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The axial penalties depend on the ring position. An axial methyl at C2 encounters the axial hydrogens on C4 and C6. The short C–O links and dioxane geometry make these 1,3-Diaxial interactions particularly unfavorable. In contrast, an axial substituent at C5 points toward ring oxygens O1 and O3 rather than toward axial C–H groups at those positions. Those oxygen sites present much less steric obstruction than axial hydrogens on carbon.

Avoiding the expensive axial C2 methyl outweighs the cost of axial tert-butyl at C5. Transferring cyclohexane tert-butyl preferences unchanged would therefore give the wrong answer. This is primarily a position-dependent steric comparison, not the usual anomeric preference for an axial electronegative substituent at an acetal carbon. Primary NMR work on substituted dioxanes documents this axial C5 tert-butyl conformer.^[ref]

Number the dioxane O1–C2–O3–C4–C5–C6, with methyl at C2 and tert-butyl at C5. Their cis relationship is maintained on ring inversion, but because they occupy opposite ring positions, one is axial and the other equatorial in either chair. The two alternatives are C2–Me equatorial/C5–tert-butyl axial, and C2–Me axial/C5–tert-butyl equatorial.

The key signal is U, the C5 methine, with only small vicinal couplings, 3.7 and 1.3 Hz. An axial H5 would have large trans-diaxial couplings to axial hydrogens on C4 and C6. Their absence identifies H5 as equatorial, hence the preferred conformer has axial tert-butyl and equatorial methyl. It is not sufficient to guess from tert-butyl bulk. The tert-butyl singlet itself supplies no vicinal H–H coupling information about its orientation.

The cis fusion admits two chair–chair conformers, related by simultaneous ring inversion. The bridgehead hydrogens remain cis, while their axial/equatorial descriptions exchange. In one fold the oxygen next to the fusion occupies the concave, inside position; in the other, the corresponding inside position is occupied by a methylene group. Both chair-only structures are shown in the lower panels of the dioxane diagram.

The O-inside chair–chair conformer is strongly favored. In the unfavorable fold, the inward methylene and its C–H bonds create close nonbonded contacts with the other ring, analogous to costly 1,3-diaxial/gauche contacts in cis-decalin. Placing oxygen in that location removes the corresponding carbon-bound hydrogens and reduces the obstruction. The comparison is between the two conformers of the same cis configuration; making the fusion trans would be a configurational change, not a chair flip. Nor is it appropriate to flatten both six-membered rings into regular hexagons, which would hide the decisive inside contacts.

The alkene undergoes concerted oxygen transfer from the terminal peroxide oxygen of mCPBA. In the cyclic electron-flow description, the alkene pi bond begins one C–O bond, the peroxide oxygen supplies the second, O–O breaks, and proton transfer/carboxyl rearrangement gives m-chlorobenzoic acid. Both C–O bonds are installed from the same alkene face, so the alkene's relative substituent geometry is retained in the epoxide. There is no freely rotating carbocation intermediate.

The neighboring free OH hydrogen-bonds to the approaching peracid and favors a cyclic, organized transition-state geometry. oxygen is delivered from the OH-directed face, giving the indicated syn relationship between epoxide oxygen and the allylic hydroxyl. In the supplied projection, the epoxide C–O bonds and the original C–OH bond are on the rear/hashed face. Steric interactions still influence the preferred hydrogen-bonded conformer, but “attack from the least hindered face” alone misses the positive directing role of OH. The distant benzyl ether is not the directing free hydroxyl.

The classical experimental comparison of allylic alcohols and protected derivatives established this directing principle; it supports the mechanistic interpretation without itself measuring the selectivity of this acyclic substrate.^[ref]

The free propargylic OH first reacts with an aluminium hydride, releasing hydrogen and forming an O–Al bond. Coordination of the nearby alkyne to this aluminium center organizes a five-membered O–C–C–C–Al unit. Hydride addition from the opposite side of the activated alkyne gives a cyclic alkenyl aluminate: aluminium bonds to the distal alkyne carbon, while hydride is delivered to the carbon nearer $\mathrm{C}{\mathrm{H}}_2\mathrm{O}$.

The key step is directed trans hydroalumination. The new C–H and C–Al bonds are anti. Aqueous workup protonolyzes C–Al with retention of its alkene geometry and hydrolyzes the O–Al attachment. Replacing that C–Al bond by C–H therefore produces the E allylic alcohol, with $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ and the rest of the chain on opposite alkene sides. A generic syn hydroalumination drawing followed by simple protonation would predict the wrong geometry.

The free OH provides the directing tether; the silyl-protected oxygen is not equivalently available. Primary electrophile-trapping studies of Red-Al-treated propargylic alcohols support the alkenyl-alanate description.^[ref]

The alcohol oxygen attacks sulfur of methanesulfonic anhydride. Cleavage of the anhydride S–O bond releases methanesulfonate, and the amine base removes the O–H proton, giving the corresponding mesylate. This activation preserves the substrate C–O bond while replacing the poor hydroxide leaving group by $\mathrm{OS}{\mathrm{O}}_2\mathrm{Me}$.

The base then removes a hydrogen from the adjacent methylene on the side toward the vinyl chloride. In a favorable antiperiplanar arrangement, C–H electrons form the new C=C bond while the C–OMs bond breaks. An E2 elimination creates the alkene between the former alcohol carbon and that methylene. It links the two neighboring unsaturated regions into the extended conjugated triene segment, while the pre-existing carbon skeleton and vinyl chloride are retained.The left-hand polyenoate portion is unchanged. The base is diisopropylethylamine; it accepts acid during mesylation and promotes elimination. The product's E/Z mixture concerns the newly formed alkene, and no single geometry needs to be assigned.

C is the best geometrical match for a strongly bound 1:1 complex with E. Both have an adjacent amide donor–acceptor edge. Arrange them head-to-tail so that the N–H of E donates to the carbonyl oxygen of C, and the N–H of C donates to the carbonyl oxygen of E. This closes the familiar two-hydrogen-bond cyclic amide motif, illustrated below.

C combines a strong acceptor with a useful donor in the correct spacing. B has a stronger individual donor, but its pyridine N and OH are separated around the ring and do not reproduce the compact adjacent amide edge of E. A lacks a matching donor, and D's N–H is weak and has the wrong geometry for the same amide–amide motif. Site strength and spatial complementarity must be considered together; multiplying two isolated probe constants is not a reliable prediction of the complete association equilibrium. Self-association of C or E can compete and must be included in a quantitative binding analysis.

4-Fluorophenol probes acceptor strength because it donates its OH hydrogen. The qualitative order is C $>$ D $>$ B $>$ A: the electron-rich lactam oxygen is the strongest acceptor here, the secondary amine is next, pyridine nitrogen is intermediate, and aldehyde oxygen is weakest. Resonance donation from amide nitrogen increases the lactam oxygen's basicity relative to an ordinary aldehyde or ketone oxygen. The N-methylpyrrole site in A does not add an independent amine-like acceptor.

N-Methylpyrrolidin-2-one is a carbonyl acceptor without N–H, so it instead probes the solute's donor ability. B $>$ C $\gg$ A and D: B's phenolic OH is a strong donor, the lactam N–H of C is appreciable, A lacks a conventional donor, and D's neutral secondary-amine N–H is weak. The amine can be a good acceptor and a poor donor at the same time; the two properties are not opposites on one scale.

These are association equilibria in cyclohexane, where solvent competition for hydrogen bonds is weak. The quoted constants should be compared within each probe series, not treated as solvent-independent measures. At 298 K, as an illustrative common temperature, a tenfold increase in $K_a$ corresponds to a binding free-energy advantage $RT\ln 10\simeq 5.7\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. This does not imply the experiment's temperature was specified, nor that every difference is the energy of one isolated hydrogen bond.

View the backbone along N–C${}_{\alpha }$ to display $\varphi =\angle (C_{i-1}^{\prime },N_i,C_{\alpha i},C_i^{\prime })$ in a Newman projection. The front amide nitrogen carries the preceding carbonyl and H, while the back C${}_{\alpha }$ carries the next carbonyl, H and side chain. Rotation changes the contacts among these groups. A corresponding view along C${}_{\alpha }$–C${}^{\prime }$ displays $\psi$. The peptide C${}^{\prime }$–N bond itself is approximately planar and is not the freely rotating bond in either definition.

glycine replaces the side chain by a second hydrogen. Removing C${}_{\beta }$ removes many contacts that exclude backbone arrangements for other residues, so its allowed $\varphi ,\psi$ region is much broader and includes substantial positive-$\varphi$ space. Its alpha carbon is also achiral, so the steric map is approximately mirror-symmetric between positive and negative torsions in the simple isolated-residue model.

proline closes a five-membered ring from C${}_{\alpha }$ through three methylenes back to nitrogen. In the N–C${}_{\alpha }$ projection, the nitrogen's side-chain arm and C${}_{\alpha }$'s C${}_{\beta }$ arm are tied together, preventing unrestricted relative rotation. Its $\varphi$ is consequently confined near the usual negative-$60^{\circ }$ region rather than exploring the broad range of an acyclic amino acid. proline also lacks a backbone N–H donor when incorporated into an ordinary internal peptide. Ring closure restricts $\varphi$ strongly, but does not freeze every backbone or ring degree of freedom. These differences are naturally represented by distinct Ramachandran plots; the original torsion/steric analysis established the basis for such comparisons.^[ref]

The product is N-[5-(2-chloroacetyl)-2,3-dihydrobenzofuran-6-yl]acetamide, with an aromatic $-\mathrm{COC}{\mathrm{H}}_2\mathrm{Cl}$ substituent at position 5. Substitution of one aromatic hydrogen by ${\mathrm{C}}_2{\mathrm{H}}_2\mathrm{ClO}$ gives the stated molecular formula. This positional assignment also agrees with the isolated intermediate reported in the original process study.^[ref]

GaCl3 activates chloroacetyl chloride by Lewis-acid coordination and chloride abstraction, represented by the acylium/contact-ion-pair limit $\mathrm{ClC}{\mathrm{H}}_2\mathrm{C}{\mathrm{O}}^{+}$. The position-5 aromatic $\pi$ bond attacks the acyl carbon, giving a Wheland intermediate. Removal of the proton at that carbon restores aromaticity. Product–Lewis-acid coordination is reversed on work-up. The carbon-bound chlorine survives; the chlorine on the acyl carbon is the leaving group.

Hydride removes the remaining alpha proton at the substituted benzylic carbon, releasing hydrogen and forming the conjugated enolate. Its oxygen attacks the carbonyl carbon of allyl chloroformate. Collapse of the tetrahedral intermediate expels chloride and gives the allyl enol carbonate, with the double bond between the original carbonyl carbon and the cyanomethyl-bearing carbon.

This is O-acylation of an enolate: the carbonyl oxygen is trapped as $\mathrm{OC}{\mathrm{O}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, retaining the enol double bond. The competing carbon-centred resonance representation explains possible C-acylation in other circumstances, but it would give a different connectivity from E. The indicated product selects the oxygen-trapping pathway and the conjugated enolate regioisomer.

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.

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 diagnostic feature is the large vicinal coupling characteristic of an axial–axial relationship in a chair. The td patterns for $H_a$ and $H_c$ contain two large couplings to neighbouring axial hydrogens, so both are axial. The ddd patterns for $H_b$ and $H_d$ also include a large ring vicinal coupling, with the smaller values reflecting the other ring hydrogen and the side chain. The assignments areThe chemical shifts alone do not determine this assignment. In particular, three unequal splittings of a methine are not a geminal-coupling pattern.

A is 3-Chlorobutan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(=\mathrm{O})\mathrm{CH}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3$. Counting chlorine as one hydrogen in the index of index of hydrogen deficiency gives $(2\times 4+2-7-1)/2=1$. A single carbonyl accounts for that unsaturation; the resonance at $203.2$ ppm identifies a ketone, while the $59.0$ ppm methine bears chlorine.

The carbon-13 NMR assignments are ${\delta }_C=203.2$ for $\mathrm{C}=\mathrm{O}$, $25.7$ for $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}$, $59.0$ for $\mathrm{CHCl}$ and $20.1$ ppm for $\mathrm{CH}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3$. The four distinct carbon environments agree with the unsymmetrical structure. The small signal near 77 ppm belongs to the usual deuterated chloroform solvent, rather than to a fifth carbon of A.

Take the bond axis as $z$. The heavy-atom $2s$ and $2p_z$ atomic orbitals can mix with hydrogen $1s$ to make three sigma-symmetry molecular orbitals. In the qualitative valence ordering, these are a low predominantly heavy-atom $2s$ level, a bonding $\sigma$ level and a higher ${\sigma }^{\ast }$ level. The perpendicular $2p_x,2p_y$ orbitals form a degenerate approximately nonbonding $\pi$ pair because hydrogen has no valence orbitals of the matching sideways symmetry. The filled heavy-atom $1s$ core is omitted from the diagram.

Counting valence electrons givesThus both exhibit paramagnetism: CH has one unpaired electron and NH has two in its ground state. For NH, Hund's rules favour separate parallel-spin occupation of the two degenerate $\pi$ orbitals. The corresponding observed ground terms are $X{}^2\Pi$ and $X{}^3{\Sigma }^{-}$; the total bond order is approximately one in this simple sigma-bond/nonbonding-pi description.^[ref]

The nitrogen-derived orbital energies lie lower than the corresponding carbon-derived ones on an approximate common energy scale, while the isolated hydrogen $1s$ reference is unchanged. The bonding orbitals consequently have different polarizations and mixing coefficients; drawing identical energies for C and N would miss the screening trend. The figure specifies ordering and occupancy, not measured gaps or a numerical prediction from (a).

First, proton transfer from acetic acid to the carbon of diazomethane gives methyldiazonium and an acetate ion:Draw an electron-pair arrow from carbon, using the carbon-negative resonance contributor, to the acidic hydrogen, and another from the acid O–H bond to oxygen. Second, an acetate oxygen attacks the diazonium methyl carbon by an SN2 reaction; an arrow from the C–N bond to nitrogen expels neutral ${\mathrm{N}}_2$:The acid's hydroxyl hydrogen has become the third hydrogen of the newly installed methyl group. This is a methyl-transfer sequence driven by formation of molecular nitrogen, without forming a free methyl carbocation or invoking Fischer esterification.

The simplest assignment is two phenylethynyl bridges and four equivalent terminal methyl groups, represented as $\mathrm{A}{\mathrm{l}}_2\mathrm{M}{\mathrm{e}}_4(\mu -\mathrm{C}\equiv \mathrm{CPh}{)}_2$. All 12 methyl protons then occupy the same class of environment, accounting for the single nonaromatic singlet without requiring methyl bridge/terminal averaging. The phenyl protons account for the aromatic-region signals; the metal-bound acetylide has no terminal acetylenic hydrogen.

An alkynyl bridge has electronic possibilities unavailable to a saturated methyl: in addition to sigma attachment from its carbon end to one Lewis acid centre, its occupied C≡C pi system can donate to the other aluminium. A delocalized sigma/pi description stabilizes the phenylethynyl-bridged arrangement while keeping the methyl groups terminal. One must not draw an unchanged localized triple bond plus two independent full ordinary Al–C bonds at the same carbon and count all of them as separate two-electron bonds. The exact distribution of sigma and pi donation requires structural or orbital evidence; this NMR observation alone does not supply a numerical bridge bond order.

The absence of concentration or temperature dependence is consistent with this preferred arrangement throughout the measured range. It is not proof that the molecule has no motion: exchange between symmetry-equivalent terminal positions, or exchange remaining fast throughout that range, could be spectroscopically invisible. Phenylethynyl bridging in this particular dimethylaluminium dimer has independent primary NMR precedent.^[ref]

A is pentane-2,3-dione, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(=\mathrm{O})\mathrm{C}(=\mathrm{O})\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. Its index of hydrogen deficiency is $(2\times 5+2-8)/2=2$, accounted for by two ketone carbonyls. The two resonances near 200 ppm, together with the three saturated carbon signals and the proton integrals, identify an unsymmetrical adjacent diketone rather than an ester or an alkene-containing structure.

The assignments are the carbonyl-adjacent methyl at ${\delta }_C=23.7$, the methylene at $29.2$, and the terminal ethyl methyl at $7.0$ ppm. The two carbonyl carbons give $197.5$ and $199.8$ ppm. Their individual ordering is not uniquely established by the supplied one-dimensional spectra; either ordering between the methyl-side and ethyl-side carbonyls should be kept tentative without a correlation or another independent assignment. In the displayed attached proton test phase convention, the two methyls have one sign, while the methylene and the nonprotonated carbonyls have the opposite sign. The solvent signal near 77 ppm is not a sixth carbon of A.

The third electron must enter the higher, degenerate and partly antibonding orbital pair rather than the already filled fully bonding orbital. At an equilateral geometry the configuration becomes $(a_1^{\prime }{)}^2(e^{\prime }{)}^1$. This weakens the collective bonding and creates orbital degeneracy, so a distortion can lower the singly occupied component and localize bonding. Along the ground-state reaction coordinate, formation of a normal two-electron H–H bond with a separated hydrogen atom becomes favourable.

The result is ground-state dissociation to ${\mathrm{H}}_2+\mathrm{H}$, not loss of a proton from the cation or formation of three isolated atoms. One should not claim that adding one electron exactly cancels two bonding electrons: a simple three-orbital count alone does not predict a zero bond order or prove dissociation. The important qualitative changes are occupancy of an antibonding/degenerate level and access to a lower-energy dimer-plus-atom arrangement. A complete dissociation statement requires the potential-energy surface, including nuclear repulsion; electronically excited neutral H${}_3$ states are a separate issue.