1.build the nucleosides (sugar and base) & …...pyridine: h n n h n less stable than d)...

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

1

1.Build the nucleosides (sugar and base) & nucleotides (sugar, base, and 5’ phosphate).

Nucleobase Nucleoside

(DNA) - deoxyribose Nucleotide

(RNA) - ribose

X

X

N

NNH

NNH2

Adenine

N

NN

NNH2

O

OH

HO

N

NN

NNH2

O

OHOH

OP-OO-

O

NH

NNH

NO

NH2

Guanine

NH

N

NO

NH2N

O

OH

HO

NH

N

NO

NH2N

O

OHOH

OP-OO

O-

N

NH

NH2

O

CytosineO

OH

HON

N

NH2

O

N

NH2

ON

O

OHOH

OP-OO-

O

NH

NH

O

O

ThymineO

OH

HON

NH

O

O

Uracil

NH

NH

O

O

NH

O

ON

O

OHOH

OP-OO-

O


2

2. Heterocycles in drugs (refer to table in Lecture 13 notes)

Nucleic Acids Components Daniel Palleros

Additional Problems. Answers 1

Nucleic Acids Components

Answers to Additional Problems

Problem 1

a) Loratadine b) Rosuvastatin

N

N

Cl

OO

pyridine

piperidine

N

N

F

N

O

OH

OHOH

S

H3C

CH3O

O

pyrimidine

c) Atorvastatin d) Cimetidine

N

F

O

OH

OHOHO

N

H

pyrrole

N

N

S NH

HN

CH3

NH3C

H

N

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.

N

O O

OH

F

N

NH

piperazine

N

O O

OH

F

N

NH

quinoline

piperazine



f) Tioconazole

N

N

O

SCl

Cl

Cl

imidazole

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; sp

2 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

N

H

+ H2ON

+ H3O+

sp2 Pyrrolidine

N

H

pKa ! 35

sp3

+ H2O

N+ H3O+

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.

N

HH

pKa = 0.4

conjugate acid of pyrrole:

+ H2ON

H

+ H3O

aromatic


3

3. Basicity in heterocycles…



f) Tioconazole

N

N

O

SCl

Cl

Cl

imidazole

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; sp

2 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

N

H

+ H2ON

+ H3O+

sp2 Pyrrolidine

N

H

pKa ! 35

sp3

+ H2O

N+ H3O+

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.

N

HH

pKa = 0.4

conjugate acid of pyrrole:

+ H2ON

H

+ H3O

aromatic



N

HH

pKa = 11.3

conjugate acid of pyrrolidine:

+ H2ON

H

+ H3O

aliphatic

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.

N

N

H

N

N+ H2O + H3O

N

H

N

+ H2O + H3O

pKb = 12.7

pKb = 8.75

pyrimidine:

pyridine:

N

N

H

N

H

less stable than

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.

N

pKb = 8.75

NH

pKb = 2.8

pyridine piperidine

sp2 sp3

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.


4

3 con’td…



N

HH

pKa = 11.3

conjugate acid of pyrrolidine:

+ H2ON

H

+ H3O

aliphatic

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.

N

N

H

N

N+ H2O + H3O

N

H

N

+ H2O + H3O

pKb = 12.7

pKb = 8.75

pyrimidine:

pyridine:

N

N

H

N

H

less stable than

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.

N

pKb = 8.75

NH

pKb = 2.8

pyridine piperidine

sp2 sp3

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.



N

NN

N

H

1

3

7

9

less basic thanimidazole

N

N

H

Purine Imidazole

1

2

34

5

take electron densityaway 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.

N

NN

N

1

3

7

9

N

N

Purine anion

1

2

34

5

take electron densityaway from adjacent ring

Imidazole anion

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.

N

N

H

1

2

34

5

pKa conj. acid = 6.95; pKb = 7.05

electron donor

Imidazole

N

N

pKa conj. acid = 1.3

pKb = 12.7

both N are e-withdrawing

Pyrimidine Problem 3 1,4-Dihydro-1,4-diazine (also known as 1,4-dihydropyrazine) has 8 π electrons (which is

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).


5

4. Risperidone

LECTURE 14 HW KEY 1. Draw structures for four different enol forms of uracil.

2. Propose a mechanism for the deamination of cytosine catalyzed by acid (+ charges):

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



NH

HN

1,4-dihydro-1,4-diazine

8 ! electrons

Problem 4

N

N

N

ON

F

CH3

O

most basic: aliphatic amine

(sp3 hybridized N)

less basic: sp2 hybridized N

less basic: sp2 hybridized N

nonbasic: amide

Problem 5 In the following molecules the N electron pair forms part of the π system. They are π-

excessive.

N

H

N

H

In the following molecules the N electron pair is outside the ring and does not form part

of the π system. They are π-deficient.

N

N

N

Pyrazole has one electron pair outside the ring and the other forming part of the π system.

The compound is both π-deficient and π-excessive.



double charge separation which is energetically unfavorable, and thus their contribution

to the stability of the molecule is small.

NH

NH

O

O

N

NH

O

O

H

N

N

O

O

H

H

NH

N

O

O

H

N

NH

O

O

H

N

N

O

O

H

H

N

N

O

O

H

H

aromatic aromatic

b)

NH

NH

O

O

N

NH

OH

O

NH

N

O

OH

N

N

OH

OH

II IIII IV

minor tautomers

major tautomer

N

NH

O

OH

V

Problem 9 The four tautomers of purine are:

N

NN

N

1

3

7

9N

NN

N

H

1

3

7

9N

NN

N

1

3

7

9 N

NN

N

1

3

7

9

H

H

H

I II III IV

Notice that in each of them the H is attached to a different N. Tautomers I and II are more

stable than tautomers III and IV because in I and II each ring has 6 π electrons and is

fully aromatic. In the resonance forms shown above for tautomers III and IV each ring

does not have a Hückel number of electrons (the pyrimidine has 7 and the imidazole has

5 π electrons), and therefore, each one separately cannot be considered aromatic,

although the whole molecule is aromatic because it has 10 π electrons). There are other

resonance forms for tautomers III and IV (shown below) in which each ring has 6 π



electrons and, thus, are aromatic but these structures involve charge separation between

atoms of identical electronegativity (two N), which has a high energy cost and, therefore,

do not contribute much to the stability of the molecule. Experimentally it was found that

only tautomers I and II occur in measurable amounts in water (both in similar amounts).

N

NN

N

H

III

N

NN

N

H

N

NN

N

H

IV

N

NN

N

H Problem 10

N

NNH

N

NH2

deaminationNH

NNH

N

O

hypoxanthineadenine

NH

NNH

N

O

NH2

deaminationNH

NH

NH

N

O

O

xanthineguanine

Problem 11

N

N

NH2

O

H3O+

H2O

N

N

NH2

O

H H2O N

N

H2N

O

H

OH2

proton transfer N

N

H3N

O

H

OH

N

N O

H

OH

- NH3

H2ON

N O

H

O

H3O+

cytosine

uracil


6

3. Mutation of methyl cytosine…

4. At pH 7.4, the main species in equilibrium are HPO4

2- (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).


7

6. Dinucleotide synthesis

**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

O

O

OPOO-

H

O

OH

OPOO-

O-

O-

OH

OH

B:

N

NN

NNH2

NH

NN

NO

NH2

O

O

OPOO-

O

OH

OP

-O

O-

OH

OH

A

G

Adenosine

Guanosine A-G Dinucleotide

HO-O

H+

H+

O

O

OPOO-

O

OH

OP

O-

OH

OH

A

GO

O-

Pentacoordinatephosphate intermediate

A-G Dinucleotide

O

O

OPOO-

O

OH

OP

O-

OH

OH

A

GO

O-

O

O

OPOO-

O

OH

OP

-O

O-

OH

OH

A

GO-O

Pentacoordinatephosphate intermediate

O

O

OPOO-

H

O

OH

OPOO-

O-

O-

OH

OH

N

NN

NNH2

NH

NN

NO

NH2

Adenosine

Guanosine

OHH

:B

B: HH+

1.build the nucleosides (sugar and base) & …...pyridine: h n n h n less stable than d)...

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