IChO 2024 · Theory problem 8

Safflower

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In Saudi culture, safflower drink is a traditional remedy used for its calming effects.

Figure 1

One ingredient with health benefits is kaempferol. The following scheme shows its total synthesis.

Figure 2

Hints:

  • C has a tricyclic system. Also, it has the same number of hydroxyl groups as the compound preceding it.
  • The step C to D is necessary to avoid any unwanted reactivity with DMDO.
  • DMDO is a hydroxylating agent.

8.1. Draw the structures of compounds A–E. 15 pt

A unique pentacyclic indole alkaloid, serotobenine found in safflower has antimicrobial properties. Its total synthesis is shown below:

Figure 3

Hints:

  • G is a bicyclic compound with one of the rings being a heterocycle.
  • H isomerizes upon heating in the presence of the amine base.
  • K is a product of an oxidative cleavage of the newly formed functional group in J. K has a \({}^{1}\mathrm{H}\) NMR signal at 12 ppm.

8.2. Draw the structures of compounds F–K. 18 pt

Carthamin gives safflower its red color.

Figure 4

Carthamin

In the total synthesis of carthamin the assembly was tested on a model substrate L. Note that M resembles carthamin’s core structure.

Figure 5

8.3. Draw the structure of compound M including its stereochemistry. (Hint: In methanol-\(d_4\) L has 11 and M has 10 distinct proton signals in their \({}^{1}\mathrm{H}\) NMR spectra). 5 pt

Official English text of the problem.

Answers

8.1.

  • A — 1-(2,4,6-Trimethoxyphenyl)ethan-1-one, or 2,4,6-trimethoxyacetophenone.
    SMILES: CC(=O)c1c(OC)cc(OC)cc1OC

  • B — (2E)-3-(4-Methoxyphenyl)-1-(2,4,6-trimethoxyphenyl)prop-2-en-1-one.
    SMILES: COc1ccc(/C=C/C(=O)c2c(OC)cc(OC)cc2OC)cc1

  • C — 5-Hydroxy-7,4′-dimethoxyflavone.
    SMILES: O=c1cc(-c2ccc(OC)cc2)oc2cc(OC)cc(O)c12

  • D — 5,7,4′-Trimethoxyflavone.
    SMILES: O=c1cc(-c2ccc(OC)cc2)oc2cc(OC)cc(OC)c12

  • E — 3-Hydroxy-5,7,4′-trimethoxyflavone, or kaempferol 5,7,4′-trimethyl ether.
    SMILES: O=c1c(O)c(-c2ccc(OC)cc2)oc2cc(OC)cc(OC)c12

8.2.

  • F — Allyl 3-methyl-4-nitrophenyl ether.
    SMILES: C=CCOc1cc(C)c([N+](=O)[O-])cc1

  • G — 5-(Allyloxy)-1H-indole.
    SMILES: C=CCOc1ccc2[nH]ccc2c1

  • H — 5-(Allyloxy)-1-tosyl-1H-indole.
    SMILES: C=CCOc1ccc2n(S(=O)(=O)c3ccc(C)cc3)ccc2c1

  • I — 4-Allyl-5-[(4-benzyloxy-3-methoxybenzyl)oxy]-1-tosyl-1H-indole.
    SMILES: C=CCc1c(OCc2cc(OC)c(OCc3ccccc3)cc2)ccc2n(S(=O)(=O)c3ccc(C)cc3)ccc12

  • J — Racemic 3-{5-[(4-benzyloxy-3-methoxybenzyl)oxy]-1-tosyl-1H-indol-4-yl}propane-1,2-diol.
    SMILES: OCC(O)Cc1c(OCc2cc(OC)c(OCc3ccccc3)cc2)ccc2n(S(=O)(=O)c3ccc(C)cc3)ccc12
    The unspecified alcohol stereocentre represents the racemic product.

  • K — 2-{5-[(4-Benzyloxy-3-methoxybenzyl)oxy]-1-tosyl-1H-indol-4-yl}acetic acid.
    SMILES: O=C(O)Cc1c(OCc2cc(OC)c(OCc3ccccc3)cc2)ccc2n(S(=O)(=O)c3ccc(C)cc3)ccc12

8.3.

  • M — The homochiral, methine-linked carthamin-model dimer of enol-O-demethylated (S)-L, with molecular formula \(\ce{C41H38O14}\). Both original stereocentres retain their configurations.

One neutral keto–enol tautomer, with both stereocentres S in this representation, is specified by:

CC[C@@]1(OC(C)=O)C(=O)/C(=C(/O)\C=C\c2ccc(OC)cc2)C(=O)/C(=C/C2=C(O)[C@](CC)(OC(C)=O)C(=O)/C(=C(/O)\C=C\c3ccc(OC)cc3)C2=O)C1=O

To draw it, demethylate the ring enol ether of each of two copies of L, then connect their originally proton-bearing ring alkene carbons through one methine carbon. Represent the central conjugated segment as \(\ce{C_{\text{ring}}=CH-C_{\text{ring}}}\), with the adjacent carbonyl/enol arrangement encoded above. Retain both acetoxy groups, both ethyl groups and both aryl methoxy groups.

Solution

8.1. Identification of A–E

A: Friedel–Crafts acylation.
The three unsubstituted positions of the starting 1,3,5-trimethoxybenzene are equivalent. Acetylation introduces an acetyl group at one of them, giving A, 2,4,6-trimethoxyacetophenone.

The balanced organic transformation is

\[ \ce{C9H12O3 + CH3COCl ->[ZnCl2] C11H14O4 + HCl}. \]

B: Claisen–Schmidt condensation.
Base forms the enolate of A, which adds to the aldehyde carbonyl. Dehydration gives the conjugated chalcone B, predominantly with the \(E\) configuration:

\[ \ce{C11H14O4 + C8H8O2 ->[KOH] C19H20O5 + H2O}. \]

The carbonyl remains attached to the trimethoxyphenyl ring.

C: Cyclization accompanied by another demethylation.
The displayed product of the \(\ce{AlCl3}\) step is (2E)-1-(2-hydroxy-4,6-dimethoxyphenyl)-3-(4-methoxyphenyl)prop-2-en-1-one:

COc1ccc(/C=C/C(=O)c2c(O)cc(OC)cc2OC)cc1

Its ortho phenolic oxygen becomes the heterocyclic oxygen during oxidative cyclization. The chalcone’s carbonyl, α-carbon and β-carbon become flavone C4, C3 and C2, respectively.

Cyclization alone would give a trimethoxyflavone with formula \(\ce{C18H16O5}\), containing no free hydroxyl group. However, C must have formula \(\ce{C17H14O5}\) and one hydroxyl group. Therefore, one further methoxy group must be demethylated.

The remaining methoxy group ortho to the carbonyl becomes flavone’s 5-hydroxyl group. Thus C is 5-hydroxy-7,4′-dimethoxyflavone. This accounts for its formula and all three rings.

D: Protection of the phenolic hydroxyl group.
Dimethyl sulfate methylates the 5-hydroxyl group, giving D, 5,7,4′-trimethoxyflavone. Protection prevents unwanted oxidation of the free phenol during the next step.

E: Hydroxylation at C3.
DMDO introduces the hydroxyl group at flavone C3, giving E, 3-hydroxy-5,7,4′-trimethoxyflavone.

The balanced oxygen-transfer equation is

\[ \ce{C18H16O5 + C3H6O2 -> C18H16O6 + C3H6O}. \]

Here the oxygen donor is dimethyldioxirane, CC1(C)OO1, and its reduction product is acetone, CC(=O)C.

Finally, \(\ce{BBr3}\) cleaves the three methyl ethers. The existing 3-hydroxyl group remains, producing kaempferol, or 3,5,7,4′-tetrahydroxyflavone:

O=c1c(O)c(-c2ccc(O)cc2)oc2cc(O)cc(O)c12

8.2. Identification of F–K

F: Phenolic O-allylation.
Potassium carbonate generates the phenoxide, which substitutes bromide from allyl bromide. This gives the allyl ether F. The methyl and nitro groups remain unchanged.

G: Leimgruber–Batcho indole synthesis.
The adjacent methyl and nitro groups provide the required ortho-nitrotoluene arrangement.

The acetal/pyrrolidine treatment converts the methyl substituent into an enamine-containing two-carbon side chain. Zinc in acetic acid then reduces the nitro group, enabling intramolecular ring closure and formation of the aromatic indole.

The original nitro-bearing aromatic carbon becomes indole C7a, while the original methyl-bearing aromatic carbon becomes C3a. Consequently, the allyloxy substituent is at C5, identifying G as 5-(allyloxy)-1H-indole.

H: Indole N-protection.
Tosyl chloride replaces the indole N–H hydrogen with a tosyl group. The oxygen is already protected as an allyl ether, so H is the N-tosyl derivative of G.

Formation of I: Claisen rearrangement followed by O-benzylation.
Heating H causes an aromatic Claisen rearrangement: the oxygen-bound allyl group migrates to the adjacent C4 position, and the phenolic hydroxyl group is restored.

The depicted rearrangement product is 4-allyl-5-hydroxy-1-tosyl-1H-indole:

C=CCc1c(O)ccc2n(S(=O)(=O)c3ccc(C)cc3)ccc12

Its phenolic oxygen then substitutes bromide from the supplied substituted benzyl bromide. The benzyl ring bears methoxy at position 3 and benzyloxy at position 4, giving I.

J: Alkene dihydroxylation.
The \(\ce{OsO4}\)/NMO system converts the terminal allyl double bond into a vicinal diol. Thus the substituent changes from an allyl group to a 2,3-dihydroxypropyl group:

  • In I, the side chain is ring–\(\ce{CH2CH=CH2}\).
  • In J, it is ring–\(\ce{CH2CH(OH)CH2OH}\).

Addition is syn. Since the substrate and reagents are achiral and only one stereocentre is created, J is racemic.

K: Oxidative cleavage followed by aldehyde oxidation.
Lead tetraacetate cleaves the bond between the two hydroxyl-bearing carbons of J. The terminal carbon is removed from the retained side chain, leaving an aldehyde attached through the original benzylic methylene.

Sodium chlorite oxidizes this aldehyde to the carboxylic acid K. Therefore, the substituent in K is ring–\(\ce{CH2CO2H}\), rather than a carboxyl group directly attached to the ring.

The signal at 12 ppm is consistent with a carboxylic-acid proton.

8.3. Structure and stereochemistry of M

For this explanation, number each ring of L locally from the ethyl/acetoxy-bearing stereocentre as C1, proceeding clockwise in the supplied drawing. Then:

  • C2 and C4 are carbonyl carbons.
  • C3 carries the exocyclic enol side chain.
  • The left-hand ring double bond is C5=C6, with methoxy at C6.

Demethylation.
Heating with \(\ce{LiCl}\) cleaves the methyl group from the activated C6 enol ether. The aryl methoxy group and the acetoxy group are retained.

The resulting 6-hydroxy analogue of (S)-L, after protonation, has SMILES:

CC[C@@]1(OC(C)=O)C(=O)/C(=C(/O)\C=C\c2ccc(OC)cc2)C(=O)C=C1O

Its formula is \(\ce{C20H20O7}\).

Methine-bridge formation.
Triethyl orthoformate supplies one carbon that links C5 of two such units. Elimination of ethanol produces the conjugated methine-linked dimer:

\[ \ce{2 C20H20O7 + HC(OC2H5)3 -> C41H38O14 + 3 C2H5OH}. \]

In the tautomer specified in the answer:

  • The first ring has a C5=CH bond to the bridge and a C6 carbonyl.
  • The second ring is singly bonded from C5′ to the bridge and retains C5′=C6′–OH.
  • Both original exocyclic enol side chains remain represented as in L.
  • The central double bond has the \(Z\) arrangement encoded in the SMILES.

This produces the same methine-linked, conjugated ring framework found in carthamin.

Stereochemistry.
Neither transformation breaks a bond at either acetoxy/ethyl-bearing stereocentre. Both centres therefore retain their absolute configurations. In the particular tautomer used in the answer, both are S. The original \(E\) side-chain alkene geometries are also retained.

Check against the NMR hint.
Exchangeable hydroxyl protons are excluded in methanol-\(d_4\). The ethyl methylene hydrogens are diastereotopic because of the neighbouring stereocentre.

For M, rapid keto–enol exchange makes the corresponding environments of its two halves equivalent on the NMR timescale.

Proton environment Signals in L Signals in M
Para-disubstituted phenyl ring 2 2
Side-chain alkene hydrogens 2 2
Aryl methoxy group 1 1
Ring enol-ether methoxy group 1 0
Acetoxy methyl group 1 1
Ethyl methyl group 1 1
Diastereotopic ethyl methylene hydrogens 2 2
Original ring C5–H 1 0
New bridging methine hydrogen 0 1
Total 11 10

The predicted counts agree with both values supplied in the problem.

The solution was checked against the official solution.