IChO 2022 · Theory problem 8

Amazing Chiral Spiro Catalyst

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Chiral compounds are important for human health, for example, more than 50% of the medicines currently in clinical use are single enantiomers of chiral molecules. However, synthesizing chiral molecules in enantio-enriched form is a great challenge. Professor Qilin Zhou's team in Nankai University in China developed a series of chiral spiro catalysts with high activity, raising the efficiency of asymmetric synthesis to a new height, and being widely used in the pharmaceutical industry. These catalysts can give up to (reach up to) 99.9% ee and can be used down to 0.00002 mol% loading. This research result won the first prize of the 2019 National Natural Science Award of China.

Figure 1

Note:

\(\ce{Ar} = 3,\,5\text{-}(\ce{^{t}Bu})_2\ce{C6H3}\)

Part A

The synthetic route of the chiral ligand SpiroPAP is shown in the following scheme.

Figure 2

Note:

eq = equivalent

\(\ce{Ar} = 3,\,5\text{-}(\ce{^{t}Bu})_2\ce{C6H3}\)

8.1 Draw the structures of 1, 2, 4, 5. (Stereochemistry is not considered).

8.2 From the following reagents, which reagent A can NOT be used for the transformation of 8 to 9?

(a) \(\ce{NaBH(OAc)3}\)

(b) \(\ce{NaBH3CN}\)

(c) \(\ce{NH2NH2}\), \(\ce{NaOH}\)

(d) \(\ce{NaBH4}\)

The racemic spiro compound 6 reacts with (–)-menthyl chloroformate (10) to generate compounds 11a and 11b, which can be separated by column chromatography, followed by hydrazinolysis to obtain optically pure (+)-6 and (–)-6.

Figure 3

Note:

eq = equivalent

8.3 Choose the correct statement below.

(a) Compounds 11a and 11b are a pair of enantiomers.

(b) Compounds 11a and 11b are a pair of diastereomers.

(c) Compounds 11a and 11b are a pair of cis-trans isomers.

(d) Compounds 11a and 11b are a pair of conformational isomers.

Part B

The Ir-SpiroPAP was prepared by reaction of SpiroPAP with \(\ce{[Ir(cod)Cl]2}\) (cod is cycloocta-1,5-diene) in MeOH under hydrogen as shown below.

Figure 4

Note:

\(\ce{Ar} = 3,\,5\text{-}(\ce{^{t}Bu})_2\ce{C6H3}\)

8.4 Write the oxidation number of Ir in the catalyst Ir-SpiroPAP.

8.5 Write the number of valence electrons of d orbitals for metal Ir in the catalyst Ir-SpiroPAP.

The chiral spiro catalyst Ir-SpiroPAP was used for the asymmetric total synthesis of Mulinane-type diterpene 22 as shown below.

Figure 5

8.6 Draw the structures of 15 and 16, including the appropriate stereochemistry.

8.7 From the following reagents, choose the best for B.

(a) \(\ce{MeLi/CeCl3}\)

(b) \(\ce{MeLi/CuI}\)

(c) \(\ce{MePh3P+ I-}\), \(\ce{^{n}BuLi}\)

(d) \(\ce{Me3S+ I-}\), \(\ce{NaH}\)

8.8 From the following reagents, choose the best for C.

(a) \(\ce{MeLi/CeCl3}\)

(b) \(\ce{MeLi/CuI}\)

(c) \(\ce{MePh3P+ I-}\), \(\ce{^{n}BuLi}\)

(d) \(\ce{Me3S+ I-}\), \(\ce{NaH}\)

Official English text of the problem.

Answers

  • 8.1
  • 1: 1,5-Bis(3-methoxyphenyl)penta-1,4-dien-3-one.
    SMILES: COc1cccc(C=CC(=O)C=Cc2cccc(OC)c2)c1
  • 2: 1,5-Bis(3-methoxyphenyl)pentan-3-one.
    SMILES: COc1cccc(CCC(=O)CCc2cccc(OC)c2)c1
  • 4: 4,4′-Dibromo-7,7′-dimethoxy-1,1′-spirobiindane.
    SMILES: COc1ccc(Br)c2c1C1(CCc3c(Br)ccc(OC)c31)CC2

  • 5: 7,7′-Dimethoxy-1,1′-spirobiindane.
    SMILES: COc1cccc2c1C1(CCc3cccc(OC)c31)CC2

  • 8.2: (c) Hydrazine, NN, with sodium hydroxide, [Na+].[OH-].

  • 8.3: (b) Diastereomers.

  • 8.4: The oxidation number of Ir is \(+3\).

  • 8.5: Six \(d\) electrons, i.e. \(5d^6\).

  • 8.6

  • 15: Ethyl (1S,5R)-1-[2-(3-methoxyphenyl)ethyl]-2-oxo-5-(propan-2-yl)cyclopentane-1-carboxylate.
    SMILES: CCOC(=O)[C@]1(CCc2cccc(OC)c2)C(=O)CC[C@@H]1C(C)C

  • 16: Ethyl (3R,3aS)-7-methoxy-3-(propan-2-yl)-2,3,4,5-tetrahydro-3aH-cyclopenta[a]naphthalene-3a-carboxylate.
    SMILES: CCOC(=O)[C@]12CCc3cc(OC)ccc3C1=CC[C@@H]2C(C)C
    In both 15 and 16, the ethoxycarbonyl and isopropyl groups are cis. In 16, the nonaromatic double bond lies in the five-membered ring.

  • 8.7: (b) Methyllithium, [Li+].[CH3-], with copper(I) iodide, [Cu+].[I-].

  • 8.8: (a) Methyllithium, [Li+].[CH3-], with cerium(III) chloride, [Ce+3].[Cl-].[Cl-].[Cl-].

Solution

8.1 — Structures of 1, 2, 4 and 5

The starting aldehyde is 3-methoxybenzaldehyde, COc1cccc(C=O)c1, and the ketone reagent is acetone, CC(=O)C.

Formation of 1. Acetone has two enolizable methyl groups. The stated 0.5 equivalent of acetone therefore allows two aldehyde molecules to condense with one acetone molecule. Two aldol additions, each followed by dehydration, give the conjugated dienone 1:

\[ \ce{2 C8H8O2 + C3H6O ->[NaOH] C19H18O3 + 2 H2O}. \]

Thus, draw a central ketone bearing a \(\ce{-CH=CH-(3-methoxyphenyl)}\) group on each side. Its formula agrees with the supplied \(\ce{C19H18O3}\). Alkene stereochemistry is omitted as instructed.

Formation of 2. Hydrogenation reduces both carbon–carbon double bonds. The ketone remains, as confirmed by the structure of 3:

\[ \ce{C19H18O3 + 2 H2 ->[Raney\ Ni] C19H22O3}. \]

Thus, 2 has a central ketone bearing two \(\ce{-CH2CH2-(3-methoxyphenyl)}\) groups.

Formation of 4. In the supplied dibrominated ketone 3, bromine blocks one ortho position of each aryl ring. Titanium tetrachloride activates the carbonyl group, permitting successive intramolecular Friedel–Crafts cyclizations at the two unblocked ortho positions.

The original carbonyl carbon becomes the common spiro carbon of two five-membered rings. The net organic balance is:

\[ \ce{C19H20Br2O3 ->[TiCl4] C19H18Br2O2 + H2O}. \]

Draw two indane units sharing only their C1 atom. Each aromatic ring carries bromine at C4 and methoxy at C7. This gives 4, 4,4′-dibromo-7,7′-dimethoxy-1,1′-spirobiindane.

Formation of 5. Butyllithium effects bromine–lithium exchange at both aryl bromides. Ethanol then protonates the two carbon–lithium sites, replacing both bromines with hydrogen. An overall balanced equation is:

\[ \ce{ C19H18Br2O2 + 2 C4H9Li + 2 C2H5OH -> C19H20O2 + 2 C4H9Br + 2 C2H5OLi }. \]

The spiro skeleton and both methoxy groups remain intact, giving 5. Subsequent boron tribromide cleavage of its methyl ethers produces the two phenolic groups of 6, confirming the methoxy positions.

8.2 — Reagent A that cannot be used

The transformation is reductive amination. The primary amino group of 8 first condenses with 3-methylpyridine-2-carbaldehyde, Cc1cccnc1C=O, forming an imine. Reduction then gives the secondary amine 9.

Writing \(R\) for the spiroaryl portion and \(R'\) for the 3-methylpyridin-2-yl group:

\[ \ce{RNH2 + R'CHO <=> RN=CHR' + H2O}, \]

\[ \ce{RN=CHR' + 2[H] -> RNHCH2R'}. \]

Here, \(2[\ce{H}]\) denotes the formal hydrogen equivalents supplied by reduction and protonation.

The following reagents can reduce the imine:

  • Sodium triacetoxyborohydride: [Na+].[BH-](OC(C)=O)(OC(C)=O)OC(C)=O
  • Sodium cyanoborohydride: [Na+].[BH3-]C#N
  • Sodium borohydride: [Na+].[BH4-]

Sodium borohydride is suitable because the scheme specifies condensation first and reduction second.

Hydrazine with sodium hydroxide supplies Wolff–Kishner conditions. These convert a carbonyl group into a methylene group through a hydrazone; they do not perform the required reduction of the imine while retaining the linkage to the amino group of 8.

Therefore, A cannot be (c).

8.3 — Relationship between 11a and 11b

The two derivatives have:

  • Opposite configurations of the spiro framework.
  • The same configurations at every stereocentre of their menthyl groups.

A mirror image would invert both the spiro configuration and all the menthyl stereocentres. Since only the spiro configuration differs, 11a and 11b are diastereomers.

Their different physical properties allow chromatographic separation. Removal of the menthyl-derived carbonate groups then releases the separate enantiomers of 6.

Thus, (b) is correct.

8.4 — Oxidation number of Ir

The depicted complex is neutral. Its ligands comprise:

  • One neutral tridentate phosphine–amine–pyridine ligand.
  • One chloride ligand, [Cl-], of charge \(-1\).
  • Two hydride ligands, [H-], each of charge \(-1\).

The hydrogen attached to the amine nitrogen belongs to the neutral amine; it is not an additional hydride ligand.

If the oxidation number of Ir is \(x\),

\[ x+0-1-1-1=0, \]

so

\[ \boxed{x=+3}. \]

The three donor atoms of SpiroPAP, chloride and two hydrides give a six-coordinate, approximately octahedral complex.

8.5 — Number of metal \(d\) electrons

Iridium belongs to group 9. The formal \(d\)-electron count is therefore

\[ n_d=9-\text{oxidation number}=9-3=6. \]

Hence the metal has the configuration

\[ \boxed{5d^6}. \]

As a consistency check, the six \(d\) electrons plus twelve electrons donated by the six donor sites give an 18-electron complex.

8.6 — Structures and stereochemistry of 15 and 16

For tracing the atoms, number the cyclopentanone ring of 14 as follows:

  • C1: the carbon bearing the ethoxycarbonyl group.
  • C2: the ketone carbonyl carbon.
  • C3 and C4: the two methylene carbons.
  • C5: the carbon bearing the isopropyl group.

For the incoming aromatic ring, let C1′ bear the ethyl tether and C3′ bear methoxy.

Formation of 15. The alkylating reagent is 1-(2-iodoethyl)-3-methoxybenzene, COc1cccc(CCI)c1.

The hydrogen at C1 of 14 lies between the ketone and ester carbonyl groups. Carbonate-assisted enolate formation followed by \(S_\mathrm{N}2\) alkylation introduces a \(\ce{-CH2CH2-(3-methoxyphenyl)}\) group at C1.

An overall balance including carbonate neutralization is:

\[ \ce{ 2 C11H18O3 + 2 C9H11IO + Cs2CO3 -> 2 C20H28O4 + 2 CsI + CO2 + H2O }. \]

The resulting formula is the supplied \(\ce{C20H28O4}\).

Stereochemistry of 15. Enolate formation removes the original configuration at C1; that stereocentre is established again during alkylation. The configuration at C5 is retained.

The subsequent structure 17 fixes the required relative configuration: its ethoxycarbonyl and isopropyl groups are both shown as solid wedges. Therefore, these groups must already be cis in 15. The new phenethyl substituent is trans to the isopropyl group.

To draw 15 in the orientation of 14, retain the solid-wedge isopropyl group and show the ethoxycarbonyl group on the same face, with the new phenethyl group on the opposite face. The absolute configuration is (1S,5R), as encoded in the answer SMILES.

Formation of 16. Methanesulfonic acid activates the ketone for intramolecular Friedel–Crafts attack. Aromatic C6′ attacks carbonyl carbon C2, forming a new six-membered ring. C6′ is ortho to the tether and para to methoxy.

The resulting alcohol loses water. C1 has no hydrogen after alkylation, so dehydration forms the C2=C3 double bond in the five-membered ring:

\[ \ce{C20H28O4 ->[MsOH] C20H26O3 + H2O}. \]

This matches the formula supplied for 16.

The configurations at the original C1 and C5 remain unchanged. In the fused-ring nomenclature used in the answer, these correspond to 3aS and 3R, respectively.

A precise way to draw 16 is to start from the supplied structure 17:

  1. Retain its complete carbon skeleton, methoxy group, ester and isopropyl group.
  2. Retain the solid-wedge ester and isopropyl substituents.
  3. Remove the dashed-wedge hydrogen from the upper five-/six-membered-ring junction.
  4. Make the bond from that junction to the adjacent nonjunction carbon of the five-membered ring a double bond.

Hydrogenation of this alkene then gives 17:

\[ \ce{C20H26O3 + H2 ->[Pd/C] C20H28O3}. \]

8.7 — Best reagent for B

Comparing 18 with 19, the required transformation:

  • Adds a methyl group to the β-carbon of the enone.
  • Consumes the enone carbon–carbon double bond.
  • Retains the ketone.

This is 1,4-conjugate addition.

Methyllithium with copper(I) iodide supplies an organocopper methyl reagent. A representative species is lithium dimethylcuprate, [Li+].C[Cu-]C, formed formally by:

\[ \ce{2 CH3Li + CuI -> LiCu(CH3)2 + LiI}. \]

The formal conjugate-addition balance, including protonation, is:

\[ \ce{ RCOCH=CR'R'' + CH3^- + H+ -> RCOCH2C(CH3)R'R'' }. \]

Organocuprates favour this conjugate addition, producing the methyl-substituted saturated ketone 19.

Therefore, B is (b), \(\ce{MeLi/CuI}\).

8.8 — Best reagent for C

Comparing 20 with 21, the required transformation:

  • Adds methyl directly to the ketone carbonyl carbon.
  • Converts that carbonyl into a tertiary alcohol.
  • Retains the pre-existing carbon–carbon double bond.

This is 1,2-addition to the carbonyl, with the formal balance:

\[ \ce{R2C=O + CH3^- + H+ -> R2C(OH)CH3}. \]

Methyllithium with cerium(III) chloride generates an organocerium reagent that favours carbonyl addition and reduces competing strongly basic behaviour. Protonation of the alkoxide gives 21.

The copper-containing reagent instead favours conjugate addition. The phosphorus- and sulfur-ylide options do not give the required methyl-substituted tertiary alcohol.

Therefore, C is (a), \(\ce{MeLi/CeCl3}\).

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