IChO 2023 · Theory problem 8

Nazarov Reaction

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The Nazarov reaction is a frequently used reaction of divinyl ketones to give cyclopentenones. It proceeds either photochemically or via acid catalysis and is an electrocyclization, followed by a proton transfer.

Figure 1

8.1 Draw the π molecular orbitals to describe the Nazarov reaction. Fill in the electrons into the respective energy levels. Mark with an x the (i) HOMO (highest occupied molecular orbital) and (ii) LUMO (lowest unoccupied molecular orbital). For this exercise, you can consider the divinyl ketone as a pentadienyl cation with five p-orbitals.

Figure 2

8.2 From the π molecular orbitals you derived in Task 8.1, predict under which conditions the Nazarov reaction of the divinyl ketone will proceed in a disrotatory or conrotatory fashion. In the table mark with an x the conditions under which the reaction is allowed.

Disrotatory Conrotatory
Thermal
Photochemical

8.3 The Nazarov reaction was used as key reaction in a synthesis of Farnesin. For both conditions below, draw one possible structure for each of A and B, including stereochemistry. Note that the products of both reactions show \(\ce{^1H}\) NMR a signal at 6.70–6.73 ppm.

Figure 3

The synthesis of Capnellene commences with unsaturated aldehyde C shown below. Treatment with D, followed by reaction with \(\ce{MnO2}\) supported on carbon gave divinyl ketone E shown below. Exposure to a mixture of \(\ce{P2O5}\) and MsOH yielded F, which was elaborated via a sequence of reactions to the unsaturated ketone I.

Figure 4

8.4 Choose the reagent(s) from the list that would be suitable as D.

(a) \(\ce{H2C=CHMgBr}\)  (c) \(\ce{H2C=CHBr}\), \(\ce{Pd(PPh3)4}\)

(b) 1. \(\ce{NaBH4}\) 2. \(\ce{H2C=CHLi}\)  (d) \(\ce{H2C=CHMgBr}\), \(\ce{CuI}\)

8.5 Give the structures of intermediates F, G, and H, including their stereochemistry.

Enone I was then subjected to \(\ce{H2C=CHMgBr}\) and \(\ce{CuI}\) in THF to give intermediate J, followed by ozonolysis to yield intermediate K, which shows a signal at 9.61 ppm in the \(\ce{^1H}\) NMR. Treatment with 5-% \(\ce{KOH}\) in a mixture of THF and ether yielded intermediate L. Hydrogenation with a Pt catalyst and under an atmosphere of \(\ce{H2}\) yielded M, which finally gave rise to Capnellene.

Figure 5

8.6 Give the structures J, K, L, and M, including their stereochemistry.

Official English text of the problem.

Answers

  • 8.1. The pentadienyl cation has four π electrons, giving \[ (\psi_1)^2(\psi_2)^2. \] Thus HOMO \(=\psi_2\) and LUMO \(=\psi_3\). The upper-lobe phase patterns, in increasing energy, are \[ +++++,\qquad ++0--,\qquad +0-0+,\qquad +-0+-,\qquad +-+-+. \] The coefficients and drawing instructions are given in the solution.

  • 8.2.

Disrotatory Conrotatory
Thermal x
Photochemical x
  • 8.3. One possible pair consists of the following stereoisomers of 13-hydroxy-2-methyltricyclo[7.4.0.0²,⁷]tridec-6-en-8-one:
Product Configuration Isomeric SMILES
A \((1R,2S,9S,13R)\) C[C@@]12CCCC=C1C(=O)[C@H]1CCC[C@@H](O)[C@H]12
B \((1S,2S,9R,13R)\) C[C@@]12CCCC=C1C(=O)[C@@H]1CCC[C@@H](O)[C@@H]12

These representatives have an enone double bond in the original left six-membered ring. The original terminal methyl and terminal hydrogen are trans in A, but cis in B.

  • 8.4. (a) only: vinylmagnesium bromide, C=C[Mg]Br.

  • 8.5. The structures below use the enantiomer corresponding to the supplied drawing of I.

Intermediate Name Isomeric SMILES
F 8,8-Dimethylbicyclo[3.3.0]oct-1(5)-en-2-one O=C1CCC2=C1C(C)(C)CC2
G (1S,5S)-5,8,8-Trimethylbicyclo[3.3.0]octan-2-one CC1(C)CC[C@@]2(C)CCC(=O)[C@@H]12
H (1S,2R,5S)-2-Ethynyl-5,8,8-trimethylbicyclo[3.3.0]octan-2-ol C#C[C@@]1(O)CC[C@]2(C)CCC(C)(C)[C@@H]21

H is shown with ethynyl on the same face as the bridgehead methyl. Its alcohol-center epimer is also compatible with the subsequent rearrangement; see the solution.

  • 8.6.
Intermediate Name Isomeric SMILES
J 1-[(1S,2S,3R,5S)-3-Ethenyl-5,8,8-trimethylbicyclo[3.3.0]octan-2-yl]ethan-1-one C=C[C@H]1C[C@]2(C)CCC(C)(C)[C@@H]2[C@H]1C(C)=O
K (1S,3S,4R,5S)-4-Acetyl-1,6,6-trimethylbicyclo[3.3.0]octane-3-carbaldehyde CC(=O)[C@H]1[C@@H](C=O)C[C@]2(C)CCC(C)(C)[C@H]12
L (1S,2S,6R,8S)-8,11,11-Trimethyltricyclo[6.3.0.0²,⁶]undec-4-en-3-one CC1(C)CC[C@@]2(C)C[C@@H]3C=CC(=O)[C@@H]3[C@@H]12
M (1S,2S,6S,8S)-8,11,11-Trimethyltricyclo[6.3.0.0²,⁶]undecan-3-one CC1(C)CC[C@@]2(C)C[C@@H]3CCC(=O)[C@@H]3[C@@H]12

J and K are represented in the productive cis configuration leading to the drawn capnellene stereochemistry. In L and M, both five-membered-ring junctions are cis, with the junction hydrogens on opposite faces at the two junctions. The achiral synthesis also gives the corresponding mirror-image series.

Solution

8.1 — Molecular orbitals and electron occupation

The model species is the pentadienyl cation, C=C[CH+]C=C. Its five contiguous p orbitals contain four electrons: the two double bonds supply two electrons each, while the cationic center contributes an empty p orbital.

Number the atomic p orbitals consecutively along the chain, \(p_1,\ldots,p_5\). In the simple Hückel model,

\[ \psi_k=\sum_{j=1}^{5}c_{jk}p_j,\qquad c_{jk}=\sqrt{\frac13}\sin\left(\frac{jk\pi}{6}\right), \]

with energies

\[ E_k=\alpha+2\beta\cos\left(\frac{k\pi}{6}\right), \qquad \beta<0. \]

The following table specifies the orbital drawings completely. Coefficients in the third column are proportional to the orbital amplitudes.

Orbital, highest first Energy Coefficients on \(p_1,p_2,p_3,p_4,p_5\) Nodes along the chain Electrons Mark
\(\psi_5\) \(\alpha-\sqrt3\beta\) \(1,-\sqrt3,2,-\sqrt3,1\) 4 Empty
\(\psi_4\) \(\alpha-\beta\) \(1,-1,0,1,-1\) 3 Empty
\(\psi_3\) \(\alpha\) \(1,0,-1,0,1\) 2 Empty x under LUMO
\(\psi_2\) \(\alpha+\beta\) \(1,1,0,-1,-1\) 1 \(\uparrow\downarrow\) x under HOMO
\(\psi_1\) \(\alpha+\sqrt3\beta\) \(1,\sqrt3,2,\sqrt3,1\) 0 \(\uparrow\downarrow\)

To draw these orbitals:

  • Draw a p-orbital pair at each carbon with nonzero coefficient.
  • Give the upper lobe the phase indicated by the coefficient and the lower lobe the opposite phase.
  • Make lobe sizes proportional to the coefficient magnitudes.
  • At a zero coefficient, indicate a node through that atom. The atomic p orbital still exists, but its contribution to that particular MO is zero.

Changing every sign in a row gives the same physical orbital.

8.2 — Electrocyclic selection rules

This is a four-electron, or \(4n\), electrocyclization.

In the ground-state HOMO, \(\psi_2\), the terminal coefficients have opposite signs. Conrotation brings matching phases together to form the new terminal carbon–carbon σ bond.

Photoexcitation promotes an electron from \(\psi_2\) into \(\psi_3\). The terminal coefficients of \(\psi_3\) have the same sign, and the photochemically allowed mode is consequently disrotation.

Thus:

\[ \boxed{\text{Thermal: conrotatory}} \qquad \boxed{\text{Photochemical: disrotatory}}. \]

8.3 — Structures and stereochemistry of A and B

The starting material is (3R)-3-hydroxycyclohex-1-en-1-ylmethanone,
O=C(C1=C(C)CCCC1)C1=C[C@H](O)CCC1.

The new σ bond joins the two lower alkene carbons in the supplied drawing. One bears the methyl substituent; the other originally bears a hydrogen.

For the products selected in the Answers, the atom numbering is:

  • C2: original methyl-bearing terminal carbon;
  • C1: original hydrogen-bearing terminal carbon;
  • C7 and C9: original left and right carbonyl-adjacent alkene carbons;
  • C8: carbonyl carbon;
  • C13: original alcohol-bearing carbon.

The product double bond is C6=C7, in the original left six-membered ring. C6 bears a vinylic hydrogen, providing the proton consistent with the signal at 6.70–6.73 ppm.

This NMR clue matters because an alternative product with the double bond between the original right-hand alkene carbons would have a fully substituted enone double bond and no corresponding vinylic proton.

The stereochemical drawing instructions for the selected pair are:

Bond or hydrogen, with the ring skeleton oriented as in the question A B
C2–methyl Solid wedge Solid wedge
C1–H Dashed wedge Solid wedge
C9–H Dashed wedge Solid wedge
C13–OH Solid wedge Solid wedge

The C13 alcohol configuration is retained. The right-hand ring junction is chosen cis in both representatives.

Under Lewis-acid activation, conrotation places the original terminal methyl and terminal hydrogen on opposite faces: A. Under irradiation, disrotation places them on the same face: B.

Both transformations are atom-conserving isomerizations:

\[ \ce{C14H20O2 ->[BF3.OEt2] C14H20O2} \qquad(\mathbf A) \]

\[ \ce{C14H20O2 ->[h\nu,\ 366\ nm] C14H20O2} \qquad(\mathbf B). \]

The different possible proton-transfer pathways and facial choices explain why the question asks for one possible structure under each condition.

8.4 — Identification of D

The starting aldehyde C is 5,5-dimethylcyclopent-1-ene-1-carbaldehyde,
O=CC1=CCCC1(C)C.

The required product E is 1-(5,5-dimethylcyclopent-1-en-1-yl)prop-2-en-1-one,
C=CC(=O)C1=CCCC1(C)C.

To obtain E, a vinyl group must first add to the aldehyde carbonyl carbon, giving a secondary allylic alcohol. Oxidation then converts that alcohol into the divinyl ketone.

Therefore D is vinylmagnesium bromide, C=C[Mg]Br, option (a).

The alcohol intermediate is 1-(5,5-dimethylcyclopent-1-en-1-yl)prop-2-en-1-ol,
C=CC(O)C1=CCCC1(C)C. It is formed as a racemate; its alcohol-center configuration is erased during oxidation.

Including aqueous work-up, the addition is balanced by

\[ \ce{C8H12O + C2H3MgBr + H2O -> C10H16O + MgBrOH}. \]

Using \(\ce{MnO}\) as the formal reduced manganese product, the oxidation balance is

\[ \ce{C10H16O + MnO2 -> C10H14O + MnO + H2O}. \]

The other choices do not give the required transformation:

  • (b): \(\ce{NaBH4}\) first reduces the aldehyde; the subsequent organolithium encounters an alcohol rather than the required carbonyl electrophile.
  • (c): vinyl bromide and the palladium catalyst do not provide the required nucleophilic vinyl addition.
  • (d): copper promotes conjugate addition to the unsaturated aldehyde, giving the wrong carbon connectivity.

8.5 — F, G and H

For discussing this sequence, retain the bicyclo[3.3.0]octane numbering used in the name of F: the bridgeheads are C1 and C5, the ketone is at C2, and the geminal methyl groups are at C8.

Formation of F. Acid activation of E produces the pentadienyl-cation system. Thermal conrotatory closure joins the cyclopentene’s β carbon to the terminal vinyl carbon, creating the second five-membered ring.

Proton transfer gives F with its C1=C5 double bond at the shared ring junction. This placement is independently required by the next step: conjugate methyl addition must install the angular methyl observed in I.

The cyclization conserves the molecular formula:

\[ \ce{C10H14O ->[P2O5,\ MsOH] C10H14O}. \]

Formation of G. Lithium dimethylcuprate performs 1,4-addition to F. Methyl attaches at the β carbon, C5, and protonation gives the C1 hydrogen.

The configuration carried forward to the supplied I is cis: C1–H and C5–methyl are on the same face. G is therefore the cis-fused trimethyl bicyclic ketone listed above.

A formal balance including aqueous work-up is

\[ \ce{C10H14O + (CH3)2CuLi + H2O -> C11H18O + CH3Cu + LiOH}. \]

Formation of H. Lithium acetylide adds 1,2 to G’s ketone. Protonation of the alkoxide gives the tertiary propargylic alcohol H:

\[ \ce{C11H18O + LiC#CH + H2O -> C13H20O + LiOH}. \]

The molecular formula agrees with the supplied \(\ce{C13H20O}\). Its degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(13)+2-20}{2}=4, \]

accounted for by two rings and one triple bond.

The representative H in the Answers has ethynyl on the same face as the bridgehead methyl and OH on the opposite face. Its C2 epimer is (1S,2S,5S)-2-ethynyl-5,8,8-trimethylbicyclo[3.3.0]octan-2-ol,
C#C[C@]1(O)CC[C@]2(C)CCC(C)(C)[C@@H]21.

Consistency with I. Under the acidic conditions, the tertiary ethynyl alcohol undergoes a Rupe rearrangement: dehydration and hydration produce an α,β-unsaturated methyl ketone. The ring double bond becomes C2=C3, and the ethynyl group becomes the acetyl substituent.

The product is 1-[(1S,5S)-5,8,8-trimethylbicyclo[3.3.0]oct-2-en-2-yl]ethan-1-one, I,
CC(=O)C1=CC[C@]2(C)CCC(C)(C)[C@H]12.

Overall,

\[ \ce{C13H20O ->[HCOOH,\ H2SO4] C13H20O}. \]

The alcohol-bearing carbon becomes trigonal in I, so this rearrangement removes the stereochemical distinction between the two C2 epimers of H.

8.6 — J, K, L and M

I → J: conjugate vinyl addition. The copper-containing vinyl reagent adds to the β carbon of I’s enone. The acetyl carbonyl remains intact, while a vinyl group attaches at the adjacent ring carbon.

In the productive stereoisomer shown, the acetyl and vinyl substituents are cis, on the face bearing the angular methyl. Their corresponding ring hydrogens are on the opposite face.

Including aqueous work-up,

\[ \ce{C13H20O + C2H3MgBr + H2O -> [CuI] C15H24O + MgBrOH}. \]

J → K: reductive ozonolysis. Ozonolysis cleaves the terminal vinyl group. Its ring-attached carbon becomes an aldehyde carbon, while its terminal carbon gives methanal (formaldehyde), C=O.

The aldehyde explains the 9.61 ppm signal. The pre-existing methyl ketone is retained, giving K with formula \(\ce{C14H22O2}\).

The atom-balanced ozonolysis and reductive work-up are

\[ \ce{C15H24O + O3 + (CH3)2S -> C14H22O2 + CH2O + (CH3)2SO}. \]

K → L: intramolecular aldol condensation. Base removes a proton from the acetyl methyl group. Its enolate attacks the aldehyde intramolecularly, closing a five-membered ring.

The new ring contains, consecutively:

  1. the ketone-bearing original ring carbon;
  2. the ketone carbonyl carbon;
  3. the former acetyl methyl carbon;
  4. the former aldehyde carbon;
  5. the adjacent original ring carbon.

Dehydration gives the conjugated cyclopentenone L:

\[ \ce{C14H22O2 ->[KOH] C14H20O + H2O}. \]

Thus L has three rings, one carbonyl and one alkene:

\[ \mathrm{DBE}(L)=\frac{2(14)+2-20}{2}=5. \]

For its drawing, retain the original bridgehead H and angular methyl as solid wedges. At the newly formed right-hand ring junction, draw both junction hydrogens dashed, matching the final capnellene drawing.

L → M: hydrogenation. Hydrogenation saturates the enone carbon–carbon double bond while retaining the ketone:

\[ \ce{C14H20O + H2 ->[Pt] C14H22O}. \]

Both alkene carbons become methylene groups, so no additional stereocenters are created. The existing junction geometry is retained.

The formula gives

\[ \mathrm{DBE}(M)=\frac{2(14)+2-22}{2}=4, \]

corresponding to three rings and one carbonyl. The change from \(6R\) in L’s name to \(6S\) in M’s name reflects a change in CIP priorities, not inversion of that center.

Consistency with capnellene. Methyl lithium adds to M’s ketone. Subsequent dehydration gives the exocyclic methylene group of capnellene,
CC1(C)CC[C@@]2(C)C[C@@H]3CCC(=C)[C@@H]3[C@@H]12.

A formal overall balance, omitting pyridine’s trapping of the acid, is

\[ \ce{C14H22O + CH3Li + SOCl2 -> C15H24 + LiCl + SO2 + HCl}. \]

This preserves all four ring-junction configurations shown in the final target.

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