In the Classroom Resources for Student Assessment
Solving the Mathematics of a Kinetic and Regiochemical Problem: The Dehydration Reaction of a cis–trans Mixture of 2-Methylcycloalkanols John J. Cawley Department of Chemistry, Villanova University, Villanova, PA 19085
Background In a recent paper in this Journal (1), an example of the titled reaction was presented as a physical organic experiment for undergraduates. A general methodology for solving the kinetic and regiochemistry problem was given in the data treatment section, but no mathematics; only the mathematical results were presented. We discovered that students grasp the logic of what has to be done to solve the problem but are unable to express that logic in appropriate mathematical equations. It seemed reasonable to present the problem of the mathematics as a take-home exam in the Journal ’s Resources for Student Assessment column. The problem is idealized in that the experimental methodology for garnering the data is ignored. This simplification helps students focus on (i) manipulating the derived kinetic data to obtain the two individual rate constants for dehydration of the cis and trans alcohols and (ii) manipulating the derived relative percent isomeric alkene product data to obtain the mole fraction of each isomeric alkene resulting from the dehydration of each isomeric alcohol. The students may then easily solve the problem of assigning the six specific rate constants involved in this dehydration reaction. For the Instructor In preparing students to take the exam it may be useful to assign them the task of studying the cited experiment (1). Questions from students may be answered in a general way, addressing the logic involved; the mathematics should not be discussed, so that the students may search out the mathematical solution in the take-home exam itself. The exam is in two parts. In Part A the kinetics (including the possibility for correcting to a different time zero, t0) are to be solved to obtain the individual rate constants for dehydration; the relative percent of product alkene formed from each of the two reactant isomeric alcohols is to be calculated for each time increment. Part A is then handed in. The instructor should discuss this part of the problem with each student before allowing the students to continue. Only then should the students undertake Part B of the exam, mathematically manipulating the alkene product data to obtain the mole fraction of each isomeric alkene resulting from the dehydration of each isomeric alcohol. Lastly the students calculate the six specific rate constants involved in this dehydration reaction.
alkenes. Inevitably one alcohol will produce relative amounts of the three alkenes different from the other. Part A of the problem is to calculate the individual rate constants for dehydration of the two isomeric alcohols for these kinetically first-order reactions, checking to see if the t0 assigned experimentally is the true time zero and correcting it to a new t0, if necessary. The relative percent of product alkene formed from each of the two reactant isomeric alcohols at each time point is to be calculated as well. Part B of the problem is to calculate the mole fraction of the three alkenes that come from each alcohol. Finally, the specific rate constants for the production of the individual alkenes from each reacting isomeric alcohol are calculated. It may be expedient for students to “spot check” their answers, using their calculated specific rate constants to regenerate the given experimental regiochemical data before handing in Part B of the exam. Exam
Part A Given: The kinetic data are presented in Table 1. It was determined that the rel % trans alcohol/rel % cis alcohol before reaction was 55.4/44.6. Table 1. Kinetic Data Concentrations are based on hydroxyl NMR integrals relative to that of the internal standard cyclohexanol.
a t (NMR intregral)
ac
"t 0"
30.00
20.00
"t 4"
25.36
3.73
"t 8"
21.44
0.6946
"t 12"
18.12
0.1282
"t 20"
12.95
time/min
4.497 × 10 ᎑3
Solve the following: 1. Calculate k t and k c. 2. Calculate k c – k t. 3. If “t 0” is not the real time zero, generate an equation to find the correction for the real time zero needed.
For the Student
4. Solve for a t0 and a c0 at the real time zero, if necessary.
Consider two isomeric alcohols that dehydrate at different rates with the potential of giving the same three isomeric
5. Solve for rel % Pt and rel % Pc at the various times, where P = sum of product alkenes.
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Journal of Chemical Education • Vol. 76 No. 1 January 1999 • JChemEd.chem.wisc.edu
In the Classroom
Solution:
Solution: kt
kc
1. Plotting ln at versus time gives = 0.042 and = 0.420. 2. k c – k t = 0.378. 3. Since the rel % trans alcohol:rel % cis alcohol is 60.00: 40.00 at “t0 ”, it is not the real time zero; plotting ln (a tt /a ct ) vs (k c – k t ) t + ln (30.00/20.00 ) yields the real time zero, t 0 = ᎑ 0.5 min, when ln (a tt/a ct) = ln 55.4/44.6; therefore all the times have to be increased by 0.5 min.
1. Using two-by-two determinants and making use of Cramer’s rule (3), solutions are found to χ texo and χ cexo as follows: at t4.5 20.14 χ texo + 79.86 χ cexo = 2.01 at t 8.5 27.83 χ texo + 72.17 χ cexo = 2.78 therefore
χtexo =
4. ln (a t0 /30.00) = 0.042 (0.5); ln a t0 = 0.021 + ln 30.00; a t0 = 30.64; and (ln
= 0.42 (0.5);
χcexo =
a c0 = 24.67. 5. Using ∆ a lost in ∆ t yields the relative percentages of products formed at time t. At t 4.5 , a t4.5 lost is 5.28 and a c4.5 lost is 20.94; thus (5.28/26.22) × 100% = 20.14% of product from the trans alcohol, and therefore 79.86% of product from the cis alcohol. In similar fashion the following percentages are found: at t8.5 , 27.83% of product from trans alcohol and 72.17% of product from cis alcohol at t 12.5 , 33.79% of product from trans alcohol and 66.21% of product from cis alcohol at t 20.5 , 41.77% of product from trans alcohol and 58.23% of product from cis alcohol
Part B Given: The summarized regiochemical data are presented in Table 2. Table 2. Regiochemical Data
4.5
= 0.10
20.14 79.86 27.83 72.17
and
a c0/20.00)
ln a c0 = 0.21 + ln 20.00;
Time/min
2.01 79.86 2.78 72.17
Relative Percent exo-ene
1-ene
3-ene
2.01
81.94
16.05
8.5
2.78
78.87
18.35
12.5
3.38
76.48
20.14
20.5
4.18
73.29
22.53
Solve the following: 1. Using appropriate equations calculate the mole fraction, χ, for each of the three alkenes that are produced by each of the two isomeric alcohols. 2. Using the mole fraction data calculate the six specific rate constants for the production of the individual alkenes from each reacting isomeric alcohol. Are your results consistent with typical literature results (2)?
20.14 2.01 27.83 2.78
= 0.00
20.14 79.86 27.83 72.17
In similar fashion it is calculated that χ t1-ene is 0.50 and χ c1-ene is 0.90; then it is calculated that χ t3-ene is 0.40 and χ c3-ene is 0.10 . 2. Multiplying the individual rate constants by the appropriate mole fraction yields the six specific rate constants. For example, k c1-ene is 0.378. The summarized data given in Table 3 show clearly that the Saytzeff rule (4 ) is being followed for the dehydration of each of the two isomeric alcohols. Table 3. Specific Rate Constants 1-ene
3-ene
exo-ene
cis alcohol
0.378
0.042
0.00
trans alcohol
0.021
0.0168
0.0042
Conclusion A cis–trans isomeric mixture of alcohols, as reactants, is dehydrated to an isomeric mixture of alkenes, a situation which may seem to complicate any kinetic and regiochemical analysis. When students are given a take-home exam in which they are asked to find the six specific rate constants involved from the experimentally obtained kinetic and regiochemical data, they are stymied not by the logic, but by the mathematics involved. Eventually, with a little coaching and encouragement, they are happily surprised to find they can solve such a problem with a little ingenuity and some basic mathematical skills, which they may not have used in a while. Literature Cited 1. Cawley, J. J.; Lindner, P. E. J. Chem. Educ. 1997, 74, 102. 2. Reinecke, M. G.; Smith, W. B. J. Chem. Educ. 1995, 72, 541. 3. Kaplan, W.; Lewis, D. J. Calculus and Linear Algebra; Wiley: New York, 1970; Chapter 1, p 16. 4. Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, 3rd ed.; Plenum: New York, 1990; Part A, p 374.
JChemEd.chem.wisc.edu • Vol. 76 No. 1 January 1999 • Journal of Chemical Education
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