Solucionario Modern Drug Discovery

Erland Stevens Medicinal Chemistry and Drug Discovery – Solutions 1

Solutions to End-of-Chapter Questions

Chapter 2Question 1

Approving raloxifene for a second use would very likely require less than 7-8 years. The safety data for the original Phase I trials (osteoporosis) would not need to be repeated. Since Phase II trails involve affected (diseased) patients, a new round of Phase II trails for the breast cancer treatment would be unavoidable. The same would be true for all Phase III trials.

Question 2

Extensive liver damage : This problem would likely be discovered in pre-clinical (animal) trials.

Insufficient efficacy : For most drugs, efficacy is confirmed in Phase II trials.

Causes arthritis after several years with chronic use : Even Phase III trials normally run a few years at the longest, so very long term problems likely would not be found until some years after the drug had been launched.

Interacts with certain cardiovascular drugs : Drug-drug interactions are normally uncovered during Phase III trials.

Question 3

For this problem it is helpful to determine the cumulative costs for each stage of development before answering the separate questions. So, here are the development costs up to and including different stages of a drug candidate’s advancement. The numbers are based on a total cost of US$800 million and the percentages in Figure 2.16.

Pre-clinical (26%) $208 million

Phase I (33%) $264 million

Phase II (43%) $344 million

Phase III (69%) $552 million

Market (100%) $800 million

At this point, just add up the numbers. There are five compounds total. One compound dies in Phase I ($264 million), two compounds die in Phase II ($344 million × 2), one compound dies in Phase III ($552 million), and one compound reaches the market ($800 million). Those numbers give a grand total of approximately $2.3 billion as the development costs for the five compounds.

Question 4

This question builds off the previous one, so most of the work is already done. A blockbuster drug with 12 years of sales will bring in US$12 billion in sales, and therefore $2.4 billion in profit. A compound that only advances to Phase I costs $264 million, so a blockbuster can cover the losses incurred by approximately 9 Phase I busts. Since the blockbuster cost $800 million to develop, its true profit would be closer to $1.6 billion – enough profit to cover 6 compounds that failed in Phase I.


Similarly, compounds that advance to Phase III cost $552 million. A blockbuster drug with $2.4 billion in profit can offset between 4 and 5 Phase III failures. Accounting for the costs of the blockbuster drug allows only about 3 Phase III failures to be covered.

Question 5

True category creators can be hard to conceive because they are, by definition, innovative. Innovation requires one to know something about the existing pharmaceutical market and deficiencies in available products.

One example of a category creator is orally-available insulin. Insulin taken in a pill form would immediately become the preferred method for diabetics to receive insulin. Note that patients can get insulin now, but the method of administration is less than satisfactory.

Other conditions that are currently manageable to some degree but not to a satisfactory level for some or advanced patients include asthma, Alzheimer’s disease, depression, multiple sclerosis, Crohn’s disease, Parkinson’s disease, and many types of addictive behavior.

Chapter 3Question 1

As it is eliminated, methotrexate (3.45) concentrates in the urine and can precipitate out of solution. Methotrexate contains two carboxylic acids. If the pH of the blood, and subsequently the urine, is drops to low, then a larger fraction of the acid groups in methotrexate will be protonated, and the molecule will be neutralized. The neutral form of methotrexate is less soluble and at risk for precipitating from solution. Raising the pH of the urine with bicarbonate shifts the equilibrium toward the deprotonated form(s) of the drug. Deprotonation places negative charges on the structure, increases water solubility, and minimizes the chance of drug precipitation in the urine.

Question 2

Remember that plasma is the non-cellular fraction of whole blood. With a blood-to-plasma ratio of less than 1 (0.86), alfentanil has a lower concentration in whole blood than plasma. When the cells in whole blood are opened and their contents mix with the plasma, the drug concentration drops from a relative value to 1.0 to 0.86. This process is shown below.


Based upon Figure 3.15, cells constitute 46% of whole blood by volume, and plasma makes up 54% of the volume of blood. This is all the necessary information to solve what amounts to a M1V1 = M2V2 dilution problem in a general chemistry class. A key point is to remember that the amount of drug in the volume is constant whether it is divided between the cells and plasma or mixed uniformly within entire sample volume.

One can determine the amount of drug in a volume by the product of the drug’s concentration (C) and its volume of occupancy (V). For this problem, all variables are known except Ccells, the concentration of the drug within the cells.

The relative concentration of alfentanil within the cells of blood to plasma is 0.70.

Question 3

Drugs enter cells of the blood the same way that drugs enter other cells of the body – primarily through passive diffusion. Epoetin alpha is a massive drug with a MW of 18,000. Molecules of this size do diffuse across the membrane. Epoetin alpha would therefore not be expected to enter the cells of the blood. Since the cellular concentration of epotein alpha should be near 0, the blood-to-plasma ratio of epoetin alpha should be less than 1, likely close to 0.5.

Question 4

Figure 3.1 lists the amount of water in the body as 60% of total body mass. For a 70-kg person, that works out to 42 kg. That 60% is split among oxygen and hydrogen based upon the respective mass and number of the atoms in water.


If the calculated masses of oxygen and hydrogen from water are removed from the masses in Table 3.1, then the remaining masses can be used to calculate the water-free mass percentages for oxygen, carbon, hydrogen, and nitrogen. The other elements in the body must be kept in the calculation in some form (even just in an “other” category) so that the final percentages work out properly.

O, C, H, and N content of a 70-kg person

Iincluding Water Water Excluding Water

element mass (kg)

% mass (kg)

mass (kg) %

Oxygen (O) 43 61.4 37.3 5.7 20.4

Carbon (C) 16 22.9 16 57.1

Hydrogen (H)

7.0 10.0 4.7 2.3 8.2

Nitrogen (N) 1.8 2.6 1.8 6.4

Others 2.2 3.1 2.2 7.9

total 70 100.0 42 28 100.0

When water is excluded, carbon is the most predominant element in the body as one might expect from an organism that consists largely of fats, carbohydrates, and proteins.

Question 5

The mass balance of the cited study is 84.8% and calculated by adding all the listed percentages. The drug in question would appear to be well absorbed. Any drug found in the urine (65.9%) and tissues (8.8%) must have been absorbed, so the minimal absorption percentage must be nearly 75%. Absorption of 75% is a reasonably good value for a drug. Furthermore, the amount of drug found in the cage rise (4.9%) could be from any number of elimination pathways such as hair, sweat, or saliva. For a drug to be eliminated through any of these pathways, it must first be absorbed. The absorption of the drug is therefore likely nearly 80%.

The drug found in the feces (5.5%) may have been absorbed and the excreted in the bile or simply passed through the rat. Detection of an unchanged drug in the feces is typically a sign that the drug was never absorbed.

Question 6

Polar surface area is indeed raised by the presence of oxygen and nitrogen atoms (the common hydrogen bond acceptors) and O-H and N-H bonds (the common hydrogen bond donors). In addition to hydrogen bond donors and acceptors, polar surface area includes all polarized bonds, including carbon-halogen bonds. Perhaps surprisingly, carbon-carbon π-bonds can be very highly polarized if they bear strong enough electron-donating and/or electron-withdrawing groups.

Question 7


The kidneys filter waste from the blood that passes through the kidneys. Proteins in the blood are too large to pass through the initial filter in the glomulerus and therefore are not generally removed by kidneys or observed in urine. The presence of protein in the urine is associated with a number of physiological conditions including various kidney diseases, dehydration, and even very vigorous exercise.

Question 8

Here is how the three compound stack up to the Lipinski’s Rules for CNS drugs in Table 3.11.

Donezepil MW 379 g/molH-bond donors 0H-bond acceptors 4 (one for each oxygen and nitrogen)

Galantamine MW 287 g/molH-bond donors 1 (alcohol)H-bond acceptors 4 (one for each oxygen and nitrogen)

Rivastigmine MW 250 g/molH-bond donors 0H-bond acceptors 2 (one for each oxygen and nitrogen, excluding the amide nitrogen)

All three Alzheimer’s drugs are (not surprisingly) within the specifications of Lipinski. Some common software packages, including more recent versions of ChemDraw, can provide clog P values. The clog P values for the three compounds are 4.01, 1.41, and 2.36, also all in the expected ranges for CNS drugs.

Compounds that violate Lipinski’s Rules for CNS drugs are expected not to diffuse passively across the blood-brain barrier. Therefore, if a drug has non-ideal properties but still crosses into the brain, then the compound likely hitches a ride into the CNS through a facilitated or active transport process. Lipinski’s Rules only predict passive diffusion.

Question 9

The upper portions of the small intestine are neutral to slightly alkaline. This environment is idea for many oral drugs, which often contain weakly basic amines. Under these conditions, a reasonable fraction of the amines are present in their neutral, basic form and more readily enter into lipophilic membranes for absorption from the intestinal lumen. A neutral-to-acidic environment such as the rectum is more ideally suited for weakly acidic drugs. Under weakly acidic conditions, weak acids tend to be present in their neutral, acid form for improved absorption.

In the small intestine with its higher pH, acidic drugs (drugs containing acidic functional groups such as carboxylic acids) will be deprotonated to a greater degree than in the rectum.

Question 10

With a density near 1 g/mL, a 70-kg patient has a volume of approximately 70 L. According to Table 3.1, the amount of iron in the body is only 4.2 g, corresponding to 0.075 mol (Fe MW = 55.85 g/mol). Therefore, the concentration of iron in the body is approximately 1.1×10-3 M (0.075 mol/70 L), or 1.1


mM. On a purely mass basis, the concentration of iron in the body is approximately 60 ppm (1,000,000 × 4.2 g / 70,000 g).

The body contains only 15 mg (0.015 g) of selenium (MW 78.96 g/mol), which is just 1.9×10-4 mol for a concentration of 2.7×10-6 M, or 2.7 μM. On a mass basis, selenium is present in the body at a concentration of only 0.21 ppm.

Chapter 4Question 1

Scheme 4.7 is shown below.

First step, forward reaction:

First step, reverse reaction:

Second step, forward reaction:

Units of Km:

(4.12)

Question 2

The key to this question is to remember the impact of each inhibitor upon an enzyme. A competitive, reversible inhibitor decreases Km without affecting Vmax of an enzyme-substrate combination. The inhibited and uninhibited reach the same Vmax, but the inhibited system requires a higher [S] to get there.


[S]

VV max

1/2 V max

uninhibitedinhibited

K m (uninhibited)

K m (inhibited)

A pure non-competitive, reversible inhibitor does the opposite – Km stays constant while Vmax decreases. Both systems reach ½Vmax at the same [S].

[S]

V

1/2 V max

(uninh)

1/2 V max

(inh)

K m

uninhibited

inhibited

An uncompetitive, reversible inhibitor decreases both Km and Vmax by the same factor. That is, if the inhibitor decreases the Km by half, then Vmax likewise decreases by half. The graph below is not extended to high [S] values so that the key [S] region near Km is clearer.


[S]

V

1/2 V max

(uninh)

1/2 V max

(inh)

K m

(uninh)

K m

(inh)

inhibited

uninhibited

Question 3

This question is just an algebraic exercise on the Michaelis-Menten equation. For the first part, replace V with 0.1Vmax and solve for [S]. To reach 0.1Vmax, [S] must be equal to 1/9 of Km (or 0.11Km).

(4.11)

For the second part, replace V with 0.95Vmax and solve for [S]. To reach 0.95Vmax, [S] must be equal to 19Km.


Question 4

The Lineweaver-Burk equation is given below.

(4.13)

This equation has an x-intercept of −1/Km and a y-intercept of 1/Vmax. Based on these two points, the slope of the line can be calculated.

Question 5

As a competitive inhibitor of fumarase, compound 4.c binds in the active site of the enzyme. Structurally 4.c resembles both the substrate (fumaric acid, 4.a) and the product ((S)-malic acid, 4.b). If all three compounds are viewed along side each other in a series of Newman projections, the overlapping of key functional groups is apparent in two ways. First, aziridine 4.c nearly preserves the trans orientation of the carboxylic acids in both 4.a and 4.b. Second, the NH in the aziridine ring matches the position of the OH in (S)-malic acid when the acid groups are held in a trans orientation. Compound 4.c resembles both the product and substrate and is able to bind the active site of fumarase and inhibit its reactivity.

Predicting the biological activity of stereoisomers is very difficult. Stereoisomers are clearly different based on configurational assignments of chiral centers, but they are also similar – same molecular formula, functional groups, and connectivity. So, should stereoisomers have different or similar activity? In the end, the only way to be sure is through testing. Regardless, since 4.c aligns so closely with both the substrate and product of fumarase, it would be reasonable to expect the enantiomer of 4.c to match less well and therefore serve as a weaker inhibitor.

Examination of two different views of the Newman projections of the enantiomer of 4.c shows the impact of the stereochemical changes. In the first Newman projection, the enantiomer still has nearly trans carboxylic acids, but the front acid is on the left instead of the right. Likewise, the back acid group is on the right instead of the left. In the second Newman projection, the acid groups are in the right position, but now the NH no longer has the same orientation as the OH in (S)-malic acid. In the enantiomer of 4.c, some spatial elements can be preserved at the cost of others. Therefore, 4.c would likely not be as potent of an inhibitor of fumarase. The only way to know the activity of the enantiomer with certainty would be to make and test the compound.


Question 6

This is a simple “plug and chug” problem. The only trick is to make sure the concentration of Ki is used in molar units (not millimolar).

(4.a)

The energy of dissociation has a positive sign, so dissociation is disfavored under the tested conditions (i.e. binding is favored).

Question 7

Dissociation of an inhibitor-enzyme complex is entropically favored because the free enzyme and inhibitor are more disordered than the complex. ΔS for dissociation is therefore a positive number. The −TΔS term, however, would be negative. Therefore, in order for ΔG to be positive, ΔH must have an even greater positive magnitude than ΔG to offset the impact of a negative −TΔS term.

None of these ideas is a surprise. If an inhibitor binds strongly, then it can form strong intermolecular forces with the enzyme. Entropy always favors dissociation and disfavors binding. Therefore, in order for binding to be favored overall, the intermolecular forces (enthalpic contributors) must be strong enough to offset entropic factors.

Before finishing this question, a word of warning must be shared. Entropy can be very important in the association and dissociation of drugs with their targets. This question ignores the impact of solvent (water). Lots of water molecules can be involved during binding and dissociation, and those many water molecules can collectively have a big impact on entropy. This idea gets fleshed out more completely in Chapter 9.

Question 8

Before answering this question, one must convert all the [S] and V data into 1/[S] and 1/V values for the Lineweaver-Burk plot. The Lineweaver-Burk graph of the [S] data is shown below with the best fit line.


0

0.05

0.1

0.15

0.2

0.25

0.3

-0.05 0 0.05 0.1 0.15 0.2 0.25

1/[S] (mM-1)

1/V

(s/

mM

)

The r2 of the line is 0.99. Based on the y-intercept of the line, Vmax is 66 s-1. Km works out to be 76 mM. The high linearity of the data seems to indicate that the enzyme follows Michaelis-Menten kinetics.

Question 9

Based upon the induced fit model, when a compound binds an enzyme, the conformation of the enzyme changes. When a non-competitive inhibitor binds an enzyme at an allosteric site, the shape of active site of the enzyme will likely be affected. It therefore seems unlikely that the substrate will have the same affinity for both the free enzyme and the enzyme-inhibitor complex.

Similarly, when the substrate binds the active site, the shape of the allosteric site will probably change. It therefore seems improbable that the inhibitor will have the same affinity for both the free enzyme and the enzyme-substrate complex.

Independence of binding of both the inhibitor and substrate to the enzyme is a requirement for a pure, non-competitive inhibitor. Because true independence of binding is rare, most non-competitive inhibitors are mixed, not pure.

Question 10

For indole to show substrate-dependent behavior, it likely causes changes to the active site that affect some substrates but not others. In the original Scheme 4.4, the substrate is depicted as a hexagon, which is too large to fit into the active site if the inhibitor is bound. This would be an example of a mixed, non-competitive inhibitor. The presence of the inhibitor affects the affinity of the enzyme for its substrate. If the substrate is smaller, like the trapezoid below, then changes to the active site should not impact substrate affinity. In this case, the inhibitor would possibly be a pure, non-competitive inhibitor.


What is not clear in the simple diagram is the assumption that the inhibitor must somehow be preventing the conversion of the substrate to the product.

Question 11

The addition of inhibitor to an enzyme-substrate system gives an inverted V vs. [S] plot. Incremental increases in [I] have a decreasing impact on the rate of substrate conversion.

[I]

V

IC 50

Question 12

The name “glutamate dehydrogenase” certainly gives the impression that the enzyme is designed to dehydrogenate glutamate to ketoglutaric acid. The Km values give a different impression. The enzyme’s affinity for glutamate and NADP+ is less than ketoglutaric acid and NADPH. (The affinity for NH4

+ is lowest of all, but the concentration of NH4+ is higher than the other substrates/products.)

Based on Km values, the enzyme may run backwards better than forwards.


Of course, enzymes are catalysts that help a system achieve equilibrium faster than without the enzyme. Whether the enzyme runs net forward or backward depends the position of the system relative to its equilibrium.

In practice, biological systems often do not ever reach equilibrium. As soon as a product is formed, the product is used in a subsequent reaction. The rate of the reverse reaction is inconsequential because the product never accumulates.

Question 13

Enzymes, and catalysts in general, undergo changes – chemical and/or conformational – during a reaction and are reformed in their original state by the end of the process. A shovel, as a rigid object, never changes. An archery bow might be a better illustration. The bow flexes through the process of shooting an arrow, but it returns to its original state once the arrow has been launched.

Question 14

Competitive inhibitors of the same enzyme all bind the active site of the enzyme. They therefore tend to share similar structural features. Non-competitive inhibitors, however, do not necessarily bind the same position of an enzyme. Enzymes can have multiple allosteric sites, and inhibitors that bind different sites will have different structures.

Question 15

Binding an enzyme at an allosteric site can cause conformational changes that may close the active site and shut down an enzyme. If the conformational changes open an otherwise restricted active site, then the binding agent would turn on the enzyme.

Chapter 5Question 1

A smoothed curve of response vs. log [L] is shown below. The horizontal line is at 50% response, and the vertical line is at log 1.63. This corresponds to log KD. KD = 43 μM. Just glancing at the data points, KD should fall somewhere between 30 and 100 μM.


0

20

40

60

80

100

0 0.5 1 1.5 2 2.5 3

log [L]

% r

esp

on

se

Question 2

The data set gives a fairly normal looking, hyperbolic saturation curve.

0

10

20

30

40

50

60

70

80

90

100

0 100 200 300 400 500 600 700 800 900 1000

[L]

% r

esp

on

se

When forced into a Lineweaver-Burk plot, the data give somewhat scattered line. Note that one data point must be ignored (E = 0%) because it cannot be validly used in the Lineweaver-Burk equation. The equation of the best-fit line is shown below.


0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 0.02 0.04 0.06 0.08 0.1 0.12

1/[L]

1/%

res

po

nse

Based on the y-intercept, Emax is approximately 530%. KD comes out to be 630 μM. These values do not match at all with the values seen in Question 1 or what one would expect from cursory inspection of the data.

The Lineweaver-Burk approach to analyze receptor data is valid, but data in this example are not of high enough quality to give good results. The analysis is not helped by having to discard one of the data points because it cannot be inverted.

Question 3

A simple plot of the data points provided gives the V vs. [S] graph below.

0

5

10

15

20

25

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

[S]

V

Plotting 1/V against 1/[S] gives a Lineweaver-Burk plot.


0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

-40 -20 0 20 40 60 80 100

1/[S]

1/V

The equation for the line is

.

Vmax is 26. Km is 0.026.

When V/Vmax is plotted against log [S], the following scatter plot is formed. This plot would normally look like a sigmoid curve and be useful for determining log KD. In this case, the points seem to give a nearly straight line. The best-fit line is shown, and the equation of the line is provided.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-2.5 -2 -1.5 -1 -0.5 0

log [S]

V/V

max

Solving this equation for [S] at V/Vmax of 0.5 gives a value of 27. This is analogous to KD and is not terribly far off from the Km determined from the Lineweaver-Burk plot. V/Vmax vs. log [S], however, is


not a linear plot, and forcing it into a line is not a valid treatment of the data (regardless of how close the number is to the desired outcome). What is missing in this log [S]-response plot is a full range of log [S] values. A larger data set would properly show the regions in which the sigmoid has flattened out. With the provided data, only the nearly linear region of the graph near the point of inflection can be plotted.

The lesson is that most enzymatic studies are performed with a narrower range of concentration points than receptor studies.

Question 4

To be completely accurate, in the presence of a receptor, some of the ligand is free [L] and the rest of the ligand is bound to the receptor [LR]. The equation for total receptor is [Lt] = [L] + [LR]. Since the concentration of receptor in any binding assay is very small relative to the concentration of the ligand, the impact of the receptor on ligand concentration is generally ignored.

Question 5

The key to this question is to recognize that that y = 0 (bound/free = 0) at the x-intercept. Just set bound/free to 0 and solve for bound.

(5.12)

(at the x-intercept)

Bmax is reached only when y = 0, and y = 0 when the concentration of free ligand reaches infinity. For this reason, Bmax is graphically a theoretical value.

Question 6

The new line will look very much like the line of the partial agonist alone. The only difference is that a higher concentration of the partial agonist will be needed to affect a response. At high concentrations the partial agonist will completely overwhelm the competitive antagonist.


0

0.2

0.4

0.6

0.8

1

log [partial agonist]

resp

on

se

partial agonist alone

partial agonist with full agonist

partial agonist with antagonist

Question 7

There is nothing tricky about this question. Under Clark’s occupancy theory, the response is directly proportional to the percentage of the receptors that are occupied. Therefore, both the percentage of response and occupancy are the same. See Figure 5.18.

% Response % Receptor occupancy

20 20

40 40

60 60

80 80

Question 8

The first part of this question is just an application of Equation 5.8. For the full agonist questions, Emax = 100%. For the very first case, set [L] equal to 0.01KD and solve for E.

(5.8)

In a similar fashion, the percent responses for the other [L] values are 9.1, 91, and 99, respectively.

For the partial agonist questions, Emax = 35%. So, if [L] = 0.01KD…


For [L] = 10KD, E = 32%.

Question 9

Equation 5.28 is best equation for expressing both upregulation and downregulation. Upregulation and downregulation involve changes in receptor concentration based on previous ligand exposure. Only Equation 5.18 includes a term for receptor concentration.

(5.18)

Question 10

The allosteric model can handle both competitive and non-competitive antagonists. For competitive antagonists, a sufficient concentration of agonist can overcome the effect of the antagonist on the equilibrium of the tensed and relaxed states. For non-competitive antagonists, since the antagonist binds the ligand at a site away from that of the agonist, the antagonist’s impact on the tensed-relaxed equilibrium cannot be completely overcome by an agonist. The agonist may be able to cause a response, but the noncompetitive antagonist, no matter what the agonist concentration might be, will diminish the response. Note that this discussion is consistent with Figures 5.13 and 5.14.

Question 11

Agonist binding causes a physiological response. An antagonist decreases the response caused by an agonist. The molecule is therefore neither an agonist nor antagonist. The molecule must bind an inconsequential part of the receptor that has no influence on the receptor’s function. Just because a molecule happens to bind a receptor does not mean that the molecule has an influence upon the receptor.

Question 12

If the intrinsic activity (e) of a partial agonist is close to zero, the partial agonist effectively acts an antagonist.

Question 13

Tyrosine kinase linked receptors are receptors that catalyze phosphorylations. Since the receptors have catalytic activity, enzyme terminology is used on molecules that block the action of these receptors.

Question 14

Under rate theory, a full agonist must have a high koff value so that the compound may quickly re-bind with the receptor. A partial agonist would have a lower koff value, which would in turn limit how frequently the partial agonist can re-bind and trigger a response. In aggregate, the partial agonist causes a sub-maximal response because binding occurs less frequently than with a full agonist.

Question 15

For this question, one needs Equation 5.13 (for the simple case of E = S) as well as a combination of both Equations 5.14 and 5.15 (for E = 2×S).


(5.13)

(Derived from Eq. 5.14 and 5.15)

All systems shown in Figure 5.23 achieve 100% response, so all involve full agonists (e = 1.0).

The table below gives ligand concentration and response data for a hypothetic receptor with a KD of 10 nM.

[L] (nM) log [L] %E (if E = S) %E (if E = 2×S)

0.1 −1.0 0.99 1.98

0.3 −0.52 2.9 5.8

1.0 0.0 9.1 18

3.0 0.48 23 46

10 1.0 50 100

30 1.48 75 100 (150)

100 2.0 91 100 (182)

300 2.48 97 100 (194)

1,000 3.0 99 100 (198)

Note that if E = 2×S, then E calculates to be greater than 100%. This is impossible, so the %E is capped at 100. These data points demonstrate the idea of spare receptors. At 50% occupation of the receptor population ([L] = KD), the response for the E = 2×S is already at 100%. The response pathway is fully saturated, and half of the receptors are still unbound. These unbound receptors represent spare receptors.

Plotting the two response columns against log [L] gives two lines. (For complete disclosure, some data points were inserted between 3.0 and 10 nM to allow the E = 2×S line to smooth properly.)


0

20

40

60

80

100

-1.4 -0.9 -0.4 0.1 0.6 1.1 1.6 2.1 2.6 3.1

log [L]

resp

on

se

E = S

E = 2xS

The E = 2×S line climbs more steeply and abruptly flattens out at 100% response. The standard E = S line has a smooth sigmoid shape.

The question has been fully answered, but the shape of the E = 2×S line deserves more comment. Why is the transition at 100% response so marked? Remember that the line for E = 2×S is not a proper function. It is actually two functions. The front half of the line, up to log [L] = 1.0, is simply E = 2×S as defined by the equation below.

As log [L] continues to climb, the percent response rises beyond 100%. This rise is very smooth if the data is plotted. In our case, we declared any response greater than 100% to be impossible and capped the response at 100% at higher log [L] values. The E = 2×S line appears disjointed because it is actually a merger of two functions.

0

50

100

150

200

-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3

log [L]

resp

on

se

E = 2xS

E = 100


Chapter 6Question 1

The cytosine-guanine interaction involves three hydrogen bonds, while adenine-thymine has just two. Because of this difference in number of hydrogen bonds, C-G interactions receive a greater weight in determining Tm, as can be seen in the Wallace Rule (Equation 6.1).

(6.1)

Question 2

This problem requires assumptions on the duplex strand in terms of C-G and A-T composition. The easiest assumption is to make the C-G equal to 50%, and A-T is 50%. In the case of odd numbers of base pairs in the duplex, just split the difference and use a fractional number (e.g., 15 nucleotides = 7.5 C-G and 7.5 A-T).

For 10 base pairs…

(6.1)

For 15 base pairs, Tm = 45 ºC. So the impact of lengthening the strand by five base pairs is an increase in Tm by 15 ºC. Regardless of the length of the shorter strand, adding five base pairs of length is predicted by Equation 6.1 to raise Tm by 15 ºC. That corresponds to a 3 ºC increase in Tm per additional base pair.

Tm cannot scale linearly and indefinitely with duplex length. A duplex strand of 1,000 base pairs would have a Tm of 3,000 ºC. Thus, the Wallace Rule has its limitations.

Question 3

The challenge to this question is to tautomerize compound 6.a in a manner that gives the right orientation of hydrogen bond donors and acceptors to interact with guanine.

Question 4

In order to get a 1:1 mixture of products, the reaction must go through a symmetrical intermediate. Like all mustards, the reaction starts with an intramolecular ring closure by the nitrogen on the alkyl halide. The aziridine intermediate is then opened through attack by the thiosulfate anion. The two carbons of the aziridine ring are virtually equally reactive. They differ only in isotope of hydrogen that they bear (2H vs. 1H). The impact of the hydrogen isotopes is very small, so the ring can open either by attack on the CH2 or the CD2 group. The products are formed in an approximately 1:1 ratio.


Question 5

The structure of bizelesin (6.e) is very similar to the product shown in Scheme 6.11, which is based on the activity of adozelesin (6.44). Adozelesin starts with a reactive, three-membered ring that opens when attacked by a nucleophile. Bizelesin does the same, but it must first form its three-membered ring. The ring forms through attack of the 4-carbon of the phenol ring onto the carbon bearing the chlorine atom. The ring is highly reaction. Opening the ring with a nucleophile both releases ring strain and restores the aromaticity in the ring. Only one side of bizelesin is shown. Both sides must react for bis-alkylation to occur.

Question 6

Equation 6.2 is undefined under two circumstances. One, if L (length of the duplex strand) is 0, then Tm cannot be calculated because of division by 0. Since a duplex strand must have a non-zero length, this circumstance is irrelevant. Two, if [M+] (concentration of monovalent cations) is 0, then log [M+] is negative infinity. DNA is a negatively charged polymer and must include positive counterions to maintain charge balance. Determining Tm with [M+] = 0 may be theoretically possible, but the experimentally determined value would provide no understanding for the properties of DNA in a living cell.

(6.2)

Question 7


As the duplex strand length (L) approaches infinity, the last term in Equation 6.2 approaches a value of 0. From the text, the concentration of monovalent metal ions ([M+]) is 0.15 M. Finally, the question tells us that the combined mole fraction of guanosine and cytidine (XG + XC) is 0.40. With this information, we can solve for Tm. The answer is 89.8 ºC.

(6.2)

(6.2)

Question 8

The “sticky ends” consist of four base pairs, and all four are A-T. Based on Equation 6.1, the Tm of this strand should be just 8 °C. That assumes that the presence of the additional base pairs do not affect the Tm value.

(6.1)

Question 9

Both lysine and arginine are basic amino acids (see Table 4.1). Basic amino acids are protonated and positively-charged at physiological pH. The high density of positive changes in histones allows the proteins to bind especially strongly to the negatively charged backbone of DNA. Histones are therefore ideal structures for the tight packing of DNA in the nucleus.

Question 10

Below is a graph of the Rosenthal type plot (r/Dr vs. r).

0

0.5

1

1.5

2

2.5

0 0.2 0.4 0.6 0.8 1

r

r/D

f (10

-6)

The best-fit line for this data is shown, and the formula for the line is provided below.

From this equation, replacing r/Dr with 0 (for the x-intercept) and solving for r provides a value of 0.12 netropsin molecules per DNA base pair. This is Bmax for netropsin.

Question 11


Below is a plot of the data from the table.

0

2

4

6

8

10

12

14

16

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

N /nt

DT

m

Just by looking at the data points in the table, one might estimate that the curve is approaching a maximal ΔTm of nearly 15º C. That would put ½ΔTm at about 7.5 º C. The value on the x-axis at this point is approximately N/nt = 0.02. Note that entry 3 in the table corresponds almost exactly to these values.

Question 12

Below is the plot of 1/ΔTm vs. 1/(N/nt) as inspired by the Lineweaver-Burk equation with a best-fit trendline.

0

0.05

0.1

0.15

0.2

0.25

0 10 20 30 40 50 60 70 80 90 100

1/(N /nt )

1/D

Tm

The equation for the best-fit line is provided below.


Based on the y-intercept value of 0.060, ΔTmmax is 17º C. That is not too far off our simple

interpretation of the data table in Question 11. If the slope of the line is 0.0018, we can use the calculated ΔTm

max to determine that KD is 0.03. Again, this value is not far from the estimate in Question 11. The point of this question is that reading values off a graph is not as accurate as letting the data set speak for itself through a best-fit line. Of course, just because the data form a nice line does not mean that the data are of high quality.

Question 13

The spectroscopic data be interpreted to show that the favored tautomer of 2-pyridone is 6.h because of its λmax is closer to 230 nM than 270 nM. While this is not a true apples-to-apples comparison, if 2-pyridone is in the keto form, then guanine may also favor the keto tautomer (6.f.).

Question 14

The full equilibrium of interest is shown below.

From this equilibrium, just use 7.4 for the pH and 6.6 for the pKa in the Henderson-Hasselbalch equation. The concentration is the conjugate base (dibasic phosphate) is indeed greater than the acid (monobasic phosphate) at physiological pH.

Henderson-Hasselbalch eq.

Question 15

Antisense drugs act by binding a complementary oligonucleotide strand. Since DNA exists in a duplex form, DNA is not free to bind antisense drugs.

Chapter 7Question 1


Vd can be quickly determined with Equation 7.13. Do is 350 mg. Cmax is approximately equivalent to Cp

o. As always, watch the units.

(7.13)

With a Vd of only approximately 3 L, infliximab is restricted to a very small volume of distribution. A value of 3 L is comparable to just the plasma of a 70-kg patient. Considering the large molecular weight of the drug, one should not be surprised that the drug does not cross the pores in the capillaries and enter the interstitial fluid. Furthermore, the high molecular weight of infliximab prevents the drug from diffusing across the cell membranes of the cellular components of whole blood. The drug should be exclusively confined to the plasma.

Question 2

One approach to this question is through Equation 7.12.

(7.12)

If the half-life for a drug is very long, then kel must be very small. By Equation 7.12, small kel values arise from a low clearance, a high volume of distribution, or both. Based on Question 1, the Vd of infliximab is extremely small, so the drug’s long half-life (small kel) is indeed surprising. The only possibility is that clearance for infliximab must be remarkably small. From Equation 7.8, clearance is a function of both blood flow (Q) and extraction ratio (E).

(7.8)

Because blood flow to an organ is not a variable, the low clearance of infliximab must be attributable to a low extraction ratio of the drug. A low hepatic extraction value (EH) indicates that the drug is apparently inert to the metabolic enzymes of the liver. A low renal extraction value (ER) indicates that the drug is either not filtered by the kidneys or it undergoes very extensive tubular reabsorption.

Infliximab is a massive drug – 144,000 g/mol. Compounds of this size cannot undergo glomerular filtration. If the kidneys cannot filter a compound, its renal extraction ratio should be essentially 0. Consequently, renal clearance is 0, and kel is small despite a small value for Vd.

Question 3

As a drug more extensively distributes throughout the body and out of the central compartment (plasma), its Vd increases. Regardless of a drug’s distribution, renal elimination is restricted to the amount of drug that is present in the plasma. Very little of an extensively distributed drug can be found in the plasma at any given time, so the elimination rate constant due to renal filtration is very small (half-life is large) for such a drug. Furthermore, hepatic clearance also only applies to any drug that is found in the central compartment and passes through the liver. An extensively distributed drug would have a low hepatic clearance as well.

Question 4


If a drug has a much greater kel than its metabolite, then the drug eliminates much more quickly than its metabolite. A Cp-time plot for such a drug and metabolite is shown below.

time

Cp

Note that the drug has essentially completely eliminated while the metabolite is still present in the plasma in appreciable quantities. With the drug virtually at Cp = 0, the metabolite is no longer forming and is only undergoing elimination. At this point, the metabolite should follow first-order elimination kinetics. A log [drug]-time plot for the late data points should give a line with a slope equal to the −kel of the metabolite.

Question 5

This question starts with just a plug-and-chug with a rearranged Equation 7.18. Concerns include deciding on a mass of the patient (70 kg) and selecting reasonable units for the rate of infusion. Because the oral dosing schedule is provided as mg/day, these would be good units for Rinf. A value for kel must also be determined from the provided half-life.

(7.18)

Mathematically, determining the time required for the infusion to reach a target Cp is much messier than the previous question. It requires rearranging Equation 7.17 to solve for time.

(7.17)


The time to reach just half of the desired Css is equal to the half-life of the drug – 53 days. This is clearly not acceptable; a patient should not need to wait months (literally) for a drug to reach therapeutic levels. The solution to the problem would be to administer some sort of loading dose either orally or by IV bolus.

This question is interesting mathematically. It demonstrates the fact that drugs with a long half-life are impractical to administer by an infusion alone. From a practical perspective, one rarely administers long half-life drugs by infusion. First, because the drug has a very long half-life, its Cp is fairly constant and does not need to be continually replenished by an infusion. Second, very few patients are connected to an IV line for months at a time to permit very long-term infusions.

Question 6

Plotting the provided data in the form of log Cp vs. time provides the following graph with trendline. The equation of the best-fit line is ln Cp = −0.61t + 1.9. Since the slope is −kel, kel must be 0.61 h-1. Half-life is therefore 1.1 h.

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 1 2 3 4 5

time (h)

ln C

p

Determing Vd requires just a little more work. The y-intercept of the line (1.9) is ln Cpo, so Cp

o is 6.7 μg/mL. This brings us to Equation 7.13, which calculates the Vd of the drug in this hypothetical 70-kg patient to be 15 L. On a per mass basis, the Vd is 0.21 L/kg.


(7.13)

Question 7

This question is more challenging because we need to fill in some gaps and make assumptions. The chapter provides only a small number of methods for determining kab and F. F can be determined through comparisons of AUC information between an IV bolus and an oral dose (Equations 7.22 and 7.23). These calculations require kab. kab can be calculated directly through Equation 7.28, but this requires F. The only equation that isolates kab from F is the calculation of tmax (Equation 7.30). Determining kab from tmax does require Cp-time data that hit very close to Cp

max. The data points for this problem do show a clear tmax at 0.75 h, but the true tmax could easily be a little earlier or later than 0.75 h. Regardless, this is the best we can do with the data at hand. Therefore, we will assume that a tmax of 0.75 h is accurate for this drug.

(7.30)

Solving functions like this is not trivial. My approach is crude at best, but I normally use a spreadsheet program to test a range of values for kab until the equation is solved. As a hint, kab values for most drugs are larger than kel. This approach for determining kab gives a value of 2.8 h-1.

The last parameter to determine is F. We can use the theoretical y-intercept of post-absorption oral data provided in the question. A plot of ln Cp versus time affords the following graph. Overall, the data points form a typical curve for an oral drug and demonstrate both the absorption and elimination phases. Only the very last points occur at a time during which absorption is essentially complete, and only elimination is in effect. The last three points can be fit well to a line:ln Cp = −0.752t + 2.13. (Note that if the data were perfect, the slope of this line would be identical to the slope of the line in the previous question – 0.61. Both lines should have the same slope since both demonstrate first-order elimination of the same compound.)


-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 1 2 3 4 5

time (h)

ln C

p

The y-intercept of the line from the elimination data points is equal to ln Cpint. Cp

int is theoretical variable that is equal to a complex term that includes F (Equation 7.27). Based on the y-intercept of 2.13, Cp

int is 8.4 μg/mL. Use Equation 7.27 to determine F.

(7.27)

F works out to 0.98, so the drug has a bioavailability of 98%. Remember that the accuracy of this calculation is limited by the accuracy of our estimate of kab. If kab is somewhere between 0.5 h and 0.75 h (perhaps 0.6 h), then our value of 0.75 h has a 25% error. That could explain why the slope of our line in this question is so far off from the slope we observed in the previous question.

Question 8

This is a direct application of Equation 7.30.

(7.30)

It is difficult to solve this kind of equation for kab. Perhaps the easiest way to determine kab is by trial-and-error. Just try different values for kab with the known kel (0.693/2 h) until tmax works out to approximately 1.6 h. kab is approximately 1.0 h-1.

Question 9

Equation 7.21 fails if kel = kab, which results in division by zero.


(7.21)

Question 10

As kab becomes very large Equation 7.23 simplifies significantly. The 1/kab term approaches 0, and the kab/(kab − kel) approaches 1.

(7.23)

(if kab = ∞ and F = 1)

This simplified form of Equation 7.23 is equivalent to Equation 7.10, which is the formula for AUC of an IV bolus.

(7.10)

If kab is infinite and F is 1, then the drug is instantaneously and completely absorbed into the central compartment (the bloodstream). This exactly describes the situation of an IV bolus – complete and instantaneous absorption.

Question 11

The bioavailability of iron is likely close to 0.12 (1.0/8.0).

Question 12

A little thinking can greatly simplify this problem; refer to Figure 7.16. Each dose gives a higher peak in Cp than the previous one until the system reaches a pseudo Cp

ss. Therefore, the Cpmax for the

azithromycin will occur after the oral dose on Day 5. The exact time on Day 5 can be determined through Equation 7.30.

(7.30)

If Day 1 starts with the first oral dose (t = 0 h), then Day 2 begins at t = 24 h, Day 3 starts at t = 48 h, Day 4 starts at t = 72 h, and Day 5 begins at t = 96 h. The overall tmax should occur 2.80 h into Day 5 or at 98.8 h.

To determine Cp at 98.8 h, the Cp contributions for each of the five oral doses need to be calculated separately at the correct time interval with Equation 7.21. Do not forget that the dose on Day 1 is 500 mg.

(7.21)


Day 1 dose (t =

98.8 h)

Day 2 dose (t =

74.8 h)

Day 3 dose (t =

50.8 h)

Day 4 dose (t =

26.8 h)

Day 5 dose (t = 2.8

h)

Adding all five Cp values together gives a Cpmax of 0.156 mg/L (or 0.156 μg/mL or 156 ng/mL).

Question 13

It is certainly possible that the concentration of azithromycin is greater in a patient’s tissues than in the plasma. With a Vd of 31.1 L/kg, the Vd in a 70-kg patient is over 2,000 L. Azithromycin extensively distributes into tissues and out of the central compartment. As azithromycin concentrates in tissues, it is very possible for it to reach its minimum inhibitory concentration in an infected region.

Question 14

The scheme below depicts three compartments, and the third compartment is only accessible through the peripheral compartment.


The scheme below shows three compartments. The third compartment is in equilibrium with both the central and peripheral compartments.

Question 15

To do this problem, we need to assume a mass of the patient. As usual, 70 kg is a standard number. Rearrangement of Equation 7.13 allows direct calculation of the required dose to reach the specified LC50.

(7.13)

(7.13)

An IV bolus with 3.5-g dose is sufficient to reach the LC50 in a 70-kg patient. Remember that immediately following an IV bolus, the dose is largely contained in the central compartment. The central compartment certainly has a smaller volume than the full Vd of this drug. Therefore, immediately following an IV bolus, the true Cp

o of this drug is likely many times higher than the calculated Cp

o (which assumes instantaneous distribution – see Figure 7.7).

Chapter 8Question 1


If dextrorphan (8.b) is the primary metabolite, then the first metabolic reaction is a Phase I oxidative dealkylation. Metabolite 8.b then apparently undergoes either Phase II sulfonylation to form 8.c or another Phase I oxidative dealkylation (N-demethylation) to form 8.d.

Question 2

Below are examples from all five Phase I reactions.


Question 3

Below are examples from all five Phase II reactions.


Question 4

The liver and kidneys often compete for elimination of a drug. If the liver metabolizes any drug, then less of that drug is eliminated by the kidneys. On the other hand, if the rate of metabolism by the liver is slowed, then the kidneys will eliminate a larger fraction of the drug.

Question 5


The glucuronidation product is fairly straightforward. Glucuronidation tends to occur on alcohols, and bicalutamide only has one OH group. The oxidation is harder to predict. Bicalutamide has several potential sites for oxidation: either of the benzene rings, the NH group of the amide, the methyl group, and the CH2 group next to the sulfone. Physiologically, the oxidation occurs adjacent to the fluorine on the benzene ring.

Question 6

The data in Table 8.2 are not completely conclusive. None of the conditions can be directly tied to the action of the kidneys. The liver is featured prominently in the table, and all liver problems significantly lengthen the half-life of theophylline. Therefore, the liver likely plays a more important role in clearing theophylline than the kidneys.

Question 7

Many of the Phase I redox reactions should be reversible. As an example, a ketone may be reduced to an alcohol, and the same alcohol may be oxidized back to the original ketone. Acetylation (Phase II) of an amine to form an amide can also occur in the opposite direction as the hydrolysis (Phase I) of an amide to an amine.

Question 8

A compound with a higher polarity should concentrate in the parts of the body that are more polar, such as the blood, and distribute less into lipophilic tissues. Since filtration by the kidneys occurs exclusively on the blood, concentrating a compound in the blood (the central compartment) makes it more subject to elimination by the kidneys.

In the context of Vd, a more polar drug should have a smaller Vd. Recalling Equation 7.12, if Vd decreases, then the value for kel should increase. A higher value for kel indicates faster elimination and a shorter half-life.

(7.12)

Question 9

Based on Equation 7.9, CL is a function of blood flow (Q) and extraction ratio (E).

(7.9)

Blood flow through the kidneys (QR) is constant, but the extraction ratio (ER) can vary with the properties of a drug. Raising a molecule’s polarity may minimize tubular reabsorption because the more polar compound is unable to cross the tubule wall after it has been filtered by the glomerulus. A drop in tubular reabsorption will increase ER and subsequently increase renal clearance (CLR) and total


clearance (CLT). A compound with higher CLT has a higher value for kel and shorter half-life according to Equation 7.12.

(7.12)

Question 10

The liver can metabolize any drug that is transported by the blood. Since drug metabolites can sometimes cause liver damage, any drug can potentially put the liver at risk. The full dose of an oral drug, however, must pass through the liver to reach the general circulatory system. Therefore, the exposure of the liver to orally-administered drugs is especially high. This answer follows the intent of the question.

It is possible to undermine this question. Oral drugs tend to be fairly chemically stable because they must survive the acidic environment of the stomach, enter the hepatic portal system, and pass through the liver in concentrations high enough to reach therapeutic Cp levels. More highly reactive drugs, especially alkylating agents used as cancer drugs (see Chapter 6), often cannot be administered orally and are instead delivered by IV. The reactivity required for a DNA alkylating agent makes the compound too unstable to survive oral administration. Highly reactive compounds, especially electrophilic alkylating agents, are exactly the type of compounds that deplete glutathione reserves in the liver and leave the liver vulnerable to damage.

Chapter 9Question 1

The PDB data can be fit with Excel. A third-order polynomial fits the data considerably better than a power, exponential, or logarithmic function. Excel is able to plot the original data and graph the best-fit trendline. (The most common trendlines − linear, power, exponential, logarithmic − poorly fit the PDB data.) Excel, however, does not provide the coefficients in equation of the best-fit line with enough significant figures to answer the question. Another application, like an applet from the web, is needed. Regardless, the best-fit trendline and its equation as determined by Excel are shown below.


structures = 4.6694yr3 - 27804yr2 + 6E+07yr - 4E+10

r2 = 0.9993

0

50000

100000

150000

200000

250000

1976 1981 1986 1991 1996 2001 2006 2011 2016 2021

year

tota

l p

rote

in s

tru

ctu

res

in P

DB

The best-fit line determined by the recommended applet (third-order polynomial) is provided below.

Based on this equation, at the end of 2015 the number of structures in the PDB can be estimated to be 121,323.

Similarly, in 2020 the number of structures in the PDB is predicted to be 193,520.

Question 2

The first character can be selected from 10 possible values, and the other three characters can be selected from 36 possible values (10+26). The total number of possible structures codes with the PDB format is 466,560.

While some may be able to solve the cubic equation below for the variable year, I cannot.


An alternative is to set up an Excel spreadsheet and calculate the number of structures predicted in various future years. One can quickly hone in on the correct year through trial-and-error. Surveying five-year intervals puts the target structure value between 2030 and 2035. A year-by-year analysis indicates that the PDB will reach 466,560 structures sometime during the year of 2031. At that time, all the PDB codes under the current labeling convention will be exhausted. There remains plenty of time to work out a solution to this impending crisis.

year structures

2025 289,486

2030 412,562

2031 440,725

2032 470,132

2035 566,091

Question 3

Below are the Ramachandran plots for 2HHB, 3HHB, and 4HHB.

2HHB


3HHB


4HHB


According to the output text that accompanies the Ramachandran plots, structure 2HHB and 3HHB have only 1.0% of their residues that fall outside the standard plot regions. Structure 4HHB, however, has 5.1% outliers. Furthermore, 4HHB includes a D-amino acid. Natural amino acids have an L configuration. Therefore, structure 4HHB contains more incorrect assignments than 2HHB and 3HHB.

Question 4

The ω dihedral angle is describes rotation about the C-N amide bond. The C-N bond of an amide has two predominant conformations – s-cis and s-trans. In these conformations, the C-N sigma bond gains pi-bond character because of conjugation of the nitrogen lone pair with the neighboring carbonyl pi-


bond. The possibility of resonance with the lone pair stabilizes both the s-cis and s-trans conformations in comparison to all other possible conformations. The s-cis conformation has a dihedral angle of 0º, and the s-trans conformation has a dihedral angle of 180º.

Question 5

Two Newman projections showing the and dihedral angles are shown below. The magnitudes of the angles are typical for an amino acid contained in a -helix. A value of −50º for gives high separation between the R-group of the amino and the carbonyl oxygen and NH on the adjacent carbon. Separation of the R-group is not as ideal for = −50º, but the R-group is eclipsed (or nearly eclipsed) with a very small hydrogen atom.

Glycine is a special amino acid because R = H. If R = H, the spatial, steric demands of the R-group are dramatically decreased. The energetic benefit of minimizing the steric effects of a hydrogen atom is small, so the idealized and values are not observed for glycine residues.

In the Ramachandran plots of Question 4, glycines (shown as □) have approximate dihedral values of = 90º and = −120º. The corresponding Newman projections are shown below.

Both Newman projections place the R-group (R = H for glycine) in a relatively crowded position. These placements are only reasonable because R is small for glycine.

Question 6

Below is the equilibrium of interest – the deprotonation of salicylic acid to form its conjugate base.


With the Henderson-Hasselbalch equation (Equation 9.1), biological pH (7.4), and the pKa of salicylic acid (3.0), the ratio of the conjugate base to the acid can be readily calculated.

(9.1)

At pH 7.4, the conjugate base is far and away the major form of the compound.

The three possible scenarios mentioned in the question are whether the carboxylic acid (actually in its carboxylate form) is least likely to interact with a target through an ionic bond, hydrogen bond acceptor, or hydrogen bond donor.

Ionic bond : The carboxylate is an ion and therefore could act through an ionic bond.

Hydrogen bond acceptor : The oxygens of the carboxylate group could accept a hydrogen bond from a suitable hydrogen bond donor.

Hydrogen bond donor : The anionic carboxylate does not have an intact OH group and therefore cannot interact as a hydrogen bond donor.

Question 7

As with Question 6, here is the key equilibrium. Note that at pH 7.4, salicylic acid (9.a) exists in the carboxylate form. The equilibrium considering the phenol must start with the carboxylic acid already deprotonated.

Following the same process as Question 6, one can calculate the ratio of carboxylate (phenol) to its conjugate base.

(9.1)


The phenol is essentially completely in its acid form.

The three possible scenarios mentioned in the question are whether the phenol is least likely to interact with a target through an ionic bond, hydrogen bond acceptor, or hydrogen bond donor.

Ionic bond : The phenol is neutral, so an ionic bond is highly unlikely.

Hydrogen bond acceptor : The oxygen of the phenol could accept a hydrogen bond from a suitable hydrogen bond donor.

Hydrogen bond donor : The phenol could also act as a hydrogen bond donor.

The only improbable option is an ionic bond.

Question 8

Below is a Newman projection of 9.c.

The Newman projection of 9.c shows an anticlinal relationship between the NH2 group and the aromatic ring. The corresponding Newman projection of dopamine is shown below.

The anticlinal conformation of dopamine is not a low energy conformation. It is a high-energy, eclipsed conformation. Energetically, dopamine is more stable in other conformations, including any staggered conformations. Dopamine therefore has a low probability of existing in the anticlinal, agonist conformation.


Question 9

The table below shows the molecular formula breakdown of the nine compounds in the question.

Entry Compound Formula Non-hydrogen atoms

1 9.d C18H20N3O3S 25

2 9.e C35H36ClNO3S 41

3 9.f C16H16ClNO2S 21

4 9.g C22H28FN3O6S 33

5 9.h C23H27Cl2N3O2 30

6 9.i C18H21NO4 23

7 9.j C33H35FN2O5 41

8 9.k C21H25N3O2S 27

9 9.l C18H19NOS 21

The average total number of carbon, nitrogen, oxygen, and sulfur atoms in this group of nine compounds is 29.1. Thirty carbon, nitrogen, oxygen, and sulfur atoms does seem to be a fair number for a “typical” drug. The range is values is broad with the highest having 41 atoms and the lowest having 21. A “typical” drug size must be interpreted openly.

Question 10

Even a small fraction (0.1%) of all of drug space is still an inconceivable number of 1060. The number of possible drugs from this pool of 1060 hits could very well run into the billions of compounds.

For students who are new to drug discovery, the search for a drug seems to be a hunt for the best compound. While finding the best molecule would be ideal, the true goal of a drug discovery program is to find a sufficiently active and safe drug. While there is only one best compound, there may exist millions or billions or even more compounds that have the appropriate activity and safety to become an effective and profitable drug.

Question 11

Multi-component reactions can very quickly generate large molecular libraries. The calculation for potential Ugi products from building blocks available from the Sigma-Aldrich catalog is shown below.

In theory, reagents available from the Sigma-Aldrich catalog could be used to prepare a molecular library containing approximately 3.2 billion compounds. That number is far greater than the current number of molecules in the CAS registry. Even though this would be a massive library, its value for


drug discovery may be poor. The molecular diversity of the library would be severely limited. It would resemble the library with many members but low diversity in Figure 9.18.

Question 12

Scheme 9.7 is reproduced below.

The functional group that connects the molecule of interest to the resin is an ester. Esters can undergo reduction with hydride reagents, especially LiAlH4 and to a lesser degree NaBH4. Hydride reagents transfer hydride (H−) to the carbonyl of the ester and fragment the linker as shown. Once the molecule of interest has been separated from the resin bead, the molecule is lost.

Questions 13

The mechanism is shown below starting with the complete structure of 9.65. The key feature in the resin linker that allows the benzodiazepine to be cleaved is the substitution of three electron-donating OR groups on the aromatic ring. These three groups give the ring enough electron density to eject the protonated benzodiazepine as a leaving group.


Question 14

Compound D3 should have three tags, tw, tx, and tz, on its resin. Compound B2 should have just two tags, tw and ty, on its resin. These tags should be unique to each of these compounds.

Question 15

Deconvolution of a hit from the 27-member library would require the synthesis of six compounds. The identity of the reagents used in the last step of making the hit is known. Three cataloged resins from the second step – one resin that last reacted with 1, one from 2, and one from 3 – would need to be carried through the last step to determine whether 1, 2, or 3 gave the hit. Three cataloged resins from the first step would need to be carried through the second and third step to fully characterize the hit. As a result, a total of six final compounds are prepared.

In general, deconvolution of a multi-step synthesis will require the synthesis of a number of compounds equal to the number of building blocks used in the all steps but the last. For a three-step synthesis with x, y, and z buildings blocks, x + y compounds would need to be prepared for deconvolution of a hit. For a four-step synthesis with m, n, o, and p building blocks, m + n + o compounds would need to be prepared for deconvolution of a hit.


Chapter 10Question 1

The phrase “essentially one product” leaves room for interpretation, but here is an analysis of the three reactions.

E2 elimination of HBr with a base This reaction will likely give a mixture of products. The elimination involves removal of a β-hydrogen. The compound contains two different β-hydrogens, each of which will give a different product. Furthermore, one of the elimination products can form as either an E- or Z-isomer. In total, three products may form in significant quantities.

OsO 4-catalyzed dihydroxylation of an alkeneThis reaction should form just one product. It will be a mixture of enantiomers, but almost all libraries consider racemic mixtures to be a single compound for screening purposes. If a racemic mixture gives high activity in a screen, confirmation screening and deconvolution will require single enantiomer testing.

Diels-Alder cycloaddition of a diene with acrylonitrile Diels-Alder cycloadditions tend to be highly regioselective (1,2- vs. 1,3-product) and diastereoselective (endo vs. exo). This reaction is no exception. Predominantly one product will be formed. As with the dihydroxylation in the previous reaction, the cycloaddition will form a racemic mixture.

Question 2


Both compounds have complicated, polycyclic structures with multiple stereocenters. Furthermore, compound 10.b has a fairly high MW (425 g/mol) for a lead compound. It would be difficult to synthesize analogues. This can be a significant challenge for natural products. While these structures might be too complex to serve as good leads, they may still be useful for drug discovery. Medicinal chemists can sometimes reduce an active molecule down to its minimal core. This core is called a pharmacophore and normally has a much simpler structure than an original hit or lead. Pharmacophores and their use in drug discovery are covered in Chapter 11.

Question 3

Below are some possible consequences of problems in storing, handling, and dispensing compounds in a molecular library.

Evaporation of the stock solution of a library member If a solution has evaporated to a smaller volume, then the concentration of the compound in the solution will be higher than expected. The compound will erroneously be recorded at its original, lower concentration, but its activity will reflect its elevated concentration. Evaporation of the solvent in the stock solution will make the library member seem higher in activity than it actually is. Note that precautions are taken to minimize evaporation. Storage solvents tend to have a low vapor pressure, and samples can be stored at low temperatures.

Degradation of the compound during long-term storage If a compound degrades, then its concentration will be lower than expected. The assay will underreport the activity of the molecule. If the compound barely misses the threshold activity level of a hit, then the underestimation of activity may cause the degraded compound to be missed altogether in the assay. Degradation of compounds during storage can be a serious problem even though compounds are stored at low temperature.

Random error in dispensing of small volumes Very small volumes of liquids can be difficult to dispense reproducibly with robotic liquid handlers. If the inaccuracies are random, then the impact cannot be predicted. In practice, dispensing errors tend to provide too small of a volume of liquid than too large a volume.

Impure sample in the library If a compound is impure (and yet assumed to be impure), then its actual concentration is lower than what is expected. This results in the activity of the compound being underreported in the assay. Therefore, impure compounds run the risk of being missed in a screen. An impure sample is very similar to a degraded sample.

Question 4

This is a high school chemistry question.


This number is miniscule compared to the 30 large atom library. Why skimp at just one molecule of each library member? Go wild and have 10 molecules of each. The main difference is the number of members in the library – 107 vs. 1063.

Question 5

Here is Equation 10.1.

(10.1)

Equation 10.1 evaluates to a negative number for ΔGºbind because log KD gives a negative value. This occurs because KD normally has a value of less than 1. Remember that KD for a receptor is determined by the equation below.

KD evaluates to less than one because the equilibrium favors the receptor-ligand complex over the free receptor and unbound ligand. If the equilibrium favored the free receptor and ligand, KD would become greater than 1, log KD would be positive, and ΔGºbind would be positive.

Of course, good drugs tend to bind strongly, not weakly. The target KD (same as EC50 under Clark’s occupancy theory) for a drug that binds a receptor is in the range of 10 nM. So, the binding energy for such a compound is…

(10.1)

If KD is greater than 1, then EC50 is also greater than 1. The units on EC50 are mol/L. In other words, the concentration of such a compound (KD > 1) to achieve a 50% response from a receptor would need to be greater than 1 M. That would be a preposterous concentration for a drug.

The reason that positive binding energies are not encountered in the discussion of hits, leads, and drugs is because compounds with positive binding energies have no possible value to a drug discovery program.

Question 6

The breakdown of the compounds is below. Entries 2 and 7 both violate Lipinski’s rules with regard to MW and log P.

Entry Compound Formula MW log PH-bonddonors

H-bondacceptors

1 10.c C18H20N3O3S 345 0.6 1 5

2 10.d C34H34ClNO3S 586 7.9 2 4


3 10.e C16H16ClNO2S 322 2.5 0 2

4 10.f C22H28FN3O6S 482 2.4 3 8

5 10.g C23H27Cl2N3O2 448 4.5 1 4

6 10.h C18H21NO4 315 0.3 1 5

7 10.i C33H35FN2O5 559 5.7 4 5

8 10.j C21H25N3O2S 384 2.8 1 5

9 10.k C18H19NOS 297 4.0 1 2

Comments

Esomeprazole – The NH nitrogen does not count as a hydrogen bond acceptor because that nitrogen’s lone pair is involved in the aromaticity of the ring.

Clopidogrel – The OR group of esters is not considered to be a hydrogen bond acceptor because it contributes electron density to the carbonyl.

Rosuvastatin – The NMe nitrogen is part of a sulfonamide and is not included in the hydrogen bond acceptor tally.

Aripiprazole – The NH nitrogen, which is part of an amide, is not considered to be a hydrogen bond acceptor.

Atorvastatin – The ring nitrogen and NH of the amide are not counted toward the number of hydrogen bond acceptors.

Question 7

Lipinski’s rules predict cell membrane permeability and oral bioavailability of a drug. If a drug is administered by inhalation, then its oral bioavailability is not of any consequence.

Question 8

The possible explanations attributed the extra binding energy to either the binding of the CH2CH2 to the enzyme or some other impact of changes introduced by the tether. In Chapter 9, the potential van der Waals attraction from a single CH2 group is 0.8 kcal/mol. Therefore, two CH2 groups would provide 1.6 kcal/mol of binding energy. Based on this information, the simplest explanation for addition binding energy seems to be van der Waals attraction between the CH2CH2 tether and the enzyme.

Chapter 9

The formula for ligand lipophilicity efficiency is shown below.

(10.2)

The four compounds compare as follows…

Entry Number log ED50 clog P LLE


1 10.l −6.96 2.97 3.99

2 10.m −4.49 0.08 4.41

3 10.n −6.14 2.76 3.38

4 10.o −7.02 3.35 3.67

Despite being the least potent based on its ED50 value, compound 10.m is the most attractive hit based on its LLE score.

Question 10

Privileged structures can be very fruitful areas for searching for biologically active compounds. From a drug discovery perspective, however, it can be very difficult to find areas within privileged structures that have not been thoroughly explored by other companies and researchers. Finding undiscovered activity from privileged structures is a challenge, and any new activity can be hard to protect with patents.

Question 11

The four compounds compare as follows…

Entry Number log ED50ΔGºbind

(kcal/mol)n

(atoms)LE

(kcal/mol/atom)

1 10.l −6.96 −9.49 20 −0.47

2 10.m −4.49 −6.12 19 −0.32

3 10.n −6.14 −8.37 23 −0.27

4 10.o −7.02 −9.57 19 −0.50

Compounds 10.l and 10.o stand out as having particularly large LE values.

Question 12

For a typical drug…

(10.a)

If this is the LE value for a typical drug, then 10.l and 10.o stand out even more as very strongly binding compounds based on their size (number of non-hydrogen atoms).

Question 13

For a typical drug…

(10.2)


A LLE value of 3 should be seen as a minimum value since log P should ideally be less than 5.

Question 14

The structure of compound 9.74 is shown below.

This compound is fantastically complex – polycyclic with several stereocenters. Furthermore, the molecule has a MW of 430 g/mol. This compound is somewhat larger than an ideal lead (MW ≤ 350 g/mol), and its complexity would hinder analogue synthesis.

Question 15

Polar surface area is exposed molecular surface that should interact favorably with a polar solvent, most notably water. As a compound crosses a cell membrane, the availability of water will be negligible. The polar surface area of the molecule will need to interact with the lipophilic tails of the phosphodiesters of the lipid bilayer. These interactions will be weak in comparison to salvation in water, so a compound with a large amount of polar surface area will favor being dissolved in a polar medium.


MPPP does fit the morphine rule. It has (a) a phenyl ring next to (b) a quaternary carbon bearing (c) a tertiary aminoethyl chain. When MPPP is redrawn to resemble the compounds in Figure 11.1, the structural analogies between the morphine pharmacophore and MPPP are clear. Indeed, MPPP is nearly identical to meperidine except that the carbonyl and oxygen of the ester have been swapped. This type of reverse ester analogue is another example of a bioisostere, a few examples of which can be seen in Table 11.5.


Question 2

trans-2-(3,4-Methylenedioxyphenyl)cyclopropylamine (11.c) preserves very roughly an anti orientation of the amino and the aromatic ring. Highlighting the key elements of dopamine within the structures of 11.d and 11.e reveals that the amine and hydroxylated ring are held roughly syn (lie on the same side of the plane defined by the carbons and hydrogens of the ethylene group) to one another in 11.d. The amine and hydrolyated ring are approximately anti to one another in 11.e. If this observation is correct, then the anti orientation of 11.c is more closely approximated by 11.d than 11.e. Compound 11.d would therefore more likely have agonist activity than 11.e.

Question 3

Although compounds from Figure 11.5 are mentioned in the question, their structures are not relevant to the conformational restrictions in 11.f. All of the structures that have an impact on this question are shown below.

From 9.28 and 9.29 it would seem that the amino group must be able to position itself to the left of the structure in order for the compound to be active. Compound 11.f shows that forcing the amino group


too far to the left is not favorable. Therefore, the problem is how to restrict the amino group to the left without forcing it too far.

The two compounds below try to address this problem. Both keep the general features of 9.27-9.29. The amino group is connected to the ring nitrogen through a tether of three carbons. Rings are used to hold the amino group in a restricted volume. The positions of the amino groups in both proposed compounds are consistent with available conformations of the aminopropyl chain of 9.28, the active analogue of chlorpromazine.

A challenge for both of these compounds is that the additional ring would likely make them much more difficult to synthesize.

Question 4

The compounds in the table reveal a few trends. The acid is vital for activity (entry 1 vs. 2). The length of the tether to the SH cannot be changed (entry 1 vs. 4). The SH group is needed for activity (entry 1 vs. 7). Based on the table data, the likely pharmacophore of captopril has the structure below.

From this core structure, a large number of reasonable analogues are possible. The table below shows some examples with explanatory text.

Entry Structure Rationale

1Adding N-substitution may restore some lost activity without introducing any more bulk than was present in the original compound.

2Changing the size of the ring may improve binding geometry.

3

Connecting the methyl group to the ring may hold the compound in an ideal geometry for binding (or it might make the ideal conformation inaccessible). The new connection introduces a new stereocenter, and both stereochemical configurations would likely be explored.


4

In theory, this is a reasonable compound to try. It is a simple isosteric replacement of the Me with a Cl atom. In reality, this compound would almost certainly be too unstable chemically to be a drug candidate. Analogues must be chemically feasible.

5

This is a bioisosteric replacement of the carboxylic acid with a tetrazole ring.

6This compound is a combination of entries 1 and 3 – add a ring and cap off the nitrogen.

7

This very compact structure challenges the idea that the SH should be extended away from the carbonyl. While the structure looks very different from captopril, the SH group is still removed from the carbonyl by just two carbons. As with a previous entry, this change introduces a new stereocenter, and both configurations should be tested.

Many other reasonable analogues are possible.

Question 5

The six isosteric analogues of morphine are shown below. Other possibilities exist. The replacement atoms are bolded to make them more obvious.


This morphine example is not the best to demonstrate univalent isosteres. Replacing an N-CH3 with an N-NH2 or (far worse) an N-Cl would be an uncommon and unstable substitution. In this example, however, these are the only possible replacements. A more representative univalent isosteric replacement would be to substitute a CH3 group on an aromatic ring with an NH2 or Cl.

Question 6

Everything behaves normally until OH and NH2 groups are placed in the ortho position. What is unique about OH and NH2? – Both are hydrogen bond donors and acceptors. Furthermore, the CH2CH2OH side chain on the ring of 11.g can undergo hydrogen bonding. If the side chain undergoes hydrogen bonding with an ortho substituent, then the chain will be folded back toward the ring. This could drop the activity of the analogue if strong binding requires the chain to be fully extended away from the ring.

Question 7

Sulfur is different because it is on the third row of the periodic table. Atoms get much larger in terms of atomic radius down a column of the periodic table. The atomic radii (Å) of carbon (0.77), nitrogen (0.70), and oxygen (0.66) are fairly close in magnitude and all considerably smaller than sulfur (1.04). A larger atomic radius gives rise to larger bonds. Since sulfur forms significantly longer bonds than carbon, nitrogen, and oxygen, sulfur is a poor isosteric replacement for the other three atoms.


Question 8

The key to this question is lining up the functional groups on the two molecules – ritonavir (11.75) in Figure 11.19 and A-74704 (11.74) in Figure 11.18. If you inspect the left sides of the two molecules, both contain an aromatic ring connected to a CH2 group and then a carbamate (NH-CO-O). The next group is a non-polar side chain. In 11.74 this group (isopropyl) fills pocket P2'. Presumably, in 11.75 this group (benzyl) will also fill pocket P2'. At this point, all the other non-polar groups can be aligned. Moving right-to-left on compound 11.75, the next benzyl group fills P1', the isopropyl fills P1, and the isopropyl on the thiazole ring fills P2.

Only the hydrogen bonds remain. Again, moving right-to-left on 11.75, the first hydrogen bond acceptor after the P2' group is the OH. This likely forms a hydrogen bond to the free water molecule in the HIV1 protease-binding pocket. Next down the line is an amide which can act as either a hydrogen bond acceptor or donor. Lastly, the urea carbonyl can form the second hydrogen bond with the active site water molecule.

Just because the groups can be aligned in this manner does not mean that this is indeed how the molecule interacts with the binding pocket of the enzyme. Regardless, this is a reasonable guess as to how ritonavir binds HIV1 protease.

Question 9

The binding energy for the 100 nM compound is −9.55 kcal/mol.


(10.1)

The binding energy for the 10 nM compound is −10.9 kcal/mol.

The impact of a magic methyl would need to be 1.35 kcal/mol.

Question 10

Instead of just adding a methyl to steer the sulfide chain into the correct conformation, the entire system could be locked into place with a ring. Either a five- or six-membered ring would be reasonable. The carefully planned pH effects from the additional alkyl substitution in cimetidine should be replicated in the new analogue.

A problem with adding additional substituents to restrict conformations is that the new groups create steric bulk that can prevent proper binding.

Question 11

The effect of the magic methyl is indeed magical. Getting a full 1.35 kcal/mol for a single methyl group is very impressive and requires the methyl group to virtually perfectly fit within a pocket.

Question 12

The equilibrium is shown below.

The ratio of the conjugate base (CH3NH2) to acid (CH3NH3+) is about 1 to 1,600.

(9.1)


That corresponds to approximately 0.06% in the conjugate base form.

Question 13

Scheme 11.3 is reproduced below. For a substituted imidazole (like histamine) only 3% of the mixture is in the protonated (acid) form, so 97% is the conjugate base. For burimamide, 40% is protonated as the acid and only 60% is the conjugate base. For metiamide, 20% is protonated and 80% is not. This is all the necessary information for the calculation.

For histamine…

(9.1)

For burimamide…

For metiamide…

Question 14

The pKa of dicyanomethane, CH2(CN)2, is approximately 12. With an alkyl substituent off the CH2 group, that value should increase a small amount. The pKa of this group is very close to that of an alcohol (16). In drugs, alcohols often act as hydrogen bond donors or acceptors. With acidity so close to an alcohol, the dicyanomethyl group should have some ability to donate a proton. Furthermore, partial deprotonation will allow the dicyanomethyl group to act as a proton acceptor as well.

Question 15

The isosteres are shown below. Exact modifications for the C- and N-termini can vary. Furthermore, an exact retroinverso isostere for the proline residue is impossible.



If a Hansch analysis minimally requires five compounds per molecular descriptor, then Equation 12.a with three descriptors should require 15 compounds in the training set. Only nine compounds were used to construct the Hansch equation.

Question 2

Lipophilicity often follows a parabolic trend as shown in Figure 12.3 (below). A molecule needs to have sufficient polarity to dissolve in aqueous media and be transported by the blood, but it also needs sufficient lipophilicity to diffuse across membranes. This polarity/lipophilicity balance can be approximated mathematically by including both a first power and second power term for lipophilicity.


1

1.5

2

2.5

3

3.5

-2.7 -2.2 -1.7 -1.2 -0.7 -0.2

log P

exp

tl.

log

1/C

Question 3

QSAR data is impacted in two important ways by accuracy of screening data. First, trends generated by a Hansch equation can only be as accurate as the activity information in the training set. If the training set contains much error, then the resulting Hansch equation will be fit to the erroneous data. The fit will likely be poorer. Second, use of an error-filled Hansch equation to predict the activity of previously unknown compounds will also be of lower quality. The old adage – “Garbage in; garbage out.” – definitely holds in Hansch equations.

Question 4

There are many ways to approach this problem. One is to consider how a result with a 4-Cl analogue might be misleading and steer the user to the wrong branch. Looking at the different outcomes with a 4-Cl group, there are three options.

4-Cl analogue is less potent The Cl either introduces an unfavorable steric interaction or acts as an EWG when an EDG is better. Analogues down this branch test either replacing the 4-Cl with an EDG group (4-OMe or 4-NMe2) or moving the Cl atom to the 3-position to avoid steric problems at the 4-position.

4-Cl analogue is equipotent This branch seems to emphasize steric bulk. The options on this branch either maintain bulk at the 4-position (4-Me or 4-t-Bu) or move the substitution to the 3-position.

4-Cl analogue is more potent This branch emphasizes the electron-withdrawing nature of the Cl group. All analogues in this branch place strong EWGs or multiple EWGs on the ring.

One possibility for a misstep would be if the 4-Cl analogue showed higher activity because the 4-Cl was filling a binding pocket, not because it was acting as an EWG. With higher activity the tree will steer the chemist to explore more and stronger EWGs when perhaps larger substituents (4-t-Bu) would provide the most active compound.


Question 5

Here are the two key equations.

(12.c)

(12.17)

Equation 12.17 can be rewritten to work with natural logarithms by adding a factor of 2.3.

Substitute Equation 12.c into Equation 12.17.

The value for A, the pre-exponential factor, is different for every reaction. In this case, however, all the reactions are very similar, so the A values for each are nearly identical. Therefore we can drop this term from the discussion. Isolating the Ea terms reveals that Es is indeed a function of Ea and directly proportional to ΔEa.

Question 6

The NH2 group is a very strong electron-donating group as reflected by its large and negative -value of –0.660. In order for the -value to become positive, the NH2 group must somehow become electron-withdrawing. The easiest way for this to happen would be for the NH2 group to be protonated to an NH3

+ group.

In general, -values are used to examine groups on an aromatic ring. Aromatic amines tend to have a pKa around 5. Therefore, if the compound is in a weakly acidic environment with a pH of 5 or less, the most common form of the amine will be as the protonated conjugate acid (R- NH3

+).


Question 7

The Hansch equation based on just log P is given below.

n = 8, r = 0.95

The Hansch equation based on both log P and (log P)2 is not meaningfully improved.

n = 8, r = 0.95

Addition of the (log P)2 term is not helpful. This is very surprising since lipophilicity normally shows a parabolic relationship with activity. This study, however, is not about biological activity. It is about how the insecticides concentrate into trout tissue. The more lipophilic a compound is, the more it should concentrate. That is exactly what the first equation (log P alone) states.

There is another possibility. It is possible that the training set of insecticides in this study do not have a broad enough range of log P values to allow the parabolic relationship of lipophilicity to bioconcentration to be seen. For example, if an insecticide were so highly lipophilic that it had zero solubility in lake and stream water, then it might be reasonable to see very low levels of bioconcentration of that insecticide in wildlife.

Question 8

Below is the Hansch equation from the provide parameters.

n = 9, r = 0.90

Because the sign on the ΔG term is negative, compounds that have a larger free energy of activation also have a smaller antimicrobial activity. With a positive sign to the coefficient on log P, more lipophilic compounds have a higher antimicrobial activity.

Question 9

The Hansch equation including log P and (log P)2 is shown below.

n = 7, r = 0.96

The value for log Pº can be determined by setting the first derivative of the Hansch equation with respect to log P to 0 and then solving for log P.

A plot of calculated log BR vs. log P (with the experimental data points) demonstrates the parabolic nature of the relationship of activity and log P.


2.5

2.6

2.7

2.8

2.9

3

3.1

3.2

3.3

3.4

3.5

4.5 4.7 4.9 5.1 5.3 5.5 5.7 5.9 6.1 6.3 6.5

log P

log

BR

log P o

It is unlikely that electronics or sterics play a very significant role in the activity of this series of compounds. The correlation coefficient for the Hansch equation is 0.96, and r2 is 0.93. The parameters in this equation account for 93% of the variability in the data set. That does not leave much room for the importance of electronic and/or steric parameters.

Question 10

While predicting activity for any series of compounds can be a significant challenge, if the compounds are very similar in structure, then the prediction is greatly simplified. To be able to predict activity trends for a diverse grouping of structures is a far harder task. The fact that CoMFA models can accommodate compounds with dissimilar structures and determine trends in biological activity attests to the power of the CoMFA method.

Question 11

Questions 7 (log P), 8 (ΔG and log P), and 9 (log P and (log P)2) all generated new Hansch equations. The plots of calculated vs. experimental activities for the three questions are shown below. In all cases, the trendlines for the calculated vs. experimental data plots have a correlation coefficient that is equivalent to the fit of the Hansch equation. This method, plotting calculated vs. experimental activities and determining the correlation of the best-fit line, is how the fit of the Hansch equation is determined.

n = 8, r = 0.95


0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

exptl. log BC

calc

. lo

g B

C

r = 0.95

n = 9, r = 0.90

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

1.00 1.50 2.00 2.50 3.00 3.50 4.00

exptl. log AA

calc

. lo

g A

A

r = 0.90

n = 7, r = 0.96


2.5

2.6

2.7

2.8

2.9

3

3.1

3.2

3.3

3.4

3.5

2.7 2.8 2.9 3 3.1 3.2 3.3 3.4

exptl. log BR

calc

. lo

g B

R r = 0.96

Question 12

A Craig Plot helps ensure that R-groups with a variety of properties are included in the training set of a Hansch analysis. Craig Plots work well for Hammett (electronics) and Hansch (lipophilicity) constants. Analogues with R-groups that reside in each of the four quadrants of the Craig Plot are readily available.

A Craig Plot for Hansch (lipophilicity) and Taft (sterics) is not as straightforward. The two parameters are not independent. Groups that provide high steric bulk tend to be more lipophilic. Therefore, they do not scatter among the four quadrants of a Craig Plot. Performing a Hansch analysis with parameters for both lipophilicity and sterics is difficult to accomplish meaningfully. Very careful attention must be paid to the training set selection.

Question 13

The pKa value is an electronic parameter in the sense that it measures how easily a structure can accept electron density from the breaking of a C-H bond. Stronger acids (lower pKa values) accept electron density well, while weaker acids (higher pKa values) accept electron density poorly.

The pKa value is also lipophilicity parameter since on one side of the acid-base equilibrium the compound will be neutral (i.e. more lipophilic) and on the other side it will be charged (i.e. less lipophilic). Most commonly, acids (HCl) are neutral and their conjugate bases are anions (Cl−). Other charge possibilities can and do occur. Rarely, both an acid and conjugate base may have a charge, such as with bicarbonate (HCO3

−) and carbonate (CO3−2). These are exceptions. Since the lipophilicity

of the species will change dramatically with the charge state of the structure, pKa has a huge influence on lipophilicity. Indeed, this is the reason that the partition coefficient (P) has been elaborated to the distribution coefficient (D) with accommodation of a pKa term.

Therefore, while pKa may be a common parameter in Hansch analyses, it is not as pure of a molecular property term as the Hammett or Hansch constant.


Question 14

Nitrogen mustards are DNA alkylating agents (Chapter 6). These are toxic compounds. It is therefore no great surprise that the Hansch equation for biological activity is nearly identical to the Hansch equation for toxicity. Finding a compound with a meaningful difference between efficacy and toxicity would be a challenge.

The only room for hope is in the correlation coefficients. The Hansch equation for activity has a somewhat lower correlation coefficient. There may be third property (sterics?) that is tied to activity but not toxicity. Through this third parameter, a potent and safe analogue may be developed.

Question 15

This question is just like the Sample Calculation in the section on the Taft parameter except it is a comparison for ethyl formate and ethyl propionate instead of ethyl formate and ethyl acetate. The set up of the calculation is identical. It begins with Equation 12.17.

(12.17)

We already know that ethyl formate hydrolyzes 17.4 times faster than ethyl acetate. If we can calculate how much faster ethyl acetate is than ethyl propionate, we can multiple the two factors together and relate the rate of hydrolysis of ethyl formate to ethyl propionate.

(12.17)

If ethyl formate hydrolyzes 17.4 times faster than ethyl acetate and ethyl acetate hydrolyzes 1.17 times faster than ethyl propionate, the ethyl formate hydrolyzes 20.4 (17.4×1.17) times faster than ethyl propionate.


First, one assumption needs to be made clear – the product has an e.e. of 100%. With this assumption one can then state that the product does not contain any of the undesired enantiomer.

In order to solve this question, one must remember that the starting material was racemic – half one enantiomer and half the other enantiomer. Although Scheme 13.3 starts with 67.1 g of material, the math is simpler if we start with 100 g. So, in the 100 g of starting material, 50 g are the desired


enantiomer, and 50 g are the other enantiomer. The resolution affords 39 g (39%) of the desired enantiomer. Therefore, the remaining material must contain 11 g (50 − 39) of the desired enantiomer and 50 g of the undesired enantiomer. The e.e. of the remaining 61% of the starting material can be readily calculated with Equation 13.1.

(13.1)

Question 2

The first step in this problem is to determine the molecular formula and then molecular weight of all the starting materials and the desired product. It will not hurt to also calculate the formula and molecular weight of all the waste products and make sure the starting materials and products balance.

Atom economy for the reaction can be quickly calculated with Equation 13.2.

Question 3

Parenteral drugs are administered by injection (see Chapter 3 for a review). This question is ultimately about bioavailability. When a drug is injected, the entire dose (with all impurities) is placed in the body. In contrast, impurities in an oral drug may never be absorbed from the digestive system. Because the bioavailability of metals from the digestive system is typically well under 100%, drug regulations allow higher levels of metal impurities in oral drugs than injectable drugs.

Question 4


In a kinetic resolution, the enzyme acts upon two different substrates. One is preferred and the other is not. As the resolution progresses, the rate of the reaction slows for two reasons. One, the concentration of the substrate decreases over time. With a lower substrate concentration, the rate of conversion slows. The Michaelis-Menten equation shown below (Equation 4.11) reflects the dependence of the rate of conversion upon the substrate concentration.

(4.11)

Two, not only is the substrate concentration decreasing, but the concentration of the preferred substrate accounts for the vast majority of the decrease early in the reaction. The two substrates have different affinities for the enzyme. In other words, each substrate has its own Km value. Km of the preferred substrate is lower (higher affinity) than the Km of the not preferred substrate. As the preferred substrate is consumed, the enzyme more and more begins to act on the not preferred substrate. Since the rate of conversion of the not preferred substrate is low, the observed overall rate of the reaction slows with percent conversion.

Question 5

The seven methods fall under two larger categories: resolutions and asymmetric synthesis. Examples of resolutions include (1) kinetic resolutions, (2) separations of diastereomeric salts, and (3) chiral chromatography. Examples of asymmetric synthesis include (4) purchase of enantiomerically pure materials, (5) use of a chiral auxiliary, (6) asymmetric catalysis, and (7) microbial processes.

The reaction associated with this question is depicted below. A key feature is that the synthesis begins with a racemic material. This fact alone makes the reaction a resolution. It may then be subcategorized as a kinetic resolution, diastereomeric salt separation, or application of chiral chromatography. In the reaction, enzymes within a microbe (Candida antarctica) preferentially perform a hydrolysis on one enantiomer over the other. This is a kinetic resolution.

Question 6

Maximizing atom economy is Principle #7 of green chemistry, but it is just one measure of “greenness”. Although having a high atom economy does not necessarily merit the “green” label for a synthesis, it does mean that the synthesis is possibly “greener” than routes with a lower atom economy.

Question 7

This question is an application of Equation 13.5.

(13.5)


This is probably an overestimation of the waste since corticoids have been widely used for decades. Their synthesis and isolation have received much attention and are likely optimized well beyond the typical pharmaceutical synthesis.

Question 8

This question is very similar to Question 2. The first step is to determine the molecule formula and molecular weight of all the starting materials and the product.

Question 9

Calculating the E-factor of the reaction requires the masses of each reagent or solvent used. If the volume is provided for a reagent, the density of the material must be used to convert volume to mass. All masses are reported in grams to keep the conversions to a minimum in the final calculation.

Reagent name Density (g/mL) Amount used Mass used

Salicylic acid (11.e) n/a 100 g 100 g

Acetic anhydride 1.082 200 mL 216 g

Phosphoric acid (85%) 1.685 1 mL 2 g

Water 1.000 2.8 L 2,800 g

Ethanol 0.789 700 mL 552 g

(13.5)

Note that the mass of the product must be subtracted from the masses of the starting materials in the calculation.


Question 10

Effective mass yield is very similar to E-factor with the exclusion of benign starting materials. All the required masses have already been calculated in Equation 9. Both water and ethanol can be excluded from the waste calculation.

(13.7)

(13.7)

Question 11

Acetamidine and acetic acid are ideally suited to form strong hydrogen bonds that are often found in cocrystals (see Figure 13.4). While the complex may be interpreted as a cocrystal, the pKa values of acetamidine and acetic acid indicate that the two compounds will undergo a strong acid-base reaction. The pKa difference is 6.7, well beyond the rule-of-thumb difference of 3 to qualify as a salt. The crystal is almost certainly a salt, not a cocrystal.

Question 12

Below is a version of Scheme 13.16 that reflects changing both reagents to their methyl analogues.

Using methyl chloroacetate and sodium methoxide raises the atom economy to 42%, a boost of only 2%.


Question 13

The best solvent is the mixture of water and N,N-dimethylformamide. On first glance, the key factor might be seen as solubility at 20 ºC. A lower solubility at low temperature translates to less lost material in the recrystallization. The more important factor, however, is the percent change between the high and low temperature values.

Solvent(s) Solubility at 20 ºC

(g/L)

Solubility at 60 ºC

(g/L)

Percent

recovery

Water 10 55 18

Methanol 5 15 33

N,N-Dimethylformamide 10 20 50

Water + Methanol 20 100 20

Water + Acetone 175 350 50

Water + Acetonitrile 400 1200 33

Water + N,N-Dimethylformamide 35 510 7

Question 14

In a resolution, one of the diastereomers must precipitate from solution. Larger structures, including those that include benzene rings, tend to have lower solubility than smaller structures. It is only natural to use large acids/bases in resolutions to boost the molecular weight of the salt to induce precipitation.

Question 15

Below are the graphs in Figure 13.11.


0

20

40

60

80

100

0 20 40 60 80 100

percent conversion

e.e.

product

startingmaterial

0

20

40

60

80

100

0 20 40 60 80 100

percent conversione.

e.

product

startingmaterial

In the graph on the left, the product initially forms with a 60% e.e. With Equation 13.1, we know the number for e.e. Rearrange the equation for major/minor, and that is the E-value of the enzyme. In this case, the E-value is 4.

(13.1)

In the graph on the right, the product initially forms with an 85% e.e. (It is hard to read precisely, but the actual number is 87.5%.) Following the same logic as the previous calculation, the E-value works out to be 12. (With 87.5% e.e., the E-value is 15.)


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Solucionario Modern Drug Discovery