For a carbonyl oxygen of residue $i$ accepting from the NH of a later residue, the possibilities are:

The alpha helix provides a favourable combination of allowed backbone torsions, reasonably linear hydrogen bonds, and side-chain packing. Its successive side chains advance by about $100^{\circ }$, rather than lining up in the three columns of the narrower 3-10 helix. Smaller repeating rings impose more severe backbone constraints; the wider pi helix gives a different, generally less favourable packing arrangement. Short 3-10 and pi-helical segments can still occur in proteins, so this is a preference for a common extended secondary structure, not a prohibition of the alternatives. Proline and sequence-specific interactions can disrupt the idealized backbone donor pattern.

The functional group interaction profile compares donor strength $\alpha$ and acceptor strength $\beta$. In water, strong–strong contacts compensate for desolvation, giving the upper-right black region. Strong–weak contacts lose favourable solvent interactions without adequate replacement, giving the two grey regions. Weak–weak association gives the lower-left black region through the hydrophobic effect.

Ethanol's ethyl groups preferentially solvate weakly polar sites, removing this incentive for their association: the lower-left region becomes approximately neutral. Its hydroxyl group retains competition for hydrogen bonds elsewhere in the profile.^[ref]

For hydrogen-bond acceptance, the three strongest groups are the terminal carboxylate, the Cys–CO–N(Pro) tertiary-amide carbonyl, and the Pro–CO–NH(Tyr) secondary-amide carbonyl, in that qualitative order. Negative charge makes carboxylate a particularly strong acceptor. Both amide oxygens accept strongly; greater nitrogen alkyl donation generally makes a tertiary amide oxygen more basic than the comparable secondary-amide oxygen. This last ordering is a functional-group estimate, not a precise calculation of a folded tripeptide's site parameters.

For donation, the order is terminal ammonium $>$ tyrosine phenolic OH $>$ the peptide NH between Pro and Tyr. The thiol is a weaker donor. Protonated ammonium has no available lone pair to accept; the proline amide nitrogen has no N–H bond to donate, and amide nitrogen lone pairs are delocalized into carbonyl groups. The carboxylate has no O–H donor in the drawn protonation state.

Apply the Felkin-Anh model to the stereocentre alpha to the ketone. Its large group is the chain continuing through the phenyl-bearing carbon, its medium group is methyl, and its small group is hydrogen. In the preferred transition-state arrangement, put the large group approximately perpendicular to C=O and approach from the opposite face, passing the small hydrogen along a Bürgi-Dunitz trajectory. The remote ether cannot form the usual small alpha-alkoxy chelate, and the coordinating solvent further supports using an open-addition prediction.^[ref]

The predicted new stereocentre is $S$. In the question's zigzag projection, this corresponds to the new OH bond directed behind the page and the newly introduced methyl toward the reader, with the original stereocentres unchanged. The new-centre priority order is OH $>$ phenyl $>$ the substituted alpha-carbon chain $>$ methyl. This is a model-based major-product prediction, rather than a quantitative diastereomeric ratio.

The products are the two mirror-image 2,3-epoxy-2,3-diphenylindan-1-ones, shown in the bottom row of the stereoisomer diagram. The original carbonyl remains, and an additional oxygen bridges the former double bond, giving the stated formula.

In the Weitz-Scheffer epoxidation, hydroperoxide anion adds conjugatively to the enone. The resulting enolate closes onto the attached peroxide oxygen with O–O bond cleavage and loss of hydroxide. Attack on the two faces of the achiral planar substrate gives the two enantiomers in equal amounts. This is a racemic epoxide pair, rather than a one-to-one mixture of different diastereomeric epoxide connectivities.

Reduction makes stereogenic centres at C2 and C3 of the indan-1-one framework. The four products are the cis pair $(2R,3S)$ and $(2S,3R)$, and the trans pair $(2R,3R)$ and $(2S,3S)$. Each pair consists of enantiomers; any cis member is diastereomeric to either trans member. There is no meso product because the carbonyl-containing and aromatic sides of the ring do not make C2 and C3 symmetry-equivalent.

For checking the Cahn-Ingold-Prelog priority rules, at C2 the order is carbonyl-side carbon $>$ phenyl $>$ C3 $>$ H. At C3 it is fused aromatic carbon $>$ phenyl $>$ C2 $>$ H. The fused aromatic path outranks the separate phenyl path because it reaches the carbonyl-containing part of the framework.

The relevant quantity is the H–C–C–H dihedral angle, through the Karplus relation, rather than the labels cis and trans alone. Couplings tend to be larger near eclipsed or antiperiplanar arrangements and smaller nearer a right angle. The observed value averages the accessible conformations.

In A, the saturated five-membered ring is puckered. A preferred envelope-like conformation can place the trans hydrogens closer to an antiperiplanar arrangement, giving a slightly larger coupling than the cis pair. In B, the double bond restricts ring geometry toward a flatter arrangement: the cis hydrogens are more nearly eclipsed, whereas the trans pair has an oblique dihedral angle. The cis coupling is then larger. Approximate Newman projections illustrate these trends; exact angles cannot be recovered from four coupling constants without a calibrated relation and conformational populations.

Base deprotonates the sulfonamide. Its nitrogen attacks one epoxide carbon from the back, breaking the C–O bond at that carbon and producing an adjacent alkoxide. The nitrogen and oxygen attachments on the cyclopentane are therefore trans. The alkoxide then attacks the aromatic carbon bearing the fluorine ortho to the sulfonyl group. Intramolecular nucleophilic aromatic substitution expels fluoride and closes a seven-membered heterocycle with the path $\mathrm{Ar}-\mathrm{S}-\mathrm{N}-{\mathrm{C}}_1-{\mathrm{C}}_2-\mathrm{O}-\mathrm{Ar}$. The para fluorine remains.

This tandem epoxide-opening/ring-closing pathway is required by the formula of F: simple epoxide addition would retain both fluorines, whereas F contains only one. In the second stage, pyrrolidine replaces the remaining para fluorine by another nucleophilic aromatic substitution, producing G. The mass balance is F plus pyrrolidine minus HF, as required by the given formula of G.

The meso epoxide has enantiotopic ends. With achiral reagents, opening at either end gives an enantiomeric product, so F is a racemic pair with the same trans relative configuration; G retains that configuration.