Useful Material Efficiency Green Metrics Problem Set Exercises for

Article pubs.acs.org/jchemeduc

Useful Material Efficiency Green Metrics Problem Set Exercises for Lecture and Laboratory John Andraos* CareerChem, 504-1129 Don Mills Road, Toronto, Ontario M3B 2W4, Canada

Andrei Hent Department of Chemistry, University of Toronto, 80 St. George Street, Toronto, Ontario M5S 3H6, Canada

Downloaded by TEXAS A&M INTL UNIV on August 27, 2015 | http://pubs.acs.org Publication Date (Web): August 26, 2015 | doi: 10.1021/acs.jchemed.5b00059

S Supporting Information *

ABSTRACT: A series of pedagogical problem set exercises are posed that illustrate the principles behind material efficiency green metrics and their application in developing a deeper understanding of reaction and synthesis plan analysis and strategies to optimize them. Rigorous, yet simple, mathematical proofs are given for some of the fundamental concepts, particularly how metrics for overall plan material performance are related to their composite counterparts for individual reactions. Throughout this exposition, whenever a synthesis scheme is examined, it is converted into a compact tree diagram that is used to depict plans of any degree of complexity (linear or convergent) as a means to conveniently keep track of all reagents, intermediates, reaction yields, stoichiometric coefficients, number of branches, number of reaction steps, and convergent steps. We demonstrate that such tree diagrams facilitate the computation of material efficiency metrics for any individual reaction in a plan as well as for the entire plan. We also show how such diagrams may be used to plan schedules for reaction operations when multiple linear branches are run concurrently. For brevity the Supporting Information contains full solutions to posed problems. KEYWORDS: Upper-Division Undergraduate, Organic Chemistry, Problem Solving/Decision Making, Green Chemistry, Synthesis

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INTRODUCTION Following the previous paper,1 which highlighted the computation of kernel and complete material efficiency metrics for synthesis plans based on mass balances using intuitive spreadsheets, we now present key insights about reaction and synthesis plan analysis. We remind the reader that the term “kernel” signifies that associated metrics parameters do not include auxiliary materials and the term “complete” signifies that they do. These insights are best understood after working through posed problem set questions that can also be used in classroom and laboratory settings when teaching green chemistry principles. Some problems are theoretically based and involve abstract thinking; however, all problems are grounded in concrete examples taken from the literature. The structure of the paper is as follows. We begin with a pedagogically useful representation of traditional synthesis schemes by means of a tree diagram2 that depicts them compactly to facilitate the computation of fundamental material efficiency metrics such as reaction yields (RY), atom economy (AE), global reaction mass efficiency (gRME), and process mass intensity (PMI). We assume from the outset that all chemical equations in any given synthesis scheme are appropriately balanced. These parameters may be determined directly by tracing the connectivity between intermediates and their progenitor reagents in each synthesis tree branch of a given plan. This allows for easy determination of basic material © XXXX American Chemical Society and Division of Chemical Education, Inc.

efficiency green metrics performances for individual reactions and for the entire plan, thus avoiding tedious tracking tasks. The tree diagram also guides the construction of the SYNTHESIS spreadsheet discussed in the previous paper,1 particularly for convergent synthesis plans composed of multiple branches. Moreover, students are able to discover fundamental insights about the basic structure of convergent plans versus linear ones. The use of tree diagrams as a powerful tool for planning schedules of reaction operations for convergent plans is also discussed. Having introduced this diagram and given examples of its use for linear and convergent plans, we next address several questions related to how material efficiency metrics for an entire plan are related to those for its composite individual reactions. As a consequence of these analyses students of green chemistry gain deep insights about the choice of reaction types in a plan that impact overall material efficiency. We also tackle the fundamental question of whether convergent plans are always more efficient than linear ones to the same target product. The Supporting Information (Part 1) contains a complete problem set of illustrative questions with worked-out solutions based on the ideas presented here that instructors can use and modify as required for their own classroom and laboratory instruction.

A

DOI: 10.1021/acs.jchemed.5b00059 J. Chem. Educ. XXXX, XXX, XXX−XXX

Journal of Chemical Education


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SYNTHESIS TREE DIAGRAMS In order to better keep track of reactions in complex syntheses involving multiple branches, a tree diagram2 was developed to convert a traditionally written synthesis scheme into a compact diagram that could be used to facilitate computation of basic material efficiency metrics by simple inspection. From these diagrams, the following parameters are easily determined by counting dots or squares: number of reaction steps, number of isolated intermediates along a synthesis pathway, number of input reagents required for a plan, number of branches, number of convergent steps, and which reactions were running simultaneously in parallel branches. All stoichiometric coefficients for input reagents, molecular weights of all materials, and reaction yields for each step in all branches were also incorporated. Essentially, the diagram’s main strength was that formulas for AE and kernel global RME3 for any reaction and, more importantly, for the entire plan could be written down by following the connectivity of dots representing output and input nodes. It is obvious that the longer a scheme is and the more branches it contains, the more complex the resulting formulas for overall AE and overall RME become. These diagrams also function as a good proofreading tool for ensuring that synthesis schemes contain no transcription errors. Here we illustrate in turn the utility of such diagrams for a multistep linear plan and a multibranch convergent plan taken from the database of examples compiled from Organic Syntheses given in the Supporting Information of the previous paper.1

Figure 1. Synthesis tree diagram of six-step linear sequence given in Scheme 1. Legend: black circle = input reagent; open circle = isolated intermediate; gray dot = final product; bold numbers = molecular weights of isolated intermediates; numbers in parentheses = molecular weights of input reagents; dotted line = connections between input materials and associated isolated intermediate for a given reaction.

Linear Plans

representing the preceding input materials and reaction intermediate.2 The centroid of a set of dots is determined by taking the midpoint of consecutive pairs of dots in a successive manner until one dot remains. For example, to find the centroid of a set of three dots with ordinate values 2, 5, and 6, we first take the midpoint of 2 and 5, which is (2 + 5)/2 = 7/2; then we take the midpoint of 5 and 6, which is (5 + 6)/2 = 11/ 2. Next, we take the midpoint of 7/2 and 11/2, which is [(7/2) + (11/2)]/2 = 18/4 = 4.5. The bolded numbers represent the molecular weights of all chemical species. The dotted lines connect all nodes so that any chemical species can be directly traced to its origin. Linear plans are characterized as having one branch containing N sequential steps and N − 1 isolated intermediates along the way before reaching the final target product. If we wish to determine the atom economy for the fourth step, for example, we see that the fourth open dot has a molecular weight equal to 174 and it is connected to the dots for monoperphthalic acid and the third intermediate with molecular weights 142 and 364, respectively. Hence, AE for the fourth step is 174/(142 + 364) = 0.344 (34.4%). If we want to know the kernel RME for that step, then we multiply the AE for that step by its reaction yield. Hence, the kernel RME = 0.344 × 0.92 = 0.316 (31.6%). The kernel RME represents material consumption that is restricted to the components required by the balanced chemical equation (i.e., excess reagents and auxiliary materials are omitted from the calculation of kernel RME). Hence, the waste that is counted consists of unreacted starting materials used in stoichiometric amounts and reaction byproducts in each reaction step. If we want to determine the stoichiometric mass of lithium aluminum hydride (LAH) required in this plan, we first note from the tree diagram that this reagent appears in two places; namely, steps 1 and 6. We set a basis scale for the final product (say, 1 mol) and then determine the mole scales of LAH by tracing the lines that connect the target product dot to each of the LAH dots and

Scheme 1 shows a 6-step linear sequence starting from cis-1,2cyclohexanedicarboxylic anhydride to cis-7,8-dimethylbicycloScheme 1. Six-Step Linear Plan from cis-1,2Cyclohexanedicarboxylic Anhydride to cis-7,8Dimethylbicyclo[4.2.0]oct-7-ene

[4.2.0]oct-7-ene4 written in a conventional manner. The corresponding synthesis tree diagram is given in Figure 1, which is based on fully balanced chemical equations for each step. From the diagram we see immediately that there are 6 steps along the x-axis and that 12 input reagents are required along the y-axis. The first input reagent begins at the origin. The reaction yields are displayed below the x-axis, and the counter for the input materials appears along the y-axis. The dark dots represent the input reagents including stoichiometric coefficients, the open dots are the isolated intermediates, and the last gray dot represents the final target product. Open dots are numbered sequentially to correspond to the reaction step numbers. The ordinate of any reaction intermediate dot is found by taking the centroid of the ordinates of the dots B



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Scheme 2. Convergent Synthesis Plan for (2R,3S,4S)-1-(tert-Butyldiphenylsilyloxy)-2,4-dimethyl-5-hexyn-3-ol

note the appropriate chains of reaction yields connecting the terminal dots. Hence, to produce 1 mol of P we need 1/0.29 = 3.45 mol (for step 6) and 1/(0.98 × 0.96 × 0.68 × 0.92 × 0.87 × 0.29) = 6.73 mol (for step 1), respectively, of LAH for a total of 10.18 mol or 38 × 10.18 = 386.9 g. The overall AE is found by tallying up the molecular weights of all input materials including stoichiometric coefficients, taking the reciprocal of the sum, and multiplying the result by the molecular weight of the target product. In this example, AE = 136/1604.9 = 0.085 (8.5%). The expression for the overall kernel RME is given by eq 1, again found by tracing appropriate nodes and noting the corresponding reaction yields that connect those nodes. The result is that the kernel RME is equal to 1.6%. (RME)overall =

136 18 + 38 + 64 0.29

364

78

+ 0.29 × 0.87 × 0.92 + 0.29 × 0.87 × 0.92 × 0.68 158 + 228.9 + 0.29 × 0.87 × 0.92 × 0.68 × 0.96 154 + 38 + 54 + 0.29 × 0.87 × 0.92 × 0.68 × 0.96 × 0.98 =

+

284 + 128 0.29 × 0.87

136 406.9 + 1633.0 + 1568.8 + 494.2 + 2553.4 + 1656.6

136 8312.2 = 0.016 =

(1)

Convergent Plans Figure 2. Synthesis tree diagram of the three-branch convergent plan given in Scheme 2. Legend: black circle = input reagent; open circle = isolated intermediate; gray dot = final product; bold numbers = molecular weights of isolated intermediates; numbers in parentheses = molecular weights of input reagents; dotted line = connections between input materials and associated isolated intermediate for a given reaction.

The versatility of tree diagrams is exemplified when they are applied to multiconvergent plans. Scheme 2 shows a convergent plan involving 3 synthesis branches and two points of convergence for the synthesis of (2R,3S,4S)-1-(tertbutyldiphenylsilyloxy)-2,4-dimethyl-5-hexyn-3-ol.5 The construction of the diagram is the same as that for a linear plan except that there is more than one synthesis branch to consider. Figure 2 shows the accompanying tree diagram for Scheme 2. We note that there are three branches in the plan that are color coded for clarity: black (first), red (second), and blue (third). We use the convention to call the first branch the main branch since this is the longest in the plan. The reaction yield labels are also color coded so that tracking of input materials in each branch is facilitated. Asterisk labels are used to designate intermediate nodes for the second and upper branches. Convergent plans are characterized as having B branches with at most B − 1 points of convergence. In the present example, B

= 3 and there is only one point of convergence which occurs in the last step. Unlike for linear plans, the metric of overall yield needs to be specified with respect to which branch it refers to. The usual reference branch for convergent plans is the one with the longest number of consecutive reaction steps, that is, the main branch. The set of reaction yields pertaining to that branch always appear in the first tier immediately below the xaxis. For example, if we wish to determine the stoichiometric mass of triethylamine (TEA) required in the plan, then we set a basis scale for the target product as before, say 1 mol, and use C



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The database of linear and convergent plans found in the Supporting Information of the previous paper1 contains fertile examples for students to practice drawing synthesis tree diagrams from traditional synthesis schemes and vice versa, as well as determining basic material efficiency metrics for individual reactions and for entire plans both at the kernel level and at the global level when all auxiliary materials are included in the calculations.

the set of reaction yields 0.76 (black) and 0.95 (blue) to work out the mole scale of TEA, 1/(0.76 × 0.95) = 1.39 mol. The required mass is then 101 × 1.39 = 139.9 g. Repeating the process for indium, we note from the tree diagram two input nodes for this material along the red branch. The mole scale of indium is [(2/3)/(0.76 × 0.92)] + [(4/3)/(0.76 × 0.92 × 0.85)] = 3.20 mol, and the required mass is then 114.8 × 3.20 = 367.0 g. A second example of a convergent plan is the type shown in Scheme 3 for the synthesis of the dyestuff aniline yellow6 from

Reaction Scheduling

Tree diagrams pertaining to convergent plans containing multiple branches may also be modified to devise a scheduling protocol for carrying out all reactions in the plan. It is readily observed that, with multiple branches, reactions in each branch will be run in parallel in simultaneous tracks. Question 12 in the Supporting Information (Part 1) illustrates this possibility with a numerical example. Instead of labeling the bottom of the grid with reaction yields, the labels now pertain to the length of time (i.e., hours) needed to carry out all operations for each reaction step (reaction time, isolation time, and purification time). Specifically, one can ask and answer a key question such as when should experimental work begin on a given arterial synthesis branch relative to the main branch so that production of the terminal intermediate product from the arterial branch will be ready at the same time as the production of an intermediate in the main branch so that the subsequent convergent reaction between the two branches can occur uninterrupted. Another question of interest is the determination of the overlap time between multiple branches. Such issues are important in project management and operations research. In fact, the modified tree diagrams are reminiscent of Gantt charts first developed in 19039−14 to optimize the scheduling of tasks in manufacturing processes. Chemical engineers use these tools routinely to schedule batch reactors in complex manufacturing networks.


Scheme 3. Convergent Synthesis Plan for Aniline Yellow

aniline which in turn is made from benzene.7,8 The types of reactions in the plan are familiar to students studying organic chemistry at a second year undergraduate level, making this an ideal example for students to gain a deeper understanding of synthesis performance and optimization with material efficiency metrics. This plan has three places where aniline is used as a reagent. The material efficiency for producing this hub intermediate is therefore central to the overall performance of the plan. Question 13 in the Supporting Information (Part 1) involves the determination of the mass split ratio of aniline in these three locations in the plan and an investigation of the effect of reaction yield improvement in each of the branches in which it appears on the overall kernel E-factor. Since the diazotization and subsequent coupling reactions both have yields of 100%, the mass split ratio of aniline is 1:1:1 corresponding to reaction stages 2, 3*, and 4**. This tells us that the mass of aniline produced is the same in all three stages in the plan. However, if the respective yields were to drop to 90 and 80% respectively, this ratio would change to 1.4:1.3:1, implying that a greater mass of benzene and nitric acid is needed in stage 1 compared to stage 3**. The greatest impact on minimizing waste occurs at the beginning of the synthesis, where the mole scale of aniline is highest, compared with the late stage appearance, where the mole scale is lowest. This key insight can be nicely reinforced by exploring what happens to the kernel E-factor for the entire plan to produce aniline yellow if the nitration reaction yield performance drops by 10% to 87% instead of the reported 97% under the same diazotization and coupling yields of 90% and 80%, respectively. In this case we find that the total mass of waste produced in the nitration reaction doubles from 118.9 to 204.3 g over the three reaction stages. A 10% decrease in the nitration yield translates into an increase of overall kernel E-factor by 0.74 units. When the diazotization and coupling reactions are fully optimized to 100% yields, this same 10% yield decrease in the nitration reaction translates into an increase of overall kernel E-factor by 0.36 units, half of the previous amount.

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EXAMPLE PROBLEMS This section is a series of worked-out problems pertaining to the determination of connecting relationships between overall material efficiency parameters for entire synthesis plans composed of two steps and their constituent individual reaction parameters. Results can be generalized to multistep plans (see Supporting Information Parts 1 and 2). Problem 1

Scheme 4 shows a two-step linear sequence with accompanying tree diagram where A, B, and C are reagents; I1 is the reaction Scheme 4. Two-Step Linear Plan

product from step 1; P is the reaction product from step 2; and Q1 and Q2 are reaction byproducts from both steps. The molecular weights of the chemical species are a, b, c, i1, p, q1, and q2, respectively. Make the following assumptions: (a) Both chemical equations are stoichiometrically balanced. (b) No excess reagents are used. D



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(c) Ignore all auxiliary materials such as reaction solvents, catalysts, workup materials, and purification materials. Determine the atom economy for each reaction step and for the overall two-step reaction sequence. Write out an algebraic expression for overall AE in terms of (AE)1 and (AE)2.

Scheme 5. Synthesis Plans for Citral, Bicyclo[4.2.0]oct-7ene, and Bullvalene Which Involve Rearrangement and Elimination Type Reactions

Solution to Problem 1

The relevant atom economy expressions are given by eqs 2 to 4. Step 1:

(AE)1 =

i1 a+b

(2)

Step 2:

(AE)2 =

p i1 + c

(3)


Overall:

(AE)overall =

p a+b+c

(4)

Beginning with eq 4 and making appropriate substitutions for p and a + b we find that p (AE)overall = a+b+c (i + c)(AE)2 = 1 i 1 +c (AE) 1

⎡ i +c ⎤ = (AE)1(AE)2 ⎢ 1 ⎥ ⎣ i1 + c(AE)1 ⎦

atom economy is analogous to that shown in eq 5. Using this property, one can examine any named rearrangement or elimination reaction in organic chemistry and construct a twostep sequence analogous to those shown in Scheme 4 where the first step pertains to the synthesis of the precursor in the following rearrangement or elimination reaction. Scheme 6 shows some well-known examples that students encounter in their traditional organic chemistry course where they can determine atom economies for each step and for the entire sequence. The Supporting Information (Part 2) contains an extensive compilation of more complex sequences taken from the recent literature which involve consecutive combinations of rearrangements and eliminations.

(5)

We note from eq 5 that the overall AE, in general, is not the multiplicative product of the individual reaction atom economies. In this example, however, equality is satisfied if and only if the molecular weight of C is zero; that is, if the second reaction does not involve any other added reagent besides intermediate I1. The existence of a simple multiplicative relationship between individual step atom economies and overall atom economy requires the second reaction to be either a rearrangement, if the molecular weights of I1 and P are identical, or an elimination reaction if the molecular weight of I1 exceeds that of P. Real example syntheses documented in the literature, shown in Scheme 5, which have this unique property are the syntheses of citral,15−17 bicyclo[4.2.0]oct-7-ene,18,19 and bullvalene.20−22 The two-step citral synthesis first involves an acid catalyzed condensation between 2 equiv of prenol and 1 equiv of prenal. The second step involves a complex sequence of transformations: elimination of one prenol unit followed by a Claisen rearrangement, followed by a bond rotation, and finally followed by a final oxy-Cope rearrangement. The synthesis of bicyclo[4.2.0]oct-7-ene involves an oxidation in the first step, a thermal elimination in the second step, and an intramolecular [2 + 2] photochemical cyclization in the third step. The synthesis of bullvalene begins with a multicomponent reaction between 4 acetylene molecules producing cyclooctatetraene, followed by thermal dimerization, and followed by a photochemical retro-[2 + 2] cyclization reaction. In all three cases, the overall atom economy for each synthesis plan is the multiplicative product of the atom economies for each step. Essentially this property is true for a given synthesis plan only when the first step can be any kind of reaction and when all reactions following it must be either a series of rearrangements or eliminations, or combination of these two kinds of reactions: no other kinds of reactions are allowed. The moment another reagent is introduced in any step following the first one, this property no longer holds and the general expression for overall

Problem 2

From Scheme 4, suppose that 1 mol of final product P is desired from this two-step synthesis. Assume that all the mass of the intermediate I1 collected from step 1 is committed as reactant in step 2. Determine the masses of reagents A, B, and C that are required to carry this out. Based on these results determine the kernel reaction mass efficiency (RME), kernel Efactor, and kernel process mass intensity (PMI) for each reaction step and for the overall two-step sequence. Solution to Problem 2

If 1 mol of P is required, then from the synthesis tree diagram we find that 1/ε2 moles of C, 1/ε1ε2 moles of A, and 1/ε1ε2 moles of B are needed. The respective masses of the reagents are therefore mA = a/ε1ε2, mB = b/ε1ε2, and mC = c/ε2. The mole scale of intermediate I1 is 1/ε2 moles, and its mass is i1/ε2. Table 1 summarizes the relationships for the kernel RME, kernel E-factor, and kernel PMI for each step and for the overall plan. We can demonstrate how the overall RME is related to the individual reaction RME expressions according to eqs 6 to 8. iε Step 1: (RME)1 = 1 1 (6) a+b E



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We observe that the overall RME is the multiplicative product of the individual reaction RMEs if and only if the molecular weight of C is zero. Hence, the special condition discussed above in Problem 1 is the only scenario that makes this statement true. In fact, under this condition the overall RME can also be written as the product of the overall atom economy and overall yield. Similarly, we can demonstrate how the overall PMI is related to the individual reaction PMIs according to eqs 10 to 12.

Scheme 6. Example Two-Step Reaction Sequences Where in Each Case Step 1 Pertains to the Synthesis of a Precursor and Step 2 Pertains to a Named Rearrangement or Elimination Reaction

Step 1:

(PMI)1 =

a+b i1ε1

(10)

Step 2:

(PMI)2 =

i1 + c pε2

(11)

(PMI)overall =


Overall:

a + b + cε1 pε1ε2

(12)

Since PMI is the reciprocal of RME, the overall PMI in terms of the reaction PMIs can also be written at once from eq 9 according to eq 13. ⎡ i + c(PMI) −1 ⎤ 1 ⎥ (PMI)overall = (PMI)1(PMI)2 ⎢ 1 i1 + c ⎦ ⎣

(13)

By analogy with RME, overall PMI is the multiplicative product of the individual reaction PMIs if and only if the molecular weight of C is zero. The respective kernel E-factor expressions are given by eqs 14 to 16.

Step 2:

Overall:

pε2 (RME)2 = i1 + c

(RME)overall =

E1 =

a+b −1 i1ε1

(14)

Step 2:

E2 =

i1 + c −1 pε2

(15)

(7)

p a+b ε1ε2

Step 1:

+

c ε2

pε1ε2 a + b + cε1

=

(8)

Eoverall =

Overall:

After making appropriate substitutions, we can rewrite overall RME in terms of the individual reaction RMEs according to eq 9, which is analogous to the form of eq 5 for overall atom economy. ⎤ ⎡ i1 + c (RME)overall = (RME)1(RME)2 ⎢ ⎥ ⎣ i1 + c(RME)1 ⎦

a+b ε1ε2

+ p

c ε2

−1=

a + b + cε1 −1 pε1ε2 (16)

It can be shown (see Supporting Information, Part 1, Q1-part 6) that the overall kernel E-factor is related to the individual kernel E-factors according to eq 17. Eoverall =

(9)

(E1 + 1)(E2 + 1)i1 + c(E2 + 1) −1 i1 + c

(17)

Table 1. Summary of RME, E-Factors, and PMI Kernel Parameters Pertaining to Scheme 4 E

Step 1

mA + mB − mI1 mI1

2

mI1 + mC − mP mP

Overall

=

=

a+b ε1ε2 i1 ε2 i1 + c ε2

p

mA + mB + mC − mP = mP

PMI = E + 1

−1=

a+b −1 i1ε1

mA + mB = mI1

a+b ε1ε2 i1 ε2

−1=

i1 + c −1 pε2

mI1 + mC

i1 + c ε2

a+b ε1ε2

+ p

c ε2

mP

=

p

mA + mB + mC mP

−1

=

F

a+b ε1ε2

+ p

c ε2

=

a+b i1ε1

=

i1 + c pε2

RME = 1/PMI mI1 iε = 11 mA + mB a+b

pε2 mP = mI1 + mC i1 + c

mP mA + mB + mC p = a+b c + ε εε 1 2

2



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Table 2. Summary of General and Reduced Relationships for Overall AE, RME, PMI, and E-Factor for a Two-Step Sequencea General Relation

Condition

Chemical Implication

⎡ i +c ⎤ (AE)1(AE)2 ⎢ 1 ⎥ ⎣ i1 + c(AE)1 ⎦

Reduced Relation (AE)1(AE)2

c=0

Step 1: any reaction Step 2: rearrangement or elimination

⎡ ⎤ i1 + c (RME)1(RME)2 ⎢ ⎥ ⎣ i1 + c(RME)1 ⎦

(RME)1(RME)2

c=0

As above

⎡ i + c(PMI) −1 ⎤ 1 ⎥ (PMI)1(PMI)2 ⎢ 1 i1 + c ⎣ ⎦

(PMI)1(PMI)2

c=0

As above

(E1 + 1)(E2 + 1)i1 + c(E2 + 1) −1 i1 + c

E1 =

c=0 E2 = 0 ε1 = 1 ε2 = 1

Step 1: any reaction Step 2: rearrangement only

a+b −1 p

or a + b + cε1 −1 pε1ε2


a

Pertaining to kernel conditions where excess reagent and auxiliary material consumption are neglected.

Moreover, we can also show that the sum of the kernel Efactors for steps 1 and 2 is not equal to the overall kernel Efactor as is readily observed in eq 18. E1 + E2 = =

Scheme 7. (A) Linear Plan Involving 3 Steps. (B) Convergent Plan Involving 3 Steps and One Point of Convergence

i +c a+b −1+ 1 −1 i1ε1 pε2 (a + b)pε2 + (i1 + c)i1ε1 −2 pε1ε2i1

≠ Eoverall

(18)

An elegant mathematical proof by contradiction is given in the Supporting Information (Part 1, Q1-part 6). Equality is satisfied, however, under a very strict set of conditions; namely, both reaction yields must be 100% and the second reaction must be a catalytic, thermal, or photochemical rearrangement. Effectively, E2 and c in eq 15 are set equal to zero so that Eoverall in eq 17 reduces to E1. These conditions are even more restrictive than those for the validity of multiplicative products for AE, RME, and PMI discussed above for a two-step sequence. Table 2 summarizes all of the key results in compact form. In the Supporting Information the results of Problems 1 and 2 are generalized for an N-step linear plan (see Q1-part 7).

Problem 3 (Part B)

For each case make a plot of the function containing the reaction yield variables versus reaction yield (0 < ε < 1). Investigate conditions when the function for the linear plan exceeds that for the convergent plan and vice versa. What about conditions when the value of the function is the same for both plans?

Problem 3 (Part A)

Solution to Problem 3 (Part B)

The material efficiency performance of a 3-step linear plan is compared against a 3-step convergent plan containing one point of convergence. Each plan involves 4 reagents. For simplicity all reaction yields are equal and all molecular weights of reagents are equal. The corresponding synthesis tree diagrams for both plans are shown in Scheme 7. If the molecular weights of A and P are a and p, respectively, and we neglect auxiliary materials and excess reagent consumption, write out the corresponding expressions for the overall RME for each plan (assume 1 mol of P is produced in each plan).

Figure 3 shows a plot of the functions f(ε) and g(ε) given by eqs 21 and 22. It appears that the function for the convergent plan is always greater than that for the linear plan over the interval 0 < ε < 1. The two functions are equal for the limits ε = 0 and ε = 1.

Solution to Problem 3 (Part A)

The RME expressions are given by eqs 19 and 20, respectively. Linear plan:

RME =

Convergent plan:

p 2a ε3

+

RME =

a ε2

+

a ε

(19)

2a ε2

(20)

p 2a ε2

+

Figure 3. Plots of functions f(ε) and g(ε) according to eqs 21 and 22, respectively. G


Journal of Chemical Education ⎡ ⎛ p ⎞⎢ 1 ⎜ ⎟ RME = ⎝ a ⎠⎢ 23 + 12 + ⎣ε ε


RME =

⎡ ⎛ p ⎞⎢ 1 ⎜ ⎟ ⎝ a ⎠⎢ 22 + ⎣ε

⎤ ⎥ = ⎛⎜ p ⎞⎟f (ε) 1 ⎥ ⎝a⎠ ε ⎦

⎤ ⎥ = ⎛⎜ p ⎞⎟g (ε) 2 ⎥ ⎝a⎠ ε2 ⎦

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plans23,24 for the pharmaceutical bortezomib using the REACTION and SYNTHESIS spreadsheets described in the previous paper.1 The analysis takes into account all auxiliary materials and excess reagent consumption. Both plans begin from (S)-pinanediol-2-methylpropane-1-boronate. The key results are summarized in Table 3. The linear plan involves 4

(21)

(22)

Table 3. Summary of Material Efficiency Metrics for Linear and Convergent Plans to Bortezomib23,24

The statement that convergent plans are always more material efficient than linear ones is true under a very strict set of conditions; namely, that both plans have the same number of steps, all reaction yields are identical in magnitude, and all molecular weights of reagents are identical in magnitude. Clearly, these conditions do not correspond to a realistic and practical scenario. In general, reaction yields will be different for each step, and more importantly the molecular weights of input reagents will definitely be different. This implies that no general statement can be made about the performances of linear versus convergent plans to a common target molecule with any degree of definiteness. One needs to evaluate the material efficiency metrics of both plans before drawing any meaningful conclusions. A simple numerical supplementary exercise may be posed as in Problem 3 (Part C) to illustrate this point more concretely. Suppose the magnitudes of the molecular weights for A and P are 50 and 100 respectively. For the convergent plan, if the reaction yields for steps 1, 1*, and 2 are 80, 90, and 80% respectively, what is the overall RME? Find a combination of reaction yields for the linear plan, other than the trivial combination of 100% in each step, that would make it have a higher overall RME than that calculated for the convergent plan. Solution to Problem 3 (Part C)

According to eq 22 the overall RME of the convergent plan is 0.3388 (33.88%). The expression for overall RME of the linear plan is given by eq 23. +

50 ε2ε3

+

50 ε3

Convergent

Winner

4 35% 15% 524 523 12.1 18.5 492.2 129.6 1.8 297.6 63.2

4 27% 18% >644 >643 14.8 42.2 >585.9 157.9 13.6 >329.4 >85.1

Same Linear Convergent Linear Linear Linear Linear Linear Linear Linear Linear Linear

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100 50 + 50 ε1ε2ε3

Linear

reaction steps and 17 input materials whereas the convergent plan involves the same number of reaction steps over two branches and 15 input materials. The kernel overall RME which takes into account only reaction yield and molecular weight parameters is readily determined from the overall E-kernel values according to RME = 1/(E + 1). The resulting overall kernel RME values for the linear and convergent plans are 7.8% and 6.3%, respectively. In terms of overall AE, the convergent plan wins by about 3%. In terms of overall yield, the linear plan wins by 8% yet the linear plan has 4 consecutive steps versus the convergent plan which has 3 consecutive steps. Over the 12 key material efficiency metrics, the linear plan prevails over the convergent one in all but two parameters, which is a very strong endorsement for that plan.

Problem 3 (Part C)

RME =

Parameter Number of reaction steps Overall yield (longest branch) Overall AE Overall PMI Overall E-factor E-kernel E-excess E-auxiliary E-solvents E-catalyst E-workup E-purification

CONCLUDING REMARKS We have presented in this paper a series of pedagogically important exercises that showcase the versatility and power of using material efficiency metrics to critique and analyze synthesis plan performance. These ideas may be applied to any kind of synthesis plan regardless of complexity. We have also filled a gap in the literature on providing rigorous proofs for connecting relationships between individual reaction metrics and their global counterparts. The teaching of green chemistry principles may be significantly enhanced and made more objective as a consequence of incorporating quantitative material efficiency metrics calculations in green chemistry courses. The visual aid of a synthesis tree diagram also facilitates this deeper understanding and gives students the confidence to tackle more complex synthesis plans to interesting and important target molecules that are relevant to industrial chemical processes, natural products, pharmacology, and biology. Upon mastery of the concepts highlighted by the problem set exercises, students will be able to apply metrics analyses to any literature plan from a journal article or patent and then identify which reactions should be targeted for optimization. Once a particular optimization strategy is chosen, again metrics analyses can be used to determine the impact of that strategy on individual reaction performance as well as on

(23)

The task is to find a combination of three reaction yield values that make the RME function for the linear plan exceed 0.3388. Some representative combinations that satisfy this criterion are as follows: ε1 = 0.9, ε2 = 0.9, ε3 = 0.9 so that RME = 0.3930; or ε1 = 0.8, ε2 = 0.9, ε3 = 0.9 so that RME = 0.3590; or ε1 = 0.8, ε2 = 0.85, ε3 = 0.95 so that RME = 0.3581. To be absolutely fair, head-to-head evaluations of linear versus convergent synthesis plans to a common target molecule should begin from a common starting material as well. Such an analysis eliminates inevitable biases due to so-called apples to oranges comparisons; however, the purity of this comparison may not be practically feasible if there exists no documented chemical pathway that connects the structures of starting materials in the first steps of both plans. Nevertheless, there are plenty of examples in the literature of linear synthesis plans that outperform convergent ones to the same target molecule. Both instructors and students are encouraged to discover such examples by performing material efficiency metrics analyses using the methods described in the previous paper.1 One such example is given in the Supporting Information (Part 1, Q17), which contains complete metrics evaluations of two synthesis H



Article

(18) Cope, A. C.; Bumgardner, C. L. Cyclic Polyolefins. XL. cis-cisand cis-trans-l,3-Cyclooctadiene from Cycloocten-3-yldimethylamine. J. Am. Chem. Soc. 1956, 78, 2812−2815. (19) Moore, W. R.; Moser, W. R. The Reaction of 6,6Dibromobicyclo[3.l.0]hexane with Methyllithium. Efficient Trapping of 1,2-Cyclohexadiene by Styrene. J. Org. Chem. 1970, 35, 908−912. (20) Reppe, W.; Schlichting, O.; Klager, K.; Toepel, T. Cyclisierende Polymerisation von Acetylen I. Ann. Chem. 1948, 560, 1−92. (21) Schrö der, G. Die Eigenschaften zweier dimerer Cyclooctatetraene vom Schmp. 53 und 76°. Chem. Ber. 1964, 97, 3131− 3139. (22) Schröder, G. Synthese und Eigenschaften von Tricyclo[3.3.2.0 4.6 ]decatrien-(2.7.9). Chem. Ber. 1964, 97, 3140−3149. (23) Ivanov, A. S.; Zhainina, A. A.; Shishkov, S. V. A convergent approach to synthesis of bortezomib: the use of TBTU suppresses racemization in the fragment condensation. Tetrahedron 2009, 65, 7105−7108. (24) Pickersgill, I.; Bishop, J.; Koellner, C.; Gomez, J. M.; Geiser, A.; Hett, R.; Ammoscato, V.; Munk, S.; Lo, Y.; Chiu, F. T.; Kulkarni, V. R. Synthesis of boronic ester and acid compounds. WO 2005097809 (Millenium Pharmaceuticals, 2005).

the entire synthesis plan. We hope to increase our repertoire of examples and exercises and report on these in due course. We also encourage feedback from instructors and students on their use in classroom and laboratory instruction.

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ASSOCIATED CONTENT

S Supporting Information *

The Supporting Information is available on the ACS Publications website at DOI: 10.1021/acs.jchemed.5b00059. Part 1, problem set with worked solutions on various green metrics exercises; Part 2, summary of literature examples of sequential rearrangement and elimination reactions (ZIP)

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AUTHOR INFORMATION


Corresponding Author

*E-mail: [email protected]. Notes

The authors declare no competing financial interest.

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REFERENCES

(1) Andraos, J.; Hent, A. Simplified Application of Material Efficiency Green Metrics to Synthesis Plans: Pedagogical Case Studies Selected from Organic Syntheses. J. Chem. Educ. 2015, DOI: 10.1021/ acs.jchemed.5b00058. (2) Andraos, J. On Using Tree Analysis to Quantify the Material, Input Energy, and Cost Throughput Efficiencies of Simple and Complex Synthesis Plans and Networks: Towards a Blueprint for Quantitative Total Synthesis and Green Chemistry. Org. Process Res. Dev. 2006, 10, 212−240. (3) Andraos, J. Unification of reaction metrics for green chemistry: application to reaction analysis. Org. Process Res. Dev. 2005, 9, 149− 163. (4) Photis, J. M.; Paquette, L. A. 1,2-Dimethylcyclobutenes by reductive ring contraction of sulfolanes. Organic Syntheses 1977, 57, 53. (5) Johns, B. A.; Grant, C. M.; Marshall, J. A. Synthesis and utilization of indium (I) iodide for in situ formation of enantioenriched allenylindium reagents and their addition to aldehydes. Organic Syntheses 2002, 79, 59. (6) Vogel, A. Textbook of Practical Organic Chemistry; Longman: London, 1948; p 599. (7) Faith, W. L.; Keyes, D. B.; Clark, R. L. Industrial Chemicals, 3rd ed.; Wiley: New York, 1966; pp 101, 541. (8) Shreve, R. N. Chemical Process Industries, 3rd ed.; McGraw-Hill Book Co.: New York, 1967; p 812. (9) Gantt, H. L. A graphical daily balance in manufacture. Trans. Am. Soc. Mech. Eng. 1903, 24, 1322−1336. (10) Gantt, H. L. Organizing for Work; Harcourt, Brace, and Howe: New York, 1919. (11) Clark, W. The Gantt chart: a working tool of management; The Ronald Press Company: New York, 1923. (12) Ku, H.; Rajagopalan, D.; Karimi, I. Scheduling in batch processes. Chem. Eng. Prog. 1987, 83 (8), 35−45. (13) Musier, R. F. H.; Evans, L. B. Batch process management. Chem. Eng. Prog. 1990, 86 (6), 66−77. (14) Wilson, J. M. Gantt charts: A centenary appreciation. Eur. J. Oper. Res. 2003, 149, 430−437. (15) Leimgruber, W.; Valentine, D. H., Jr. Verfahren zur Herstellung von Aldehyden. DE2411530 (Hoffmann-LaRoche, 1974). (16) Wegner, G.; Keibel, G.; Therre, J.; Aquilia, W.; Fuchs, H. Continuous Method for Producing Citral. WO 2008037693 (BASF, 2008). (17) Teissere, P. J. Chemistry of Fragrant Substances; VCH: New York, 1994; p 133. I


Useful Material Efficiency Green Metrics Problem Set Exercises for

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