this from experimental melting point–percentage composition data for 2-furaldehyde semicarbazone and cyclohexanone semicarbazone. From this diagram and your own experimental data you will determine the composition of the mixtures you obtain from parts 3 and 4. It is important to note something here. On the diagram above, between the melting point of the eutectic mixture and
125ºC, each temperature corresponds to two different compositions of an X-Y mixture. You will have to take this into account when determining the compositions of your semicarbazone mixtures from parts 3 and 4 of the experiment. In some cases, it will not be possible to do more than narrow down to two possibilities the composition based on the melting point of your sample and the melting point diagram you will generate in the pre-lab work.


1.
Read thoroughly the theory section of this experiment and the section in your lecture textbook dealing with kinetic and thermodynamic control.
2.
Carefully go over the procedure for this experiment, especially parts 3 and 4, to be sure that you understand the purpose of each procedural step. Since you will be working with a partner for this experiment, determine before you come to lab, which parts of the experiment each of you will do. A logical breakdown of responsibilities would be for one of you to do part 1 and either part 3 or 4 with the other of you doing the remaining parts.
3.
Answer the following questions about the AS and BS reactions discussed in the background portion of this experiment. a.
When a competitive reaction occurs between A and B and a limited amount of S, which product, AS or BS, would you expect to form in largest yield at
0ºC? Briefly explain. b.
Which product would you expect to form in largest yield at high temperatures? Briefly explain. c.
If BS and A were mixed together with the appropriate solvents used throughout the experiment and heated to the same high temperature as in “b” above, what would you expect to happen? Briefly explain.
4.
Using the following melting point-percentage composition data, draw a melting point diagram like the one drawn for X and Y in the background section of this experiment.
Use the average melting point data for your graph, not the melting point range. Graph to show the best fit line. Do not just “connect the dots”. To draw the best fit lines, do not attempt to begin and end at data points. Instead, draw a line interpolating between all points. To get a good graph, the lowest temperature on your y-axis should be 100ºC and the graph should be sufficiently large to allow you to use it to answer post-lab questions.
% by mass of CS
100
90
80
70
50
20
0
Melting point range (ºC) Average melting point (ºC)
165.2 - 166.8 166.0
144.9 - 160.0 152.5
140.0 - 144.0
143.7 - 148.4
142.0
146.1
143.0 -160.0
171.0 - 180.0
201.6 - 202.4
151.0
175.5
202.0

! Safety Considerations
! Semicarbazide hydrochloride is carcinogenic. Use gloves when working with this compound.
If you get any on your skin, wash it thoroughly with soap and hot water.
1. Preparing Cyclohexanone Semicarbazone
In a 50-mL Erlenmeyer flask, dissolve 1.0 g of semicarbazide hydrochloride and 2.0 g of dibasic potassium phosphate (K
2
HPO
4
) in 25 mL of water. Using a 10-mL graduated cylinder, measure out 1.0 mL of cyclohexanone and pour it into a test tube containing
5 mL of 95% ethanol. Pour this ethanol solution into the solution containing the semicarbazide and swirl the mixture thoroughly to mix the reactants. Let the mixture sit for 5-10 minutes to allow crystals of cyclohexanone semicarbazone to form.

Note: The reactions of aldehydes and ketones with semicarbazide hydrochloride require that the pH of the solution be acidic but not too much so. The function of the dibasic potassium phosphate is to assure the optimum pH for the reaction.

Note: Throughout this experiment, you will be allowing solutions to sit, waiting for crystals of the semicarbazones to form. Sometimes these crystals form readily, and sometimes they do not. Crystallization is a tricky thing in organic lab. When crystals are slow in forming, cooling in an ice bath can help. So can scratching the sides of the container with a glass stirring rod. The scratch mark made by the stirring rod can provide a spot just right for molecules to collect and start developing a crystal. Once crystals start to form, other crystals will form more rapidly.
Filter the crystals with suction using a Hirsch funnel. (A Hirsch funnel is a small, conical funnel designed for suction filtration. It works exactly like a Buchner funnel except it is used for filtering small amounts of solid. Use small pieces of filter paper that just cover the holes at the base of the funnel.) Wash the crystals with 2-3 mL of cold water.
Remove the crystals, place them between two pieces of filter paper folded on the sides to keep the crystals from falling out, and label them. Note the color of the crystals.
Set this solid aside for use in part 4.
The melting point of pure cyclohexanone semicarbazone is 166 o
C.

2.
3.
Preparing 2-Furaldehyde Semicarbazone
Repeat exactly the procedure for the preparation of cyclohexanone semicarbazone, using
0.8 mL of 2-furaldehyde instead of 1.0 mL of cyclohexanone. The melting point of 2furaldehyde semicarbazone is 202 o
C.
Competitive Reactions of Cyclohexanone and 2-Furaldehyde with Semicarbazide
Prepare the following solutions in appropriately sized Erlenmeyer flasks and stopper them.
Solution S (semicarbazide solution)
Dissolve 2.0 g of semicarbazide hydrochloride and 4.0 g of dibasic potassium phosphate in 50 mL of water.
Solution FC (furaldehyde and cyclohexanone mixture)
Dissolve 2.0 mL of cyclohexanone and 1.6 mL of 2-furaldehyde in 10 mL of 95% ethanol. a.
In separate Erlenmeyer flasks, cool 5 mL of solution FC and 25 mL of solution S in an ice-water bath at 0-2 o
C. Then add the cold solution FC to the cold solution
S, swirling well to mix the reactants. Crystals should form quickly. Keep the reaction mixture at 0-2 o
C for 3-5 minutes after the crystals have formed. Vacuum filter the crystals and wash them with 2-3 mL of cold water. Allow the crystals to dry for at least 24 hours before taking their melting point. b.
In separate Erlenmeyer flasks, warm 5 mL of solution FC and 25 mL of solution S in an 80-85 o
C water bath. After 5 minutes of warming, add the warm FC solution to the warm solution of S and swirl well to insure thorough mixing of the reactants. Heat the mixture in the hot-water bath for 15 minutes, cool it to room temperature and then in an ice-water bath until crystals are formed. Scratch the sides of the flask with a glass stirring rod if necessary to induce crystallization.
Vacuum filter the crystals, wash them with 2-3 mL of cold water, and let them dry for at least 24 hours before taking their melting point.

Note: Students sometimes are confused in this experiment when they cool the reaction mixture that has been heated at 85ºC for 15 minutes. They sometimes think that cooling the reaction in ice water that they are changing the nature of the products that were formed at 85ºC. The composition of the products will not change upon cooling. The cooling is just to speed up the formation of crystals.

4. Testing for Reversibility: reacting semicarbazones with cyclohexanone and furaldehyde
5. a.
Cyclohexanone Semicarbazone with 2-Furaldehyde
In a 25-mL Erlenmeyer flask, put 0.3 g of the cyclohexanone semicarbazone prepared in part 1, 0.3 mL of 2-furaldehyde, 2 mL of 95% ethanol, and 10 mL of water. Warm the mixture with swirling in an 80-85 o
C water bath until the solids dissolve and then warm the mixture for 3 minutes more. Cool the mixture to room temperature and then in an ice-water bath. Vacuum filter the crystals, wash the crystals with 2-3 mL of cold water and allow them to dry for at least 24 hours before taking their melting point. b.
2-Furaldehyde Semicarbazone with Cyclohexanone
In a 25-mL Erlenmeyer flask, put 0.3 g of the 2-furaldehyde semicarbazone prepared in part 2, 0.3 mL of cyclohexanone, 2 mL of 95% ethanol, and 10 mL of water. Warm the mixture with swirling in an 80-85 o
C water bath until the solids dissolve and then warm the mixture for 3 minutes more. Cool the mixture to room temperature and then in an ice-water bath. Vacuum filter the crystals, wash the crystals with 2-3 mL of cold water and allow them to dry for at least 24 hours before taking their melting point.
Dispose of all filtrates, the liquids left over from crystal filtration, down the sink.
Leftover CF and S solutions will be disposed of in waste bottles in the hood
6. After 24 hours, take the melting points of all of your samples and record them in your notebook. See Appendix 4 for details on how to correctly measure melting point. Your instructor will show you how to crush your crystals, put them in a capillary tube and pack the crystals in the tube. Your instructor will also talk to you about how to use the melting point apparatuses in the lab and when to record the beginning and end of the melting point range. To get an accurate and meaningful melting point range, the most important thing is to heat slowly, at a rate of no more that 2ºC/minute. Heating any faster than that near the melting point will cause your melting points to be too high and the ranges to be too broad to be of value. When you have finished with your melting points, dispose of the solid products in a waste jar in the hood.

1. Write equations for the reactions of cyclohexanone and 2-furaldehyde with semicarbazide in the presence of an HCl catalyst.
2. Give a theoretical discussion of the difference between a kinetically controlled product and a thermodynamically controlled product. Be sure to explain how temperature affects the product mixture in reactions such as the ones we studied in this experiment in which

4.
3. the kinetically controlled product and the thermodynamically controlled product are not the same. Include a labeled energy diagram in your answer.
Give the experimental melting point ranges for each of the solids that you and your partner made. For each of the sample mixtures of parts 3 and 4, average the beginning and ending temperatures of the range and use this average temperature to determine the approximate composition of the sample from the melting point diagram that you made for your pre-lab assignment.
State which semicarbazone is the kinetically controlled product and which is the thermodynamically controlled product. a.
Explain in detail how the results from the competitive reaction experiments in experimental procedures 3a and b support this conclusion. b. Explain in detail how the results from the experiments in experimental procedures 4a and b support this conclusion