Heterocycles 2 Daniel Palleros

Heterocycles

1. Structures 2. Aromaticity and Basicity 2.1 Pyrrole 2.2 Imidazole 2.3 Pyridine 2.4 Pyrimidine 2.5 Purine 3. Π-excessive and Π-deficient Heterocycles 4. Electrophilic Aromatic Substitution 5. Oxidation-Reduction 6. DNA and RNA Bases 7. Tautomers 8. H-bond Formation 9. Absorption of UV Radiation 10. Reactions and Mutations

Heterocycles 3 Daniel Palleros

Heterocycles

Heterocycles are cyclic compounds in which one or more atoms of the ring are heteroatoms: O, N, S, P, etc. They are present in many biologically important molecules such as amino acids, nucleic acids and hormones. They are also indispensable components of pharmaceuticals and therapeutic drugs. Caffeine, sildenafil (the active ingredient in Viagra), acyclovir (an antiviral agent), clopidogrel (an antiplatelet agent) and nicotine, they all have heterocyclic systems.

O CH3 N HN O O N O CH 3 N H3C N N HN N OH O S O H N N N 2 N O N N O

CH3 N

CH3 caffeine sildenafil acyclovir

Cl S N

CH3 N N

H COOCH3 nicotine (S)-clopidogrel

Here we will discuss the chemistry of this important group of compounds beginning with the simplest rings and continuing to more complex systems such as those present in nucleic acids.

Heterocycles 4 Daniel Palleros

1. Structures Some of the most important heterocycles are shown below. Note that they have five or six-membered rings such as pyrrole and pyridine or polycyclic ring systems such as quinoline and purine. Imidazole, pyrimidine and purine play a very important role in the chemistry of nucleic acids and are highlighted.

Five-membered 3 N 4 3 N 4 N 5 5 2 2 O N S N O S 1 1 H H Furan Pyrrole Thiophene Imidazole Oxazole Thiazole

O N S H Tetrahydrofuran Pyrrolidine Thiolane (THF)

Six-membered

N N

N N N N N

Pyridine Pyrimidine Pyridazine Pyrazine

O

N N H H Piperidine Morpholine

Polycyclic 5 4 5 4 4 6 3 7 N 5 6 3 6 3 5 N 1 2 8 2 7 7 N 6 N N 2 N 9 4 N 2 1 3 8 1 8 1 7 H H Quinoline Isoquinoline Indole Purine

2. Aromaticity and Basicity Heterocycles are aromatic if they obey Hückel’s rules of aromaticity: a) The ring is planar. b) There is a continuous conjugated system (all atoms of the ring are sp2 hybridized). c) The number of pi electrons is equal to 4 n +2, where n = 0; 1; 2; 3; etc.

According to these rules, pyrrole, imidazole, pyridine, pyrimidine and purine, just to mention a few, are aromatic heterocycles. We will discuss these five systems in some detail because they play an important role in the chemistry of nucleic acids and Heterocycles 5 Daniel Palleros pharmaceuticals. We will pay special attention to the hybridization and basicity of the nitrogen atoms.

2.1 Pyrrole The nitrogen atom is sp2 hybridized. The N electron pair is on a p orbital, perpendicular to the plane of the ring and overlapping with the other p orbitals. There are a total of 6 π electrons, which is a Hückel number (Hückel numbers are 2, 6, 10, etc.) and, therefore, the molecule is aromatic.

sp2

N H

H

H N H

nonbasic H H

6 pi electrons

You may wonder why the hybridization of the nitrogen atom is sp2 and not sp3 as expected for a nitrogen attached to three other atoms. The answer to this question can be found in the fact that the sp2 hybridization makes the nitrogen coplanar with the rest of the molecule and places the lone electron pair on a p orbital parallel to the other p orbitals. This leads to aromaticity and further stability. If the nitrogen were sp3 hybridized, it wouldn’t be coplanar and the molecule couldn’t be aromatic.

H

H N

sp3 H H H

nonaromatic

Heterocycles 6 Daniel Palleros

Because the N electron pair is forming part of the aromatic system, it is not available for protonation (protonation would destroy the aromaticity), thus pyrrole is nonbasic.

not available for protonation

N H

nonbasic

This is demonstrated by an unusually low pKa value for the conjugate acid of pyrrole, which is 11 orders of magnitude more acidic than the conjugate acid of the nonaromatic counterpart, pyrrolidine.

conjugate acid of pyrrole: conjugate acid of pyrrolidine:

pKa = 11.3 pKa = 0.4 N N H H H H

Let’s consider now the dissociation of pyrrole itself.

pKa = 17.5 + H2O + H O+ N N 3 H sp2 The pKa of pyrrole (the dissociation of the H on the nitrogen) is 17.5. Compared to the pKa of pyrrolidine (≈ 35), it is about 20 orders of magnitude lower.

Pyrrolidine

pKa 35 + H O 2 + H O+ N N 3 H sp3 The higher acidity of pyrrole is due to the sp2 hybridization of its N; sp2 hybridized atoms have more s-character, hold electrons tighter and, in general, are more tolerant of negative charges than sp3 hybridized atoms and thus yield more stable anions.

2.2 Imidazole The imidazole ring is aromatic. Both N atoms are sp2 hybridized. The N attached to H (N1) has the electron pair on a p orbital, perpendicular to the plane of the ring. This electron pair forms part of the six-electron aromatic cloud and, like in the case of pyrrole, Heterocycles 7 Daniel Palleros is not available for protonation. The double-bonded nitrogen (N3), on the other hand, has the electron pair on an sp2 hybrid orbital which lies outside the ring and, thus, does not form part of the aromatic system. This electron pair is available for protonation.

2 outside the ring sp available for protonation (basic N) 2 N 3 N H 1 4 5 inside the ring; aromatic not available for protonation (nonbasic N) available for protonation H not available for protonation N N H

H H

6 pi electrons

It is worth mentioning that once the basic N gets protonated, both N atoms become indistinguishable because of resonance:

H H N N HA N

A- N N N

H H H

Protonated imidazole is often represented by the following structure that clearly shows that both N atoms are identical. The pKa of the conjugate acid of imidazole is 6.95.

H N + pKa = 6.95 N H Heterocycles 8 Daniel Palleros

The pKa of imidazole itself is 14.2, about three order of magnitude lower than the pKa of pyrrole (pKa = 17.5). This is due to the extra N atom in imidazole which is electron- withdrawing and thus stabilizes the negative charge in the conjugate base.

electron withdrawing: stabilizes the negative charge N N pKa = 14.2 + H O 2 + H O+ N N 3 H

The imidazole ring is present in histidine, a basic amino acid, and histamine, a biologically active amine involved in immune responses.

O H N N O NH2 NH3 N N H H histidine histamine

2.3 Pyridine The N atom is sp2 hybridized. The N electron pair lies outside the ring on an sp2 hybrid orbital and is available for protonation, making pyridine a basic heterocycle. The pKa of the conjugate acid of pyridine is 5.25.

N pKa = 5.25 N

outside the ring H available for protonation (basic N)

2.4 Pyrimidine Both N atoms are equivalent and sp2 hybridized. Both electron pairs lie outside the aromatic ring on sp2 hybrid orbitals. Both N are slightly basic. Pyrimidine is less basic than pyridine because of the inductive, electron-withdrawing effect of the second N atom. The pKa of the conjugate acid of pyrimidine is 1.3.

Heterocycles 9 Daniel Palleros

N N

N N H pKa = 1.3

Note that pyrimidine is about six orders of magnitude less basic than imidazole.

2.5 Purine Purine consists of a pyrimidine ring fused with an imidazole ring. All four N atoms are sp2 hybridized. N1, N3 and N7 are basic but N9 is not. N7, being on the imidazole side of the molecule, is more basic than N1 and N3 on the pyrimidine side.

less basic than imidazole 7 3 N 4 N 1 N 5 2 9 N N N 3 1 more acidic than H H imidazole nonbasic

Purine Imidazole

However, N7 less basic than the corresponding N in imidazole (N3) because the neighboring pyrimidine ring with two N atoms, which are electron-withdrawing, takes electron density away from the imidazole ring. Once N7 gets protonated both N7 and N9 become almost indistinguishable; they both have a pKa of 2.4, which makes them about four orders of magnitude more acidic than the protonated nitrogens in imidazole.

H N H 7 N 1 N

pKa = 2.4 9 N N N 3 pK = 6.95 H a H

The pyrimidine ring, because of the electron-withdrawing effect of the nitrogen atoms, also makes the H on N9 more acidic than the corresponding H in imidazole (on N1).

7 3 N N N 1

N N 1 9 N H 3 H pK = 14.2 pKa = 8.9 a

It should be remarked that there are two important tautomers of purine. As you may recall, tautomers are constitutional isomers in rapid equilibrium with one another. Usually, they differ in the position of a hydrogen atom (tautomers are also treated in section 7). The only difference between the two tautomers of purine is in the position of the hydrogen on the imidazole ring. Heterocycles 10 Daniel Palleros

7 H N 7 N N N

9 N N 9 N N H H9 H7 In solution, both tautomers exist in equilibrium in equal amounts. However, from a biological standpoint, the tautomer with hydrogen on N9, called H9, is more important that the one with the hydrogen on N7 (H7), because in biological systems N9 is usually the most reactive nitrogen and gets attached to other molecules by losing its hydrogen.

A summary of pKa values for all the heterocycles discussed so far is presented in the table below.

Table of pKa values for selected heterocycles Pyrrole Pyrrolidine

N N N N H H H H H H 17.5 0.4 35 11.3

Imidazole H N N + N N H H 14.2 6.95

Pyridine Pyrimidine

N

N N H H 5.25 1.3

Purine H H N N N N N N + N N N N N N H H 8.9 8.9 2.4

3. Π−excessive and Π-deficient Heterocycles The presence of a nitrogen atom in an aromatic ring affects the electron density of the rest of the atoms. If the nitrogen atom acts as an electron-donor, there is a net gain in electron density in the ring and the ring is called π-excessive. If, on the other hand, the nitrogen atom acts as an electron-withdrawing group, the aromatic ring loses electron density and is called π-deficient.

Whether the N atom donates or takes electron density away depends on the location of its electron pair. There are two possibilities. Possibility one: the electron pair is on a p orbital, perpendicular to the plane of the ring. The electron pair forms part of the π system and is delocalized around the ring, increasing its electron density. In such cases, the nitrogen atom acts as an electron donor and the ring is π-excessive. An example of a Heterocycles 11 Daniel Palleros

π-excessive heterocycle is pyrrole. In the figure below the resonance forms show a negative charge delocalized in the pyrrole ring (note that the nitrogen atom is positive). As a result, the four carbon atoms have an excess of electron density, which can be calculated with the tools of quantum mechanics and it’s called the effective π charge. In general, the effective π charge ranges from -1 to +1. A negative value indicates an excess of π electron density and a positive value a deficiency. The results for pyrrole are shown below. It can be observed that all four carbon atoms have a negative effective π charge (and the nitrogen atom a positive value).

Pyrrole, a π-excessive heterocycle

N N N N N H H H H H

-0.105 -0.105

-0.035 -0.035

N + 0.28 H

The second possibility is that the N electron pair is on an sp2 hybrid orbital that lies outside the ring, does not form part of the aromatic ring and cannot be donated to the ring. Under these circumstances the nitrogen atom, due to its electronegativity, acts solely as an electron-withdrawing group and the ring is π-deficient. Pyridine and pyrimidine are examples of π-deficient rings.

The resonance forms for pyridine show a positive charge spread on the aromatic ring while the nitrogen remains negative. This is also reflected by the effective π charge values.

Pyridine, a π-deficient heterocycle

N N N N

+0.050

0 0

+0.077 +0.077 N -0.195

Heterocycles 12 Daniel Palleros

Imidazole with two N atoms, one with the electron pair on a p orbital (electron-donor N) and the other with the electron pair on an sp2 hybrid orbital (electron-withdrawing N), is both π-excessive and π-deficient. This is illustrated by the positive and negative effective π-charge values for the ring carbons.

-0.287 N -0.068

+0.094 -0.037 N+0.298 H

4. Electrophilic Aromatic Substitution Aromatic heterocycles undergo electrophilic aromatic substitutions such as bromination, and nitration like benzene rings do. If the heterocycle is π-excessive, the excess of electron density activates the ring for such reactions. Pyrrole, a π-excessive ring, undergoes bromination so easily that the reaction must be controlled by carrying it out at low temperature.

3 0oC + Br 2 2 + HBr Br N 1 N H H

The bromination takes place preferentially at C2 rather than C3. This can be explained by the more effective charge delocalization for the transition state and the intermediate that results when the bromine gets attached to C2. As shown below, there are three resonance forms for the intermediate with bromine at C2 but only two with bromine at C3.

Reaction on C2:

+ Br Br Br Br H N N N H N H H H H H

three resonance forms

Reaction on C3: Br Br Br+ H H

N N N H H H

two resonance forms

Heterocycles 13 Daniel Palleros

Contrary to pyrrole, pyridine is a π-deficient heterocycle and thus, undergoes electrophilic aromatic substitution with difficulty. The bromination of pyridine occurs only at high temperature (300oC) and yields exclusively 3-bromopyridine.

4 3 Br 300oC + Br 2 2 + HBr N 1 N

Pyridines are so unreactive in electrophilic aromatic substitutions in part because of the electron-withdrawing effect of the nitrogen atom itself, but also because the N electron pair easily reacts with the electrophiles needed for the reaction resulting in a net positive charge that deactivates the ring even further.

+ E+

N N strong electron-withdrawing effect E (inductive and resonance): the ring is strongly deactivated

5. Oxidation-Reduction In general, the presence of an electron-withdrawing group in an organic molecule lowers the energy of its molecular orbitals. The electrons located in these low-lying energy levels are more stabilized and are less reactive as they are less prone to be donated. As a result, such molecules have little tendency to lose their electrons or get oxidized. Instead, they have a greater tendency to gain electrons and get reduced. Examples of such compounds are pyridine and pyridinium ions and other π-deficient heterocycles.

In the case of pyridinium ions, the nitrogen atom with a positive charge acts as a very strong electron-withdrawing group, making these compounds easy targets for electron- rich reagents such as the hydride anion and other nucleophiles. An example of reduction by hydride is shown below with NAD+.

O H H O - NH2 H NH2

N N R R + NAD NADH

Nature uses this reaction to drive many biological oxidation-reduction processes.

Contrary to π-deficient heterocycles, π-excessive heterocycles have their π electrons in high energy molecular orbitals. These compounds have higher tendency to lose their electrons and get oxidized. At the same time, they have low tendency to gain more Heterocycles 14 Daniel Palleros electrons and thus, are difficult to reduce. An example of such compounds is pyrrole. The reaction below shows the oxidation of pyrrole with a mild oxidizing agent, 0.3% H2O2. It yields 2-hydroxypyrrole which tautomerizes to a mixture of pyrrolinones.

0.3% H2O2

OH O + O N N N N H H H H 2-hydroxypyrrole mixture of tautomeric pyrrolinones

6. DNA and RNA Bases The bases present in both nucleic acids (DNA and RNA) are adenine, guanine and cytosine. Thymine is exclusively present in DNA and uracil exclusively present in RNA. The structures of these nucleobases are shown below.

DNA and RNA

NH NH 2 O 2 7 6 N 5 N N N 1 NH 8

N 2 N O H 4 N N H 9 3 H N NH2

Adenine Guanine Cytosine

DNA RNA O O

H3C NH NH

N O N O H H

Thymine Uracil

These compounds are planar and fully conjugated. Adenine has 6 π electrons in each ring and is aromatic.

Heterocycles 15 Daniel Palleros

Adenine

NH2

N N

N H N

imidazole ring: pyrimidine ring: 6 ! electrons 6 ! electrons aromatic aromatic

Several important resonance forms can be drawn for each one of the nucleobases. Some of the resonance forms for guanine, cytosine, thymine and uracil are also aromatic. However, their contribution to the total stability of these molecules is small because these aromatic resonance forms involve separation of charges which is energetically unfavorable.

Guanine

O O

N NH N H N etc

N H N NH2 N H N NH2

6 ! electrons 6 ! electrons aromatic aromatic

Cytosine

NH 2 NH2

N N etc

N O N O H H

6 ! electrons aromatic

Heterocycles 16 Daniel Palleros

Uracil and Thymine

O O

R R H NH N etc

N O N O H H R = H : Uracil 6 ! electrons aromatic R= CH3: Thymine

7. Tautomers The five nucleobases: adenine, guanine, cytosine, thymine and uracil can exits in more than one tautomeric form. As mentioned earlier (section 2), tautomers are constitutional isomers in rapid equilibrium with one another. Usually they differ in the position of a hydrogen atom. A typical example of a tautomeric reaction is the keto-enol equilibrium. Since a carbonyl group is more stable than a carbon-carbon double bond, the equilibrium normally favors the keto form.

O OH

C H C C C

keto enol

The reaction is catalyzed by acids and bases.

O H H O H-A - O C H A C C H C - A C H-A C

Guanine, thymine and uracil undergo a keto-enol-like tautomeric equilibrium. At physiological pH they exist mainly in the keto form.

Guanine O OH

N NH N N

N H N NH2 N H N NH2

99.99 % 0.01 % Heterocycles 17 Daniel Palleros

Thymine and Uracil

OH O O OH 4 3 R R N R R 5 NH NH N 2 6 N O N O N OH 1 H H N OH I II III IV R = H : Uracil R= CH3: Thymine

For uracil and thymine only tautomers I and II are important in biological systems. Tautomer I is the predominant form at physiological pH values. Tautomer II exists only at low pH.

Adenine and cytosine undergo an amino-imino-like tautomeric equilibrium in which the amino form is favored.

NH2 NH

H

amino imino Cytosine

NH2 NH

4 5 N 3 NH

2 6 N O N O H 1 H Adenine

NH H 2 N

N N N NH

N N N H H N

8. H-bond Formation All five nucleobases can form H-bonds. The NH and NH2 groups can act as H-bond donors and the N and O atoms as H-bond acceptors. It should be noticed, however, that only the nitrogen atoms whose electron pairs are outside the ring can act as H-bond acceptors. If the electron pair is inside the ring, forming part of the aromatic π system, is not available to accept H-bonds. In the structures below the nucleobases are shown as they appear in nucleic acids, bound to the rest of the molecule through one of the N atoms (shown as a squiggly bond). Their H-bond donor (d) and acceptor (a) capabilities are indicated. Heterocycles 18 Daniel Palleros

NH NH2 d 2 d O a a a N H a N a N N N d

N a N N N O N NH2 d a a Adenine Guanine Cytosine

O a O a

H3C H H N N d d

N O a N O a

Thymine Uracil

You may wonder why the NH2 group in adenine and cytosine, with a free electron pair on the nitrogen atom, acts exclusively as an H-bond donor and not as an H-bond acceptor. The answer can be found in the resonance forms of adenine and cytosine (shown below for cytosine). In one of the resonance forms, there is positive charge on that N atom making it an unlikely acceptor of H-bonds.

unable to accept H-bonds

NH2 NH2

N N etc.

N O N O H H

Once the nucleobase is forming part of a nucleic acid molecule, not all of the atoms that can form H-bonds actually do. Only those indicated by stars () and moons () form H- bonds in normal DNA. In the figure below, stars indicate H-bond donors and moons, acceptors.

! NH O " 2

N H ! N " N

N ! N O " N NH2

Guanine Cytosine

Heterocycles 19 Daniel Palleros

! NH2 O " O "

N ! ! N " H3C H H N N

N N N O N O

Adenine Thymine Uracil

Guanine and cytosine form three H-bonds each and adenine, thymine and uracil two each. Guanine has a pattern of acceptor-donor-donor (from top to bottom) that is complementary to the pattern of cytosine: donor-acceptor-acceptor. In DNA these two bases are associated through H-bonds. Adenine having a donor and an acceptor is complementary to thymine and uracil, each with an acceptor and a donor.

H R N O H H HN O N N N H N H N N N N O O H N N N N N Cytosine H R = CH3 : Thymine R = H : Uracil Guanine Adenine

In normal DNA guanine is paired with cytosine (G≡C) and adenine to thymine (A=T).

A mismatch between bases can occur if the less stable tautomers of the nucleobases are present at the time of DNA replication because the less stable tautomers have different H- bond donor-acceptor pattern. This mismatch may lead to mutations. Let’s examine for example, cytosine. In the stable tautomer, the NH2 group is an H-bond donor, while in the less stable tautomer the same N (now double-bonded to the adjacent carbon) is an H- bond acceptor. In the stable tautomer, N3 is an H-bond acceptor while in the less stable tautomer this nitrogen (now with an H attached) is an H-bond donor.

Cytosine " NH2 NH ! 4 3 5 N ! NH "

2 6 N O N O H 1 ! H ! amino imino Heterocycles 20 Daniel Palleros

The H-bond donor-acceptor capabilities of the less stable tautomer of cytosine can be explained by its resonance forms. In one of them there is a negative charge on the nitrogen atom at C4 and a positive charge on N3, making them an acceptor and a donor of H-bonds, respectively.

H NH N 4 4 3 3 H H 5 N 5 N etc 2 2 6 6 N O N O H 1 1 H

This tautomer of cytosine forms hydrogen bonds with adenine instead of guanine, resulting in the abnormal A=C pairing.

Abnormal pairing H N

H HN N N H N N O

N N Cytosine (minor tautomer) Adenine

Similarly, the minor tautomer of guanine has a different H-bond acceptor-donor pattern than the major tautomer and this may lead to mutations through abnormal pairing. Guanine can form abnormal pairing to thymine.

Guanine

O ! OH"

N NH" N N !

N H N NH2" N " H N NH2

minor tautomer

Heterocycles 21 Daniel Palleros

Abnormal pairing: CH3 O

H O N H N N N O

H N Thymine N N H

Guanine (minor tautomer)

Notice that in normal DNA, thymine forms only two H-bonds but in abnormal pairing to guanine it forms three.

9. Absorption of UV Radiation Depending on its wavelength, UV radiation can be divided into three regions: UVA, UVB and UVC. UVA radiation is between 400 and 320 nm (called “black light”) and is the least energetic and least damaging of all UV radiation. UVB radiation is between 320 and 280 nm and UVC between 280 and 100 nm. Most of the UVC radiation is absorbed by the ozone layer in the atmosphere but the fraction that makes it through is the most damaging to life.

Because of the extended π system of the purine and pyrimidine bases, nucleosides, nucleotides and nucleic acids all absorb UV radiation. They show maxima around 185, 205 and 260 nm due to π  π* transitions. The free base and the corresponding nucleosides and nucleotides all absorb in the same region and have nearly the same molar absorptivities and λmax. The absorption of UV radiation around 260 nm (UVC) is responsible for UV-induced DNA damage that may result in cancer and other genetic aberrations.

UV spectra of nucleosides adenosine (A); guanosine (G); cytidine (C); uridine (U) and free base thymine (T); pH 7.2. From Experimental Organic Chemistry, D. Palleros. Heterocycles 22 Daniel Palleros

10. Reactions and Mutations The absorption of UV radiation may result in photochemical reactions which in turn may lead to mutations. For example, the thymine residue attached to DNA undergoes a dimerization reaction upon interaction with UV light. The reaction is a cycloaddition (you may recall that the Diels-Alder reaction is also a cycloaddition) involving the π electrons of the carbon-carbon double bonds.

O O O O CH3 CH3 H CH3 H3C H H H N N UV light N N + O N N O O N N O

Thymine Thymine Thymine dimer

A cyclobutane ring forms and this creates a kink or bend in the DNA molecule; this precludes proper base pairing which may lead to mutations.

The purine and pyrimidine rings can also react with nucleophiles and electrophiles. An inspection of the electrostatic potential maps for these molecules (see Fig. 28.2 in McMurry and the figure below) shows that there are atoms with positive charge density (shown in blue – online- or bold black in print) and atoms with negative charge density (shown in red – online - or gray in print). The atoms with negative charge density act as nucleophiles (indicated by a thick arrow ) and react with electrophiles, while the atoms with positive charge density are electrophilic (indicated by a thin arrow ) and are attacked by nucleophiles. Notice that the atoms with negative charge density are the same atoms that can act as H-bond acceptors.

NH NH2 O 2 C C N N C H N N N C N C N N N O N NH2

Adenine Guanine Cytosine

O O

H3C C H C H N N

C C N O N O

Thymine Uracil

Heterocycles 23 Daniel Palleros

An example of a nucleophilic attack on a nucleobase is the reaction of cytosine with water. The overall reaction leads to the deamination of the cytosine residue and its transformation to uracil. The reaction begins with the protonation of N3 followed by the nucleophilic attack of water on C4. After a proton transfer, NH3 leaves.

NH2 NH2 H2N OH2 H3N OH 4 + 3 H3O H H H 5 N N H2O N proton transfer N 2 H2O 6 N 1 O N O N O N O

cytosine - NH3

H O O H H H O N 2 N

+ H3O N O N O

uracil

This deamination reaction takes place at very low rate but environmental factors such as pH and the presence of certain chemicals such as nitrites may accelerate it. Nitrites - (NO2 ), present in most cure meats, can form nitrous acid that may lead to the deamination of cytosine via the formation of unstable diazonium salts.

N NH2 2 OH O

H O N HNO2 N 2 N NH

several N steps 2 N O N O N O N O

Uracil Cytosine a diazonium salt

If cytosine is chemically transformed into uracil, it may lead to a mutation as cytosine pairs up with guanine but uracil pairs up with adenine. This reaction may be the reason why DNA does not contain uracil but has thymine instead. Any uracil that is formed as a result of the deamination of cytosine is recognized by the DNA repair enzymes as a foreign base and rapidly removed. If uracil was a native base of DNA then the repair enzymes wouldn’t be able to tell it apart from the uracil formed as a result of the deamination reaction, the deamination would go unnoticed and unrepaired and soon the genetic information contained in DNA would vanish.

Purine and pyrimidine bases also react with electrophiles. Some of the most powerful electrophiles are alkylating agents such as methyl bromide (largely used as a pesticide), dimethyl sulfate and mustard gas (sadly used as a chemical weapon). They are mutation- inducing agents.

Heterocycles 24 Daniel Palleros

O - + - ! !- !+ !- ! !- ! !+ H C O S O CH Cl S Cl H3C Br 3 3 !+ !+ O methyl bromide dimethyl sulfate mustard gas

They react with centers of negative charge density (nitrogen and oxygen atoms; see figure on the previous page) and prevent them from participating in H-bonding.

The reaction between adenine and methyl bromide is shown below. N1 gets methylated and, as a result, it is unable to form H-bonds because it no longer has a free electron pair. This may lead to a mutation. N3 and N7 can also react with methyl bromide but since they are not directly involved in base pairing, their methylation may have less important consequences.

NH2 NH2 7 6 5 1 N CH3 N N N H3C Br 8 Br

2 N N N 9 4 N 3 adenine methylated adenine