Page 166 - Medicinal Chemistry Self Assessment
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2.12  Chlorpropamide and Other Sulfonylureas        155



                    of chlorpropamide would be expected to be more acidic (pK =4.9) than the one present within the
                                                                           a
                    structure of tolbutamide (pK =5.4) due to the electron withdrawing character of the halogenated
                                              a
                    aromatic ring.

                 3.  Using your answer from question 2, calculate the percent of tolbutamide that will be ionized at an
                    intestinal pH=6.1.

                    Answer
                    As determined in question 2, the pK  of the sulfonylurea groups of tolbutamide is 5.4. To solve this
                                                     a
                    problem, we need to use the Henderson-Hasselbalch equation. Because the functional group is acidic,
                                           –
                    the ionized form (R-SO N CO-R) is the base form and the unionized form (R-SO NHCO-R) is the acid
                                         2
                                                                                            2
                    form.
                                  [Base Form]
                    pH= pK +log
                            a
                                  [Acid Form]
                                  [Base Form]
                    6.1= 5.4+log
                                  [Acid Form]
                             [Base Form]
                    0.7 = log
                              [Acid Form]

                          [Base Form ]   5.01   [Base Form ]
                    5.01=              or     =
                           [Acid Form ]    1    [Acid Form ]


                    This ratio indicates that for every one molecule that contains the functional group in the acid (or
                    unionized) form, there are 5.01 molecules that contain the functional group in the base (or ionized)
                    form. The following equations can then be used to correctly calculate the percent of the molecules
                    that are ionized and the percent that are unionized.

                    5.01 molecules in base form + 1.0 molecule in acid form = 6.01 total molecules

                    Base form=Ionized form and Acid form=Unionized form

                                            5.01 Molecules in Ionized Form
                    Percent in Ionized Form=                             ×100%= 83.4%
                                                 6.01 Total Molecules
                                               1 Molecule in Unionized Form
                    Percent in Unionized Form=                             ×100%=16.6%
                                                   6.01 Total Molecules

                    The question asks for the percent that will be ionized, so the correct answer is 83.4%.


                 4.  The mechanism of action of this class of drugs involves the ability to interact with the ATP-sensitive
                    potassium channels in the pancreas. Using the structure of tolbutamide, identify the types of binding
                    interactions that would be possible between its functional groups and a protein ion channel. Also
                    identify amino acids present within this ion channel whose side chains could participate in the inter-
                    actions identified. Assume a plasma pH=7.4 for all ionizable functional groups.
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