Lectures 13 & 14 Key, with contributions from Palleros CHEM 109 1.Build the nucleosides (sugar and base) & nucleotides (sugar, base, and 5’ phosphate). Nucleoside Nucleotide Nucleobase (DNA) - deoxyribose (RNA) - ribose NH2 NH2 NH 2 N N N N N N N N O N N HO N N -O P O H O O O- Adenine OH OH OH O O O N NH N NH N NH N N NH O N N NH HO 2 2 N N NH -O P O H 2 O O O- Guanine OH OH OH NH2 NH2 NH 2 N N N N O O N O N O HO -O P O H O O - Cytosine O OH OH OH O O NH NH N O N O HO H O Thymine OH X O O NH NH O N O N O - H O P O O Uracil - O X OH OH 1 Nucleic Acids Components 1 Daniel Palleros Additional Problems. Answers Nucleic Acids Components Answers to Additional Problems Lectures 13 & 14 Key, with contributions from Palleros CHEM 109 2. HeterocyclesProblem 1 in drugs (refer to table in Lecture 13 notes) a) Loratadine b) Rosuvastatin Cl pyridine OH pyrimidine OH O N N OH H C piperidine 3 N N N O S F CH3 O O O c) Atorvastatin d) Cimetidine N S NH OH O OH O N N H3C N OH N H HN N CH3 H pyrrole F imidazole e) Ciprofloxacin You may not recognize the quinoline ring system in the structure on the left, but it’s easier to see it in the resonance structure on the right. O O O O F OH F OH N N N N quinoline Nucleic NAcids Components Daniel Palleros H N 2 Additional Problems. Answers H piperazine piperazine f) Tioconazole N imidazole N Cl O S Cl Cl thiophene Problem 2 a) Pyrrole (pKa = 17.5) is more acidic than pyrrolidine (pKa ≈ 35) because the anion of pyrrole is more stable than the anion of pyrrolidine. The higher acidity of pyrrole is due to the sp2 hybridization of the N; sp2 hybridized atoms have more s-character, hold electrons tighter and, in general, are more electronegative and more tolerant of negative charges than sp3 hybridized atoms and, thus, yield more stable anions. 2 Pyrrole pKa = 17.5 + H2O + H O+ N N 3 H sp2 Pyrrolidine pKa ! 35 + H O 2 + H O+ N N 3 H 3 sp b) The conjugate acid of pyrrole (pKa = 0.4) is more acidic than the conjugate acid of pyrrolidine (pKa = 11.3) because pyrrole is an aromatic compound while its conjugate acid is not. Deprotonation of the conjugate acid of pyrrole restores the aromaticity to the pyrrole ring with a significant increase in stability. Such gain in stability does not occur in the deprotonation of pyrrolidine, an aliphatic amine. conjugate acid of pyrrole: + H O + H3O N 2 N H H H pKa = 0.4 aromatic Nucleic Acids Components 2 Daniel Palleros Additional Problems. Answers f) Tioconazole N imidazole N Cl O S Cl Cl Lecturesthiophen e13 & 14 Key , with contributions from Palleros CHEM 109 3.P rBasicityoblem 2 in heterocycles… a) Pyrrole (pKa = 17.5) is more acidic than pyrrolidine (pKa ≈ 35) because the anion of pyrrole is more stable than the anion of pyrrolidine. The higher acidity of pyrrole is due to the sp2 hybridization of the N; sp2 hybridized atoms have more s-character, hold electrons tighter and, in general, are more electronegative and more tolerant of negative charges than sp3 hybridized atoms and, thus, yield more stable anions. Pyrrole pKa = 17.5 + H2O + H O+ N N 3 H sp2 Pyrrolidine pKa ! 35 + H O 2 + H O+ N N 3 H 3 sp b) The conjugate acid of pyrrole (pKa = 0.4) is more acidic than the conjugate acid of pyrrolidine (pKa = 11.3) because pyrrole is an aromatic compound while its conjugate acid is not. Deprotonation of the conjugate acid of pyrrole restores the aromaticity to the pyrrole ring with a significant increase in stability. Such gain in stability does not occur in the deprotonation of pyrrolidine, an aliphatic amine. conjugate acid of pyrrole: + H O + H3O N 2 N Nucleic Acids Components H H 3 H Daniel Palleros Additional Problems. Answers pK = 0.4 aromatic a conjugate acid of pyrrolidine: + H O N + H2O N 3 H H H pK = 11.3 aliphatic a c) Pyrimidine (pKa conj. acid = 1.3; pKb = 12.7) is less basic than pyridine (pKa conj. acid = 5.25; pKb = 8.75) because the second N atom in pyrimidine is electron- withdrawing and destabilizes the positively charged conjugate acid. It can also be explained by saying that the electron-withdrawing effect of the second N makes the electron pair less prone to protonation. pyrimidine: N N + H2O + H3O N N H pKb = 12.7 pyridine: + H2O + H3O N N H pKb = 8.75 N less stable than N N H H d) Pyridine is less basic than piperidine because in pyridine the lone electron pair is on an sp2 hybridized orbital which, having more s-character, is more electronegative and more difficult to get protonated than the electron pair in piperidine which is located on an sp3 orbital. 3 pyridine piperidine N N 3 sp2 H sp pKb = 8.75 pKb = 2.8 e) N7 in purine is less basic than N3 in imidazole because the pyrimidine ring, present in purine but not in imidazole, due to its two electron-withdrawing N atoms takes electron density away from the adjacent ring, making it less prone to protonation. Nucleic Acids Components 3 Daniel Palleros Additional Problems. Answers conjugate acid of pyrrolidine: + H O N + H2O N 3 H H H pK = 11.3 aliphatic a c) Pyrimidine (pKa conj. acid = 1.3; pKb = 12.7) is less basic than pyridine (pKa conj. acid = 5.25; pKb = 8.75) because the second N atom in pyrimidine is electron- withdrawing and destabilizes the positively charged conjugate acid. It can also be explained by saying that the electron-withdrawing effect of the second N makes the electron pair less prone to protonation. pyrimidine: N N + H2O + H3O N N H pKb = 12.7 pyridine: + H2O + H3O N N H pKb = 8.75 N less stable than Lectures 13 & 14 Key, with contributionsN from PallerosN CHEM 109 H H 3 con’td… d) Pyridine is less basic than piperidine because in pyridine the lone electron pair is on an sp2 hybridized orbital which, having more s-character, is more electronegative and more difficult to get protonated than the electron pair in piperidine which is located on an sp3 orbital. pyridine piperidine N N 3 sp2 H sp pKb = 8.75 pKb = 2.8 e) N7 in purine is less basic than N3 in imidazole because the pyrimidine ring, present in purine but not in imidazole, due to its two electron-withdrawing N atoms takes electron Nucleic Acids Components 4 Daniel Palleros deAnsddiititoyn aal wPraoybl efmroms. An stwheer sa djacent ring, making it less prone to protonation. Purine Imidazole less basic than imidazole 3 7 N 4 N N 1 5 2 N N 9 N 1 H 3 H take electron density away from adjacent ring f) The H on purine’s N9 is more acidic (pKa = 8.9) than the H on imidazole’s N1 (pKa = 14.2), because the conjugate base of purine (anion on N9) is more stabilized than the conjugate base of imidazole (anion on N1) due to the presence of the electron- withdrawing pyrimidine ring. Purine anion Imidazole anion 7 N 3 4 N 1 N 5 2 9 N N N 3 1 take electron density away from adjacent ring g) Both heterocycles have two N atoms. In pyrimidine, a π-deficient ring, both N atoms with their electron pairs outside the ring, are electron withdrawing and they mutually decrease their electron densities, making themselves less basic. In imidazole, on the other hand, N1 with an electron pair inside the ring, donates electron density and increases the electron density on the other nitrogen, N3, making it more basic. pKa conj. acid = 6.95; pKb = 7.05 3 4 N N 2 5 N 1 N H pK conj. acid = 1.3 electron donor a pKb = 12.7 both N are e-withdrawing Imidazole Pyrimidine Problem 3 1,4-Dihydro-1,4-diazine (also known as 1,4-dihydropyrazine) has 8 π electrons (which is 4 not a Hückel number; Hückel numbers are (4n + 2), where n = 0, 1, 2, 3, etc.). The molecule is antiaromatic and has not been isolated (although some derivatives of it have). Nucleic Acids Components 5 Daniel Palleros Additional Problems. Answers Nucleic Acids Components 8 Daniel Palleros H Additional Problems. Answers N 8 ! electrons electrons and, thus, are aromatic but these structures involve charge separation between N atoms of identical electroneH gativity (two N), which has a high energy cost and, therefore, Nucleic Acids Components Daniel Palleros 1,4-dihydro-1,4-diazine 7 do not Lectures c Aont d d i t ri oi nbut a l 13 P e r o mb &le uc14ms.h AKey ntsow ,et rhewiths st contributionsability of the from mol Pallerosecule. E xperimentally it was CHEMfound t109hat only tautomers I and II occur in measurable amounts in water (both in similar amounts). P roblem4.doubl Risperidone4 e charge separation which is energetically unfavorable, and thus their contribution to the stabilityII Iof the molecule is small. less basic: sp2 hybridized N O O O O N H N H N N H H NH NH N N N N N N NO nonbasic: amN ide O N O N O N O H H H IV H O N N N F N N N N N N N O H COH3 O N H H H Problem 10 H N most basic: alipNhatic amine N 3 NH2 (sp hybridized N) O 2 N O less basic: sp hybridized N N O N O H N NH deamination H N NH ProblemLECTURE 5 14 HW KEYaro matic aromatic N In the fol1low.