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Only the low-coupling pair remains, so the detected product is the trans diastereomer, with approximate coupling 4 Hz. The reaction is highly diastereoselective within the spectrum's detection limit. Its possible absolute configurations are $(2R,3R)$ and $(2S,3S)$.

A conventional proton NMR spectrum in an achiral environment cannot distinguish those enantiomers: they have the same shifts and couplings. Thus a chiral catalyst plus one set of resonances does not establish an enantiomeric excess or an absolute configuration. Chiral chromatography, an appropriate chiral NMR method, or a calibrated stereochemical comparison would be needed.

Protonation of a spiroacetal oxygen makes the adjacent C–O bond easier to cleave. Reversible opening gives an oxocarbenium ion, in which the spiro carbon is locally planar. The tethered alcohol can attack either face; deprotonation closes the acetal and regenerates acid. Repetition interconverts the spiro epimers while retaining the other stereogenic centres. Methanol can participate in reversible proton transfer or trapping, but there is no need to invert the remote carbon centres.

At pH 7, a glutamate residue is normally carboxylate. Its most natural role is general-base activation of the attacking water. Draw the carboxylate oxygen lone pair toward a water H, the water O–H pair toward its oxygen, and that oxygen's lone pair toward C3 while the conjugated pi system moves toward the carbonyl oxygen:This can be concerted, avoiding a freely diffusing hydroxide intermediate. The protonated glutamate subsequently returns its proton to the enolate system, directly or through solvent, regenerating carboxylate. A favorable arrangement is needed; ordinary carboxylate does not quantitatively deprotonate bulk water in a prior equilibrium.

The side-chain amide of glutamine is neutral. Its N–H can donate a hydrogen bond to the carbonyl oxygen and stabilize developing enolate charge, while its amide oxygen can accept a hydrogen bond to orient attacking water. Draw $\mathrm{Gln}-\mathrm{CON}{\mathrm{H}}_2\cdots \mathrm{O}=C1-C2=C3$, with the dotted interaction from amide N–H to substrate oxygen. This assists polarization and transition state stabilization; the amide nitrogen is not an ordinary amine base because its lone pair is delocalized into its own carbonyl.

The guanidinium group of arginine is positively charged. It can stabilize developing negative charge by hydrogen bonding and electrostatic attraction, drawn $\mathrm{Arg}-\mathrm{guanidiniu}{\mathrm{m}}^{+}\cdots {\mathrm{O}}^{\delta -}-C1$. Its delocalized cation is not the preferred neutral general base at pH 7. Productive catalysis requires stronger stabilization of the reacting arrangement than of the unreactive ground-state complex.

Phenylalanine and leucine have hydrocarbon side chains and provide no conventional acid/base or hydrogen-bond donor/acceptor group here. They can position the enone, exclude unproductive solvent and select the attacked face through hydrophobic effect and steric hindrance. Those binding effects can assist catalysis but should not be drawn as an imaginary side-chain proton relay. The original schematic depicts the Glu water relay and the Gln/Arg oxyanion-stabilizing contacts.

The C–O-forming transition state has a partial bond from water oxygen to C3, positive charge developing on that oxygen, and negative charge delocalizing toward the carbonyl oxygen. A compact drawing isHere the two C–C dotted connections denote redistribution of pi bonding, not cleavage of either carbon–carbon sigma bond. In a bond-order drawing retain the full sigma skeleton: C1–O is changing from double toward single, C1–C2 from single toward double, C2–C3 from double toward single, and C3–O from absent toward single. The incoming water approaches a face of the approximately planar conjugated unit and C3 begins to become tetrahedral. The total electric charge remains zero. If proton transfer to another water occurs concertedly, include a partially broken incoming-water O–H bond and a partially formed relay O–H bond; that is a solvent-assisted variant, not a different regioselectivity. The original mechanism schematic identifies these bond changes without depicting a falsely isolated free hydroxide reagent.

Use C1 for the carbonyl carbon, C2 for the adjacent carbon, and C3 for the beta carbon of the reactive enone fragment $\mathrm{R}-C1(=\mathrm{O})-C2H=C3H-{\mathrm{R}}^{\prime }$. water adds conjugately at C3, giving a beta-hydroxy ketone. The oxygen lone pair of neutral water attacks C3; move the C2=C3 pi pair toward C1–C2 and the carbonyl pi pair onto oxygen. This produces the charge-separated enolate–oxonium intermediateA second water molecule can relay a proton from the newly attached oxonium oxygen to the enolate system. Using the carbon-centered resonance structure, reform C1=O and protonate C2 while the donating O–H bond returns its pair to the added oxygen. Equivalently, oxygen protonation first gives an enol, followed by water-mediated keto-enol tautomerism. Either path yields a neutral beta-hydroxy ketone and regenerates the relay water. There is no need to introduce externally supplied H${}^{+}$ or OH${}^{-}$; solvent molecules perform the proton transfers. The uncatalyzed charge separation has a large barrier, so this mechanism describes a slow formal pathway, not a claim of rapid hydration in pure water.

The beta-carbon bond-forming transition state retains the same basic connectivity across this pH range, but the predominant proton-transfer route and enzyme protonation change. Near the acidic end, protonated glutamate can act as a general acid, donating a proton or a strong hydrogen bond to carbonyl oxygen and making C3 more electrophilic; it is less available as the carboxylate general base that activates water. Above its local $p{K}_a$, carboxylate-mediated water activation is favored. Acid catalysis therefore shifts positive-charge stabilization toward a protonated carbonyl, whereas base-assisted attack reduces positive-charge buildup on incoming water. Neither a unique transition-state geometry nor a numerical pH optimum follows from pH alone.

Using a free-solution glutamate side-chain $p{K}_a\simeq 4.2$ only as an illustration, the deprotonated fraction is $[1+10^{p{K}_a-\mathrm{pH}}{]}^{-1}$: about 0.059 at pH 3 and 0.998 at pH 7. An enzyme environment can shift that $p{K}_a$ substantially. The glutamine amide and the hydrocarbon side chains remain in their ordinary forms from pH 3 to 10; arginine guanidinium, with a typical free side-chain $p{K}_a$ around 12.5, remains predominantly positive even at pH 10. Thus the major side-chain switch among those listed is glutamate. At the alkaline end, bulk hydroxide can also contribute to nonenzymatic conjugate addition. A measured pH-rate profile must control substrate stability and enzyme folding before assigning every loss of activity to an active-site titration.

Hydride is delivered to steroid C5 and a proton to C4; the C3 ketone is retained. Use the conjugated fragment $\mathrm{O}=C3-C4=C5$. Carbonyl hydrogen bonds from Tyr91–OH and protonated Glu57 lower the barrier to conjugate reduction. Draw a curved arrow from the nicotinamide C4–H bond of NADPH to steroid C5, shift the C4=C5 pi pair toward C3–C4, and shift C3=O pi electrons onto oxygen. The nicotinamide ring rearomatizes to NADP+; the steroid becomes a bound enolate, or its protonated enol equivalent. electron shifts across the nicotinamide ring must accompany C–H bond cleavage rather than leaving an unexplained neutral carbon radical.

Next, reform C3=O and protonate steroid C4 from an appropriately positioned acid, directly or through an enol/proton-relay route. This regenerates the catalytic acid after solvent reprotonation. The enzyme fixes the hydride-accepting face so that the new C5–H has the 5α configuration, giving dihydrotestosterone; delivery to the opposite face would give a different ring-junction configuration. This is not simple 1,2 reduction of the C3 carbonyl to an alcohol. The net local bookkeeping is one hydride plus one proton across C4=C5, with two electrons supplied by NADPH. The original reactive-fragment diagram shows the enolate-forming arrows and the separate protonation step.

Y91F implicates the Tyr hydroxyl, while E57Q implicates the Glu carboxylic-acid chemistry. Replacing tyrosine by phenylalanine retains an aromatic side chain but removes its hydroxyl, supporting a role for Tyr91 as a hydrogen bond donor that polarizes the steroid C3 carbonyl and stabilizes an enolate-like intermediate or transition state. Its hydroxyl could participate in a proton relay, but inactivity alone does not establish that it supplies the final proton.

Replacing a glutamate residue by glutamine approximately preserves side-chain size while replacing a titratable carboxylic acid by an amide. Loss of activity, combined with the acidic activity range, supports protonated Glu57 as a carbonyl-activating acid and possible proton donor. The two residues can cooperate to stabilize oxygen charge during reduction. The inhibitor-binding structure supports local contacts; it is not a direct picture of a testosterone transition state, and an N–H contact of the aza inhibitor need not occur in testosterone.

To distinguish these chemical roles from indirect effects, compare mutant and wild-type abundance, membrane insertion and folding, cofactor binding, and steroid binding. Measure initial rates with matched enzyme amounts and include no-enzyme and inhibitor-treated wild-type controls. If binding survives but turnover fails, the evidence for a catalytic defect is stronger. The published study combines structure, mutagenesis and modeled testosterone placement, so the proposed residue roles should be distinguished from direct observation of proton trajectories.^[ref]

catalysis requires a protonated group, consistent with a general-acid contribution. Loss of its proton as pH rises would remove carbonyl activation or the proton needed to finish the conjugate reduction. A glutamate residue with a shifted active-site $p{K}_a$ is a plausible candidate, but a pH cutoff alone neither identifies the residue nor measures a precise $p{K}_a$. Irreversible unfolding or changes in substrate binding must also be excluded by reversible pH jumps, protein-integrity checks and initial-rate measurements.

For the mechanistic notation below, the reduced cofactor is written NADPH, the established cofactor of human steroid 5α-reductase. The original PDF explicitly labels it NADH; its abbreviated N-substituent does not display the phosphate distinction. NADH and NADPH have the same reactive reduced nicotinamide ring, so the local hydride-transfer arrows are unchanged, although their biochemical specificity is different.^[ref]

A formal two-electron reductive elimination couples two anionic, mutually cis donors and reduces Ti(IV) to Ti(II). There are two chemically distinct couplings: Cl–Cl elimination gives $\mathrm{C}{\mathrm{l}}_2$, leaving a Ti(II) complex with the intact anionic O,N,P ligand and one chloride; O–Cl elimination gives an aryl hypochlorite group, $\mathrm{ArOCl}$, breaking the original Ti–O attachment and leaving a Ti(II) center coordinated by N,P and two chlorides. The latter product can remain tethered through the N,P part of the same ligand; it is not necessarily a free small molecule. The starting facial arrangement supplies cis pairs of both kinds.

Cl–Cl coupling is the more plausible of these formal alternatives, because it retains the particularly strong early-metal Ti–O interaction and makes the stronger Cl–Cl rather than relatively weak O–Cl bond. Both alternatives reduce the valence count by two, from 12 to 10, so the 18-electron rule does not choose between them. Neither should be asserted to occur readily: reduction of a stable, oxophilic $d^0$ Ti(IV) center is intrinsically unfavorable. Coupling a chloride to the neutral amine or phosphine is not the ordinary elimination of two X-type ligands. proton transfer between N–H and oxygen, if it occurred, would instead be acid-base chemistry, not automatically a two-electron reductive elimination.

Write $L^{-}$ for the cyclopentadienyl ligand bearing its $\mathrm{C}{\mathrm{H}}_2$–2-pyridyl arm; ${\eta }^5:\kappa N$ means that its ring and pyridyl nitrogen both bind the same metal. $D=[\mathrm{Mo}({\eta }^5:\kappa N\text{-}L)(\mathrm{CO}{)}_2(\mathrm{NCMe}){]}^{+}$. The base removes the pyridinium N–H proton in $C$, restoring a pyridine lone pair. Intramolecular nitrogen coordination then replaces one acetonitrile ligand and closes a chelate ring. Both CO ligands are retained; the molybdenum remains Mo(II), $d^4$.

This is an proton transfer followed by chelation, with metal electron count $4+6+2_N+4_{\mathrm{CO}}+2_{\mathrm{NCMe}}=18$. The chelate effect, favorable formation of the Mo–N bond, and release of a solvent molecule explain the driving force. Merely invoking a 16-to-18-electron change would be misleading because $C$ is already 18-electron. In the preceding acid treatment, protonation of the ${\eta }^3$ allyl ligand allows release of propene; the other proton resides on the pendant pyridine, accounting for $C$'s overall $2+$ charge. One proton is removed here; the cyclopentadienyl tether is not deprotonated or detached.

Secondary amines normally favor non-geminal disubstitution, putting the second amino substituent on a different phosphorus. The first installed $\mathrm{N}{\mathrm{R}}_2$ group donates electron density to the P–N framework and deactivates its own phosphorus toward further attack. It is also bulkier than an initially installed primary-amino group. Crucially, $\mathrm{P}-\mathrm{N}{\mathrm{R}}_2$ has no N–H proton, so the proton-abstraction/chloride-elimination pathway that can direct a primary amine geminally is unavailable. Direct attack at a less-deactivated $\mathrm{PC}{\mathrm{l}}_2$ site is favored. The resulting non-geminal products can have cis or trans relative orientations around the ring; non-geminal substitution alone does not select one of these stereoisomers. These are useful trends, with known substituent-dependent exceptions in experimentally characterized mixed derivatives.^[ref]

Primary amines can favor geminal disubstitution through a proton-abstraction route. The first substitution leaves an exocyclic $\mathrm{PCl}(\mathrm{NHR})$ group. Its remaining N–H proton provides a pathway unavailable to a fully substituted amino group: base removes H, chloride elimination gives a reactive low-coordinate phosphorus intermediate with increased exocyclic P–N bonding, and a second amine attacks that same phosphorus. proton transfer then gives $\mathrm{P}(\mathrm{NHR}{)}_2$. Thus the directing effect is not simply the inductive effect of an amino group.

This needs a qualification: “primary amine” alone does not specify a universal product distribution. Substituent, solvent and base concentration affect whether the proton-abstraction route outcompetes direct nucleophilic substitution. Kinetic and trapping experiments with aromatic primary amines support the base-catalyzed route to geminal products, while also finding direct substitution pathways under other conditions.^[ref]

$B={\mathrm{N}}_3{\mathrm{P}}_3(\mathrm{NHCy}{)}_6$, with two cyclohexylamino substituents on each original phosphorus. All six P–Cl bonds have been replaced; the six N–H bonds of those substituents remain. Each equivalent of triethylamine traps one equivalent of HCl.

$C$ contains three four-membered ${\mathrm{P}}^{\mathrm{V}}-\mathrm{N}(\mathrm{Cy})-{\mathrm{P}}^{\mathrm{III}}(\mathrm{Cl})-\mathrm{N}(\mathrm{Cy})$ spiro rings, one at each original ring phosphorus. Each PCl${}_3$ joins the two geminal NHCy groups: replacement of two of its chlorides forms two new P–N bonds and releases two HCl, consuming both N–H protons. The new phosphorus retains one chloride and a lone pair, making it three-coordinate P(III); the original six-membered ${\mathrm{P}}_3{\mathrm{N}}_3$ ring retains P(V). The product formula is ${\mathrm{N}}_3{\mathrm{P}}_3\{(\mathrm{NCy}{)}_2\mathrm{PCl}{\}}_3$. For each spiro ring, the original phosphorus is bonded to both exocyclic nitrogens, and each of those nitrogens is bonded to the same new PCl center; the braces do not indicate a simple linear substituent.

The three new P–Cl directions permit an all-cis isomer and a cis,trans isomer. Relative to the central ring, write these as $(+,+,+)$ and $(+,+,-)$, with cyclic relabeling giving equivalent choices. Reversing all signs does not create a third relative configuration. This describes the configuration pattern; puckering and substituent conformations are additional structural details.

A: ethoxy CH${}_2$ ↔ CH${}_3$, $J\simeq 7\mathrm{Hz}$; vinyl CH ↔ each alkene methyl, small ${}^{4}J\simeq 1$–$1.5\mathrm{Hz}$.

C: ethoxy CH${}_2$ ↔ CH${}_3$, about $7\mathrm{Hz}$; the two OCH${}_2$ proton types have a geminal coupling of about $9$–$10\mathrm{Hz}$; acetal CH ↔ C≡CH has ${}^{4}J\simeq 2\mathrm{Hz}$.

B: internal vinyl CH ↔ trans terminal H, about $17\mathrm{Hz}$; ↔ cis terminal H, about $10\mathrm{Hz}$; terminal CH${}_2$ geminal coupling about $1$–$2\mathrm{Hz}$. Vinyl CH ↔ OCH${}_2$ is about $6\mathrm{Hz}$. Both vicinal links along the butanoate alkyl chain are about $7$–$8\mathrm{Hz}$.

Use $J=300\Delta \delta$ when a separation is measured in ppm. Thus $0.057\mathrm{ppm}$ corresponds to about $17\mathrm{Hz}$, whereas $0.023\mathrm{ppm}$ corresponds to about $7\mathrm{Hz}$. These are approximate readings, not a fitted spin simulation.

The small four-bond splittings illustrate allylic scalar coupling and acetal-alkyne long-range coupling.

A, spectrum 4: ethoxy CH${}_3$ at about $1.3$ is a NMR triplet, and OCH${}_2$ at about $4.15$ is a NMR quartet. The two alkene methyls at about $1.9$ and $2.2$ each integrate to three protons; they are apparent NMR singlets on the overview but have small allylic NMR doublet splitting. The vinyl CH at about $5.7$ is correspondingly a narrow NMR multiplet, formally a quartet of quartets when its two methyl couplings are unequal, or approximately a NMR septet if they coincide. A broadened overview must not be used to declare all those small couplings exactly zero.

C, spectrum 5: the acetal CH at about $5.25$ and terminal C≡CH at about $2.55$ each integrate to one proton and show their small mutual long-range NMR doublet splitting. The ethoxy methyls give a six-proton NMR triplet near $1.25$. The OCH${}_2$ region has two sets near $3.6$ and $3.75$, each integrating to two protons. These are an ABX3 spin system: geminal coupling and coupling to CH${}_3$ produce approximately one doublet of quartets for each proton type, with second-order distortion in the expanded trace. Treating the whole region as one ideal four-proton NMR quartet misses the printed structure.

B, spectrum 6: the internal alkene CH near $5.95$ is a doublet of doublets of triplets from trans/cis coupling to CH${}_2$ and vicinal coupling to OCH${}_2$. The two terminal alkene protons near $5.2$–$5.35$ are distinct doublet of doublets-type signals, with possible additional small allylic splitting. OCH${}_2$ near $4.6$ is principally a NMR doublet. The acyl-alpha CH${}_2$ near $2.3$ is a NMR triplet, the middle CH${}_2$ near $1.65$ an approximate NMR sextet or overlapping NMR triplet of quartets, and the terminal methyl near $0.95$ a NMR triplet.

Spectrum 4 has one vinyl proton, two alkene-methyl groups and an ethoxy group. Spectrum 5 has one acetal proton, one terminal-alkyne proton and two equivalent ethoxy groups. Spectrum 6 has the three-proton terminal-alkene pattern and the allylic OCH${}_2$ group, together with the butanoate chain.

First hydrolyze the benzylidene acetal. Protonate one acetal oxygen and break its bond to the Ph-CH centre, giving one free alcohol and a benzylic oxocarbenium ion. Water attacks that centre; proton transfers give a hemiacetal. Protonate the remaining acetal-derived OR group and cleave it to release the second alcohol and benzaldehyde.

The resulting unsaturated hydroxy ester isThe OH on the allyl-substituted methine is gamma to the ester carbonyl. Protonation of that carbonyl enables intramolecular OH attack to give a five-membered tetrahedral intermediate. Proton transfer to methoxy and collapse expel methanol, followed by deprotonation to give the alpha,beta-unsaturated gamma-lactone.

The other liberated OH remains CH${}_2$OH. Both C=C units survive: one is the allyl substituent and the other becomes the lactone's conjugated double bond. Hydrolysis to a hydroxy acid followed by lactonization is an alternative sequence under the aqueous acid; direct acid-catalysed intramolecular ester exchange gives the same atom mapping.

The alkene pi bond attacks Br${}_2$ as the Br-Br bond breaks to bromide, forming a bridged, unsymmetrical bromonium ion. The carbon bearing the two methyl groups has greater carbocation-like character.

The tethered alcohol oxygen attacks that carbon from the side opposite the bromine bridge. Its lone pair forms the new O-C bond while the attacked C-Br bridge bond opens, giving a six-membered cyclic oxonium ion. Bromide or solvent removes the O-H proton to give the neutral bromoether.

This explains why O becomes attached at the more substituted alkene carbon while Br remains at the other carbon. Polar solvent assists charge separation, and intramolecular capture competes effectively with external bromide attack. Opening is anti; attack on the two faces of the achiral starting material gives a racemic product at the newly formed CHBr stereocentre.

The ring-forming step is bromoetherification.

Write the proton-labelled structure as $\mathrm{C}{\mathrm{H}}_3^{\mathrm{a}}\mathrm{C}(=\mathrm{O})\mathrm{C}{\mathrm{H}}^{\mathrm{b}}(\mathrm{Cl})\mathrm{C}{\mathrm{H}}_3^{\mathrm{c}}$. The assignments areThe NMR quartet is centred at $(4.371+4.302)/2=4.3365$ ppm; each adjacent line differs by $0.023$ ppm. The NMR doublet has the same spacing, $1.618-1.595=0.023$ ppm. At 300 MHz,Three equivalent methyl protons split $b$ into a $1:3:3:1$ NMR quartet; the single methine proton splits $c$ into a $1:1$ NMR doublet. No resolved long-range scalar coupling splits the acetyl methyl, giving its NMR singlet. The integrals total seven protons.