Lecture no 4
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Lecture no 4 Lecture no 4 Presentation Transcript

  • Medicinal Chemistry Fourth Lecture
    • Physicochemical Properties of Drugs
  • Partition Coefficient of Drugs
    • Partition Law: “ A given substance, at a given temperature, will partition itself between two immiscible solvents in a constant ratio of concentrations”
    • Consider 100 mg of a drug added to an immiscible mixture of a 50 mL organic solvent (e.g. octanol) and 50 mL of water in a separating funnel.
    • After the mixture was allowed to equilibrate, the mass of the drug in the organic phase was determined to be 66.7 mg.
    • The mass of the drug in the water phase will be
    • 100-66.7=33.3 mg
    • Concentration of the drug in the organic phase will be
    • 66.7/50 = 1.33 mg/mL
    • Concentration of the drug in the aqueous phase will be
    • 33.3/50= 0.67 mg/mL
    • P = [Organic]/[Water]= 1.33/0.67= 2
  • How Can the Partition Coefficient be Used?
    • Prediction of:
    • Absorption
    • Distribution
    • Elimination
    • of the drug in the body.
    • Is the above given value of P pH dependent?
    • To ensure that the above equation applies in the case of:
    • Acidic substances (the pH of the measurement must be low)
    • Basic Substances (the pH of the measurement must be high)
    • Why ?
    • Alternatively (in the case of drug ionization):
    • The above expression is termed the “apparent partition coefficient (P app )” while the similar expression for the unionized drug is termed “the true partition coefficient”.
    • P app is dependent on the proportion of substance present in the solution which is pH dependent.
    • P app = P X f un ionized
    • f un ionized = 1, P app = P, the drug is unionized
    • Consider the following situation:
    • The previous drug is found to be 40 mg in the organic phase, while the remaining drug in the water phase is 66.7% ionized.
    • The mass of the drug in the water phase = 100-40 = 60 mg
    • The mass of the ionized drug in water = 60 X 0.67= 40 mg
    • The mass of the unionized drug in water = 60 X 0.33 = 20 mg
    • Concentration of the drug in the organic phase = 40/50= 0.8 mg/mL
    • Concentration of the unionized drug in water = 20/50 = 0.4 mg/mL
    • Concentration of the total drug in water = 60/50 = 1.2 mg/mL
    • The percentage of the drug extracted into the organic phase = (40 mg/100 mg) X 100 = 40 %
    • The partition coefficient of the unionized drug (the true P) is given by:
    • P = [drug] in organic phase/[unionized drug] in water
    • = 0.8/0.4 = 2
    • P app = [drug] in organic phase/ total [drug] in water phase = 0.8/1.2 = 0.67
    • P app can be checked:
    • P app = P X f unionized
    • 0.67 = 2 X 0.33
    • P values can vary from small fractions to several thousends therefore the log of P is usually used especially in QSARs
    • P represents a measure of the ease of movement of drugs through biological membranes.
    • The accuracy of the correlation depends on the system used as a model for the membrane.
    • n-octanol, chloroform, and olive oil are used to model the lipid component while water and buffered solutions are used for the aqueous medium.
    • n-octanol-water is the most commonly used model.
    • Matching the organic phase with the area of biological activity leads to a better correlation:
    • n-octanol gives the best consistency for the drugs that are absorbed in the GI tract
    • Less polar solvent like olive oil give more consistent correlation for drugs crossing the blood-barrier membrane.
    • SARs and QSARs
    • Compounds that are similar to pharmacologically active drugs are most likely to be active themselves.
    • These drugs may have:
    • 1. Similar activity, different potency
    • 2. Different activity
    • Structurally related activities are referred to as “ SARs”
    • These studies on the lead and its analogues may lead to the identification of the “pharmacophore” and the part of the structure that is related to the side effects.
    • This information can in turn be used to design drugs that are
    • More potent and which have fewer side effects.
    • SARs are usually carried out by making a small change to the lead and assessing the concomitant change in the biological activity.
    • These changes can be in:
    • The size and shape of the carbon skeleton
    • The nature and degree of substitution
    • The stereochemistry of the lead
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    • The selection of the change depends on many consideration:
    • Activities of compound with similar structures
    • Biochemistry of the intended analogue.
  • Changing Size and Shape
    • Can be done by:
    • Changing the number of methylene groups
    • Increasing or decreasing the degree of unsaturation
    • Introducing or removing a ring system
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  • Introduction of New Substituents
    • These substituents will either:
    • Replace an existing one
    • Occupy a previously unsubstituted position
    • A great deal of information has been collected over the years about the change of properties that will be caused by the introduction of new substituents.
    • This fact makes it possible to make some generalizations
    • Substituent choice will depend on the desired function
  • Incorporation of a group in an unsubstituted position
    • Leads to change in the size and shape of the molecule
    • May Introduce a chiral center
    • May impose some conformation restrictions on some of the bonds of the analogue
    • May lead to:
    • increased rate of metabolism
    • Reduction in the rate of metabolism, or
    • Alternative route of metabolism
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    • The End