Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section F

A hydroxyl-directed Evans–Saksena reduction gives the required anti-1,3-diol. A suitable reagent is tetramethylammonium triacetoxyborohydride in an acidic medium, for example acetic acid/acetonitrile. Ligand exchange between the pre-existing alcohol and an acetoxyborohydride group tethers the reducing reagent to the substrate. Intramolecular hydride delivery to the ketone then proceeds through a six-membered arrangement, converting C=O to an alkoxide while conserving all existing stereocentres. Hydrolysis releases the second hydroxyl group.

The preferred directed transition-state conformation minimizes crowding by the larger carbon-chain substituents. It delivers the new C–H bond on the face that leaves the two hydroxyl groups anti, as in B: with the original hydroxyl drawn as a wedge in the printed zigzag, the new hydroxyl is dashed. The intervening methyl stereocentre is retained. Choosing an undirected borohydride reduction would not by itself rationalize this strong substrate-controlled relationship.

Dibutylboron triflate coordinates the imide carbonyl and triethylamine removes the propionyl $\alpha$ proton, giving predominantly the Z dibutylboron enolate. Its connectivity is $\mathrm{Aux}-\mathrm{N}-\mathrm{C}(\mathrm{O}-\mathrm{BB}{\mathrm{u}}_2)=\mathrm{CHMe}$; the imide auxiliary remains attached, rather than being displaced during enolization. Coordination of cinnamaldehyde's oxygen to boron organizes a six-membered Zimmerman–Traxler transition state.

A chair with the aldehyde's styryl group pseudoequatorial avoids the auxiliary's bulky benzyl side. Formation of the C–C bond and transfer of the aldehyde oxygen into the boron coordination sphere give the boron aldolate, which hydrolyses to D. A Z enolate gives the syn aldol relationship. In the printed product zigzag, the labelled $\alpha$-methyl bond should therefore be a solid wedge, on the same stereochemical face as the neighbouring solid-wedge hydroxyl bond. With the stated connectivity, the labelled carbon is R: priority is imide carbonyl, then $\mathrm{CH}(\mathrm{OH})\mathrm{CH}=\mathrm{CHPh}$, then methyl, then hydrogen. The adjacent hydroxyl centre is S. These assignments refer to the actual auxiliary and product orientation, not an arbitrary choice of an otherwise equivalent chair.

The free allylic hydroxyl can hydrogen-bond to the peracid and accelerate directed oxygen transfer to the nearby double bond. The remote styryl double bond lacks this directing contact. Epoxidation of the hydroxyl-adjacent alkene is therefore favoured, even though both alkenes are in principle oxidizable. Peracid epoxidation is a concerted oxygen transfer: alkene $\pi$ donation to terminal peroxide oxygen, O–O cleavage, proton transfer and carbonyl re-formation occur through a cyclic transition-state arrangement. The alkene's E geometry is retained as a trans relationship between its substituents in the epoxide.

The two alkene faces are diastereotopic because the molecule already contains stereocentres. A plausible substrate-controlled conformation minimizes allylic strain and presents the face opposite the bulky OTBS-bearing part of the chain to the oxidant. Predict epoxide oxygen on the face opposite OTBS, retaining the pre-existing methyl and OTBS configurations. Hydrogen-bond direction, steric approach and conformational population compete; one cannot deduce an exact diastereomeric ratio from the flat drawing alone. The same facial preference has been discussed for this synthetic series.^[ref]

The acetylide attacks the less substituted terminal epoxide carbon by SN2 ring opening. The C–O bond at that carbon breaks, leaving the oxygen attached to the original substituted stereocentre. Protonation gives $\mathrm{BnOC}{\mathrm{H}}_2\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{C}\equiv \mathrm{CMe}$. Dissolving-metal reduction then gives the E alkene and also cleaves the benzyl ether.

Thus H is $\mathrm{HOC}{\mathrm{H}}_2\mathrm{CH}(\mathrm{OH})\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{CHMe}$ with an E double bond, and its original epoxide-derived stereocentre is retained geometrically. No new tetrahedral stereocentre is formed: the attacked epoxide carbon becomes CH${}_2$. The new stereochemical information is the E alkene geometry. Calling its secondary hydroxyl carbon “newly generated” would confuse changing its functional environment with creating a new stereocentre.

The secondary alcohol of H condenses with the aldehyde under acid catalysis to form a hemiacetal, then loses water to an oxocarbenium ion. Its tethered E alkene attacks the aldehyde-derived carbon through a chair-like six-membered arrangement: the alkene carbon farther from the alcohol forms the new C–C bond, placing a carbocation at the nearer alkene carbon. Trifluoroacetate captures that cation; the basic methanolic work-up cleaves the trifluoroacetate ester to reveal the ring hydroxyl group. This is a Prins cyclization.

Number the product ring from oxygen through the aldehyde-derived carbon as C2, then the alkene-derived methyl-bearing carbon C3, the hydroxyl-bearing carbon C4, the intervening methylene C5 and the original alcohol carbon C6. A chair keeping the bulky C2 side chain and C6 hydroxymethyl group equatorial predicts 2,6-cis substitution. Suprafacial alkene capture preserves the E-alkene-derived relationship, giving 2,3-trans between the C2 side chain and C3 methyl. Preferred equatorial trapping of the C4 cation gives 3,4-trans, hence the 2,4,6-cis arrangement of the substituent-bearing bonds in the favoured chair. In the printed flat-ring orientation, draw the C2 side chain, C4 hydroxyl and C6 hydroxymethyl behind the plane, and C3 methyl in front. Existing stereocentres in the aldehyde side chain and at C6 are retained; they anchor the absolute face of this assignment.

DBU removes the proton $\alpha$ to sulfonium, generating the propargyl-substituted sulfur ylide $\mathrm{M}{\mathrm{e}}_2{\mathrm{S}}^{+}-\mathrm{C}{\mathrm{H}}^{-}\mathrm{C}\equiv \mathrm{CMe}$. Its carbon attacks the enone $\beta$ carbon bearing indolyl, shifting the conjugated electrons into an enolate. The new intermediate has the carbon sequence ${\mathrm{O}}^{-}-\mathrm{C}(\mathrm{Ph})=\mathrm{C}(\mathrm{CN})-\mathrm{CH}(\mathrm{indolyl})-\mathrm{CH}(\mathrm{SM}{\mathrm{e}}_2^{+})(\mathrm{C}\equiv \mathrm{CMe})$.

Intramolecular attack by the enolate oxygen on the sulfonium-bearing carbon displaces dimethyl sulfide and closes a five-membered ring. This conjugate addition–O-alkylation sequence gives the dihydrofuran, retaining the C≡CMe substituent and placing the original carbonyl oxygen in the ring. Enolate carbon alkylation would instead give a cyclopropane; the observed oxygen-containing ring identifies the alternative closure. An approach and betaine conformation minimizing crowding between indolyl and alkynyl substituents, followed by stereoinverting displacement at the sulfonium carbon, rationalizes the depicted trans product. The result is racemic in the absence of an external chiral influence.

The first two operations invert the allylic hydroxyl group. Triphenylphosphine adds to the azodicarboxylate, generating the reagent pair that converts the alcohol into an alkoxyphosphonium leaving group. Benzoate displaces it with inversion at carbon, giving the inverted benzoate ester. Methoxide then cleaves that ester by nucleophilic acyl substitution, retaining the newly inverted carbon–oxygen stereochemistry because cleavage occurs at the acyl group, not by substitution at the stereogenic carbon.

The inverted allylic alcohol exchanges with the orthoacetate under acid catalysis. Ethanol loss produces an allyl ketene acetal, schematically $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}(\mathrm{OEt})-\mathrm{O}-(\mathrm{allyl})$. A six-membered suprafacial Johnson-Claisen rearrangement breaks the allylic C–O bond, shifts the ring alkene by one position and forms the new bond from the original alkene terminus to the ketene-acetal CH${}_2$. The result is the $-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}_2\mathrm{Et}$ substituent at the carbon that was the remote end of the original double bond. A chair-like transition state placing the larger substituents away from axial crowding explains the stereochemical transfer; the inverted alcohol configuration selects the corresponding face.

Finally, TiCl${}_4$ activates the anomeric methoxy group and methoxide departure creates an oxocarbenium ion. Hydride transfer from triethylsilane replaces that methoxy substituent by hydrogen. The ring oxygen and pendant OTBS group remain intact. The sequence is inversion, stereospecific allylic rearrangement, then reductive removal of the anomeric methoxy group, rather than direct reduction or hydrolysis of the ester introduced by the Claisen step.

A consistent proposed sequence is desilylation, five-membered hemiketal formation, retro hetero-Diels–Alder opening, intramolecular aldol addition and hemiacetal closure. The essential carbon rearrangement is loss of the original ketone–side-chain bond and formation of a bond between its two neighbouring carbons; it is not a simple Michael addition.

For an explicit atom map, label the starting six-membered ring $A-B=C-D-E-O_r-A$, with the ketone at $D$, methyl substituents at $A$ and $B$, the hydroxy group at $A$, and the side chain $E-X-T-\mathrm{Et}$. Here $X$ is the methylene and $T$ is the carbon bearing the protected alcohol. Let $O_D$ be the original ketone oxygen and $O_T$ the protected alcohol oxygen. These labels identify atoms, not new substituents.

First, Fluoride removes the silyl protecting group from $O_T$. The resulting alcohol attacks the protonated ketone at $D$, forming the five-membered hemiketal $D-E-X-T-O_T-D$. The original oxygen $O_D$ is now a hydroxy group; the original pyran ring remains intact at this stage.

Next, The original pyran ring can undergo a retro hetero-Diels-Alder reaction. The three electron-pair movements are $B=C\to C=D$, $D-E\to E=O_r$ and $O_r-A\to A=B$: break $D-E$ and $O_r-A$, and form the three indicated double bonds. The result is the tethered aldehyde/dienol esterThe temporary $D-O_T$ bond keeps both reacting ends within the same molecule. At $D$, both oxygen substituents remain attached, and the new $D-O_T$ tether prevents separation into independent fragments.

Then The $C=D$ enol attacks the tethered aldehyde at $E$. Formation of $C-E$, restoration of $D=O_D$, and proton transfer give the six-membered lactone ring $C-D-O_T-T-X-E-C$, with $E-O_{r}H$. This is an intramolecular aldol reaction, and it supplies the crucial new carbon–carbon bond without losing any carbon atom.

Finally, Keto-enol tautomerism at $A=B$ gives the methyl ketone at $A$ and introduces the hydrogen at $B$. The $E-O_{r}H$ alcohol then attacks that ketone, closing the five-membered hemiacetal ring $A-B-C-E-O_r-A$. The resulting fused rings have exactly the product's connectivity: $D$ is the lactone carbonyl, $O_T$ is its ring oxygen, and $O_r$ is the five-membered-ring oxygen.

The stereocentre at $T$ survives throughout this sequence. The configuration initially at $E$ is erased at the aldehyde stage, and those at $A$, $B$, $C$ and $E$ arise during the later cyclization and proton-transfer steps. A folded intramolecular aldol reaction followed by reversible hemiacetal formation is consistent with the drawn cis ring junction, but connectivity and orbital bookkeeping alone do not prove that it is the unique selected diastereomer. In particular, a claim of simple retention at $E$ would contradict this atom map.

A related pyranone-to-fused-lactone rearrangement under aqueous hydrogen fluoride is experimentally reported; that account also explicitly qualifies the relative stereochemistry as assigned on mechanistic grounds. The sequence above is a proposed explanation of the examination structure, rather than a mechanism claimed to have been experimentally established in that paper.^[ref]