Codex @codex  0

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.

B is 2-Methylpropanoyl chloride, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHC}(=\mathrm{O})\mathrm{Cl}$. The single index of index of hydrogen deficiency is supplied by the carbonyl. Its relatively high-frequency infrared spectroscopy band identifies an acyl chloride, while two equivalent methyl groups explain why four carbons produce only three carbon-13 NMR resonances.

Assign ${\delta }_C=178.1$ to the acyl chloride carbonyl, $46.3$ to the methine and $18.9$ ppm to both methyl groups. Their chemical equivalence in NMR follows from the symmetry of the isopropyl fragment in this achiral environment.

C is 2-Chloroethyl vinyl ether, $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-\mathrm{O}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}$. There is one alkene and no carbonyl. Assign the carbon-13 NMR resonances asThe strongly differentiated pair of alkene shifts is characteristic of a vinyl ether; interpreting $151.1$ ppm as a carbonyl would leave the unusually shielded unsaturated $87.4$ ppm carbon and the three distinct vinylic protons unexplained.

The vinyl ether has an important resonance contributorDonation from oxygen into the conjugated system increases electron density at the terminal unsaturated carbon, helping explain its unusually shielded ${\delta }_C=87.4$ ppm and its terminal protons near 4.1–4.2 ppm. The directly oxygen-bound unsaturated carbon is strongly deshielded at $151.1$ ppm; its proton at $6.480$ ppm also experiences the oxygen substituent and alkene magnetic anisotropy. These trends are consistent with the whole delocalized electronic structure; a chemical shift is not simply a meter for an atom's partial atomic charge.

For the saturated part, the stronger local deshielding by oxygen places $\mathrm{OC}{\mathrm{H}}_2$ at ${\delta }_C=67.8$, ${\delta }_H\simeq 3.948$, while $\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}$ lies at $41.9$ and $3.708$ ppm. The two terminal vinylic protons have different geometrical relationships to the oxygen substituent and therefore different chemical shifts as well as different vicinal couplings.

The configurations are $\mathrm{C}:1s^{2}2s^{2}2p^2$ and $\mathrm{N}:1s^{2}2s^{2}2p^3$. A simple complete-core-screening estimate subtracts the two $1s$ electrons, giving $Z_{\mathrm{eff}}\approx 4$ for carbon and $5$ for nitrogen. This neglects screening by the other valence electrons, and is an explicitly rough estimate.

A more systematic estimate using Slater's rules includes $0.85$ for each $1s$ electron and $0.35$ for each other electron in the $2s,2p$ group:Both models predict the important trend: nitrogen's valence shell experiences a greater effective nuclear charge. There is no unique numerical effective nuclear charge without specifying an orbital and a screening prescription; the one-electron formula in (a) does not by itself define neutral-atom orbital energies.^[ref]

The $2s$ atomic orbital is spherically symmetric and has one radial node. A real $2p$ atomic orbital has two lobes of opposite quantum phase, one atomic-orbital nodal plane through the nucleus and no radial node. The three spatial orientations $2p_x,2p_y,2p_z$ remain degenerate in a spherically symmetric atom when external fields and fine structure are neglected.

In a many-electron atom, $2s$ penetrates the screened core more effectively than $2p$, experiences a stronger attraction and is lower in energy: ${\epsilon }_{2s}<{\epsilon }_{2p}$. The hydrogen-like ion degeneracy therefore does not survive screening. Both valence orbital types are generally more tightly bound and spatially contracted in nitrogen than in carbon, because of the larger effective nuclear charge. These are central-field orbital statements; the detailed atomic multiplet structure also involves electron repulsion and Hund's rules.

The framework has a planar ${\mathrm{H}}_2\mathrm{C}$ group attached to a straight C–N–N axis. Two useful octet resonance contributors areIn the left contributor, the terminal nitrogen has two lone pairs; in the right, the terminal nitrogen and carbon each have one. Each formula has 16 valence electrons and zero net electric charge. The double-headed resonance arrow changes the electronic allocation while retaining the same nuclear framework; these are not two different equilibrium species.

An internally consistent elementary orbital hybridisation model is $s{p}^2$ at carbon and $sp$ at both nitrogen atoms; the hydrogens use $1s$ orbitals. The three carbon hybrids form two C–H sigma bonds and one C–N sigma bond, leaving one perpendicular p orbital. The central nitrogen uses opposite $sp$ hybrids for its two collinear sigma connections, leaving two perpendicular p orbitals for the two pi systems.

At terminal nitrogen, one $sp$ hybrid points toward the central nitrogen and the other houses an axial lone pair. One perpendicular p orbital forms the second, N–N-only pi system; the other participates in the three-atom C–N–N pi system. This description remains the same across both resonance contributors; counting lone-pair domains in just one drawing and assigning a different terminal hybridisation to each contributor would obscure the delocalization. orbital hybridisation is a useful choice of localized basis, not a uniquely measurable label: advanced treatments of diazomethane require a superposition of bonding patterns.^[ref]

The four sigma bonds are the two C–H connections, C–N and N–N, containing eight electrons. A terminal-nitrogen axial lone pair contains two more. Of the remaining six electrons, two occupy an N–N pi bond lying in the molecular plane, while four occupy a perpendicular three-centre four-electron pi bond extending over C–N–N. There are therefore 16 valence electrons in total.

For orientation, a three-site model with equal on-site energy $\alpha$ and adjacent interaction $\beta <0$ gives perpendicular pi levels at $\alpha +\sqrt{2}\beta$, $\alpha$, and $\alpha -\sqrt{2}\beta$. The lower two are doubly occupied and the upper is empty. The middle level has a terminal-end combination with little central-atom amplitude; unequal C and N orbital energies change its coefficients and make it only approximately nonbonding.

The highest occupied molecular orbital is the upper occupied perpendicular pi orbital, with substantial carbon amplitude, broadly above the strongly bonding sigma framework and the lower occupied pi orbitals. It is not the terminal axial nitrogen lone pair. Real diazomethane has C–N bonding and N–N antibonding character in this frontier orbital, rather than being an exactly nonbonding equal-site model. Photoelectron and Rydberg-state experiments support this assignment.^[ref]

An intense band near $2088\mathrm{c}{\mathrm{m}}^{-1}$ is consistent with the high-frequency coupled C–N–N stretching motion of a diazo group, with substantial N–N multiple-bond character. It lies above a typical isolated C=N double-bond stretch, and supports contributions with a strong N–N connection. It does not establish that the molecule is a literal localized C–N single bond plus an unperturbed N≡N triple bond: a normal mode moves the coupled framework, and the resonance contributors describe one electronic structure.

Protonation at the nucleophilic carbon gives methyldiazonium, $\mathrm{C}{\mathrm{H}}_3-{\mathrm{N}}^{+}\equiv \mathrm{N}$. The carbon-rich highest occupied molecular orbital and the carbon-negative resonance contributor rationalize this site. Subsequent displacement of molecular nitrogen can be strongly favourable; the initial protonation product still contains both nitrogens.

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.

For the ordinary ester, the carbon nucleophile of a grignard reagent attacks the carbonyl carbon and moves the C=O pi pair onto oxygen. Collapse of the tetrahedral intermediate expels ethoxide, transiently giving ${\mathrm{R}}^1\mathrm{COR}$. This ketone is more electrophilic than the starting ester because it lacks an alkoxy group's stabilizing resonance donation. A second equivalent of the grignard reagent therefore adds to it before isolation:The negatively charged intermediates are associated with magnesium counterions. Merely reducing the amount of reagent to one equivalent does not generally trap the ketone selectively; it can leave unreacted ester together with the double-addition product.

In the first successful preparation, magnesium insertion into the primary C–Br bond gives n-butylmagnesium bromide. The carbon chain has four carbons, so its acetylation produces hexan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}(\mathrm{C}{\mathrm{H}}_2{)}_3\mathrm{C}{\mathrm{H}}_3$, not pentan-2-one. Attack on acetic anhydride forms $\mathrm{C}{\mathrm{H}}_3\mathrm{C}({\mathrm{O}}^{-})(\mathrm{nBu})\mathrm{OCOC}{\mathrm{H}}_3$. Coordination of the anhydride-derived oxygens to magnesium can stabilize this tetrahedral addition complex. Under suitably cold, controlled-addition conditions it survives sufficiently to delay release of the free ketone until aqueous work-up has destroyed the remaining grignard reagent. Protonation followed by collapse releases acetate/acetic acid and the ketone.

This interpretation has direct experimental precedent: Melvin S. Newman and William T. Booth Jr. obtained methyl ketones by cold addition of grignard reagents to acetic anhydride, attributing selectivity to a persistent addition complex. The temperature and reagent-addition regime are genuine conditions on the explanation, rather than proof that an anhydride can never undergo overaddition; the exam scheme omits them.^[ref]

For the benzonitrile route, n-butyl attacks the nitrile carbon and one C≡N pi pair moves to nitrogen, producing the magnesium ketimine salt $\mathrm{PhC}(\mathrm{nBu})={\mathrm{N}}^{-}\mathrm{MgB}{\mathrm{r}}^{+}$. This anionic nitrogen-containing intermediate resists a second nucleophilic addition much more than a neutral carbonyl does. The ketone $\mathrm{PhC}(=\mathrm{O})\mathrm{nBu}$ appears only during acidic aqueous work-up, after organomagnesium has been quenched. Protonation gives the iminium ion; water attacks its carbon, proton transfers give a carbinolamine, and C–N cleavage from its nitrogen-protonated form produces the protonated ketone and ammonia, followed by deprotonation. In the acidic mixture ammonia is present largely as ammonium. This sequence explains why nitrile addition stops at a ketone without requiring the ketone itself to be resistant to grignard reagents.

The atom-labelled hydrolysis sequence isWater's lone pair attacks the iminium carbon while its C=N pi pair moves to nitrogen; the carbinolamine oxygen then reforms C=O while the C–N pair moves to the departing protonated amino group. Final deprotonation gives the neutral ketone. The arrows involving protonation and hydrolysis also exchange protons with the acidic aqueous reservoir; the condensed sequence tracks the carbon–nitrogen and carbon–oxygen transformations rather than displaying every solvent molecule.

The oxygen substituent changes position through a halohydrin–epoxide sequence. First, hydroxide attacks the acetate carbonyl of $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2\mathrm{OAc}$, giving a tetrahedral intermediate. Collapse expels the substrate alkoxide, and proton transfer gives $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2\mathrm{OH}$ plus acetate. This is saponification, breaking the acyl–oxygen bond rather than replacing bromide directly with hydroxide.

A small equilibrium population of $\mathrm{BrC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Me}{)}_2{\mathrm{O}}^{-}$ undergoes intramolecular SN2 reaction at the adjacent primary carbon: the O lone pair attacks $\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$ and its C–Br pair leaves with bromide. The result is 2,2-Dimethyloxirane. Addition of HBr protonates the ring oxygen. bromide then attacks preferentially at the more substituted carbon, where the protonated ring has greater carbocation character, breaking that C–O bond and leaving oxygen attached to the former primary carbon:The middle formula denotes a three-membered epoxide, not an open-chain ether. Acidic opening of this highly unsymmetrical epoxide has the opposite regioselectivity to a simple basic nucleophilic opening at its less substituted carbon.

Finally, the primary alcohol oxygen attacks acetyl chloride. The tetrahedral intermediate collapses with chloride departure, and triethylamine removes the O–H proton and captures HCl. The product is $\mathrm{AcOC}{\mathrm{H}}_2\mathrm{C}(\mathrm{Br})(\mathrm{Me}{)}_2$. oxygen is carried through the epoxide to the other carbon; it is not necessary to invent a methyl migration or a carbon-skeleton rearrangement.

The two starting ester groups are adjacent on the benzene ring. Select one ester carbon as the site receiving both incoming methyl groups. A first methyl addition gives a tetrahedral intermediate; its collapse expels methoxide to give the ortho-substituted ketone $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}(=\mathrm{O})\mathrm{Me}$. A second methyl addition produces $\mathrm{o}-!\mathrm{Me}{\mathrm{O}}_2\mathrm{C}-{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}({\mathrm{O}}^{-})\mathrm{M}{\mathrm{e}}_2$, with the alkoxide bound to magnesium.

The newly generated oxygen is now positioned to attack the remaining ester intramolecularly. Draw an arrow from its lone pair to that ester carbonyl and another from the C=O pi bond to its oxygen. Collapse of the resulting tetrahedral intermediate reforms the carbonyl and expels methoxide, closing a five-membered lactone. Protonation of magnesium-bound species in the work-up gives 3,3-Dimethylphthalide. Equivalently, protonation can first give the ortho hydroxy ester, followed by acid-assisted lactonization with loss of methanol; both descriptions show the same required bond formation.

The oxygen in the new ring is the oxygen of the first ester's original carbonyl; the surviving lactone carbonyl comes from the second ester. The product retains the six ring carbons and both original acyl carbons, loses the two ester methoxy carbons, and gains two carbons from methylmagnesium bromide. The favourable intramolecular five-membered closure accounts for the displayed product. It does not make the isolated lactone universally inert to a large excess of grignard reagent; stoichiometry and trapping/work-up conditions matter.

Disconnect the new bond between the two six-carbon units. One portion of bromocyclohexane supplies a grignard reagent; another supplies cyclohexene oxide:A bulky strong base can favour E2 elimination in the alkene-forming step; a peracid such as meta-chloroperoxybenzoic acid provides epoxidation. None of its carbons enters the target.

Treat cyclohexene oxide with cyclohexylmagnesium bromide, then perform acidic aqueous work-up. Backside attack on one epoxide carbon breaks its C–O bond and installs cyclohexyl opposite the retained oxygen. Protonation of the oxygen gives trans-2-cyclohexylcyclohexan-1-ol, the required relative configuration. The two six-carbon reactant portions supply all 12 target carbons. An achiral synthesis produces a racemic mixture trans pair; obtaining only the particular enantiomer represented by the drawing would require resolution or an enantioselective step. The useful stable chair can place the two large trans-1,2 substituents approximately diequatorial.

The four terminal methyl groups form ordinary Al–C sigma bonds, with formal bond order one. Each of the two bridging methyl carbons connects the two aluminium centres through a three-centre two-electron bond. One occupied bonding combination extends over Al–C–Al; the other combinations are unoccupied in the simple electron-deficient bridge model. A single shared pair therefore supplies one bond in total across the two contacts, giving formal Al–C bridge bond order $1/2$ per contact, compared with one for terminal Al–C.

Each aluminium has approximately tetrahedral, $s{p}^3$-like orbital directions. A terminal methyl carbon uses $s{p}^3$-like orbitals for its three C–H bonds and its Al–C bond. A bridging methyl carbon still uses three directed orbitals for C–H bonding, while its remaining approximately $s{p}^3$ hybrid overlaps with orbitals on both aluminium atoms. This shared bridge avoids placing five independent ordinary two-electron bonds around carbon. The hybrid labels are idealized and the bridge geometry is distorted; the four occupied C–H/bridge orbital directions are not equivalent to five localized carbon bonds.

The total valence count provides a useful check: $2(3)+6(4)+18(1)=48$ electrons. Eighteen C–H bonds use 36, four terminal Al–C bonds use eight, and two three-centre bridges use four. A description assigning a full two-electron bond to each of the four bridging contacts would require too many electrons.

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.

B gives spectrum 4. Number the unsaturated acyl chain from its carbonyl carbon. For $\mathrm{MeO}-\mathrm{C}(=\mathrm{O})-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$, the assignments areThe strongly deshielded beta carbon and relatively shielded alpha carbon reflect polarization of the conjugated ester. All four protonated carbons here are CH or methyl environments and share one attached proton test phase, whereas the carbonyl has the opposite phase. The methoxy resonance is decisive: B has an O-bound methyl, not an O-bound methylene.

The mechanism is nucleophilic addition followed by elimination of methoxide. An ethoxide oxygen lone pair attacks the carbonyl carbon, while the C=O pi pair moves to oxygen:The tetrahedral intermediate collapses when its anionic oxygen reforms the carbonyl and the C–OMe bond breaks with its electron pair going to the methoxy oxygen:lithium is a coordinating counterion; it does not supply an electron pair to the acyl carbon. proton exchange with ethanol connects the methoxide produced to the ethoxide catalyst pool and gives methanol, so the net solvent-mediated exchange is methyl acetate plus ethanol to ethyl acetate plus methanol. The reversibility of the elementary acyl steps is important: changing an alkoxy group is not automatically an irreversible reaction.