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Section 3 General Self Assessment Answers 2.3 Solving pH/pKa Problems 1.Chapter Shown 3 below are the structures of cefotaxime, nitrofurantoin, atenolol, and ezetimibe. Each of these drug molecules contains one ionizable functional group. The pKa values have been provided. Solving pH/pKa Problems a. Match the pKa values provided with the appropriate functional groups. For each functional group, identify the name of the group and whether it is acidic or basic. 1. Shown below are the structures of cefotaxime, nitrofurantoin, atenolol, and ezetimibe. Each of these drug b. For each functional group indicate whether it would be primarily ionized or primarily unionized at a moleculesstomach contains pH one= 1.8, ionizable a urinary functional pH = 6.1, group. or aThe plasma pKa values pH = 7.4.have Provide been provided. an explanation for your responses for cefotaxime at a plasma pH = 7.4, nitrofurantoin at a urinary pH = 6.1, and atenolol at a stomach Answer:pH = 1.8. Answer: Imide Acidic Cefotaxime NiNitrofurantointrofurantion pK = 3.4 a Carboxylic acid pKa = 7.1 Acidic Aromatic hydroxyl (phenol) Secondary amine Acidic Basic Atenolol pKa = 9.6 Ezetimibe pKa = 10.2 Cefotaxime contains an acidic carboxylic acid. At a plasma pH = 7.4, the environment (i.e., the pH) is more basic than the functional group (i.e., the pH > pKa). An acidic functional group will be primarily ionized in 3. Showna basic below environment. is the structure of natamycin. It contains two functional groups that could be potentially ionized. Nitrofurantoin contains an acidic imide group. At a urinary pH = 6.1, the environment (i.e., the pH) is more The pKa values for natamycin are 4.6 and 8.4. acidic than the functional group (i.e., the pH < pKa). An acidic functional group will be primarily unionized in an acidic environment. Atenolol contains a basic secondary amine. At a stomach pH = 1.8, the environment (i.e., the pH) is more O OH acidic than the functional group (i.e., the pH < pKa). A basic functional group will be primarily ionized in an acidic environment. O OH H 3 C OOOH 1 COOH O O CH3 Natamycin H O OH NH 2 Page1of2 Chapter 3 Solving pH/pKa Problems 1. Shown below are the structures of cefotaxime, nitrofurantoin, atenolol, and ezetimibe. Each of these drug molecules contains one ionizable functional group. The pKa values have been provided. Answer: 2 Medicinal Chemistry Self Assessment Imide Acidic Drug (pKa Value) Stomach (pH = 1.8) Urine (pH = 6.1) Plasma (pH = 7.4) Cefotaxime (3.4) Primarily unionized Primarily ionized Primarily ionized Nitrofurantoin (7.1) Primarily unionized Primarily unionized Primarily ionized Atenolol (9.6)Cefotaxime Primarily ionized Primarily ionized NitrofPrimarilyuranti onionized pK = 3.4 Ezetimibe (10.2)a Primarily unionizedCarboxylic acid Primarily unionized pKPrimarilya = 7.1 unionized Acidic 2. In the previous question, we examined four pKa values in three different environments for a total of 12 different scenarios. Which of these 12 scenarios allow you to use the Rule of Nines to calculateArom atic the percent of ionization of the functional group in the specific environment? Identify the specifichydrox yl scenarios and use the Rule of Nines to calculate the percent of the functional group that would( phebe nol) Secondary amine ionized. Acidic Basic Answer: Atenolol To use the Rule of Nines, the difference between the pH and the pK must be an integer (i.e., 1, 2, pK = 9.6 Eza etimibe 3). In evaluating the above 12a scenarios, there are only two scenarios that meet this criterion, cefo- pKa = 10.2 taxime (pKa = 3.4) in a plasma pH = 7.4 and nitrofurantoin (pKa = 7.1) in a urinary pH = 6.1. For cefo- taxime, |pH – pK | is equal to 4; thus, there is a 99.99:0.01 ratio. Because the carboxylic acid (pK = 3.4) a a would be primarily ionized in a basic environment (pH = 7.4), we can use this ratio to determine that it would be 99.99% ionized. For nitrofurantoin, |pH – pK | is equal to 1; thus, there is a 90:10 ratio. a The imide (pKa = 7.1) functional group is acidic, so it would be primarily unionized in an acidic envi- 3. Shownronment below (pHis the = structure6.1). We canof natamycin. use this information It contains two to predictfunctional that groups it would that couldbe 10% be ionized.potentially ionized. The pKa values for natamycin are 4.6 and 8.4. 3. Shown below is the structure of natamycin. It contains two functional groups that could be poten- tially ionized. The pKa values for natamycin are 4.6 and 8.4. O OH O OH H 3C OOOH COOH O O CH3 Natamycin H O OH NH2 a. Match the pKa values provided to the appropriate functional groups and identify if the functional group is acidic or basic. b. Using the Henderson-Hasselbalch equation, calculate the percent ionization that occursPage for each1of 2 of these functional groups at an intestinal pH = 6.2. a. Using the Henderson-Hasselbalch equation, calculate the percent ionization that would occur for each of these functional groups at an intestinal pH of 6.2. 2.3 Solving pH/pK Problems 3 a Answer: Answer: a. O OH O OH H 3C OOOH Carboxylic acid COOH (pKa = 4.6) Acidic O O CH3 Natamycin H O OH NH2 Primary amine (pKa = 8.4) Basic b. For the carboxylic acid, the pKa = 4.6 and the pH = 6.2. Using the Henderson-Hasselbalch equation gives the following: 4. The most basic functional group present within the structure of ranitidine has a pKa value of 8.2. Identify this [Base Form] functional group and6.2 calculate = 4.6+log the pH that is necessary for this functional group to be 70% ionized. [Acid Form] Answer: Solving the equation provides a ratio of [Base Form]/[Acid Form]. [Base Form] 1.6 = log The electron withdrawing property of the nitro [Acid Form] group greatly decreases the pKa value for the [Base Form] 39.8 [Base Form] adjacent nitrogen containing group. 39.8 = or = [Acid Form] 1 [Acid Form] Tertiary amine pK = 8.2 a Ranitidine For every molecule that contains this functional group in the acid form, there are 39.8 molecules that contain this functional group in the base form. Because the functional group is acidic, the base form is equal to the ionized form, and the acid form is equal to the unionized form. 39.8 molecules in base form + 1.0 molecule in acid form = 40.8 Total Molecules Base Form = Ionized Form, and Acid Form = Unionized Form for This Functional Group 39.8 Molecules in Ionized Form Percent in Ionized Form = x 100% 40.8 Total Molecules Percent in Ionized Form= 97.5% For the primary amine, the pKa = 8.4 and the pH = 6.2. Using the Henderson-Hasselbalch equation gives the following: [Base Form] 6.2 = 8.4 +log [Acid Form] Page2of2 a. Using the Henderson-Hasselbalch equation, calculate the percent ionization that would occur for each of these functional groups at an intestinal pH of 6.2. 4 Medicinal Chemistry Self Assessment Answer: Solving the equation provides a ratio of [Base Form]/[Acid Form]. [Base Form] O OH −2.2 = log O OH [Acid Form] [Base Form]H 3 0.0063C OO [Base Form] OH Carboxylic acid 0.0063 = or = COOH [Acid Form] 1 [Acid Form] (pKa = 4.6) Acidic For every molecule that contains this functional group inO the acidO form,CH there are 0.0063 molecules that contain this functional group in the base form. Because the functional3 group is basic, the base form is equal to the unionized form, andNa thetam acidycin form is equal to the ionized form. 0.0063 molecules in base form + 1.0 molecule in acid formH O = 1.0063 TotalOH Molecules NH2 Primary amine Base Form = Unionized Form, and Acid Form = Ionized Form for This Functional Group (pKa = 8.4) 1 Molecule in Ionized Form Basic Percent in Ionized Form = x 100% 1.0063 Total Molecules Percent in Ionized Form = 99.4% 4. The most basic functional group present within the structure of ranitidine has a pKa value of 8.2. Identify this functional group and calculate the pH that is necessary for this functional group to be 70% ionized. 4. The most basic functional group present within the structure of ranitidine has a pKa value of 8.2. Answer:Identify this functional group and calculate the pH that is necessary for this functional group to be 70% ionized. Answer: The electron withdrawing property of the nitro The electron withdrawing property of the nitro group greatly decreases the pKa value for the group greatly decreases the pKa value for the adjacentadjacent nitrogen-containing nitrogen containi group.ng group. Tertiary amine pK = 8.2 a Ranitidine We can use the Henderson-Hasselbalch equation to solve this problem. Because the ionized form of a basic functional group can also be designated as its protonated form or its conjugate acid form, either of the following equations can be used. [Base Form] pH = pK +log a [Acid Form] [Unprotonated Form] or pH = pK +log a [Protonated Form] Therefore, if 70% of this functional group is ionized: [30] pH = 8.2 + log [70] [30] pH = 8.2 + log = 8.2 +(− 0.37) = 7.83 Page2of2 [70] Because a pH = 7.83 does not exist physiologically, this percent ionization could only occur in an exogenously prepared solution..
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    SYNTHESIS AND STRUCTURAL STUDIES OF NOVEL ACYCLIC UNSYMMETRIC FUNCTIONALIZED IMIDES AND MECHANISTIC STUDIES ON E. COLI ASPARAGINE SYNTHETASE B By JOSE G. ROSA RODRIGUEZ A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1996 To my wife and family. ACKNOWLEDGMENTS First I would like to thank my advisor, Dr. Nigel G. J. Richards, for his guidance, patience, support, and valuable input throughout this entire journey. Second, I would like to thank Dr. Sheldon M. Schuster for his advice, input, and useful discussions on enzyme kinetics, and for the excellent food I consumed on Friday afternoons. Third, I want to thank Dr. Susan K. Boehlein for her advice and instructive conversations and for conducting all enzyme assays in this study. I also would like to thank Dr. William Dolbier Jr., Dr. Eric Enholm, and Dr. Gus Palenik for their advice and valuable suggestions. I could not have done it without the help from these important people. They have my appreciation and best wishes for the future. Next, I would like to thank my colleagues Anthony (Tony) B. Dribben, Yannis (Janice) Karayannis, Nagraj (Nag) Bokinkere, Rajiv (Raju) Moghe, and Susan (Sue) Rasmussen for daily productive and nonproductive conversations in the topics of chemistry and life. Special thanks go to Tony Dribben for useful advice regarding computers and computational chemistry. I wish all of them the best in the future and hope to see them again. iii Special thanks go to my wife, Lymari Montalvo, for her company, love, and support throughout these years of graduate study and to her family, Jose A.
  • Photoluminescence and Electron Migration Characteristics of Aromatic Poly(Amide-Imide) Containing 1-Naphth- Aldehyde and 4,4′-Diamino Diphenyl Ether

    Photoluminescence and Electron Migration Characteristics of Aromatic Poly(Amide-Imide) Containing 1-Naphth- Aldehyde and 4,4′-Diamino Diphenyl Ether

    http://www.e-polymers.org e-Polymers 2009, no. 016 ISSN 1618-7229 Photoluminescence and electron migration characteristics of aromatic poly(amide-imide) containing 1-naphth- aldehyde and 4,4′-diamino diphenyl ether Sepideh Khoee,* Somayeh Zamani *Polymer Laboratory, Chemistry Department, School of Science, University of Tehran, P. O. Box 14155-6455, Tehran, Iran; fax: +98 21 6649 5291; e-mail: [email protected] (Received: 23 June, 2008; published: 16 February, 2009) Abstract: A new photoactive polymer poly(amide-imide), was fabricated by reacting two synthesized monomers, N,N'-(4,4'-oxy-bis(4,1-phenylene))-bis-(1,3- dioxo-1,3-dihydroisobenzofuran-5-carbox amide) and N-(4,6-diamino-1,3,5- triazine-2-yl)-1-naphthamide. Polycondensation reaction of the two monomers was performed under 10 min microwave irradiation in o-cresol as solvent for producing a photoactive poly(amide-imide) in a quantitative yield. The resulting polymer was characterized by FT-IR, 1H-NMR, TGA techniques and its solvatochromic and fluorescence properties were investigated in different solvents. The poly(amide- imide) exhibit broad fluorescent band, and the fluorescence intensity is related to the intermolecular chain-chain or chain-solvent interaction. Also the self-quenching mechanism was studied according to the specific behavior of the polymer in different solvents and the self-quenching rate constant for the association reaction in the excited state (Kq) could be measured from the Stern-Volmer equation. This kind of fused system and diamines show various electron migration mechanisms and photoluminescent properties in the singlet excited states. Introduction The application of microwave energy to organic synthesis is known for a long time [1] and in the last decade was successfully applied to radical and condensation polymerization reactions [2, 3].