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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.

protonation of the secondary OH permits water departure. The neighbouring phenylthio sulfur can assist this departure by forming a bond to the adjacent alcohol carbon, generating a thiiranium ion spanning that carbon and the original sulfur-bearing cyclohexane carbon. Label the latter Q and the former L. The temporary Q–S–L bridge transfers the phenylthio attachment from Q to L when its Q–S bond subsequently breaks. sulfur attack opposite the departing OH accounts for the new C–S stereochemistry without requiring a freely rotating carbocation.

For the protected terminal alcohol, no free tethered OH is available to trap the bridge. Removal of a beta hydrogen from a cyclohexane carbon adjacent to Q forms the ring alkene while Q–S breaks. sulfur remains bonded to L, giving the allylic phenyl thioether. acid is regenerated by the deprotonation, so the catalyst need not be consumed.

For the free terminal alcohol, its oxygen attacks Q intramolecularly, opening Q–S instead. Count the new ether ring: O–CH2–CH(Me)–L–Q, a five-membered ring sharing Q with the cyclohexane. deprotonation gives the spirocyclic ether. This is capture of the same neighbouring-group-stabilized intermediate rather than migration of the methyl group. The pre-existing methyl-bearing carbon retains its configuration; its wedge/dash appearance changes when the product is redrawn in a different ring projection. The two substrates differ in whether a suitably placed oxygen nucleophile is available for intramolecular trapping.

A is the predicted major diastereomer. At the carbon adjacent to the ketone, take phenyl as the large group, vinyl as medium, and hydrogen as small. There is no adjacent heteroatom donor that would force a five-membered chelate. The appropriate first model is therefore the non-chelating Felkin-Anh model.

View along the carbonyl-carbon–stereocentre bond. Rotate to place the large alpha phenyl group approximately perpendicular to the carbonyl plane. Of the two resulting approaches opposite that large group, the favourable Bürgi-Dunitz trajectory passes nearer hydrogen rather than vinyl. Propyl transfer to the carbonyl and protonation of its oxygen give the configuration labelled A: in the printed projection the new OH projects forward and the propyl bond backward, while the original vinyl bond remains backward. Rotation used to construct a Newman projection must not be mistaken for inversion of the original stereocentre.

This is a relative-selectivity prediction, not a numerical product ratio. Simple attack on the face that happens to look less crowded in the initial flat drawing is unreliable because the relevant C–C single bond can rotate. B is the competing product from the less favourable approach beside vinyl. The model and its preference for the trajectory beside the small group are described in the institutional treatment.^[ref]

In water, separated benzene molecules require separate hydration environments around nonpolar surface. Bringing them together buries part of that surface and reorganizes the surrounding water. The resulting solvent contribution favours association relative to cyclohexane, where benzene was already well solvated by dispersion. This solvent-mediated contribution is the hydrophobic effect, not formation of a new covalent bond or a claim that benzene has become a conventional hydrogen-bond acceptor like oxygen.

The supplied constants imply a thirtyfold enhancement. With matched standard states,at an illustrative 298 K. The water value is still slightly below $1{\mathrm{M}}^{-1}$, so enhancement relative to cyclohexane does not mean a negative standard association free energy for a 1 M standard state, nor does it imply that dilute benzene is all dimer. No universal “entropy only” explanation is needed: solvent enthalpy and entropy both contribute and depend on temperature.

Association in solution includes replacement of solute–solvent contacts, not just formation of the isolated dimer. cyclohexane solvates the benzene surface well through dispersion. Much of the London dispersion gained on bringing two benzene molecules together replaces comparable benzene–cyclohexane contacts, leaving only a modest net gain before the translational and orientational entropy penalty. cyclohexane cannot provide the complementary donor/acceptor interactions available when two waters hydrogen-bond, so water association gains a more distinctive interaction relative to its separated solvated monomers.

Using the same concentration standard state, $K^{\circ }=K_c{c}^{\circ }$ and $\Delta G_{\mathrm{assoc}}^{\circ }=-RT\ln K^{\circ }$. For illustration at 298 K and $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, the ratio of the two supplied association constants corresponds toThus the water-molecule pair can associate more favourably in this solvent even though the stipulated isolated-pair binding energies are similar. Gas-phase potential energy alone omits both differential solvation and the entropy of association.

water achieves substantial binding through a localized O–H···O hydrogen bond; benzene achieves a comparable total through a larger set of weaker contacts. In the water dimer, one water is the hydrogen-bond donor and the other presents an oxygen lone pair region as acceptor. The interaction includes electrostatics, induction interaction, dispersion and short-range exchange repulsion; a hydrogen bond is not a separate fundamental force that can simply be added to all those terms again.

For benzene, draw a T-shaped edge-to-face pair or a slipped parallel pair. Its larger, more polarizable electron clouds provide appreciable London dispersion. Its quadrupolar charge distribution also makes the energy orientation dependent, despite each molecule having no permanent dipole by symmetry. An edge C–H approaching the other ring's pi region is therefore preferable to a model of two structureless, noninteracting dipole-free objects. Dispersion attracts while Pauli repulsion limits close overlap. Several distributed contacts can sum to the scale of one water–water hydrogen bond; counting only the number of conventional hydrogen bonds would miss them.

The comparison concerns the given gas-phase pair interaction energies, not an equality of every solvent-dependent association free energy. Correlated primary calculations explicitly identify both T-shaped and parallel-displaced benzene-dimer minima, rather than a uniquely preferred perfectly eclipsed sandwich.^[ref]

Number the ring with chlorine at C1 and the methylsulfonyl group at C4. The new nitro group is at C2, equivalently C6 by the starting ring's symmetry: the product is 1-chloro-2-nitro-4-(methylsulfonyl)benzene. Both substituents favour this site. Although chlorine deactivates the ring inductively, its lone pair donation stabilizes ortho/para Wheland intermediates. The sulfone is strongly electron withdrawing and meta directing; attack ortho or para to it produces especially unfavourable cationic contributors to resonance in chemistry. C2 is ortho to chlorine and meta to sulfone, whereas C3 has the opposite, less favourable relationship.

sulfuric acid protonates nitric acid, followed by loss of water to produce the nitronium ion. A ring pi bond attacks nitrogen, giving the Wheland intermediate with the new C–N bond. deprotonation at the attacked carbon restores aromaticity. This is electrophilic aromatic substitution: the aryl chloride and the sulfone both remain attached. The formula check is ${\mathrm{C}}_7{\mathrm{H}}_7\mathrm{Cl}{\mathrm{O}}_2\mathrm{S}+\mathrm{N}{\mathrm{O}}_2-\mathrm{H}={\mathrm{C}}_7{\mathrm{H}}_6\mathrm{ClN}{\mathrm{O}}_4\mathrm{S}$. It confirms replacement of one ring hydrogen rather than chlorine, but directing effects, rather than the formula alone, establish the position.

The first base equivalent removes O–H; the second removes the alpha C–H. The first equivalent of lithium diisopropylamide converts the carboxylic acid to its lithium carboxylate. A further equivalent removes the remaining hydrogen on the carbon between the two allyl groups, giving a carboxylic acid dianion:The second negative charge has enolate-like delocalization into the carboxyl group; lithium coordination and aggregation are not represented by this simple charge-bookkeeping structure. The carbon nucleophile attacks methyl iodide with C–I bond cleavage, forming the new C–Me bond. acid workup reprotonates the carboxylate, giving $F=\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{Me})(\mathrm{L}{)}_2$. The two equivalents are therefore chemically necessary for the stated carbon alkylation rather than simply an arbitrary excess. A single equivalent is consumed predominantly by the much more acidic carboxylic proton. The product has formula ${\mathrm{C}}_9{\mathrm{H}}_{14}{\mathrm{O}}_2$ and still has two identical allyl groups, so this methylation does not create a stereocenter.

hydroxide attacks an ester carbonyl to form a tetrahedral intermediate. Collapse expels ethoxide, and acid–base transfer gives the carboxylate and ethanol. Repeating this nucleophilic acyl substitution at the other ester produces the malonate dianion; acid workup gives $\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{L}{)}_2-\mathrm{C}{\mathrm{O}}_2\mathrm{H}$, with $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ as defined above.

On heating, one carboxylic acid group is lost through a six-membered cyclic proton-transfer/bond-cleavage arrangement. A carbonyl oxygen of the retained group accepts the proton from the departing carboxyl OH as its C–C bond breaks and carbon dioxide forms; electron redistribution gives the enol-like form of the retained acid. proton transfer/tautomerization then producesThe net decarboxylation is ${\mathrm{C}}_9{\mathrm{H}}_{12}{\mathrm{O}}_4\to {\mathrm{C}}_8{\mathrm{H}}_{12}{\mathrm{O}}_2+\mathrm{C}{\mathrm{O}}_2$. The central carbon gains a hydrogen, both allyl substituents survive, and only one carboxyl group is removed. It would be incorrect to draw hydrolysis directly giving the final acid without accounting for this carbon loss. The favourable cyclic pathway is characteristic of malonic acid decarboxylation.

The malonate carbon receives two allyl groups. With $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, attached through its first carbon, the product issodium hydride removes a methylene proton from diethyl malonate; molecular hydrogen formation accompanies production of an enolate stabilized by resonance in chemistry. Its carbon attacks the bromomethylene carbon of allyl bromide in an SN2 reaction, expelling bromide. The monoalkylated malonate still has one hydrogen between the carbonyls. Excess base removes that hydrogen, and a second equivalent of allyl bromide produces a second carbon–carbon bond. The final malonate carbon has no remaining hydrogen and cannot undergo a third analogous deprotonation there.

The two alkenes are preserved because the substitution occurs at the primary carbon bearing bromine. Counting the introduced groups gives ${\mathrm{C}}_{13}{\mathrm{H}}_{20}{\mathrm{O}}_4$ for the diallylated diethyl ester. Its two identical allyl groups also mean that the central carbon is not a stereocenter. This stage is the alkylation portion of the malonic ester synthesis.

The bromines ortho and para to OH are removed, leaving the meta bromine at C3. A suitable polar mechanism begins with electrophilic ipso protonation of a brominated carbon. protonation at C4 or C6 gives a Wheland intermediate stabilized by donation of the phenolic oxygen lone pair; an important contributor has an O-protonated cyclohexadienone-like framework. Corresponding protonation at C3 lacks this direct stabilization by resonance in chemistry from OH.

iodide can capture/reduce the departing electrophilic bromine as the C–Br bond electrons restore the aromatic pi system. Thus the incoming hydrogen remains on carbon while bromine is removed, unlike ordinary electrophilic substitution in which the incoming electrophile remains and a proton leaves. bromine trapping can be represented through IBr formation followed by reaction with further iodide to give iodine and bromide. For each replacement the balanced net reduction isApplying it at C4 and C6 gives 3-bromo-2,5-dimethylphenol, with formula ${\mathrm{C}}_8{\mathrm{H}}_9\mathrm{BrO}$. This explains why the position least activated by OH is the one at which bromine survives. It is a proposed acid-assisted protodebromination mechanism consistent with the selectivity; the preparative sequence alone does not prove every proton-transfer or halogen-transfer intermediate. It should not be drawn as iodide simply displacing bromide from an unactivated aryl carbon.

Number the phenolic carbon C1, with methyl groups at C2 and C5. All three remaining aromatic hydrogens are replaced: $H$ is 3,4,6-tribromo-2,5-dimethylphenol. The starting phenol is ${\mathrm{C}}_8{\mathrm{H}}_{10}\mathrm{O}$; replacing three hydrogens gives ${\mathrm{C}}_8{\mathrm{H}}_7\mathrm{B}{\mathrm{r}}_3\mathrm{O}$, agreeing with the intermediate's composition. Bromination of methyl side chains would not follow the stated Lewis acid-promoted aromatic pathway.

aluminium tribromide polarizes Br–Br. The aromatic ring attacks the electrophilic bromine end, forming a Wheland intermediate; removal of the proton on the attacked carbon restores aromaticity. Repetition fills C4 and C6, the available para and ortho positions activated by OH, and C3, which is activated by the methyl substituents when the other positions are occupied. The formula fixes the extent of substitution without requiring an experimentally unreported order for the individual brominations. The preparative perbromination/selective-debromination sequence is documented in the primary synthetic work.^[ref]

Number the cyclopentanone ring $p_1$ at its carbonyl, then $p_2$ at the methyl-bearing carbon, $p_3$ at the mesylate-bearing carbon, $p_4$ at the carbon joined to cyclohexane $h_2$, and $p_5$ at the remaining methylene. B has a $p_2=p_3$ enone and the $h_1=h_6$ enamine. triethylamine removes the hydrogen at $p_2$ while the $p_3$–O bond breaks to release mesylate. An E2 elimination, with approximately antiperiplanar geometry in the puckered ring, gives the conjugated cyclopentenone. Elimination towards $p_4$ instead would give a nonconjugated alkene.

The enamine carbon $h_6$ now attacks enone beta carbon $p_3$ intramolecularly. nitrogen donates into $h_1$ to give the iminium ion, while the enone electrons move onto the ketone oxygen. protonation of the ketone enolate gives C. The second new bond is $h_6-p_3$; the first was $h_2-p_4$. These two connections account for the bridged skeleton without a concerted cycloaddition or an aldol step. The formula of B is ${\mathrm{C}}_{16}{\mathrm{H}}_{23}\mathrm{NO}$: relative to A, elimination removes $\mathrm{C}{\mathrm{H}}_4{\mathrm{O}}_3\mathrm{S}$ overall, namely methanesulfonic acid after proton transfers.

The bridged, sterically hindered biphenyl framework is twisted and does not rapidly interconvert the two faces of each methylene group on the NMR timescale. Label one hydrogen on each CH2 $H_A$ and the other $H_B$. Replacing $H_A$ versus $H_B$ by a test substituent gives diastereomeric environments, so these hydrogens are diastereotopic and need not have equal chemical shifts even in an achiral solvent. Molecular symmetry relates the two $H_A$ sites to each other and the two $H_B$ sites to each other.

Each methylene hydrogen undergoes geminal coupling to its partner, giving the two doublets, each accounting for two hydrogens overall. Their separation is $0.21$ ppm; at operating frequency ${\nu }_0$ in MHz it corresponds to $0.21{\nu }_0$ Hz. This separation is not the geminal coupling constant, which must be measured from the line spacing within a doublet. If the shift separation is not large compared with that coupling, an AB pattern with second-order distortion is expected rather than ideal first-order doublets. A racemic pair of twisted conformations still has these signals: enantiomers have the same spectrum in an achiral medium, while the two hydrogen environments within each molecule remain different.

The initial nucleophile attacks silicon, not the nitrile or an alkene carbon. Each trimethylsilyl group has an electrophilic Si center. A methyl group from methyllithium attacks Si, passing through a pentacoordinate silicon arrangement; Si–O cleavage releases $\mathrm{SiM}{\mathrm{e}}_4$ and the lithium salt of the oxygen-containing fragment. Cleaving the silyl enol ether generates a lithium enolate; cleaving the aryl silyl ether generates a phenoxide. Neither operation needs a new deprotonation at a carbon stereocenter.Here $R-O^{-}$ represents either the phenoxide or the oxygen-centered enolate contributor. Excess reagent permits both cleavages. The pre-existing methyl/nitrile stereocenter retains its configuration: it has no hydrogen to remove, and neither bond attached to that center is broken.

After HMPA is added, each anionic oxygen attacks the carbonyl carbon of allyl chloroformate. The C=O pi electrons move to the carbonyl oxygen, giving a tetrahedral intermediate; collapse expels chloride. The enolate becomes an allyl enol carbonate, and the phenoxide becomes an aryl allyl carbonate, each containing $\mathrm{O}-\mathrm{C}(\mathrm{O})-\mathrm{OC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$. oxygen acylation accounts for preservation of the enol double bond and for the two newly installed carbonate groups. carbon attack would give a different carbon skeleton. The two oxygens are trapped because they were selectively unmasked by desilylation, rather than because every ether oxygen in the molecule is equally reactive.

Oxazolination occurs para to oxygen on the ring that does not bear bromine. The product is 2-[4-(4-bromophenoxy)phenyl]-2-Oxazoline, shown in the final panel above. The brominated ring retains its para bromine; the other ring receives the new oxazolinyl group at its open para position.

The diaryl ether oxygen activates both rings toward electrophilic aromatic substitution. On the brominated ring, its least crowded para position is blocked, and bromine also decreases ring reactivity inductively. The unbrominated ring has an accessible para position, favored over its ortho sites by steric effects. This is substitution of an aromatic hydrogen, not displacement of the aryl bromide.

The formula provides a useful consistency check: ${\mathrm{C}}_{12}{\mathrm{H}}_9\mathrm{BrO}$ loses one aromatic H and acquires ${\mathrm{C}}_3{\mathrm{H}}_4\mathrm{NO}$, giving ${\mathrm{C}}_{15}{\mathrm{H}}_{12}\mathrm{BrN}{\mathrm{O}}_2$. It cannot by itself distinguish positional isomers; the directing and steric arguments select the connectivity.

The same base first makes $\mathrm{Ph}{\mathrm{C}}^{-}(\mathrm{OPh})\mathrm{CN}$. This time addition occurs at a carbon bearing hydrogen para to the nitro group, giving a sigma-H adduct. The attacked carbon temporarily bears both its original H and the new carbon substituent. Negative charge is again delocalized through the ring toward the nitro oxygens.

A further base removes the hydrogen on that aromatic addition site. Its C–H electrons form an exocyclic bond toward the incoming carbon, while that carbon expels phenoxide. This is a base-induced elimination across the new ring–side-chain bond. Redistribution of the nitronate/quinoid pi system restores an aromatic ring and gives the side-chain carbanion $\mathrm{p}-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4-{\mathrm{C}}^{-}(\mathrm{Ph})\mathrm{CN}$. protonation during acidic workup yields $\mathrm{p}-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{CH}(\mathrm{Ph})\mathrm{CN}$.

This is vicarious nucleophilic substitution: the ring loses hydrogen while the nucleophile sacrifices its phenoxy leaving group. hydrogen is not expelled as hydride, and the product's side-chain hydrogen can be supplied on workup rather than moving directly from the ring in a compulsory hydride shift. Base and isotope experiments in the primary mechanistic work support reversible addition followed by base-induced elimination, while not uniquely choosing every E2/E1cb limit.^[ref] The displayed para product is consistent with access to the less crowded activated position.

tert-Butoxide removes the acidic hydrogen from $\mathrm{PhCH}(\mathrm{OPh})\mathrm{CN}$. The resulting carbon anion is stabilized by the nitrile group and the phenyl substituent. It attacks the aromatic carbon bearing fluorine, giving a Meisenheimer addition intermediate. Ring pi bonds shift toward the para nitro group, where the negative charge can be delocalized onto oxygen.

Loss of fluoride restores aromaticity and gives $\mathrm{p}-\mathrm{N}{\mathrm{O}}_2{\mathrm{C}}_6{\mathrm{H}}_4-\mathrm{C}(\mathrm{Ph})(\mathrm{OPh})\mathrm{CN}$. This is an addition–elimination nucleophilic aromatic substitution. The phenoxy group remains attached to the incoming carbon. The ipso aromatic carbon already has a leaving group, so rearomatization can occur by fluoride departure without removing phenoxide. Acid workup neutralizes residual anions; it is not a separate step that replaces the aryl fluoride.

Hydroxide initially deprotonates a little thiophenol, generating the reactive thiolate $\mathrm{Ph}{\mathrm{S}}^{-}$. Number the extended conjugated system $\mathrm{O}=C1-C2=C3-C4=C5$, where C1–C2–C3 is the ring enone and C4=C5 is the attached vinyl group. sulfur attacks the terminal delta carbon C5. The pi bonds relay toward oxygen, generating an extended enolate.

proton transfer at C4, either directly from thiophenol or through solvent-mediated enolization, gives the product retaining the conjugated ring enone. The product is 3-[2-(phenylsulfanyl)ethyl]cyclohex-2-en-1-one, drawn below. Its side chain is $-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{SPh}$; the ring C2=C3 double bond remains. This is net sulfur/hydrogen addition to the terminal part of the extended system, proceeding through 1,6-Conjugate addition and protonation of the extended enolate.

The terminal site is less crowded than the ring beta carbon, and the isolated product retains favorable carbonyl–alkene conjugation. Reversible proton transfers and conjugate addition under thiolate catalysis can favor that product over an unconjugated enone-derived alternative. Transfer of a proton from another PhSH molecule regenerates PhS${}^{-}$, explaining why a catalytic quantity of hydroxide suffices. The net balance is one thiophenol added, ${\mathrm{C}}_8{\mathrm{H}}_{10}\mathrm{O}+{\mathrm{C}}_6{\mathrm{H}}_6\mathrm{S}={\mathrm{C}}_{14}{\mathrm{H}}_{16}\mathrm{OS}$; carbonyl 1,2 addition would instead produce a different functional group and is not favored for this soft sulfur nucleophile.

P is H2, the acetal methine between two oxygens; its only resolved coupling is to the attached methyl, giving the quartet. S is that C2 methyl, whose doublet has the same 6.9 Hz coupling. T is tert-butyl, integrating to nine equivalent hydrogens with no adjacent carbon-bound methine proton across its quaternary attachment.

U is equatorial H5, coupled to the four hydrogens at C4 and C6. The two equivalent equatorial methylene hydrogens are Q, each a doublet of doublets with a 12.3 Hz geminal coupling to its axial partner and a 1.3 Hz coupling to H5. The two equivalent axial methylene hydrogens are R, each with the same 12.3 Hz geminal coupling and a 3.7 Hz vicinal coupling to H5. Equivalence between C4 and C6 makes two pairs rather than four separately shifted methylene hydrogens.Coupling U to two equivalent R hydrogens and two equivalent Q hydrogens produces its triplet of triplets. The different small vicinal couplings are consistent with oblique H–C–C–H torsions for equatorial H5; the 12.3 Hz value is geminal and must not be mistaken for a large vicinal trans-diaxial coupling.