IChO 2020 · Theory problem 3

(±)-Coerulescine

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A spiro compound is typically an organic compound containing rings linked together by one common atom (spiroatom) as carbon atom with bold in figure below. The spiro[pyrrolidin-3,3′- oxindole] ring system is a structural framework incorporated in several cytostatic alkaloids and unnatural compounds. Coerulescine (1) and horsfiline are the simplest prototype members of this subfamily that can be synthesized by the route shown below.

Figure 1

Claisen rearrangement, which is a [3,3]-sigmatropic rearrangement, is a powerful carbon– carbon bond-forming reaction in which an allyl vinyl ether is converted thermally to an unsaturated carbonyl compound as shown in the Scheme below. When compound A is heated, it undergoes Claisen rearrangement to give carbonyl compound B.

For this entire task, your answers can be given without any stereochemical details.

Figure 2
Figure 3

3.1 Draw the structures of A and B.

  • A is an inseparable mixture of cis/trans isomers.
  • B has IR absorption at \(1726\ \mathrm{cm^{-1}}\).

3.2 Draw structures for C, D, E, and F.

  • D–F have a bicyclic structure.

3.3 Choose the correct order of steps for the transformation of F to G.

☐ Imine formation, then reduction, then amidation

☐ Amidation, then imine formation, then reduction

☐ Reduction, then amidation, then imine formation

3.4 Draw the structures for G and H (both spiro compounds).

3.5 Draw the structure of the intermediate produced by treatment with n-BuLi in the step H → coerulescine (1).

Coerulescine (1), on treatment with N-bromosuccinimide (NBS), gives the bromo derivative, which upon heating with sodium methoxide in the presence of cuprous iodide gives horsfiline (I) in 60% yield.

3.6 Choose the correct structure for compound I consistent with the following selected \(\ce{^1H}\)-NMR data: \(\delta\) 7.05 (d, \(J=1.4\ \mathrm{Hz}\), 1H), 6.78 (d, \(J=8.0\ \mathrm{Hz}\), 1H), 6.72 (dd, \(J=8.0,\ 1.4\ \mathrm{Hz}\), 1H) ppm. Tick the relevant box.

Figure 4

3.7 When the allyl ether of 2-naphthol is heated a sigmatropic rearrangement is initiated. Write the structure of major product isolated from this reaction.

Figure 5

Official English text of the problem.

Answers

Unprimed positions below belong to the oxindole ring; primed positions belong to the second nitrogen-containing ring. Stereochemistry is omitted where permitted by the problem.

  • 3.1
  • A: Allyl 2-(2-nitrophenyl)vinyl ether, an \(E/Z\) mixture.
    \(E\): C=CCO/C=C/c1ccccc1[N+](=O)[O-]
    \(Z\): C=CCO/C=C\c1ccccc1[N+](=O)[O-]

  • B: 2-(2-Nitrophenyl)pent-4-enal.
    O=CC(CC=C)c1ccccc1[N+](=O)[O-]

  • 3.2

  • C: Ethyl 2-(2-nitrophenyl)pent-4-enoate.
    CCOC(=O)C(CC=C)c1ccccc1[N+](=O)[O-]
  • D: 3-Allylindolin-2-one.
    O=C1Nc2ccccc2C1CC=C
  • E: 1-tert-Butyl 3-ethyl 2-oxo-3-(prop-2-en-1-yl)indoline-1,3-dicarboxylate.
    CCOC(=O)C1(CC=C)C(=O)N(C(=O)OC(C)(C)C)c2ccccc21

  • F: 1-tert-Butyl 3-ethyl 2-oxo-3-(2-oxoethyl)indoline-1,3-dicarboxylate.
    CCOC(=O)C1(CC=O)C(=O)N(C(=O)OC(C)(C)C)c2ccccc21

  • 3.3 Tick the first box: imine formation, then reduction, then amidation.

  • 3.4

  • G: The oxindole-\(N\)-Boc derivative of 1′-methylspiro[indoline-3,3′-pyrrolidine]-2,2′-dione.
    CN1CCC2(C1=O)c1ccccc1N(C(=O)OC(C)(C)C)C2=O

  • H: 1′-Methylspiro[indoline-3,3′-pyrrolidine]-2,2′-dione.
    CN1CCC2(C1=O)c1ccccc1NC2=O

  • 3.5 The lithium salt of H formed by deprotonation of its oxindole nitrogen.
    CN1CCC2(C1=O)c1ccccc1[N-]C2=O.[Li+]

  • 3.6 Tick the second structure from the left: horsfiline, the 5-methoxy derivative of coerulescine.
    CN1CCC2(C1)c1cc(OC)ccc1NC2=O

  • 3.7 1-Allylnaphthalen-2-ol, also called 1-allyl-2-naphthol.
    C=CCc1c(O)ccc2ccccc12

Solution

Throughout the solution, \(\mathrm{Ar}\) denotes the 2-nitrophenyl group. In reduction equations, \(\ce{[H]}\) denotes one formal reducing hydrogen equivalent supplied by the stated reagents.

3.1 — Wittig reaction and Claisen rearrangement

Sodium tert-butoxide removes a proton from the carbon adjacent to phosphorus in the phosphonium salt. This gives the allyloxymethylene triphenylphosphorus ylide, represented by the charge-separated SMILES:

C=CCO[CH-][P+](c1ccccc1)(c1ccccc1)c1ccccc1

The activation and Wittig reactions can be written as:

\[ \ce{[Ph3PCH2OCH2CH=CH2]Cl + NaOC(CH3)3 -> Ph3P=CH-O-CH2-CH=CH2 + HOC(CH3)3 + NaCl} \]

\[ \ce{ArCHO + Ph3P=CH-O-CH2-CH=CH2 -> A + Ph3PO} \]

The aldehyde carbon becomes the aryl-bearing carbon of the new double bond. The ylide carbon becomes the oxygen-bearing carbon. Thus A has the connectivity

\[ \ce{Ar-CH=CH-O-CH2-CH=CH2}. \]

Each alkene carbon bears two different substituents, so \(E\) and \(Z\) forms are possible, explaining the stated inseparable cis/trans mixture.

At \(139\,^\circ\mathrm{C}\), A undergoes a concerted Claisen rearrangement:

\[ \ce{Ar-CH=CH-O-CH2-CH=CH2 ->[\Delta] OHC-CH(Ar)-CH2-CH=CH2}. \]

The six-membered transition-state arrangement produces three simultaneous changes:

  1. The oxygen–allyl carbon bond breaks.
  2. A new carbon–carbon bond forms between the terminal allyl carbon and the aryl-bearing vinyl carbon.
  3. The vinyl ether becomes a carbonyl group, while the allyl double bond shifts.

The oxygen-bearing vinyl carbon originally carries hydrogen, so the carbonyl product is the aldehyde B, rather than a ketone. Its aldehyde carbonyl is consistent with the absorption at \(1726\ \mathrm{cm^{-1}}\).

Both A and B have formula \(\ce{C11H11NO3}\), as required for an isomerization.

3.2 — Identification of C, D, E, and F

Formation of C. Jones oxidation converts the aldehyde of B into 2-(2-nitrophenyl)pent-4-enoic acid:

O=C(O)C(CC=C)c1ccccc1[N+](=O)[O-]

A balanced oxidation equation is

\[ \ce{3B + 2CrO3 + 3H2SO4 -> 3ArCH(CH2CH=CH2)COOH + Cr2(SO4)3 + 3H2O}. \]

Acid-catalysed esterification with ethanol then gives C:

\[ \ce{ArCH(CH2CH=CH2)COOH + C2H5OH <=>[H+] C + H2O}. \]

The ethyl ester has formula \(\ce{C13H15NO4}\), matching the scheme.

Formation of D. Zinc/ammonium chloride reduces the ortho nitro group to an amino group. The resulting amino group attacks the ester carbonyl intramolecularly. Loss of ethanol closes the five-membered lactam ring, giving 3-allylindolin-2-one, D.

The balanced net ionic equation, combining reduction and cyclization, is

\[ \ce{C + 3Zn + 6H+ -> D + C2H5OH + 3Zn^2+ + 2H2O}. \]

The benzene ring and newly formed lactam ring constitute the required bicyclic structure. The product formula is \(\ce{C11H11NO}\).

Formation of E. First, the oxindole nitrogen is Boc-protected. Call the explicitly drawn protected intermediate P, tert-butyl 3-allyl-2-oxoindoline-1-carboxylate:

CC(C)(C)OC(=O)N1C(=O)C(CC=C)c2ccccc21

A balanced net equation for protection under the stated sodium hydride conditions is

\[ \ce{D + NaH + (Boc)2O -> P + H2 + CO2 + NaOC(CH3)3}. \]

A second treatment with sodium hydride removes the remaining hydrogen at oxindole C3. The resulting enolate attacks ethyl chloroformate, CCOC(=O)Cl, installing an ethoxycarbonyl group at C3:

\[ \ce{P + NaH + ClC(=O)OC2H5 -> E + H2 + NaCl}. \]

Thus E has both an allyl group and an ethoxycarbonyl group attached to C3, while its nitrogen carries Boc. Its formula is \(\ce{C19H23NO5}\).

Formation of F. Catalytic osmium tetroxide and sodium periodate cleave the terminal alkene:

\[ \ce{-CH2-CH=CH2 -> -CH2-CHO} \]

on the retained molecular fragment. The terminal alkene carbon becomes methanal (formaldehyde), C=O.

The balanced net oxidation is

\[ \ce{E + 2NaIO4 ->[OsO4\ (cat.)] F + HCHO + 2NaIO3}. \]

Consequently, the allyl substituent of E becomes a 2-oxoethyl substituent in F. The formula changes from \(\ce{C19H23NO5}\) to \(\ce{C18H21NO6}\), exactly as printed.

3.3 — Order of the steps from F to G

The correct sequence is imine formation → reduction → amidation.

  1. Imine formation. Methylamine reacts with the aldehyde group of F, eliminating water. The N-methyl imine of F is:

CCOC(=O)C1(CC=NC)C(=O)N(C(=O)OC(C)(C)C)c2ccccc21

For the aldehyde-containing fragment, the balanced condensation is

\[ \ce{R-CHO + CH3NH2 <=> R-CH=N-CH3 + H2O}. \]

  1. Reduction. Sodium cyanoborohydride reduces the imine/iminium functionality to a secondary amine:

\[ \ce{R-CH=N-CH3 + 2[H] -> R-CH2-NH-CH3}. \]

The resulting 2-(methylamino)ethyl derivative of F is:

CCOC(=O)C1(CCNC)C(=O)N(C(=O)OC(C)(C)C)c2ccccc21

  1. Amidation. The secondary amine attacks the ethyl ester carbonyl intramolecularly. Ethanol is eliminated, and a five-membered \(N\)-methyl lactam forms.

The complete atom-balanced net transformation is

\[ \ce{F + CH3NH2 + 2[H] -> G + H2O + C2H5OH}. \]

The aldehyde readily undergoes condensation with methylamine. Amidation first would instead produce a much less nucleophilic amide nitrogen, making the proposed subsequent imine formation unfavourable.

3.4 — Structures of G and H

The ring formed during amidation follows this sequence:

\[ \text{spiro carbon} -\ce{C(=O)} -\ce{N(CH3)} -\ce{CH2} -\ce{CH2} -\text{the same spiro carbon}. \]

It shares only the original oxindole C3 atom with the oxindole ring, so G is a spiro compound. It contains:

  • the original oxindole carbonyl;
  • the newly formed \(N\)-methyl lactam carbonyl;
  • a Boc group on the oxindole nitrogen.

Its formula is \(\ce{C17H20N2O4}\).

Treatment with \(2.5\ \mathrm{M}\) hydrochloric acid removes Boc and gives the oxindole N–H compound H. Net Boc fragmentation can be written as

\[ \ce{G ->[H+] H + CO2 + (CH3)2C=CH2}. \]

Here the hydrocarbon coproduct is 2-methylpropene, C=C(C)C. The molecular formula changes by loss of \(\ce{C5H8O2}\):

\[ \ce{C17H20N2O4 -> C12H12N2O2 + CO2 + C4H8}. \]

Both lactam carbonyls are present in H.

3.5 — Intermediate formed with n-BuLi

The oxindole nitrogen of H bears an acidic hydrogen. n-Butyllithium removes this hydrogen, producing the lithium amidate of H and butane:

\[ \ce{C12H12N2O2 + C4H9Li -> Li[C12H11N2O2] + C4H10}. \]

To draw the intermediate, replace the oxindole N–H in H with \(\ce{N^-}\) and an associated \(\ce{Li+}\). Both carbonyl groups remain present.

The ionic SMILES in the Answers section records this formal N-deprotonated structure. The amidate charge is delocalized over the nitrogen–carbonyl unit; that SMILES does not specify lithium coordination or aggregation.

Deprotonation makes the oxindole carbonyl less electrophilic. Lithium aluminium hydride then reduces the other, \(N\)-methyl lactam carbonyl to a methylene group. Work-up restores the oxindole N–H, giving coerulescine, 1′-methylspiro[indoline-3,3′-pyrrolidin]-2-one:

CN1CCC2(C1)c1ccccc1NC2=O

The overall reduction, including work-up, has the formal balance

\[ \ce{C12H12N2O2 + 4[H] -> C12H14N2O + H2O}. \]

3.6 — Identification of horsfiline

The answer is the second structure from the left, with methoxy at oxindole C5.

Its three aromatic protons are H4, H6, and H7:

Proton Given signal Coupling assignment
H4 \(\delta 7.05\), d, \(J=1.4\ \mathrm{Hz}\) Meta coupling to H6
H7 \(\delta 6.78\), d, \(J=8.0\ \mathrm{Hz}\) Ortho coupling to H6
H6 \(\delta 6.72\), dd, \(J=8.0,\ 1.4\ \mathrm{Hz}\) Ortho coupling to H7 and meta coupling to H4

The large coupling identifies one adjacent proton pair, H6–H7. The small coupling identifies the meta pair H4–H6.

The 4- and 7-methoxy alternatives each leave three consecutive aromatic protons and therefore require two substantial ortho couplings. They do not fit the observed pattern.

The coupling topology alone also permits a 6-methoxy arrangement. The chemical-shift assignments and reaction regiochemistry select C5: the lactam nitrogen directs aromatic bromination preferentially to the para position, C5. Copper-assisted methoxylation then replaces bromine at that same carbon.

The intermediate is 5-bromocoerulescine:

CN1CCC2(C1)c1cc(Br)ccc1NC2=O

The balanced bromination is

\[ \ce{C12H14N2O + C4H4BrNO2 -> C12H13BrN2O + C4H5NO2}, \]

where the brominating reagent is N-bromosuccinimide, O=C1CCC(=O)N1Br, and its coproduct is succinimide, O=C1CCC(=O)N1.

Methoxylation gives horsfiline:

\[ \ce{C12H13BrN2O + NaOCH3 ->[CuI,\ \Delta] C13H16N2O2 + NaBr}. \]

The stated 60% yield is the isolated yield; it does not change the structural assignment.

3.7 — Claisen rearrangement of the allyl ether of 2-naphthol

The starting material is 2-(allyloxy)naphthalene:

C=CCOc1ccc2ccccc2c1

Its preferred aromatic Claisen rearrangement forms the new carbon–carbon bond at naphthalene C1, adjacent to the oxygen-bearing C2. This pathway preserves a benzenoid ring during rearrangement.

The immediate keto intermediate is 1-allylnaphthalen-2(1H)-one:

O=C1C=Cc2ccccc2C1CC=C

Proton transfer and keto–enol tautomerization restore aromaticity, furnishing 1-allylnaphthalen-2-ol:

C=CCc1c(O)ccc2ccccc12

Thus the product to draw is a naphthalene ring with OH at C2 and \(\ce{-CH2-CH=CH2}\) at C1. The double bond remains terminal.

The overall reaction is an atom-conserving isomerization:

\[ \underbrace{\ce{C13H12O}}_{\text{2-(allyloxy)naphthalene}} \xrightarrow{\Delta} \underbrace{\ce{C13H12O}}_{\text{1-allylnaphthalen-2-ol}}. \]

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