Ozonolysis Problems That Promote Student Reasoning - Journal of

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Thomas A. Holme University of Wisconsin–Milwaukee Milwaukee, WI 53201

Ozonolysis Problems That Promote Student Reasoning

W

Ray A. Gross, Jr. Department of Physical Sciences & Engineering, Prince George’s Community College, Largo, MD 20774; [email protected]

This article examines how the structures of many monoterpenenes that follow the isoprene rule can be determined when their molecular formulas and those of their ozonolysis products are known.1 The result is a set of problems suitable for use in a beginning organic chemistry course or in a qualitative organic analysis course. The problems vary in difficulty, which allows students to improve their reasoning skills by working the easier problems first, followed by the more challenging ones. The problems require students to apply logic and reasoning together with their knowledge of ozonolysis and the isoprene rule; however, instructors may restate the problems to eliminate the need for the isoprene rule if desired. The problems help students develop critical thinking skills, improve their understanding of structure and bonding, and learn the subtle differences among ozonolysis products, depending on the reactant and workup conditions. They allow students to extract knowledge from raw data and apply that knowledge to structure determinations. They also introduce students to natural products vis-à-vis the isoprene rule (1), help students improve their nomenclature skills by requiring them to name a variety of unsaturated structures, and help students better understand the structural relationships among aldehydes, ketones and carboxylic acids. For a given problem, the only data given to students are the molecular formulas of a reactant and those of its products of ozonolysis. These problems differ from available textbook problems involving structure determinations by ozonolysis in two ways. These problems begin with molecular formulas; whereas, textbook problems (2) provide structural formulas or partial structures to aid students in the solutions. In addition, textbook problems give only one kind of workup, reductive or oxidative; whereas, many of these problems include both kinds of workups, affording instructors a set of rich problem-solving exercises.

units, although isoprene is not the biogenetic precursor. Isoprene is 2-methyl-1,3-butadiene. Scheme I shows the headto-tail union of two isoprene units to form 2,6-dimethyloctane, which serves as a template for the problems. Structures of compounds containing various combinations of double and triple bonds can be made by adding one or two π bonds, respectively, to existing σ bonds on the template. The molecular formulas of the resulting compounds together with those of their ozonolysis products produce a variety of structure-determination problems. Students are told the unknown follows the isoprene rule from which they must infer the presence of the template from C10 in the molecular formula and prove it is acyclic from additional data.2

Introduction

As seen in Scheme II, the reductive workup of an ozonide invariably produces two carbonyl groups from a double bond and two carboxylic acids from a triple bond. These results allow the following calculations (3) to be made from molecular formulas, leading to structural features in the reactant hydrocarbon.

Isoprene Rule The isoprene rule states that many natural products may be considered to form by the head-to-tail linkage of isoprene

Scheme I. Formation of a C10 template that obeys the isoprene rule.

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Reductive and Oxidative Workups A summary of ozonolysis reactions involving reductive (R) and oxidative (O) workups applied to alkenes and alkynes derived from the template are shown in Schemes II and III, respectively. The ozonolysis products are determined by the nature of the multiple bond (i.e., double or triple bond), the location of the multiple bond, and the kind of workup used by the experimenter. Ozonolysis reactions start by the addition of ozone. An initial ozonide or molozonide forms and rearranges into an ozonide that is hydrolyzed under either reductive or oxidative conditions to give the isolated products. The equations are not balanced,3 because they show only the organic reactants and products. However, the imbalance allows students to determine structural features present in the starting hydrocarbon by an analysis of the molecular formulas of the reactant hydrocarbon and the products obtained from both workups. Data from a Reductive Workup

• The number of degrees of unsaturation, DU4 or π + r, in the reactant and in each product can be calculated from the corresponding molecular formulas by the DU equation π + r = n + 1 − h兾2. In this equation, π is the number of π bonds, r the number of rings in the structure, n is the number of carbon atoms, and h the number of hydrogen atoms in the molecular formula (4).

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In the Classroom • The number of triple bonds (tb) in the reactant is given by the number of new hydrogen atoms appearing in the products divided by two. In Scheme II, the alkyne has 18 hydrogen atoms and the products have 20; the two-hydrogen atom difference proves the presence of a triple bond in the hydrocarbon. As seen in Scheme II, new hydrogen atoms are not produced by the cleavage of double bonds. • The number of π bonds (π) in the reactant is the sum of oxygen atoms in the products divided by two. As seen in Scheme II, the cleavage of a double bond with its one π bond produces two oxygen atoms, and the cleavage of a triple bond with its two π bonds produces four oxygen atoms. Thus, every π bond in a reactant produces two new oxygen atoms. • The number of double bonds (db) in the reactant is the number of π bonds (π) minus two times the number of triple bonds (2tb) in its structure. • The number of rings (r) in the reactant is its degrees of unsaturation (π + r) minus the number of π bonds (π).

Scheme II. Ozonolysis of three compounds by reductive workups (R = Zn dust/HOAc or Me2S).

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Data from an Oxidative Workup Scheme III shows the products obtained by oxidative workups for the same three hydrocarbon reactants of Scheme II. Products obtained during the oxidative workup of an ozonide by cleavage of a double bond may be a carboxylic acid, a ketone, or carbon dioxide, depending on the structure of the starting alkene. A terminal alkene or alkyne yields carbon dioxide, and any non-terminal alkyne yields two carboxylic acids. Additional structural information may be obtained by a comparative analysis in which each formula in the reductive workup is compared to the corresponding formula in the oxidative workup. The comparison is made as though the reductive products were isolated first and the ensuing mixture oxidized later to produce the oxidative products. All comparisons are made between a pair of molecular formulas. • A terminal alkene yields formaldehyde, CH2O, in the reductive workup and carbon dioxide, CO2, in the oxidative workup. The presence of CH2O proves the presence of a terminal alkene in the reactant. • A terminal alkyne yields formic acid, CH2O2, in the reductive workup and carbon dioxide, CO2, in the oxidative workup. The presence of CH2O2 proves the presence of a terminal alkyne in the reactant.

Scheme III. Ozonolysis of the three compounds by oxidative workups (O = Ag2O).


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In the Classroom • Except for formaldehyde, an aldehyde product in the reductive workup is isolated as a carboxylic acid in the oxidative workup. The difference is indicated by an additional oxygen atom in the acid’s formula (e.g., the C6H12O aldehyde in Scheme II becomes the C6H12O2 acid in Scheme III). • Any ketone isolated in the reductive workup is also isolated in the oxidative workup (e.g., C4H8O in Scheme II remains C4H8O in Scheme III). • Any acid isolated in the reductive workup is also isolated in the oxidative workup (e.g., C 9H 18O 2 in Scheme II remains C9H18O2 in Scheme III). • A dialdehyde in the reductive workup becomes a diacid in the oxidative workup as evidenced by a twooxygen atom increase in the diacid (O2 to O4); a ketoaldehyde becomes a ketoacid (O2 to O3), an aldehyde-acid becomes a diacid (O3 to O4), and so forth.

A summary of a systematic application of the above conclusions to structure determinations from the available molecular formulas follows. The molecular formulas of the reductive-workup products are analyzed first. This analysis provides the number of degrees of unsaturation, the number of triple bonds, π bonds, double bonds, and rings in the hydrocarbon reactant as well as the presence of any terminal alkenes or alkynes in its structure. Then structural formulas or structural features of compounds present in the reductive products are determined as allowed by the data. If the information obtained from the reductive workup is sufficient to construct the structure with the aid of the template, the analysis of the oxidative workup can be omitted. If the information is not sufficient, the oxidative products are analyzed. In this analysis, organic families in the reductive products are deduced by means of a one-to-one comparison of their molecular formulas with those of the oxidative products. Functional groups or families such as aldehydes, ketones, ketoaldehydes, and so forth, are identified and associated with a specific number of carbon atoms. For example, a reductive product might be characterized as a C6 aldehyde, a C4 ketone, a C5 ketoaldehyde, and so forth. The structural features of the compounds characterized in this manner can then be correctly placed on the template owing to its asymmetry. The last step is to name the unknown. Example Problems The kind of reasoning necessary to solve the structures in Schemes II and III is exemplified in the solutions to three example problems below. Example 1 shows that the reductive workup products alone are sometimes sufficient to determine a structure. Example 2 introduces a split-ozonide technique, which is applied when the oxidative data are required for a structure determination. Example 3 shows that two structures may satisfy the available data. The Supplemental MaterialW includes 58 unique sets of ozonolysis data, 50 of which may be solved for a single structure, exclusive of R–S or E–Z stereochemistry. Thirty data sets can be solved from the reductive workup products. The problems require students to link the concepts inherent in the content described above.

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Example 1: Structure of a Triene from Its Reductive Products A compound, C10H16, follows the isoprene rule and yields CH2O, C3H6O and 2 C3H4O2 upon ozonolysis by a reductive workup. What is the IUPAC name of the hydrocarbon, excluding stereochemistry? The systematic approach outlined above yields the solution shown in Scheme IV. First, structural information about the hydrocarbon is determined and summarized beneath its molecular formula. From the molecular formula, the number of degrees of unsaturation in the hydrocarbon is 3 (π + r = 3). The reaction produces no new hydrogen atoms (16 − 16 = 0); therefore, the number of triple bonds in the reactant is zero (tb = 0). The six oxygen atoms in the products show the presence of three π bonds in the reactant (π = 3), and all of them are in double bonds because there are no triple bonds (db = 3). A structure with 3 degrees of unsaturation and 3 double bonds has no rings (r = 0). From the foregoing analysis, three double bonds must be placed on the 2,6-dimethyloctane template, which is required by a C10 acylic hydrocarbon that follows the isoprene rule. Next, the number of π bonds in each product is calculated as shown beneath the corresponding formula. None of the products can have rings, because the reactant is acyclic. Thus, a calculation of the degrees of unsaturation in each product gives the number of π bonds in the corresponding structure. Likewise, the number of π bonds in each product equals the number of carbonyl groups in its structure, because the products cannot have any carbon–carbon multiple bonds. Carbon–carbon multiple bonds are oxidized by ozone. Thus, two compounds with one π bond and two compounds with two π bonds are produced by the cleavage of three double bonds. The template is drawn and numbered as in Scheme I. The two products with one π bond must be derived from the ends of the template, and the two products with two π bonds must be derived form the middle portion of the template. The CH2O product is formaldehyde, which must be produced from a double bond located at one of three sites C1, C6–C10 or C7. The C3H6O product can only come from a double bond at C2 on the template. The C2 double bond is indicated by a dashed line on the template in Scheme IV. This establishes acetone as the product derived from the left end of the template as drawn. Formaldehyde must then come from right end. Thus, one of the C3 dicarbonyl compounds must come from C3–C4–C5 on the template, which locates the second double bond at C5. The placement of the second double bond at C5 leaves a C7 double bond as the only location for a terminal double bond that can account for the presence of formaldehyde, because two double bonds cannot share C6. The complete structure is 3,7-dimethyl1,3,6-octatriene also known as ocimene. The numbering in ocimene differs from that of the template because of the location of the terminal double bond. Other solutions5 that give the same answer are possible, and students are encouraged to look for multiple solutions. One of the goals of administering these problems is for students to reason correctly but not necessarily by a rote pattern.


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Example 2: Structure of a Diene by a Split-Ozonide Technique An unknown C10H18 hydrocarbon reacts with ozone to give the products shown in Scheme V for the oxidative (O) and reductive (R) workups. What is the IUPAC name, excluding stereochemistry, of the hydrocarbon if its structure follows the isoprene rule? The reductive workup is analyzed as in example 1 to show that the unknown is a diene. Formaldehyde and the C4 compound (π = 1) must come from the ends of the structure. It cannot be determined from the reductive workup whether the C4 carbonyl compound is an aldehyde or ketone, because C4H8O is derivable from the left end of the template as an aldehyde (2-methylpropanal) or from the right end as a ketone (2-butanone). Thus, the oxidative workup is required. The molecular formula of the C4 compound is unchanged in the oxidative workup. Therefore, it is 2-butanone derived from the right end of the structure as indicated by the dashed line at C5 on the template. Formaldehyde must then come from a C1 double bond, and the structure is 2,6dimethyl-1,5-octadiene. If the C4H8O product of the oxidative workup were changed to C4H8O2, the answer would change to 2-ethyl-6-methyl-1,4-heptadiene. In that case, the C4 compound would be 2-methylpropanal derived from a C3 double bond on the left end of the template, and the C5 ketoaldehyde would require a C6–C10 double bond. Many structures, including all three in Scheme III, may be determined from their oxidative products alone by an acute awareness of the structural limitations imposed by the template. In example 2, C4H8O must arise from a C5 double bond, leaving only C1 for the location of the second one.

Scheme IV. Solution to example 1.

Example 3: Two Structures from the Data An unknown that follows the isoprene rule reacts with ozone to give the products shown in Scheme VI. What are the IUPAC names, excluding stereochemistry, of the two possible structures for the unknown? The reactant has three degrees of unsaturation, and the systematic analysis of its reductive workup products shown at the left in Scheme VI proves it is an acyclic enyne. The reductive products are formaldehyde, a C4 ketoacid, and a C5 acid as verified by the oxidative workup products. The C5 acid requires the triple bond to lie at C4 as shown by dashed lines on the template. However, the location of the double bond cannot be established, because a C4 ketoacid would be produced if the double bond were at either C1 or C6–C10 on the template. Thus, two structures, 2,6-dimethyl1-octen-4-yne or 2-ethyl-6-methyl-1-hepten-3-yne satisfy the data.

Scheme V. Solution to example 2.

An Alternative Approach That Eliminates the Use of the Isoprene Rule For instructors who wish to omit the isoprene rule or who do not teach it, the target molecules may be considered as C10 -enes, -ynes, -enynes, and so forth. Instead of stating the target compound follows the isoprene rule, the problem may state that exhaustive hydrogenation of it yields 2,6dimethyloctane. This change eliminates the need to prove the

Scheme VI. Solutions to example 3.

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compound is acyclic but also suggests an alternative way of finding the number of triple bonds and double bonds in the unknown’s structure. This alternative method for finding the number of tb and db works only for acyclic compounds; whereas, the method exemplified in the solutions above works equally well for cyclic compounds. The number of tb in an acyclic structure is a function of the number of degradation products P formed in the reductive workup, as shown by the following analysis. The number of π bonds in the unknown is given by eq 1, and the total number of reductive products, P, arising from an acyclic compound is one more than the sum of multiple bonds as given by eq 2. (1) π = db + 2 tb (2)

P = db + tb + 1

From eq 1, db = π − 2tb. The substitution of the latter value of db into eq 2 gives eq 3, which may be simplified and rearranged to give eq 4: P = π − 2tb + tb + 1

(3)

tb = π + 1 − P

(4)

For acyclic compounds, DU = π + r = π because r = 0. Therefore, eq 4 may be rewritten as eq 5: (5)

tb = DU + 1 − P

For example 1, DU = 3 and P = 4; therefore, tb = 0 and all three π bonds are in double bonds. For example 2, DU = 2 and P = 3; therefore, tb = 0 and db = 2. For example 3, DU = 3 and P = 3; therefore, tb = 1 and db = 1. At this point, the two methodologies merge. The alternative method for finding tb and db may be used as soon as it is determined that the compound is acyclic. Students might also arrive at the conclusion inherent in eq 5 as follows. For an acyclic unknown, the number of π bonds is given by its formula or DU, because it has no rings. The maximum number of products occurs when all of the π bonds are in double bonds and is one more than the number of π bonds. Each triple bond present decreases the maximum number of products by one. In the case of example 3, the maximum number of products is π + 1 or 4. The actual number of products is only three, so the compound contains one triple bond. Discussion and Student Results Students are provided exercises in which they determine the gross structures of hydrocarbons from their molecular formulas as exemplified above. They are not asked to determine R–S or E–Z configurations, because these structural features cannot be determined from the available data. The problems evolved from the following beliefs about students: • Students learn how to apply the concepts of organic chemistry by working problems. • Many students have poor or underdeveloped reasoning skills. • Many students do not work enough problems to learn the reasoning required to solve them.

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Instructors have several options for engaging students with exercises from this set over the course of a semester. Three or four problems might be posted on the Internet and changed frequently. Problems might be given during recitation periods or during labs, depending on the available time, or assigned as homework on a regular basis. In some instances, one of these problems might be given on an examination to assess the level of student understanding of ozonolysis reactions. In our case, the problems were assigned to students during their laboratory periods with the following thoughts: • It might be possible to improve student reasoning skills by engaging them for 15 minutes a week with individual, solvable problems. • A gradual increase in the complexity of the problems would afford students an opportunity to learn the logic required to solve the more difficult problems in stages, not all at once. • The individual and collective progress of the students could be monitored by keeping track of the number of correct solutions versus the number of trials.

Accordingly, problems were tendered to students during the first 15 minutes of their second-semester organic laboratory periods. The organic chemistry required to solve the problems was covered during the first semester. Students were introduced to the isoprene rule and given one of the problems with two solutions during their first lab period. The problem was not graded, because the students were not expecting it. Only three of 27 students (11%) correctly named the structure. During the semester, each student had a chance at 11 graded problems taken from the tables in the Supplemental Material,W including one on the midterm exam. Students were given the information deducible from Schemes II and III but were never given a specific procedure for solving the problems. After each trial, they were given the names of the structures from which they could work backwards to see how a correct solution might be obtained. Students were given graded unknowns, starting with those with a single double bond and thereafter increasing in structural complexity. For example, a solution to C10H20 + O3 (R) → C5H10O + C5H10O requires only that the student understand that the unknown is cleaved into two, five-carbon fragments; whereas, other problems, such as examples 1–3 above, require more reasoning skills. The gradual increase in the complexity of the reasoning needed to solve a problem allowed the students to build upon their previous week’s experience and not to be overwhelmed by the immediate problem. The idea was to engage students in the problems over the entire semester and allow them to discover for themselves how to solve the problems by analysis, synthesis, and evaluation (5). The results were mixed. Students solved only 36% (90兾253) of all of the graded problems, including those tested multiple times. However, they solved 88% (44兾50) of the specific problems tendered at least once. These problems accounted for less than 8% of the lab grade. The student success rate increased during the semester; 18 of the 25 students (72%) who took the final exam successfully solved the 12th problem of this type. The increase in student success from 11% to 72% from the first to last problem indicates that stu-


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dents can improve their problem-solving skills by practice and repetition. Every student who commented on the usefulness of the problems stated that the exercises helped them develop problem-solving skills in some fashion. Sample statements include: …I look at problems differently…I know what to look for when piecing fragments together… …At the end, I began to understand… …I ask myself, what combination of factors is necessary to arrive at the answer. If I know more than one way to solve a problem, I will try them all.

Conclusions The problems exemplified above are difficult for beginners, yet solvable. Instructors are afforded a set of problems that complement synthetic and spectral problems in making students think critically about structure and bonding to arrive at the solutions. The problems are formulated on a realistic natural-products scenario that chemists might encounter, and the problems supplement lectures on aldehydes, ketones, and carboxylic acids. This kind of teaching approach follows that espoused by Mohrig and others (6). Did the students improve their reasoning skills? The instructor and most of the students who participated believe they did for the kinds of problems given here. In general, an online homework system, such as the one developed by Penn et al. (7) in which the instructor can monitor student engagement in all kinds of problems seems warranted. WSupplemental

Material

A list of 66 compounds suitable for analysis, 50 problems with a single solution and 16 with dual solutions, and notes for instructors are available in this issue of JCE Online. Notes 1. Presented at the 37th Middle Atlantic Regional Meeting of the American Chemical Society, Rutgers University, New Brunswick, NJ, May 24, 2005; Paper 685. 2. The problems in this article are limited to C10 hydrocarbons (i.e., two isoprene units); however, the techniques exemplified can be extended to compounds with three or more isoprene units. 3. Students are given a complete set of ozonolysis products in these problems (i.e., the equations are balanced in carbon). Hence, CO2 is shown in the equations when it might not be de-

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tected if the experiment were actually conducted. Reference 4 shows how to account for undetected structures. 4. Degrees of unsaturation, DU, are called index of hydrogen deficiency, units of unsaturation, double bond equivalents, units of unsaturation, and so forth by various authors. By any name, DU equals the sum of the π bonds and rings in a structure, where π = the number of π bonds and r = the number of rings. The structural features (π + r) in compounds containing C and H or C, H, and O are related to the molecular formula by the DU equation π + r = n + 1 − h兾2, where n = the number of carbon atoms and h = the number of hydrogen atoms. The equation allows π + r for the reactant and every product of ozonolysis to be calculated from molecular formulas by subtracting half of the hydrogen atoms from one more than the number of carbon atoms. When a compound is acyclic (r = 0), a calculation of DU gives the number of π bonds in the compound. 5. The systematic solution presented is methodical and intended to show how to apply fundamentals. Other solutions are possible, including trial and error. One might solve example 1 by realizing the reactant is an acyclic triene, because its DU of 3 is satisfied by the four products, which must come from 3 double bonds. C3H6O can only come from the left end of the structure, and its isopropylidene group must be bonded through a C2 double bond to a 3-carbon fragment followed by a double bond at C5, making the third double bond lie at C7.

Literature Cited 1. Sebastián, A.; Colombo, M. I.; Rúveda, E. A. Chem. Educator 2003, 8, 364–367. 2. (a) McMurry, J. Organic Chemistry, 6th ed.; Brooks/Cole: Pacific Grove, CA, 2004; pp 225, 256–257. (b) Solomons, T. W. G.; Fryhle, C. B. Organic Chemistry, 8th ed.; Wiley & Sons: New York, 2004; pp 364–365, 367. 3. Gross, R. A., Jr. J. Chem. Educ. 1996, 73, 1019–1021. 4. Gross, R. A., Jr. Chem. Educator 2003, 8, 13–14. 5. Bloom, B. S. Taxonomy of Educational Objectives, Handbook I: Cognitive Domain; David McKay Company, Inc.: New York, 1956; pp 144–200. 6. (a) Mohrig, J. R. J. Chem. Educ. 2004, 81, 1083–1085. (b) Symposium on New Approaches for the Organic Chemistry Laboratory I. 18th Biennial Conference on Chemical Education, Iowa State University, Ames, IA, July 20, 2004. (c) Mohrig, J. R.; Hammond, C. N.; Schatz, P. F. In Proceedings, 228th National Meeting of the American Chemical Society, August 23, 2004; American Chemical Society: Washington, DC, 2004; Paper 77. 7. Penn, J. H.; Nedeff, V. M.; Gozdzik, G. J. Chem. Educ. 2000, 77, 227–231.


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Ozonolysis Problems That Promote Student Reasoning - Journal of

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