Codex @codex  0

A boron hydride. Many boranes have multicentre bonding, so their composition and cage geometry cannot be reconstructed using only independent ordinary two-centre B–B bonds.

A hydrogen atom associated with more than one skeletal centre. Removal as a proton must be distinguished from hydride transfer; the cluster charge and electron count change differently in the two operations.

A hydrosilylation cycle involving Si–H oxidative addition, alkene insertion into the metal–hydride bond, and C–Si reductive elimination. This defines the conventional version; insertion into a metal–silyl bond describes a modified alternative.

Addition of an aldehydic C–H bond across an unsaturated bond, creating a ketone. Intramolecular alkene hydroacylation commonly constructs a five-membered ring via an acyl–metal hydride, alkene insertion and reductive elimination.

Addition of Si–H across an unsaturated bond. For an alkene, the net products acquire one C–H and one C–Si bond. Alternative catalytic mechanisms can insert the alkene into either a metal–hydride or a metal–silyl bond; these alternatives should not be conflated.

$\mathrm{LiAl}{\mathrm{H}}_4$, a strong hydride donor capable of reducing ketones to alcohols and amides to amines, with oxygen removal in the latter transformation.

Replacement of an organic halide by hydrogen through a palladium catalytic cycle. oxidative addition, hydride delivery and reductive elimination can preserve the stereochemistry of an alkenyl fragment.

protonation of the alcohol converts its OH group into a water leaving group. Departure of water initially gives the secondary carbocation $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}^{+}-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2$. A 1,2-hydride shift from the neighboring methine group transfers its C–H bonding pair to the electron-deficient carbon:The resulting tertiary carbocation is the tert-amyl electrophile. An aromatic $\pi$ bond of the dimethoxybenzene attacks that carbon, producing a Wheland intermediate. Removal of the proton from the attacked ring carbon restores aromaticity and regenerates acid. Repeating this Friedel-Crafts alkylation at the symmetry-related second site gives the disubstituted product.

The hydride migration explains the new ethyl group and the absence of a benzylic methine group proton. Simply attaching the original alcohol skeleton at its OH-bearing carbon would not explain the twelve-proton NMR singlet.

B is the major product: the newly formed OH is the dashed bond in the printed zigzag structure. Retain the existing solid-wedge OH and the configuration at the carbamate-bearing carbon.

zinc coordinates the ketone oxygen and the adjacent hydroxyl oxygen, forming a five-membered O–Zn–O chelate. This constrains rotation around the carbonyl–alpha-carbon bond. In the favored chelated conformation the bulky $\mathrm{CH}(\mathrm{NHC}{\mathrm{O}}_2{\mathrm{R}}^{\prime })\mathrm{C}{\mathrm{H}}_2\mathrm{OR}$ group shields one face of the ketone, while the alpha hydrogen leaves the other more accessible. hydride attacks from that less crowded face, the carbonyl $\pi$ bond opens onto oxygen, and aqueous work-up gives the second alcohol.

Mapping this Cram chelation model back onto the printed zigzag gives B, not A. Merely placing a new OH “on the same side” of an unrotated page drawing is not a valid chelation argument: the reactive chelate and that drawing use different conformations. Zinc-borohydride reductions of alpha-hydroxy ketones are experimentally known to favor the corresponding anti-1,2-diol family under chelation control.^[ref]

The reagent combination can generate zinc borohydride and related zinc-associated reducing species. The stereochemical justification is coordination-controlled facial selectivity; it does not require identifying a unique solution aggregate from the reagent list. The quoted yield is not itself a measurement of a 95:5 diastereomeric ratio.

The hydroxyl group first binds aluminium, normally by deprotonation to an aluminium alkoxide with release of hydrogen. Coordination also brings aluminium close to the alkyne. directed trans-hydroalumination then puts hydride on the alkyne carbon proximal to CH2O and aluminium on the distal carbon.

One useful mechanistic description has external hydride attacking the coordinated alkyne opposite the aluminium-containing face. The resulting vinylaluminate is stabilized by a five-membered O–Al–C–C–CH2 chelate. This is not the ordinary undirected syn addition of a neutral dialkylaluminium hydride. Propargylic-alcohol-directed trans-hydroalumination with Red-Al gives vinylaluminate intermediates that can be intercepted by electrophiles.^[ref]

iodine replaces the distal C–Al bond with C–I without scrambling the alkene geometry; aqueous work-up releases the alcohol. hydride and iodine are consequently trans across the alkene, while iodine and the CH2OH group are on the same side.

By Cahn-Ingold-Prelog priority rules priority the product is therefore $Z$: CH2OH outranks H at one alkene carbon and I outranks CH2R' at the other. Describing the two carbon chains as trans would not make this an $E$ alkene, because the iodine changes which substituent has higher priority.

Number the OH-bearing carbon C1 and the adjacent brominated carbon C2. In the trans bromohydrin, base generates an alkoxide. A chair conformation can place O at C1 and Br at C2 trans-diaxial, aligning the oxygen nucleophile for backside attack at C2. Intramolecular SN2 closure expels bromide and inverts C2: the retained C1–O bond and the newly formed C2–O bond are both on the original OH face, giving the epoxide shown. ring inversion can provide the reactive chair even if it is not the dominant ground-state conformer.

The cis bromohydrin cannot place OH and Br trans-diaxial: one is axial when the other is equatorial. In the chair with Br axial, the hydrogen on C1 can instead be antiperiplanar to C2–Br. An alkoxide-assisted semipinacol pathway moves the C1–H bonding pair to C2 while the oxygen lone pair forms C1=O and bromide departs. hydride migration gives the ketone without changing the carbon skeleton. The hydrogen is transferred to C2 rather than expelled as a free hydride. The original phenyl-bearing stereocentre is not involved in either bond-reorganization sequence. Configuration controls which properly aligned displacement or migration can occur; a flat ring drawing alone conceals that distinction.

oxidative addition of an intact Si–H bond initially places H and SiR${}_3$ cis in the metal coordination sphere; this is the local arrangement generated by addition of a sigma bond. hydride and alkene must also attain adjacent sites for intramolecular insertion, and the eventual alkyl and silyl groups must be cis for direct reductive elimination. Subsequent rearrangements should not be confused with the stereochemistry of the bond-addition event itself.

For an alkyl-substituted terminal alkene, insertion puts Co on the terminal carbon and H on the substituted carbon, producing $\mathrm{Co}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{R}$. It avoids the more crowded secondary metal–alkyl arrangement. Replacement of that Co–C bond by C–Si therefore gives the linear product $\mathrm{RC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Si}{\mathrm{R}}_3$, an Anti-Markovnikov addition outcome. Hydrometallation is syn at the coordinated alkene, although this simple terminal substrate generally does not retain two substituents whose relative stereochemistry could diagnose the complete sequence.

Step 1 is oxidative addition of Si–H to Co(I), giving Co(III)–H and Co(III)–SiR${}_3$. It cleaves the Si–H bond and forms two metal–ligand bonds, increasing the formal metal oxidation state by two. The phosphine dissociation shown before this numbered step makes room for the reaction; alkene binding and phosphine exchange are ligand-substitution events rather than oxidative additions.

Step 2 is migratory insertion of the coordinated alkene into Co–H, equivalently intramolecular hydride migration to the alkene. One alkene carbon gains H and the other gains the Co–C bond. Co remains formally Co(III); a two-electron alkene ligand plus a hydride is replaced by one alkyl ligand, reducing the electron count by two before phosphine reassociation.

Step 3 is C–Si reductive elimination between the alkyl and silyl ligands. It releases the hydrosilylated alkane and regenerates Co(I). The electron count for the drawn sequence is consistent: the one-PMe${}_3$ alkene adduct is 16-electron; oxidative addition gives 18; insertion gives 16; PMe${}_3$ association gives 18; reductive elimination returns the 16-electron catalyst. Because insertion occurs into Co–H, this is the conventional Chalk-Harrod mechanism.

Write the reacting terminus as $R-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, where $R$ includes the pre-existing stereocentres and the intact $\mathrm{OC}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O})\mathrm{NM}{\mathrm{e}}_2$ substituent. ozone undergoes 1,3-dipolar cycloaddition across the terminal alkene to give a molozonide. In a resonance description of ozone as ${\mathrm{O}}^{-}-{\mathrm{O}}^{+}=\mathrm{O}$, the alkene pi pair bonds to one terminal oxygen while the other terminal oxygen forms the second C–O bond and the ozone pi bond shifts onto its central oxygen. Cleavage of this unstable primary ozonide produces a carbonyl compound and a carbonyl oxide, which recombine to form the secondary ozonide. Reductive workup breaks the peroxide bonds and gives the carbonyl fragments $R-\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ and formaldehyde.

Excess sodium borohydride also reduces the newly formed aldehyde: hydride attacks its carbonyl carbon, shifting the pi electrons onto oxygen, and the final acidic workup protonates the alkoxide. Thus the retained chain becomes $R-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. The terminal alkene carbon becomes formaldehyde and can be reduced to methanol; it is no longer part of the main product. This accounts for the one-carbon shortening of the retained skeleton.

The original amide carbonyl is present in both H and I and is not reduced under the ordinary borohydride workup. The existing stereocenters and benzyl ether are retained. Stopping the mechanism at an aldehyde, or deleting the amide C=O when reading the smaller printed scheme, would give the wrong product. The new terminal alcohol carbon is CH2 and therefore adds no stereocenter.

For an octahedral complex, strong axial sigma donation raises the metal–ligand antibonding $e_g$-derived orbitals. A hydride is a very strong sigma donor: its occupied $1s$ orbital overlaps effectively along a metal–ligand axis and has no filled perpendicular p orbitals to act as a conventional pi donor. It can therefore produce a large octahedral splitting by raising $e_g$ relative to $t_{2g}$.

Triphenylphosphine also gives sigma donation, but its useful acceptor orbitals additionally receive metal pi backbonding. This lowers the metal $t_{2g}$-derived levels and enlarges their separation from $e_g$. The acceptors have substantial P–C antibonding character; invoking a dominant phosphorus $3d$ acceptor is unnecessary. Thus hydride is strong-field principally through sigma bonding, while PPh${}_3$ combines sigma donation with pi acceptance. The spectrochemical series describes the resulting splitting, not just a ranking of ligand basicity.

The all-carbonyl hydride, HCo(CO)${}_4$, is more acidic. Loss of ${\mathrm{H}}^{+}$ gives $[\mathrm{Co}(\mathrm{CO}{)}_4{]}^{-}$, whose negative charge is stabilized by pi backbonding into four carbonyl ligands. Replacing one CO by triphenyl phosphite reduces the net electron-withdrawing character of the ligand set relative to four carbonyls. The substituted hydride is more electron rich and its conjugate-base anion is less stabilized against protonation.Both conjugate bases have a conventional 18-electron count, so the difference is not a missing-electron argument. It is the relative stabilization of acid and base by the ligands in coordination chemistry. This concerns heterolytic deprotonation, not the homolytic Co–H bond-dissociation energy.

The trityl cation acts as an electrophile and removes one metal-bound methyl by methyl abstraction. The balanced conversion isThus $C=[\mathrm{C}{\mathrm{p}}_2\mathrm{ZrMe}{]}^{+}[\mathrm{B}({\mathrm{C}}_6{\mathrm{F}}_5{)}_4{]}^{-}$ and $D=\mathrm{P}{\mathrm{h}}_3\mathrm{CMe}$, 1,1,1-triphenylethane. Transfer is formally of $\mathrm{M}{\mathrm{e}}^{-}$ to carbon, not transfer of a hydride or a methyl radical. zirconium remains Zr(IV), $d^0$. Ionic electron counting changes from $12+4=16$ in the dimethyl to $12+2=14$ at the cation, creating an accessible coordination site. The borate is a weakly coordinating anion, although ion pairing can still occur in solution.

A ring-retaining bromination replaces the three hydridic B–H positions while preserving N–H. Its net stoichiometry isUse controlled bromination and conditions allowing elimination of hydrogen bromide from addition intermediates, rather than identifying the first bromine adduct as the final ring product. Boron-bound hydrogen has hydride-like character because boron is less electronegative than hydrogen; its replacement is consistent with electrophilic attack by bromine. The primary preparative work reports B-tribromoborazines obtained by bromination of borazine and its N-methyl analogue.^[ref]

The product must retain the unsaturated B${}_3$N${}_3$ ring. Merely adding three HBr molecules to borazine initially gives a saturated hydrogen-halide addition product, not the required formula. Addition followed by suitable elimination is distinct from simply asserting direct B–H protonolysis by HBr.

Sodium borohydride is chemoselective for the ketone; lithium aluminium hydride reduces both the ketone and the lactam. In an amide, donation of the N lone pair into the carbonyl $\pi$ system reduces the carbonyl carbon's electrophilicity. It is therefore substantially less susceptible to the milder hydride donor.

Lithium aluminium hydride supplies more strongly reducing hydride equivalents, and the strong affinity of aluminium for oxygen assists removal of the amide oxygen through the iminium pathway. This difference explains both the unchanged lactam carbonyl in B and the absence of that oxygen in C. Both reactions are interpreted after the usual protonating workup.

B is 4-hydroxy-1-methylpyrrolidin-2-one; C is 1-methylpyrrolidin-3-ol. Their ring connectivities are shown below. Each ring remains five membered; the attached N-methyl group is retained.

For B, a B–H bond of sodium borohydride supplies a hydride equivalent to the ketone carbonyl carbon, and the carbonyl $\pi$ pair moves to oxygen. This gives an alkoxide, whose protonation during the reaction or workup gives the secondary alcohol. The lactam carbonyl remains. Converting one $\mathrm{C}=\mathrm{O}$ to $\mathrm{CH}(\mathrm{OH})$ adds two H atoms and gives ${\mathrm{C}}_5{\mathrm{H}}_9\mathrm{N}{\mathrm{O}}_2$.

For C, lithium aluminium hydride reduces that ketone in the same net way and also reduces the lactam. Coordination of the amide oxygen to aluminium assists hydride attack at its carbonyl carbon, giving a tetrahedral intermediate. The N lone pair then forms a C=N bond while the C–O bond breaks to an aluminium-bound oxygen group. A resulting iminium ion accepts a second hydride at carbon, yielding the cyclic tertiary amine. Aqueous workup protonates the ketone-derived alkoxide.

The lactam conversion replaces $\mathrm{N}-\mathrm{C}(=\mathrm{O})$ by $\mathrm{N}-\mathrm{C}{\mathrm{H}}_2$; its oxygen is removed rather than retained as another alcohol. Together with ketone reduction, the net change is $+4\mathrm{H}-\mathrm{O}$, giving ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{NO}$. No stereochemistry is specified: the new alcohol stereocentre is formed as a racemic mixture in an achiral environment.