Kinetic Vs Thermodynamic Control in Competing Reactions

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 x y = k[R1] [R2] , 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

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 1,2- addition, 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 H O

H2NN H H2NN H2NN NO2 H2NN C NH2 hydrazine phenylhydrazine 2,4-dinitrophenylhydrazine 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 H O H  H addition dehydration C + N N C C positive to minus H2O N  H R negative N  H N H analogous N H to loss of R water from R a gem-diol

Figure 1. General Reaction.

Mechanism

Figure 2 shows the mechanism of the reaction. In Step 1, the

Lab 4 3 H - OH O H O H O+ H Step 1 C Step 2 C C N N H fast N R N H transfer H H N H of H+ H N H Step 3 R R H O+ H Step 4 C C N N H N H H N H H O R 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 NNHR C O 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 H2NNH2 CH3CH hydrazone

NNHPh H NNH 2 CH3CH phenylhydrazone O O2N + O2N CH3CH NNH NO2 H2NNH NO2 CH3CH 2,4-dinitrophenylhydrazone O O NNHCNH2 H2NNHCNH2 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 2,4- dinitrophenylhydrazine 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 O H2NNHCNH2 C O C NNHCNH2 O O H H semicarbazone of 2-furfural 2-furfural a yellow solid

Equation 1

O H NNHCNH O O 2 2 NNHCNH2

cyclohexanone semicarbazone of cyclohexanone a white solid Equation 2

We can distinguish the products by their colors. The semicarbazone of 2- furfural 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 Br

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