IChO 2012 · Theory problem 5

Chemical modification of DNA

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Deoxyribonucleic Acid (DNA) is one of the fundamental molecules of life. This question will consider ways that DNA’s molecular structure may be modified, both naturally and in ways devised by humankind.

Consider the pyrimidine bases, cytosine (C) and thymine (T). The N-3 atom (indicated by *) of one of these bases is a common nucleophilic site in single strand DNA alkylation, while the other is not.

5.1 Select (circle) which base, C or T, has the more nucleophilic N-3 atom.

Figure 1

5.2. Draw two complementary resonance structures of the molecule you select to justify your answer.

One common modification of DNA in nature is methylation of the indicated (*) position of guanine (G) by S-adenosyl methionine (SAM).

5.3 Draw the structures of both of the products of the reaction between guanine and SAM.

Figure 2

One of the earliest man-made DNA alkylating agents was mustard gas. Mustard gas acts by first undergoing an intramolecular reaction to form intermediate A which directly alkylates DNA, to give a nucleic acid product such as that shown in the part above.

Figure 3

5.4 Draw the structure for reactive intermediate A.

The nitrogen mustards react via an analogous pathway to the sulfur mustard of previous part. The reactivity of the compound may be modified depending on the third substituent on the nitrogen atom. The reactivity of nitrogen mustards increases with increasing nucleophilicity of the central nitrogen atom.

5.5 Select the most and least reactive from each of following groups of nitrogen mustards.

Figure 4
Figure 5
Figure 6

Some classes of natural products act as DNA alkylators, and in this way, they have the potential to serve as cancer therapies due to their antitumor activity. One such class is the duocarmycins. Shown below are steps from an asymmetric total synthesis of the natural product.

5.6 Draw the structures of isolable compounds J and K in the following scheme:

Figure 7

Related small molecules were synthesized to study the way in which the duocarmycins work. One such example is the thioester shown below.

5.7 Draw the structure of reactive intermediate Z.

Figure 8

Official English text of the problem.

Answers

  • 5.1 — C, cytosine: NC1=NC(=O)NC=C1.

  • 5.2 — Two charge-separated resonance contributors of cytosine, supplementary to its neutral structure above:

  1. Exocyclic amino nitrogen positive; N-3 negative: [NH2+]=C1[N-]C(=O)NC=C1.
  2. Ring N-1 positive; N-3 negative: NC1=CC=[NH+]C(=O)[N-]1.
  • 5.3 — The products are:

\(N^2\)-Methylguanine: CNc1nc2[nH]cnc2c(=O)[nH]1.

S-Adenosyl-L-homocysteine (SAH), shown as its zwitterion: Nc1ncnc2c1ncn2[C@@H]1O[C@H](CSCC[C@H]([NH3+])C(=O)[O-])[C@@H](O)[C@H]1O.

  • 5.4 — A is the 1-(2-chloroethyl)thiiran-1-ium ion, a three-membered cyclic sulfonium ion: ClCC[S+]1CC1. Its counterion is chloride, [Cl-].

  • 5.5 — Reactivity selections:

Group Most reactive Least reactive
a II: N,N-bis(2-chloroethyl)aniline, ClCCN(CCCl)c1ccccc1 I: N,N-bis(2-chloroethyl)-4-nitroaniline, ClCCN(CCCl)c1ccc([N+](=O)[O-])cc1
b I: N,N-bis(2-chloroethyl)-4-methoxyaniline, COc1ccc(N(CCCl)CCCl)cc1 III: N,N-bis(2-chloroethyl)-4-nitroaniline, ClCCN(CCCl)c1ccc([N+](=O)[O-])cc1
c II: mechlorethamine, CN(CCCl)CCCl I: N,N-bis(2-chloroethyl)acetamide, CC(=O)N(CCCl)CCCl
  • 5.6 — J and K:

J: methyl 7-benzyloxy-1-(benzyloxycarbonyl)-5-[(benzyloxycarbonyl)amino]-4-iodo-1H-indole-2-carboxylate. It is the starting compound with iodine replacing hydrogen at indole C-4.

COC(=O)c1cc2c(I)c(NC(=O)OCc3ccccc3)cc(OCc3ccccc3)c2n1C(=O)OCc1ccccc1

K: the 5-carbamate N-[(2R)-oxiran-2-ylmethyl] derivative of J. The aryl iodine and epoxide are both retained.

COC(=O)c1cc2c(I)c(N(C[C@@H]3CO3)C(=O)OCc3ccccc3)cc(OCc3ccccc3)c2n1C(=O)OCc1ccccc1

  • 5.7 — Z: the cyclopropane-fused cyclohexadienone thioester obtained by cyclization of the depicted seco precursor, with formula \(\ce{C21H18N2O4S}\).

CSC(=O)c1cc2c([nH]1)C(=O)C=C1N(C(=O)OCc3ccccc3)C[C@H]3C[C@@]213

In Z, the pre-existing methine stereocenter is \(S\), and the newly tetrahedral carbon shared by the cyclopropane and cyclohexadienone rings is \(R\).

Solution

5.1

Cytosine has the more nucleophilic N-3 atom.

In the neutral cytosine structure, N-3 is an imine-like nitrogen: its lone pair is available for bonding to an electrophile. Electron donation from the amino group and through the ring further supports electron density at this position.

In thymine, CC1=CNC(=O)NC1=O, N-3 is an imide nitrogen situated between two carbonyl groups. Its lone pair is strongly delocalized into those carbonyl groups and is consequently much less available for alkylation.

5.2

The two charge-separated contributors listed in the Answers can be obtained from neutral cytosine without moving any atoms or protons.

For the first contributor:

  • Donate the exocyclic amino nitrogen’s lone pair into its bond to C-4.
  • Move the C-4=N-3 \(\pi\)-electron pair onto N-3.

The exocyclic nitrogen becomes positively charged, and N-3 becomes negatively charged:

[NH2+]=C1[N-]C(=O)NC=C1.

For the second contributor:

  • Donate the N-1 lone pair to form N-1=C-6.
  • Shift the C-6=C-5 electrons to form C-5=C-4.
  • Shift the C-4=N-3 electrons onto N-3.

This gives:

NC1=CC=[NH+]C(=O)[N-]1.

Both contributors place additional electron density at N-3. They are resonance contributors of the same molecule, not separate species or tautomers.

5.3

The marked atom is guanine’s exocyclic amino nitrogen at C-2, not a ring nitrogen.

Guanine is Nc1nc2[nH]cnc2c(=O)[nH]1. SAM, in the protonation state drawn in the question, is:

Nc1ncnc2c1ncn2[C@@H]1O[C@H](C[S+](C)CC[C@H](N)C(=O)[O-])[C@@H](O)[C@H]1O.

The guanine amino nitrogen attacks the methyl carbon attached to positively charged sulfur in an \(S_\mathrm{N}2\) displacement. The methyl–sulfur bond breaks, leaving a neutral thioether sulfur. Subsequent proton transfer gives \(N^2\)-methylguanine and SAH.

To draw the products:

  • Replace guanine’s exocyclic \(\ce{-NH2}\) group by \(\ce{-NHCH3}\).
  • Remove the sulfur-bound methyl group from SAM, leaving the adenosyl–sulfur–homocysteine linkage intact.
  • Preserve all amino-acid and ribose stereochemistry.

With SAH represented as the zwitterion given in the Answers, the balanced equation is

\[ \ce{C5H5N5O + C15H22N6O5S -> C6H7N5O + C14H20N6O5S}. \]

This representation accounts for the proton lost from guanine’s nitrogen: the SAH amino group is \(\ce{-NH3+}\), while its carboxylate remains negatively charged.

5.4

Sulfur mustard, or bis(2-chloroethyl) sulfide, is ClCCSCCCl.

A sulfur lone pair attacks the terminal carbon of one of its own chloroethyl groups. Chloride leaves, forming a three-membered ring consisting of sulfur and the two carbon atoms of that arm. Sulfur now has three carbon–sulfur bonds and carries a positive charge.

Thus A is ClCC[S+]1CC1, and its formation is balanced as

\[ \ce{C4H8Cl2S -> C4H8ClS+ + Cl-}. \]

For the requested drawing, make a triangle with \(\ce{S+}\) at one vertex and \(\ce{CH2}\) at each of the other two vertices. Attach the remaining \(\ce{-CH2CH2Cl}\) arm to sulfur.

Guanine can attack a ring carbon, opening the strained ring. After deprotonation, this gives the depicted \(N^2\)-{2-[(2-chloroethyl)sulfanyl]ethyl}guanine:

ClCCSCCNc1nc2[nH]cnc2c(=O)[nH]1.

The overall substitution is

\[ \ce{C4H8Cl2S + C5H5N5O -> C9H12ClN5OS + HCl}. \]

Hydrogen chloride has SMILES [H]Cl.

5.5

The central nitrogen must use its lone pair for intramolecular displacement of chloride. Substituents that make this lone pair more available increase reactivity; substituents that withdraw or delocalize it decrease reactivity.

(a) The order is

\[ \boxed{\mathrm{II>III>I}}. \]

II has no nitro substituent. In III, the meta nitro group withdraws mainly through its inductive effect. In I, the para nitro group additionally allows strong resonance withdrawal through the aromatic ring, reducing the availability of the mustard nitrogen lone pair.

The meta compound III is N,N-bis(2-chloroethyl)-3-nitroaniline, ClCCN(CCCl)c1cccc([N+](=O)[O-])c1.

Therefore II is most reactive and I least reactive.

(b) The order is

\[ \boxed{\mathrm{I>II>III}}. \]

The para methoxy group in I donates electron density by resonance, increasing nitrogen nucleophilicity. The unsubstituted compound II is intermediate. The para nitro group in III strongly withdraws electron density.

Therefore I is most reactive and III least reactive.

(c) The order is

\[ \boxed{\mathrm{II>III>I}}. \]

II is a tertiary alkylamine, so its nitrogen lone pair is readily available.

I is an amide: the nitrogen lone pair is strongly delocalized into the adjacent carbonyl group. III is methyl N,N-bis(2-chloroethyl)carbamate, COC(=O)N(CCCl)CCCl. In III, the methoxy oxygen also donates into the carbonyl system, reducing the carbonyl’s resonance withdrawal from nitrogen relative to I.

Therefore II is most reactive and I least reactive.

5.6

Use ordinary indole numbering: the methyl ester is at C-2, the protected amino group at C-5, and the benzyloxy group at C-7. Here Cbz denotes benzyloxycarbonyl, represented as the substituent *C(=O)OCc1ccccc1, where * is the attachment point.

Formation of J. N-Iodosuccinimide, O=C1CCC(=O)N1I, effects electrophilic aromatic iodination.

C-4 is ortho to the protected amino substituent and para to the benzyloxy substituent. It is also less crowded than C-6, which lies between those substituents. Iodination at C-4 supplies the aryl iodide required for the later ring closure.

Thus, to draw J, copy the starting structure and replace its C-4 hydrogen by iodine.

For reference, the uniodinated starting indole is:

COC(=O)c1cc2cc(NC(=O)OCc3ccccc3)cc(OCc3ccccc3)c2n1C(=O)OCc1ccccc1.

The other iodination product is succinimide, O=C1CCC(=O)N1. The balanced equation is

\[ \ce{C33H28N2O7 + C4H4INO2 -> C33H27IN2O7 + C4H5NO2}. \]

Formation of K. Sodium hydride removes the proton from the exocyclic carbamate nitrogen, releasing hydrogen. The resulting nitrogen nucleophile displaces the nosylate group from the primary carbon of the glycidyl reagent.

The reagent is (S)-oxiran-2-ylmethyl 2-nitrobenzenesulfonate, or (S)-glycidyl nosylate:

O=[N+]([O-])c1ccccc1S(=O)(=O)OC[C@@H]1CO1.

The epoxide remains intact. Thus the change from J to K is precisely

\[ \ce{Ar-NH-C(=O)-OBn -> Ar-N(CH2\text{-}oxiranyl)-C(=O)-OBn}, \]

with the complete connectivity and stereochemistry specified by K’s SMILES in the Answers.

The displaced group forms sodium 2-nitrobenzenesulfonate:

[Na+].[O-]S(=O)(=O)c1ccccc1[N+](=O)[O-].

The balanced overall alkylation is

\[ \ce{C33H27IN2O7 + NaH + C9H9NO6S -> C36H31IN2O8 + C6H4NNaO5S + H2}. \]

The substitution occurs at the primary carbon bearing the sulfonate, so there is no inversion at the epoxide stereocenter. Its descriptor changes from \(S\) in the reagent to \(R\) in K because replacing the oxygen-containing substituent by a nitrogen-containing substituent changes the CIP priorities.

Consistency with the final cyclization. Magnesium inserts into the aryl carbon–iodine bond. The resulting carbon nucleophile attacks the less substituted epoxide carbon intramolecularly, opening the epoxide and forming a six-membered ring fused across indole C-4 and C-5. Acid workup protonates the alkoxide.

The resulting \(S\)-configured cyclic secondary alcohol, as drawn in the question, is:

COC(=O)c1cc2c(n1C(=O)OCc1ccccc1)c(OCc1ccccc1)cc1c2C[C@H](O)CN1C(=O)OCc1ccccc1.

The bond between the stereocenter and oxygen is retained throughout ring opening. Its \(S\) descriptor in the alcohol again reflects the changed CIP priorities.

A balanced net ionic equation for this sequence is

\[ \ce{C36H31IN2O8 + Mg + H3O+ -> C36H32N2O8 + Mg^{2+} + I- + H2O}. \]

The small species used here are sodium hydride, [Na+].[H-]; hydrogen, [H][H]; magnesium, [Mg]; hydronium, [OH3+]; magnesium(II), [Mg+2]; iodide, [I-]; and water, O.

5.7

The starting chloromethyl phenol, or seco thioester precursor, is:

CSC(=O)c1cc2c([nH]1)c(O)cc1c2[C@H](CCl)CN1C(=O)OCc1ccccc1.

Use the underlying indole numbering to describe the changes. The phenolic oxygen is attached at C-7, and C-4 is para to it. Let the existing stereogenic carbon bearing the chloromethyl substituent be \(C^*\).

Bicarbonate removes the phenolic proton. Electron donation from oxygen through the aromatic ring permits intramolecular carbon alkylation at C-4:

  1. The original phenolic C-7–O bond becomes a carbonyl bond.
  2. C-4 forms a new bond to the terminal carbon of the chloromethyl substituent.
  3. Chloride leaves.
  4. The existing C-4–\(C^*\) and \(C^*\)–methylene bonds, together with the new bond, form a cyclopropane.

Consequently, Z is a cyclopropane-fused cyclohexadienone, not an oxygen-cyclized ether.

To draw Z precisely:

  • Replace the phenolic \(\ce{-OH}\) by \(\ce{=O}\).
  • Remove chlorine.
  • Join the carbon that formerly bore chlorine to C-4.
  • Make C-4 tetrahedral.
  • In a Kekulé representation, place the cyclohexadienone double bonds at C-5=C-6 and C-3a=C-7a.
  • Preserve the pre-existing stereochemistry and use the \(S,R\) configuration encoded in Z’s SMILES.

The net cyclization is

\[ \ce{C21H19ClN2O4S + NaHCO3 -> C21H18N2O4S + NaCl + CO2 + H2O}. \]

This gives exactly the molecular formula printed for Z: \(\ce{C21H18N2O4S}\).

The additional small species are sodium hydrogen carbonate, [Na+].O=C(O)[O-]; sodium chloride, [Na+].[Cl-]; and carbon dioxide, O=C=O.

Finally, guanine’s exocyclic amino nitrogen attacks the less substituted cyclopropane carbon—the carbon originally bearing chlorine. Cleavage of its bond to C-4 allows the six-membered ring to regain aromaticity. Proton transfer restores the phenolic hydroxyl group and gives the depicted \(N^2\)-guanine adduct of the seco thioester:

CSC(=O)c1cc2c([nH]1)c(O)cc1c2[C@H](CNc2nc3[nH]cnc3c(=O)[nH]2)CN1C(=O)OCc1ccccc1.

This addition is balanced without a separate coproduct:

\[ \ce{C21H18N2O4S + C5H5N5O -> C26H23N7O5S}. \]

Cyclopropane strain relief and restoration of aromaticity explain why Z is the reactive alkylating intermediate.

The answers were checked against the official answers published by the organisers.