IChO 2011 · Theory problem 8

Click chemistry

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Click chemistry is a chemical concept introduced by K. B. Sharpless in 2001 and describes a set of chemical reactions that generate substances quickly, reliably and quantitatively by joining molecules through small units under mild conditions. This methodology has recently been applied as a key step in the following synthesis of bicyclic compounds.

Mandelic acid is a versatile natural compound and widely used as a “chiral pool” in synthesis. The reduction of \((R)\)-mandelic acid with \(\ce{LiBH4}\) affords A.

Figure 1

Reaction of A with one equivalent of \(p\)-toluenesulfonyl chloride gives B (Scheme 8.1). Heating B in pyridine yields C. During this transformation, compounds B and C retain their absolute configurations.

Figure 2

Scheme 8.1:

Figure 3

8.1 Draw the structures of B and C with the correct stereochemistry. Use dashed-wedged line notation throughout this problem.

Reaction of C with sodium azide in aqueous acetonitrile gives a mixture of enantiopure regioisomers D and E in a ratio of 3 : 1 (Scheme 8.2). On the other hand, the compound B affords E as the sole product under the same condition.

Figure 4

Scheme 8.2:

8.2 Draw the structures of D and E with the correct stereochemistry.

Part I: Compounds D and E are separately subjected to \(\ce{NaH}\) mediated reaction with 3-bromoprop-1-yne to afford F and G, respectively (Scheme 8.3). Heating F and G separately in toluene gives the bicyclic products H and I, respectively.

Figure 5

Scheme 8.3:

8.3 Draw the structure of compounds F, G, H and I with the correct stereochemistry.

Part II: Reaction of D and E separately with dimethyl acetylenedicarboxylate in water at \(70\,^\circ\mathrm{C}\) forms the optically active monocyclic regioisomers J and K, respectively (Scheme 8.4). Subsequent treatment of J and K with \(\ce{NaH}\) gives final bicyclic products L and M, respectively, both having the formula \(\ce{C13H11N3O4}\).

Figure 6

Scheme 8.4:

8.4 Draw the structures of compounds J, K, L and M with the correct stereochemistry.

Official English text of the problem.

Answers

All configurations below refer to the phenyl-bearing stereocenter. The bicyclic names describe the fused heterocyclic core explicitly.

  • 8.1

B: \((2R)\)-2-Hydroxy-2-phenylethyl 4-methylbenzenesulfonate.
SMILES: Cc1ccc(S(=O)(=O)OC[C@H](O)c2ccccc2)cc1

C: \((2R)\)-2-Phenyloxirane, also called \((R)\)-styrene oxide.
SMILES: c1ccccc1[C@@H]1CO1

  • 8.2

D: \((2S)\)-2-Azido-2-phenylethan-1-ol, the major product.
SMILES: OC[C@@H](N=[N+]=[N-])c1ccccc1

E: \((1R)\)-2-Azido-1-phenylethan-1-ol, the minor product from C and sole product from B.
SMILES: [N-]=[N+]=NC[C@H](O)c1ccccc1

  • 8.3

F: 3-[(2S)-2-Azido-2-phenylethoxy]prop-1-yne.
SMILES: C#CCOC[C@@H](N=[N+]=[N-])c1ccccc1

G: 3-[(1R)-2-Azido-1-phenylethoxy]prop-1-yne.
SMILES: C#CCO[C@@H](CN=[N+]=[N-])c1ccccc1

H: \((2S)\)-2-Phenyl-4-oxa-1,8,9-triazabicyclo[4.3.0]nona-6,8-diene.
SMILES: c1nnn2[C@@H](c3ccccc3)COCc12

I: \((3R)\)-3-Phenyl-4-oxa-1,8,9-triazabicyclo[4.3.0]nona-6,8-diene.
SMILES: c1nnn2C[C@@H](c3ccccc3)OCc12

  • 8.4

J: Dimethyl 1-[(1S)-2-hydroxy-1-phenylethyl]-1H-1,2,3-triazole-4,5-dicarboxylate.
SMILES: COC(=O)c1nnn([C@H](CO)c2ccccc2)c1C(=O)OC

K: Dimethyl 1-[(2R)-2-hydroxy-2-phenylethyl]-1H-1,2,3-triazole-4,5-dicarboxylate.
SMILES: COC(=O)c1nnn(C[C@H](O)c2ccccc2)c1C(=O)OC

L: Methyl \((2S)\)-5-oxo-2-phenyl-4-oxa-1,8,9-triazabicyclo[4.3.0]nona-6,8-diene-7-carboxylate.
SMILES: COC(=O)c1nnn2[C@@H](c3ccccc3)COC(=O)c12

M: Methyl \((3R)\)-5-oxo-3-phenyl-4-oxa-1,8,9-triazabicyclo[4.3.0]nona-6,8-diene-7-carboxylate.
SMILES: COC(=O)c1nnn2C[C@@H](c3ccccc3)OC(=O)c12

Solution

Here \(\ce{Ph}\) denotes phenyl, \(\ce{Me}\) methyl, and \(\ce{Ts}\) \(p\)-toluenesulfonyl.

The starting material is \((R)\)-mandelic acid, O=C(O)[C@H](O)c1ccccc1. Reduction gives A, \((1R)\)-1-phenylethane-1,2-diol, OC[C@H](O)c1ccccc1. The formal organic reduction balance is

\[ \ce{PhCH(OH)CO2H + 4[H] -> PhCH(OH)CH2OH + H2O}. \]

Here \([H]\) represents reducing equivalents supplied by lithium borohydride, [Li+].[BH4-], followed by protonation during workup; water has SMILES O. The existing stereocenter is retained.

Wedge–dash drawing convention. For each product, place the phenyl group to the lower left and the carbon side chain to the lower right of the stereocenter, using ordinary bonds as in the drawing of A. Place the directly attached oxygen or nitrogen above that carbon:

  • For B, C, E, G, I, K and M, draw this heteroatom bond as a solid wedge, with hydrogen directed away.
  • For D, F, H, J and L, draw it as a dashed wedge, with hydrogen directed toward the viewer.

In this orientation these are, respectively, the \(R\) and \(S\) configurations. The ring connections are specified below; the drawings do not imply a fixed conformation of the flexible rings.

8.1 — Identification of B and C

One equivalent of \(p\)-toluenesulfonyl chloride, Cc1ccc(S(=O)(=O)Cl)cc1, converts the less hindered primary hydroxyl group of A into a tosylate:

\[ \ce{A + TsCl + Et3N -> B + [Et3NH]Cl}. \]

Triethylamine, CCN(CC)CC, accepts the proton, forming triethylammonium chloride, CC[NH+](CC)CC.[Cl-]. Thus B has the connectivity

\[ \ce{Ph-CH(OH)-CH2-OTs}, \]

with the original \(R\) stereocenter.

On heating, the remaining hydroxyl oxygen attacks the primary carbon bearing the tosylate group. Intramolecular displacement closes the three-membered epoxide ring, while pyridine accepts the hydroxyl proton:

\[ \ce{B + C5H5N -> C + [C5H6N]+ + TsO-}. \]

Pyridine is n1ccccc1; pyridinium is c1cc[nH+]cc1; \(p\)-toluenesulfonate is Cc1ccc(S(=O)(=O)[O-])cc1.

The substitution occurs at the terminal, nonstereogenic carbon. The benzylic carbon–oxygen bond remains intact, so C is \((R)\)-2-phenyloxirane.

To draw C, connect the hydroxyl oxygen of the original A skeleton to its adjacent terminal carbon, producing the triangle containing oxygen, the phenyl-bearing carbon and \(\ce{CH2}\). The stated retention also excludes a pathway involving displacement at the benzylic stereocenter.

8.2 — Identification of D and E

Sodium azide, [Na+].[N-]=[N+]=[N-], supplies azide ion, [N-]=[N+]=[N-]. Backside attack can open C at either carbon.

  1. Attack at the benzylic carbon: the benzylic carbon–oxygen bond breaks, and that stereocenter undergoes inversion. The original epoxide oxygen remains on the terminal carbon and is protonated. This gives D, the \(S\) azido alcohol: \[ \ce{Ph-CH(N3)-CH2OH}. \]

  2. Attack at the terminal carbon: the terminal carbon–oxygen bond breaks, while the benzylic carbon–oxygen bond is retained. This gives E, the \(R\) azido alcohol: \[ \ce{Ph-CH(OH)-CH2N3}. \]

The separately balanced opening pathways are

\[ \ce{C + NaN3 + H2O -> D + NaOH}, \]

\[ \ce{C + NaN3 + H2O -> E + NaOH}. \]

Sodium hydroxide is [Na+].[OH-]. The supplied product ratio is D : E = 3 : 1.

The reaction of B independently establishes the identity of E. Azide replaces the terminal tosylate directly:

\[ \ce{B + NaN3 -> E + NaOTs}. \]

Sodium \(p\)-toluenesulfonate is [Na+].Cc1ccc(S(=O)(=O)[O-])cc1. Since this substitution does not occur at the stereocenter, its \(R\) configuration is retained.

8.3 — Identification of F, G, H and I

Sodium hydride, [Na+].[H-], removes the alcohol proton. The resulting oxygen nucleophile displaces bromide from 3-bromoprop-1-yne, or propargyl bromide, C#CCBr. These Williamson ether syntheses give

\[ \ce{D + NaH + BrCH2C#CH -> F + NaBr + H2}, \]

\[ \ce{E + NaH + BrCH2C#CH -> G + NaBr + H2}. \]

Sodium bromide is [Na+].[Br-], and hydrogen is [H][H].

Consequently:

  • F has connectivity \(\ce{Ph-CH(N3)-CH2-O-CH2-C#CH}\), retaining \(S\).
  • G has connectivity \(\ce{Ph-CH(OCH2C#CH)-CH2N3}\), retaining \(R\).

Heating brings about an intramolecular azide–alkyne \([3+2]\) cycloaddition:

\[ \ce{F -> H}, \qquad \ce{G -> I}. \]

All three azide nitrogens and both alkyne carbons become the five atoms of a 1,2,3-triazole; no atoms are eliminated.

For the following drawing instructions, number the triazole consecutively N1–N2–N3–C4–C5–N1, with N1 being the nitrogen originally attached to the carbon chain. These are triazole positions, distinct from the bicyclic numbering used in the names above.

The tether favors closure of a six-membered second ring. The alkyne carbon carrying the \(\ce{CH2O}\) tether becomes C5, adjacent to N1; the original terminal alkyne carbon becomes C4, bearing hydrogen.

The precise fused-ring connections are:

  • H: Starting at N1, follow the six-membered ring through the \(S\) phenyl-bearing carbon, then \(\ce{CH2-O-CH2}\), then C5, and back to N1.
  • I: Starting at N1, follow the six-membered ring through \(\ce{CH2}\), then the \(R\) phenyl-bearing carbon, then \(\ce{O-CH2}\), then C5, and back to N1.

Both products therefore share the N1–C5 bond between their triazole and six-membered ether rings. One valid Kekulé drawing has double bonds between N2 and N3 and between C4 and C5.

Neither cycloaddition changes a bond at the stereocenter. Thus H retains \(S\), and I retains \(R\). Their formulas remain \(\ce{C11H11N3O}\), identical to those of F and G.

8.4 — Identification of J, K, L and M

Dimethyl acetylenedicarboxylate, COC(=O)C#CC(=O)OC, undergoes an intermolecular azide–alkyne cycloaddition with each azido alcohol:

\[ \ce{D + MeO2C-C#C-CO2Me -> J}, \]

\[ \ce{E + MeO2C-C#C-CO2Me -> K}. \]

Because the alkyne is symmetric, both triazole carbon atoms bear a methyl ester group. Using the triazole numbering defined above:

  • J has a \(\ce{CH(Ph)CH2OH}\) substituent attached through its \(S\) benzylic carbon to N1, with \(\ce{CO2Me}\) at both C4 and C5.
  • K has a \(\ce{CH2CH(OH)Ph}\) substituent attached through its terminal \(\ce{CH2}\) to N1, again with \(\ce{CO2Me}\) at both C4 and C5. Its benzylic center remains \(R\).

The regioisomerism between J and K comes from the different connectivities of D and E.

Next, sodium hydride removes the remaining alcohol proton. The oxygen nucleophile attacks the carbonyl carbon of the C5 ester, producing a six-membered cyclic ester. Collapse of the tetrahedral intermediate expels methoxide. The C4 methyl ester remains intact.

The balanced reactions under the basic conditions are

\[ \ce{J + NaH -> L + NaOCH3 + H2}, \]

\[ \ce{K + NaH -> M + NaOCH3 + H2}. \]

Sodium methoxide is [Na+].C[O-].

For precise drawings of the products:

  • L: Starting at triazole N1, follow the new six-membered ring through the \(S\) phenyl-bearing carbon, then \(\ce{CH2-O-C(=O)}\), then triazole C5, and back to N1.
  • M: Starting at triazole N1, follow the new six-membered ring through \(\ce{CH2}\), then the \(R\) phenyl-bearing carbon, then \(\ce{O-C(=O)}\), then triazole C5, and back to N1.

In both drawings, retain the \(\ce{CO2Me}\) substituent on triazole C4. No substitution occurs at the stereocenter, so L remains \(S\), and M remains \(R\).

The formula check is

\[ \underbrace{\ce{C8H9N3O}}_{\text{D or E}} + \underbrace{\ce{C6H6O4}}_{\text{dimethyl acetylenedicarboxylate}} = \underbrace{\ce{C14H15N3O5}}_{\text{J or K}}. \]

Cyclization removes from the organic molecule the elements of methanol, CO:

\[ \ce{C14H15N3O5}-\ce{CH4O} = \boxed{\ce{C13H11N3O4}}, \]

in agreement with the formula specified for both L and M.

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