IChO 2005 · Theory problem 2

Organic Synthesis and Stereochemistry

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Natural carbohydrates are generally produced by photosynthesis in plants. However, unnatural carbohydrates can be prepared by organic synthesis. The following outline is a synthetic scheme for the unnatural L-ribose (compound I).

Figure 1

2.1 Compound A has the molecular formula of \(\ce{C10H10O5}\). Draw the structural formula of A.

2.2 Given the chemistry described for reaction sequence A to C, indicate whether the following statements are true or false (Use T to represent true and F to represent false).

___ (a) \(\ce{OsO4}\) is an oxidizing agent in the reaction of A to B.

___ (b) \(\ce{MeOH}\) is generated as a by-product in the reaction of B to C.

___ (c) Protons act as the catalyst in the transformation of B to C.

___ (d) C will still be formed albeit in lower yields in the absence of \(\ce{Me2C(OMe)2}\).

Pig liver esterase is an enzyme that can hydrolyze esters to carboxylic acids. Hydrolysis of C by the pig liver esterase afforded an enantiomeric mixture of D and E, in which E was the major component. The optical rotation of the mixture was \([\alpha]_{\mathrm D}^{20}=-37.1^\circ\). Further purification by recrystallization gave pure E with the optical rotation \([\alpha]_{\mathrm D}^{20}=-49.0^\circ\).

2.3 What is the molar ratio of D/E in the product mixture before the recrystallization? Show your work.

2.4 Reaction of F with meta-chloroperbenzoic acid (MCPBA) afforded G as the product. Indicate whether the following statements are true or false (Use T to represent true and F to represent false).

___ (a) The reaction was to oxidize compound F.

___ (b) The oxygen atom inserted originated from MCPBA.

___ (c) The \(R/S\) notation of C-1 remained unchanged before and after the reaction.

The molecular formula of H is \(\ce{C9H16O5}\). Proton NMR data of H are listed as follows:

\({}^{1}\mathrm H\) NMR (\(\ce{CDCl3}\)) \(\delta\) 1.24 (s, 3H), 1.40 (s, 3H), 3.24 (m, 1 H), 3.35 (s, 3H), 3.58 (m, 2H), 4.33 (m, 1H); 4.50 (d, \(J=6\ \mathrm{Hz}\), 1H), 4.74 (d, \(J=6\ \mathrm{Hz}\), 1H), 4.89 (s, 1H).

2.5 Draw the configurational formula of H.

2.6 Assign R/S notations for compound I at C-1, C-2, C-3 and C-4. Give your answers as follows:

C-1: _; C-2: ; C-3: _; C-4: .

2.7 What are the identities of P, Q, R, S, T and U in the Fischer projection of compound I (L-ribose)?

Figure 2

Disaccharides are compounds with two monosaccharide subunits linked together by a glycosidic bond. Polysaccharides contain as few as ten, or as many as thousands, monosaccharide subunits. An example of a disaccharides is as follows:

Figure 3

2.8 How many diastereoisomers would be obtained for pentasaccharide J, if it is derived from five units of D-glucose?

Figure 4

pentasaccharide J derived from D-glucose

Official English text of the problem.

Answers

  • 2.1 A is dimethyl 7-oxabicyclo[2.2.1]hepta-2,5-diene-2,3-dicarboxylate:
    COC(=O)C1=C(C(=O)OC)[C@H]2C=C[C@@H]1O2

  • 2.2 (a) T; (b) T; (c) T; (d) T.

  • 2.3 \(\boxed{D/E=0.138}\), or \(\boxed{D:E=1:7.24}\). The mixture contains approximately 12.1 mol% D and 87.9 mol% E.

  • 2.4 (a) T; (b) T; (c) F. C-1 changes from \(S\) in F to \(R\) in G, despite retention of its spatial configuration.

  • 2.5 H is methyl 2,3-O-isopropylidene-β-L-ribofuranoside:
    CO[C@H]1O[C@@H](CO)[C@@H]2OC(C)(C)O[C@H]12
    In the orientation used for G, draw the C-1 methoxy group and C-4 hydroxymethyl group on solid wedges; retain the dashed bonds from C-2 and C-3 to the acetonide oxygens.

  • 2.6 \(\boxed{\text{C-1: }S;\quad\text{C-2: }S;\quad\text{C-3: }R;\quad\text{C-4: }S}\).
    The illustrated I is β-L-ribofuranose:
    O[C@H]1O[C@@H](CO)[C@H](O)[C@@H]1O

  • 2.7 \(\boxed{P=R=T=\ce{OH};\quad Q=S=U=\ce{H}}\). All three hydroxyl groups are on the left of the Fischer projection.

  • 2.8 \(\boxed{2^5=32}\) diastereoisomeric configurations.

Solution

2.1 — Identification of A

Furan, c1ccoc1, reacts with dimethyl acetylenedicarboxylate, COC(=O)C#CC(=O)OC, by a Diels–Alder cycloaddition. Furan supplies the four-electron diene, and the alkyne supplies the two-electron dienophile.

The atom-balanced reaction is

\[ \ce{C4H4O + C6H6O4 -> C10H10O5}. \]

No atoms are eliminated. One π bond remains between the two former alkyne carbons, and another remains between the two internal carbons of the furan diene. The original furan oxygen becomes the bridge of the bicyclic product.

To draw A, take the structure of B, remove its two hydroxyl groups, and restore a double bond between the carbons that carried them. The two methyl ester groups remain attached to the other double bond. This gives the bicyclic structure named and encoded in Answer 2.1.

2.2 — Oxidation and diol protection

(a) True. Osmium tetroxide, O=[Os](=O)(=O)=O, effects syn dihydroxylation of the unsubstituted double bond of A, producing the cis vicinal diol B. The organic substrate undergoes a two-electron oxidation.

A balanced oxidation half-equation is

\[ \ce{C10H10O5 + 2H2O -> C10H12O7 + 2H+ + 2e-}. \]

(b) True. The two hydroxyl groups of B react with 2,2-dimethoxypropane, COC(C)(C)OC, to form the cyclic isopropylidene ketal, or acetonide, of C. Both methoxy groups of the reagent can leave as methanol, CO.

The balanced net equation is

\[ \ce{C10H12O7 + C5H12O2 <=> C13H16O7 + 2CH3OH}. \]

Thus two molecules of methanol are formed per acetonide formed by this net reaction.

(c) True. Protonation facilitates cleavage of a methoxy group and subsequent substitution by the diol oxygens. Deprotonation regenerates the acid catalyst. Protons are therefore catalytic, rather than consumed stoichiometrically.

(d) True. Acetone, CC(=O)C, is also present and can itself react with the diol under acid catalysis:

\[ \ce{C10H12O7 + CH3COCH3 <=> C13H16O7 + H2O}. \]

This equilibrium produces water, O, which promotes the reverse reaction. The dimethoxy reagent helps drive acetonide formation; it can also consume water according to

\[ \ce{Me2C(OMe)2 + H2O <=> Me2CO + 2MeOH}. \]

Consequently, acetone and acid can still produce C without 2,2-dimethoxypropane, but less efficiently under the stated conditions.

2.3 — Enantiomeric composition

The esterase hydrolyzes one methyl ester group:

\[ \ce{C13H16O7 + H2O -> C12H14O7 + CH3OH}. \]

Because D and E are enantiomers, their specific rotations under identical conditions have equal magnitudes and opposite signs:

\[ [\alpha]_D=+49.0^\circ,\qquad [\alpha]_E=-49.0^\circ. \]

Let their mole fractions be \(x_D\) and \(x_E\). Their molar masses are identical, so the specific rotation is weighted by these same fractions:

\[ x_D+x_E=1, \]

\[ -37.1^\circ =x_D(49.0^\circ)+x_E(-49.0^\circ). \]

Therefore,

\[ x_E-x_D=\frac{37.1}{49.0}=0.75714. \]

The enantiomeric excess in favour of E is consequently \(75.7\%\), and

\[ x_D=\frac{1-0.75714}{2}=0.12143, \qquad x_E=\frac{1+0.75714}{2}=0.87857. \]

Hence

\[ \boxed{ \frac{n_D}{n_E} =\frac{49.0-37.1}{49.0+37.1} =\frac{11.9}{86.1} =0.138 } \]

or

\[ \boxed{D:E=1:7.24}. \]

2.4 — Baeyer–Villiger oxidation and C-1 configuration

The relevant structures are:

Compound Structural description Stereochemical SMILES
F Protected ribofuranosyl α-keto ester COC(=O)C(=O)[C@H]1O[C@@H](C(=O)OC)[C@@H]2OC(C)(C)O[C@H]12
G Corresponding protected ribofuranosyl methyl oxalate COC(=O)C(=O)O[C@H]1O[C@@H](C(=O)OC)[C@@H]2OC(C)(C)O[C@H]12

The reagent is 3-chloroperoxybenzoic acid, O=C(OO)c1cccc(Cl)c1. Its reduction product is 3-chlorobenzoic acid, O=C(O)c1cccc(Cl)c1.

Writing \(R\) for the protected sugar residue attached through C-1, the balanced transformation is

\[ \ce{ R-C(=O)-C(=O)-OMe + m-ClC6H4CO3H -> R-O-C(=O)-C(=O)-OMe + m-ClC6H4CO2H }. \]

(a) True. This is a Baeyer–Villiger oxidation: oxygen is inserted between C-1 of the sugar and the adjacent ketone carbonyl carbon.

(b) True. The inserted oxygen comes from the terminal peroxide oxygen of MCPBA. Mechanistically, peracid addition gives a Criegee intermediate; migration of the sugar group accompanies cleavage of the peroxide bond.

(c) False. Migration occurs with retention of spatial configuration at the migrating C-1, but its CIP priorities change.

Compound C-1 priorities, highest to lowest Configuration
F Ring oxygen \(>\) acyl carbon \(>\) C-2 \(>\) H \(S\)
G Exocyclic oxalate oxygen \(>\) ring oxygen \(>\) C-2 \(>\) H \(R\)

In F, the acyl carbon outranks C-2 because its next attached-atom list is \((\mathrm O,\mathrm O,\mathrm C)\), compared with \((\mathrm O,\mathrm C,\mathrm H)\) for C-2.

In G, both leading substituents begin with oxygen. The exocyclic oxygen leads to a carbonyl carbon and outranks the ring oxygen, which leads to saturated C-4.

Thus the two highest priorities exchange their spatial positions:

\[ \boxed{S\longrightarrow R} \]

without inversion of the stereocenter.

2.5 — Identification and configuration of H

The two operations convert the substituents of G as follows:

  1. Acidic methanol converts the anomeric oxalate ester into a methyl glycoside.
  2. Lithium aluminium hydride, [Li+].[AlH4-], reduces the methyl ester attached to C-4 to a hydroxymethyl group. The acetonide remains intact.

A balanced net representation of the first functional-group change is

\[ \ce{R-OCOCO2Me + MeOH -> R-OMe + HOCOCO2Me}. \]

Here the oxalate-derived coproduct is methyl hydrogen oxalate, COC(=O)C(=O)O.

The ester reduction can be represented by

\[ \ce{R'-CO2Me + 4[H] -> R'-CH2OH + MeOH}, \]

where \(\ce{[H]}\) denotes a formal reducing hydrogen equivalent.

These transformations give methyl 2,3-O-isopropylidene-β-L-ribofuranoside, the structure in Answer 2.5. Its formula follows directly from its groups:

  • Five carbons in the sugar skeleton.
  • Three carbons in the acetonide.
  • One carbon in the methoxy group.
  • Five oxygens: one ring oxygen, two acetonide oxygens, one methoxy oxygen, and one alcohol oxygen.

Thus its molecular formula is \(\ce{C9H16O5}\), as required.

The proton NMR data support this assignment:

\(\delta\) / ppm Integral and multiplicity Assignment
1.24, 1.40 Two singlets, 3H each The two nonequivalent acetonide methyl groups
3.24 Multiplet, 1H Alcohol \(\ce{OH}\) proton
3.35 Singlet, 3H Anomeric methoxy group
3.58 Multiplet, 2H \(\ce{CH2OH}\)
4.33 Multiplet, 1H H-4
4.50, 4.74 Two doublets, \(J=6\ \mathrm{Hz}\), 1H each Mutually coupled H-2 and H-3
4.89 Singlet, 1H Anomeric H-1

The integrations total \(16\) hydrogens. The anomeric singlet is consistent with the very small H-1/H-2 coupling of the β anomer in this acetonide-constrained furanose. The absolute stereochemistry is supplied by the synthetic sequence.

For the requested drawing, retain the five-membered sugar ring and the 2,3-acetonide of G. Replace the C-1 oxalate substituent by a solid-wedge methoxy group, and replace the C-4 methyl ester by a solid-wedge hydroxymethyl group. The two bonds from C-2 and C-3 to the acetonide oxygens remain dashed.

The subsequent acid hydrolysis removes both protecting groups:

\[ \ce{C9H16O5 + 2H2O -> C5H10O5 + CH3OH + CH3COCH3}. \]

2.6 — Absolute configurations of I

Apply the CIP rules to the cyclic structure actually drawn.

Center Priorities, highest to lowest Result
C-1 Ring oxygen \(>\) hydroxyl oxygen \(>\) C-2 \(>\) H \(S\)
C-2 Hydroxyl oxygen \(>\) C-1 \(>\) C-3 \(>\) H \(S\)
C-3 Hydroxyl oxygen \(>\) C-4 \(>\) C-2 \(>\) H \(R\)
C-4 Ring oxygen \(>\) C-3 \(>\) hydroxymethyl carbon \(>\) H \(S\)

The distinction at C-3 deserves attention. Both adjacent carbons initially present the list \((\mathrm O,\mathrm C,\mathrm H)\). Following their oxygen branches breaks the tie:

  • The oxygen reached through C-4 is bonded onward to carbon.
  • The hydroxyl oxygen reached through C-2 is bonded to hydrogen.

Therefore C-4 outranks C-2.

The answer for the illustrated β-L-ribofuranose is

\[ \boxed{(1S,2S,3R,4S)}. \]

Free L-ribose can equilibrate between forms in solution; these assignments refer specifically to I as drawn.

2.7 — Fischer projection of L-ribose

Open-chain L-ribose, systematically (2S,3S,4S)-2,3,4,5-tetrahydroxypentanal, has SMILES

O=C[C@@H](O)[C@@H](O)[C@@H](O)CO.

With the aldehyde group at the top and the hydroxymethyl group at the bottom, all three hydroxyl groups lie on the left, and all three hydrogens lie on the right. Consequently,

\[ \boxed{P=R=T=\ce{OH}}, \qquad \boxed{Q=S=U=\ce{H}}. \]

Here a hydroxyl substituent can be represented as [*]O, with * denoting its attachment to the carbon skeleton; hydrogen is [H].

The open-chain C-3 descriptor is \(S\), whereas cyclic I has C-3 \(R\). Ring closure changes the CIP ordering of the two carbon paths; it does not require inversion at that carbon.

2.8 — Number of pentasaccharide diastereoisomers

The D-glucose units fix the configurations of their nonanomeric stereocenters. Each of the five anomeric carbons can independently have either configuration:

  • Four occur at glycosidic linkages.
  • The fifth is the free anomeric center at the reducing end.

The linkage pattern is fixed by the drawing. The chain ends are different, so reversing the sequence does not identify otherwise distinct configurations.

Therefore,

\[ N=2^4\times2=\boxed{32}. \]

They are diastereoisomers because changing these anomeric configurations leaves the other D-glucose stereocenters unchanged.

The solution was checked against the official solution.