Lectures 13&14 HW

Home , Pyrrolidine

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 O H O O- Cytosine 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) T ioconazole

N imidazole

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 sp3

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

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 sp3 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 a aromatic conjugate acid of pyrrolidine:

+ H O N + H2O N 3 H H H aliphatic pKa = 11.3

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 aliphatic pKa = 11.3

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 CH O N H O 3 H H H Problem 10 H N most basic: alipNhatic amine N 3 NH2 (sp hybridized N) O less basic: sp2 hybridized N N O N O N O N H NH deamination H N NH ProblemLECTURE 5 14 HW KEYaro matic aromatic N In the fol1low. Drawing m structuresolHecules tNhe for N fourelec tdifferentron pair f orenolms paformsNrt of ofthe uracil. π syst e m. They are π- H N excessive .b) adenine hypoxanthine O OH O O OH NH O N NH O N N

N N N O N NHN Odeamination H H N ONH N NHOH N OH H H H I N II III IV N NH N V H 2 H N O H In the following molecules the N electron pair is outside the ring and does not form part major tautomer guanine xanthine of t he π s ys t e m . T he y are π-deficient. minor tautomers Problem 11 2.P Proposeroblem a9 mechanism for the deamination of cytosine catalyzed by acid (+ charges): The four tautomers of purine are: N

NH2 NH2 H2N OH2 H3N OH H 7 7 N 7 H O+ NN N H H 7 N H N 13 H N H1ON N N 2 N protoNn tra1nsfer N N 1 H2O 9 N N N N 3 9 N 9 N 9 N N Pyrazole haHs oneN elecOtron pair outN side tOhe3 ring and theN otheOr3 forming paN rt of tOhe 3π system. The compound is both π-deficient and π-excessive. H I II III IV cytosine - NH3 Notice that in each of them the H is attached to a different N. TH automers I and II are more O O stable than tautomers III and IV because in I and II each ring has H6 π electrons and is H H O N fully aromatic. In the resonance forms showN n abov2e for tautomers III and IV each ring + does not have a Hückel number of electrons (the Hp3yrO imidine has 7 and the imidazole has 5 π electrons), and therefore, each one NseparatOely cannot be cN onsideOred aromatic, although the whole molecule is aromatic because it has 10 π electrons). There are other uracil resonance forms for tautomers III and IV (shown below) in which each ring has 6 π

**Two arrows missing in key above (arrows moving e- back to O in 1st and last steps)

5 Lectures 13 & 14 Key, with contributions from Palleros CHEM 109

3. Mutation of methyl cytosine…

2- - 4. At pH 7.4, the main species in equilibrium are HPO4 (major) and H2PO4 (minor). The dibasic anion is the more dominant species because pH 7.4 is closer to its pKa (7.21).

5. Draw the H-Bonding patterns for both A-T and G-C. Practice this a few times on your own without looking at the key (structures of bases will be given on the final).

6 Lectures 13 & 14 Key, with contributions from Palleros CHEM 109

6. Dinucleotide synthesis

NH2 N Adenosine N A - N N - A O O O- O P O O P O O O O P O O - - O O O- O OH O OH O O OH H N H+ NH G B: HO G - P O O N - O N NH2 O O P O O O P O - O - O - O H+ O OH OH OH OH OH OH Pentacoordinate phosphate intermediate A-G Dinucleotide Guanosine

**It is still OK to use the lazy NAS mechanism on phosphorus, I just didn’t want you to forget that it is an abbreviation!

7. Dinucleotide hydrolysis NH2 Pentacoordinate N N phosphate intermediate Adenosine A N N O- A O- O- O P O O O P O O O P O O O- O- O- O O OH O OH :B O OH H N NH G + H G O P O O O- H OH O P O N N NH B: H - O 2 O- O O P O O - O- OH OH O OH OH OH OH A-G Dinucleotide Guanosine