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The radical $\mathrm{BrO}$ participates in catalytic atmospheric ozone destruction. Its measured abundance is only one partition of the total bromine inventory; precursor delivery, photodissociation and reservoir reactions determine the relation between the two.

A halohydrin with bromine on the carbon adjacent to OH. Its relative stereochemistry can select between properly aligned epoxide closure and semipinacol rearrangement.

The gaseous hydrogen halide consisting of one hydrogen atom and one bromine atom. Its molecular vibration can be studied through its vibrational fundamental and weak vibrational overtones. The aqueous acid and the isolated diatomic molecule should be distinguished when specifying a dissociation process.^[ref]

In the productive mixed-radical pathway, chlorine monoxide and BrO yield chlorine and bromine atoms; subsequent atom reactions with ozone regenerate the oxides. The net cycle consumes two ozone molecules per radical-pair event, giving $L=2k[\mathrm{ClO}][\mathrm{BrO}]$ when the subsequent steps are fast. The branching of a real chlorine monoxide–BrO reaction must be included if not already absorbed into its specified effective rate coefficient.

Number the enone carbonyl carbon C1, the adjacent alkene carbon C2 and the ethoxy-bearing alkene carbon C3. In the bromine-first pathway, the enol-ether-like $C2=C3-\mathrm{OEt}$ unit attacks $\mathrm{B}{\mathrm{r}}_2$ at C2, breaking the Br–Br bond. The intermediate has a new C2–Br bond and positive charge at C3, stabilized as an oxonium chemical resonance contributor by donation from the ethoxy oxygen.

triethylamine removes the remaining C2 hydrogen. Its C–H bonding pair reforms C2=C3 and neutralizes the oxonium system. The result is replacement of the vinylic C2–H by C2–Br, with the original enone and ethoxy group retained. If bromide first captures the cationic intermediate, the equivalent vicinal dibromide can eliminate HBr to reach the same product.

The ethoxy group makes this alkene sufficiently electron-rich for the electrophilic-addition description. It is important to place bromine at C2 and restore the double bond, rather than stop at a saturated addition product.

Label the enamine carbon bonded to nitrogen C1 and its alkene partner C2; C6 is the other carbon next to C1. B is the C-alkylated iminium ion bromide. Its new substituent at C2 is $-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}(\mathrm{C}{\mathrm{O}}_2\mathrm{Et})=\mathrm{C}{\mathrm{H}}_2$, while C1 is double-bonded to the pyrrolidine nitrogen, which bears positive charge. bromide is the counterion, accounting for Br in the empirical formula without requiring a retained C–Br bond.

The C2 end of the enamine attacks the bromomethyl carbon by $S_{\mathrm{N}}2$ displacement. Simultaneously, donation of the nitrogen lone pair creates the C1=N bond. This is ordinary enamine C-alkylation, not N-alkylation.

proton transfer removes an alpha proton at C6, converting the iminium ion into the enamine regioisomer C1=C6. That enamine attacks the terminal $\mathrm{C}{\mathrm{H}}_2$ of the tethered acrylate in an intramolecular Michael addition. nitrogen donation restores the iminium ion, and protonation of the ester enolate completes the sequence.

The new bridge is $C2-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{C}{\mathrm{O}}_2\mathrm{Et})-\mathrm{C}{\mathrm{H}}_2-C6$. Together with the original ring and its C1 bridge this produces the bicyclo$[\mathrm{3.3.1}]$nonane iminium ion skeleton in C. It retains the pyrrolidine group and the ester; hydrolysis to a ketone is not part of the displayed reaction. The second panel of the original annulation figure shows the chemical atom connectivity and distinguishes ionic bromide in B from a covalent bromine substituent.

The activated aromatic pi system polarizes bromine and attacks its electrophilic end while Br–Br cleaves. This makes a C–Br bond at C1 or C3 and gives a Wheland intermediate. Donation from phenolic oxygen stabilizes contributors with positive charge at its attached ring carbon. acetate or bromide then removes the proton from the attacked carbon; the C–H bond electrons restore the aromatic pi system. Repetition at the other activated ortho site gives the dibrominated product.

Each step replaces C–H by C–Br and restores aromaticity. No radical initiator or benzylic abstraction is needed, and no Lewis acid catalyst should be invented for this already activated phenolic substrate. The overall substitution balance is $\mathrm{Ar}{\mathrm{H}}_2+2B{\mathrm{r}}_2\to \mathrm{ArB}{\mathrm{r}}_2+2HBr$.

B is 1,3-dibromo-2-hydroxyxanthone: both bromines are on the hydroxy-substituted outer ring, at its two available positions ortho to OH. Define C2 as the OH-bearing carbon; C1 lies between it and the carbonyl-side ring junction, and C3 lies on its other side. These labels specify the substitution independently of the orientation of the printed drawing.

Phenolic lone pair donation strongly activates its ortho and para positions towards electrophilic substitution. The para position is a fused junction and has no replaceable hydrogen, leaving C1 and C3 as the two directly activated sites. C1 also benefits from the conjugative influence of the ring ether oxygen. Although other positions receive some activation from that oxygen, they lack the stronger local phenolic activation of the selected pair. Excess bromine can therefore replace both ortho hydrogens. Replacing two hydrogens in ${\mathrm{C}}_{13}{\mathrm{H}}_8{\mathrm{O}}_3$ gives ${\mathrm{C}}_{13}{\mathrm{H}}_6\mathrm{B}{\mathrm{r}}_2{\mathrm{O}}_3$, consistent with the stated composition. Addition across an outer-ring double bond would sacrifice aromaticity and would not account for the hydrogen loss.

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]

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.

For both substrates, the alkene pi bond attacks bromine and polarizes/breaks Br–Br. A bromine lone pair completes a three-membered bromonium ion, and bromide subsequently attacks a carbon from the opposite face, cleaving that C–Br bridge bond. The two new C–Br bonds are anti. A free carbocation that rotates before capture is not needed.

In trans-fused $A$, the fixed conformation makes the alternative anti-opening pathways diastereomeric. A chair-like, trans-diaxial opening is favored over a geometrically strained alternative that requires a twist/boat-like arrangement of the six-membered ring. The Fürst-Plattner rule describes this conformational preference. One depicted enantiomer of $A$ therefore gives essentially one dibromide diastereomer, with the two bromines occupying the trans-diaxial arrangement in its favored chair. If a racemic sample of $A$ were used, its other enantiomer would of course give the mirror product as well.

For cis-fused $B$, the accessible mirror-related folds and the equivalent anti pathways produce mirror-related dibromides with equal rates in an achiral environment. Adding the two bromines destroys the symmetry that made the starting configuration achiral. $B$ gives a racemic dibromide pair. The racemate follows from equivalent enantiomer-forming pathways, not from a loss of anti stereospecificity.

oxidative addition inserts Pt into one B–Br bond. All three B–Br positions of the starting ring are equivalent, so there is no distinct choice among them before the first substitution. The product is the square-planar Pt(II) boryl bromide $\mathrm{trans}-[\mathrm{Pt}(\mathrm{PC}{\mathrm{y}}_3{)}_2(\mathrm{Br})({\mathrm{B}}_3\mathrm{B}{\mathrm{r}}_2{\mathrm{N}}_3{\mathrm{H}}_3)]$, with the two phosphines trans, the Pt–B bond trans to Pt–Br, and the six-membered ring intact. The third panel shows this connectivity. This is a metal–boron sigma bond, not attachment of the whole ring as an ${\eta }^6$ pi ligand. Formal electron counting gives Pt(II), $d^8$, with $8+4_{\mathrm{P}}+2_{\mathrm{Br}}+2_{\mathrm{boryl}}=16$ electrons, appropriate for square-planar coordination. The primary synthesis establishes this trans-bromo(borazinyl) structure.^[ref]

The electron-rich Pt(0) reacts at the polarized B–Br bond, whose electrophilic boron and departing bromide provide a favorable bond-activation pathway. This accounts for B functionalization without needing cleavage of a ring B–N bond or selection of a nitrogen-bound product. It does not imply that arbitrary N–H activation is impossible under every condition.

Replacing an electronegative bromine substituent by a metal fragment changes the boron's electron demand. Relative to the B–Br sites, the metal-bound boron is less effective at accepting N-to-B pi donation, so its adjacent B–N bonds are expected to have less multiple-bond character and be longer; donation toward the remaining B–Br sites produces compensating unequal ring bonds. The original threefold equivalence is lost. The detailed lengths depend on the full electronic structure; a localized drawing with three rigid alternating double bonds would conceal the redistribution that the unequal lengths reveal.

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.

The first product is $\mathrm{BrCH}=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OC}{\mathrm{H}}_3$, with the original alkene geometry retained. methoxide attacks the primary saturated carbon bearing bromine from the backside. Its oxygen lone pair forms the new C–O bond as the C–Br bonding pair leaves with bromide. The other bromine is vinylic, directly bonded to an sp${}^2$ carbon: ordinary SN2 reaction is unfavourable there, and ionization would require an unstable vinyl cation. Thus the primary alkyl bromide is replaced selectively, rather than replacing the vinylic halide or moving the double bond. There is no observable inversion label at the attacked carbon because it is a methylene.

The second product is sodium 4-propoxybutane-1-sulfonate, $\mathrm{N}{\mathrm{a}}^{+}{}^{-}{\mathrm{O}}_3\mathrm{S}(\mathrm{C}{\mathrm{H}}_2{)}_4\mathrm{OC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. sodium hydride first removes the alcohol proton, giving propoxide and hydrogen. The propoxide oxygen then attacks the primary ring carbon bonded to the ring oxygen; simultaneous cleavage of that carbon–oxygen bond opens the sultone. The departing ring oxygen remains attached to sulfur as part of the resonance-stabilized sulfonate anion. The carbon attached directly to sulfur is not the substitution site, because cleavage there would not expel the same stable sulfonate leaving group. This is carbon-centred ring opening of a cyclic sulfonate ester, not addition across an S=O bond.

The molecule with the larger inferred dissociation energy is assigned the stronger hydrogen–halogen bond: $\mathrm{A}=\mathrm{HBr}$ and $\mathrm{B}=\mathrm{HI}$. Moving from bromine to the larger iodine atom generally lengthens the bond and reduces effective overlap with the compact hydrogen $1s$ orbital, so hydrogen iodide is more weakly bound. Independently, the tabulated ground-state harmonic constants near $2649\mathrm{c}{\mathrm{m}}^{-1}$ for hydrogen bromide and $2309\mathrm{c}{\mathrm{m}}^{-1}$ for hydrogen iodide agree with the inferred assignment.^[ref][ref] The ordering is the identification criterion; the low-level Morse oscillator extrapolations above should not be presented as measured dissociation energies.

The initial ozone concentration is $(3\times 10^{-6})[M]=3.60\times 10^{12}\mathrm{molecules}\mathrm{c}{\mathrm{m}}^{-3}$. Over fifty days with twelve active hours per day, the required combined daytime rate isThe mixed chlorine-bromine ozone-loss cycle also consumes two ozone molecules per radical-pair reaction, so $L_{\mathrm{Br}}=2k_6[\mathrm{ClO}][\mathrm{BrO}]$. Subtracting the dimer contribution yieldsThe same result follows by assigning the dimer $1.897$ ppm of the fifty-day loss and the mixed cycle the remaining $1.103$ ppm. This uses the same daylight fraction for both cycles and retains the earlier fixed-ClO approximation. It estimates a sunlit BrO abundance, not the total inventory of bromine in all reservoirs.

For fluorine, the key is chemical sequestration, not an inability of free F to react. Fluorine atoms readily form very stable hydrogen fluoride through reaction with hydrogen-containing gases. Breaking hydrogen fluoride or regenerating a sustained free-fluorine radical population under ordinary stratospheric conditions is difficult, so an efficient analogous catalytic family is not maintained.

For iodine, the traditional approximation is limited delivery to the stratosphere. Many iodine source gases are short lived in the troposphere, and their reactive products are efficiently taken up or removed before slow large-scale ascent. This historically motivated omitting iodine from the major stratospheric chlorine/bromine budget. It does not establish that iodine atoms are intrinsically ineffective catalysts, or that stratospheric iodine is exactly zero.

Indeed, the blanket omission needs qualification even for this exam's date. Aircraft observations published in 2020 detected both IO and particulate iodine in stratospheric air and inferred continuing multiphase recycling and ozone loss. Gas/particle partitioning and measurement sensitivity must be considered together; a small gas-phase concentration alone can miss an active reservoir. Thus the traditional supply argument is the intended contrast with fluorine's stable-HF sink, but zero stratospheric iodine chemistry is not supported by those observations.^[ref]

Prefer a replacement that releases little reactive chlorine or bromine into the stratosphere: ideally omit those elements, or give the parent gas a short troposphere removal time compared with transport to the stratosphere. A carbon–hydrogen bond can permit rapid attack by the hydroxyl radical; other readily oxidized functionality can also shorten the atmospheric residence time. The degradation products must themselves avoid transporting long-lived reactive halogen to altitude. Merely retaining fewer chlorine atoms is less effective than preventing their delivery altogether.

Both chlorine and bromine participate in catalytic ozone destruction. Bromine is much more effective per delivered atom, by a factor of roughly 60 in the usual lower-stratospheric comparison; the precise effectiveness depends on chemical partitioning and location. Fluorine is largely trapped in very stable hydrogen fluoride and does not sustain an analogous ozone-destroying cycle. Thus short-lived gases without chlorine or bromine have the smallest direct ozone-depletion potential. Fluorination alone does not ensure a small climate impact.^[ref]