Exam 8 505 Harrelson Medchem: Gaba

Front 1

1. List the major sedative-hypnotic drug classes.

Back 1

1. Barbiturates
2. Propanediol carbamates
3. Benzodiazepines
4. “Z drugs” (ALL start with Z except Lunesta (Eszopiclone)
5. Melatonin receptor agonist
6. Piperidinediones
7. Acetylene derivatives
8. Chloral hydrate
9. SSRIs (previous lecture)

Front 2

2. Describe the properties necessary for a barbiturate derivative to have good sedative-hypnotic activity.

Back 2

1. Be a weak acid.
2. Have the appropriate lipid/water partition coefficient (log P).

Front 3

3. Define log P and explain what high and low values of log P mean.

Back 3

1. It is a physical chemical property to estimate how these agents are going to act in the body. The log of the (concentration of the drug in octanol) / (concentration of the drug in water). Drug in octanol is acting as the lipophilic media.
2. High = lipophilic, low = hydrophilic

Front 4

4. Explain how log P is determined and what it can tell you about a drug.

Back 4

2. If it is 100/2, you would say that log p = 2. The log p scale is 1-5 (x axis vs activity on y axis) and you look for an optimal log P. Log P is physical chemical quantitative measure of lipophilicity. Negative values would by hydrophilic. Positive values are lipophilic.

Front 5

5. Explain how barbituric acid differs structurally from the barbiturate derivatives that are used as drugs.

Back 5

1. Barbituric acid is too acidic to act as a sedative-hypnotic
1. The acid-base properties can be modified via substitution at the 5a and 5b positions.

Front 6

7. Explain how resonance stabilization impacts the acid/base properties of barbituric acid (draw the resonance structures).

Back 6

The acid-base properties of barbiturates are the result of lactam-lactim and ket-enol tautomerism

Front 7

8. Describe how the pKa’s of barbituric acid derivatives are affected when substitutions are made at the 5 position and how this change affects pharmacology.

Back 7

1. When 5a and 5b = Hydrogens, PKA = 4.1 (too acidic)
2. 5a, 5b = alkyl or other hydrocarbon containing groups = pKa of 7-8. By doing this you loose the keto-enol tautomerism.

Front 8

9. Given the pKa of a barbiturate, determine the ratio of unprotonated/protonated and ionized/unionized molecules at a given pH.

Back 8

%I = 100/(1+10^(Ph-pKa)) IF a base, else swap ph/pka if acid

Front 9

10. Describe the pharmacological effects observed when the nitrogen atoms of the barbiturates are modified. Provide a chemical explanation for these effects.

Back 9

1. If we got rid of BOTH of the protons attached to the nitrogens, it will abolish activity (outside of the physiological PH range. We need at least one acidic proton. If you have both 5 positions and one of the nitrogens changed to an alkyl group – you still have activity.

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11. Describe the effect of different alkyl group substitutions at the 5 position of the barbiturates.

Back 10

1. Hypnotic activity increases with increasing lipophilicity until the total number of carbons at both the C-5 positions is between 6 and 10. Further increases lead to decreases in hypnotic activity.

2. Unsaturated allyl, alkenyl, and cycloalkenyl derivatives are more potent than the corresponding saturated analogs

Front 11

12. Describe the effect of incorporating stereochemistry into the 5 position of the barbiturates.

Back 11

1. Branched chain isomers have greater lipid solubility and hypnotic activity than straight chains
2. Stereoisomers have the same potency (approximately).

Front 12

13. Describe the impact of modifications at the carbonyl carbons of the barbiturates.

Back 12

1. The substitution of a sulfur atom for the oxygen at C-2:
1. Increases lipid solubility.
2. Produces quicker onset and short duration (thiobarbituates used for i.v. anesthesia).

Substitution at more than one carbonyl oxygen with sulfur decreases hypnotic activity (i.e., there is an optimal lipophilicity range).

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14. Describe how the barbiturates are categorized and list the categories.

Back 13

1. LONG = SEDATIVES
2. SHORT-INTERMED = HYPNOTICS
3. Ultrashort acting: i.v. ANESTHETICS

Front 14

15. Describe the factors that affect a barbiturate’s duration of action and the role that acidity/basicity and lipophilicity have in determining duration of action.

Back 14

1. Renal Excretion
2. Metabolism
3. Redistribution (for short acting agents)

That is, very little of the parent drug will be excreted unchanged in the urine. The causes are:

Poor kidney filtration: Barbiturates are lipophilic, weakly acid drugs that are extensively bound to plasma proteins. Therefore, they are poorly filtered (i.e., they have to disassociate from the proteins in order to go into the urine).

Extensive reabsorption: These are weakly acidic drugs and the urinary pH = 6. So the barbiturates are mostly unionized and, since they are also relatively lipophilic molecules, they are reabsorbed.

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16. Describe what could be done in cases of barbiturate overdose to eliminate the drug more rapidly. This involves predicting ratios of unionized to ionized drug at different pHs.

Back 15

1. Sodium bicarbonate can be administered so that the urinary pH is increased. This increases the amount of drug existing in the ionized form, increases water solubility, and excretion (less reabsorption).

Front 16

17. Explain why the duration of action of most barbiturates is determined by the rate of metabolism. Which type of barbiturates is an exception?

Back 16

1. This is true for all BUT the ultrashort acting barbiturates.
2. Occurs because metabolism = more hydrophilic = higher rate of excretion.

Front 17

18. Describe the common metabolic routes of barbiturate metabolism and how the substituents at the C-5 position impact metabolism (e.g., what happens if R = an aromatic ring?).

Back 17

Common metabolic routes include hydroxylation at 5-C or para position, followed by glucuronidation of hydroxylated metabolites

Aliphatic hydroxylation at one of the C-5 substituents (ω-1 hydroxylation)

Aromatic hydroxylation is a relatively slow process for barbiturates compared to ω-1 hydroxylation. Therefore if this is the primary route of metabolism for a specific barbiturate then it will have a longer half-life.
E.g., phenobarbital undergoes hydroxylation at the para position. Therefore it has a long half life (100 hours). Phenobarbital is a long acting sedative.

Glucuronide conjugation of hydroxylated metabolites

Some barbiturates also undergoes an unusual N-glucuronidation directly on the ring nitrogen.

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19. Describe the adverse events associated with the barbiturates.

Back 18

2. Low Therapeutic Index
3. Tolerance and Dependence
4. Respiratory and Cardiovascular Depression

Front 19

20. Draw the benzodiazepine ring and label w/descriptions of the common structural groups found at R1, R2, and R3.

Back 19

1. R1: Alkyl groups at this position are common but aren’t essential for activity.
2. R2: Most of the time these are aryl substituents. Ortho-substitution with electron-withdrawing groups (e.g., Cl or F) enhance potency.
3. R3: Electron withdrawing groups enhance potency.
4. Benzodiazepine structure can profoundly impact metabolism. This is important because rates of metabolism will impact DOA and therefore, how the particular benzodiazepine is used.

Front 20

21. Given a set of structures, recognize a long-lived active metabolite of diazepam.

Back 20

More lipophilic = short DOA but long elimination t ½ in lipid stores

Front 21

22. Explain chemically why oxazepam and lorazepam can be administered as a racemate.

Back 21

1. Equal activity, self racemizing

Front 22

23. Describe and recognize the type of substitutions at the R1 position in the benzodiazepines that lead to long-lived active metabolites (that increase the effective half-life).

Back 22

1. LONG = Alkyl group that is easily N-dealkylated
2. SHORT =
1. no alkyl group that’s easily N-dealk
2. –OH grp = easy glucuronidation = ez elimination

Front 23

24. Describe and recognize the type of substitutions at the R1 position in the benzodiazepines that lead to compounds that have relatively short effective half-lives.

Back 23

1. SHORT ACTING a group at R1 that isn’t easily N-dealkylated
2. Shorter acting agents have structural features that allow them to be eliminated
more quickly (e.g., triazole ring doesn’t allow formation of nordiazepam, α-hydroxy group allows rapid glucuronidation

Front 24

select the fastest metabolized drug(s) of the group and explain which feature is responsible and how it works.

Back 24

Triazolam and Estazolam due to triazole ring, triazole ring doesn’t allow formation of nordiazepam, α-hydroxy group
allows rapid glucuronidation

Front 25

Given a set of molecules recognize the common metabolite of many benzodiazepines and explain why it is significant.

Back 25

NORDIAZEPAM: long-lived metabolite resulting from
N-dealkylation = explains SAR for long t ½

Front 26

29. Describe the metabolic processes that benzodiazepines undergo in order to generate the long lived active metabolite.

Back 26

1. N-dealkylation

Front 27

30. Describe how metabolism and duration of action can impact the selection of a benzodiazepine as a hypnotic (e.g., compare flurazepam metabolism to triazolam metabolism).

Back 27

1. Flurazepam = long t ½  ez N-dealk
2. triazolam = short t ½  no N-dealk

Front 28

31. Identify “Z drugs” that are preferred for sleep maintenance and the one that is preferred for initiating sleep. Describe the reasoning for these preferences.

Back 28

1. INITIATE SLEEP:
a. Zaleplon (Sonata) for initiating sleep
b. Short elimination t ½ vs zolpidem (Ambien)

2. MAINT SLEEP:
a. Zolpidem (Ambien) for maint sleep, opp reasons
b. Eszopiclone (Lunesta)
1) Long t ½

Front 29

32. Describe how the pharmacokinetics for “Z drugs” differ from the benzodiazepines.

Back 29

Bind to GABA-A but the DOA not impacted by the formation of long-lived active metabolites

Front 30

33. Describe how the mechanism of action for ramelteon differs from the “Z drugs”, benzodiazepines, and barbiturates.

Back 30

1. Binds MELATONIN receptor vs GABA-A

Front 31

34. Describe how piperidinediones are similar to and different from the barbiturates.

Back 31

1. SIMILAR to BARBS
a. Tolerance and dependence
b. Induction of δ ALA enzyme involved w/heme biosynth
2. Only advantage over barbiturates is that they have a lower tendency to cause respiratory depression.

Front 32

35. Describe the chemical processes involved in the activation of chloral hydrate to trichloroethanol. (be sure to draw the structure and rxn here)

Back 32

Chloral Hydrate is REDUCED TO TRICHLOROETHANOL (think bis-alcohol to R-OH w/3 Cls)
2. Chloral Hydrate –(-OH)H2O + Cl3C-HC=O –(NAD/NADH)Cl3-CH2OH

Front 33

36. Describe the mechanism that explains how ethanol acts synergistically with chloral hydrate to generate hypnotic effects.

Back 33

1. The oxidation of ethanol to an ALDEHYDE generates NADH, which is necessary for the reduction of trichloroethanal (aldehyde w/chlorinated methyl) to trichloroethanol.

2. “Mickey Finn” is a combination of chloral hydrate and ethanol.

Front 34

examine and consider these GABA medchem structs, identify as many as possible based on SARs

Back 34

none

Front 35

examine and consider these GABA medchem structs, identify as many as possible based on SARs

Back 35

none