Aromatic nitration 2026-10-20
Aromatic nitration replaces an aromatic hydrogen with a nitro group. In a common pathway, aromatic attack on a nitronium ion forms a Wheland intermediate and deprotonation restores aromaticity.
Nitrobenzene 2026-10-20
Nitrobenzene is benzene bearing a nitro group. The nitro group withdraws electron density and stabilizes certain anionic aromatic addition intermediates.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B2 a Solution 2026-10-20
The five-carbon unit is initially an open chain, not a completed carbocycle. Label it , with phenyl-bearing , carbonyl , amide-substituted , and the two ethylthio groups on . Its double bonds are initially and . Let be the nitro-bearing carbon of nitroethane; its methyl remains a substituent throughout.
DBU removes an alpha proton from nitroethane to give a nitronate. carbon attack at the Michael acceptor forms , with electron redistribution through onto the carbonyl oxygen. proton transfer restores the ketone. Removing the remaining X–H proton gives another carbon-nucleophilic nitronate. It attacks the other conjugated end, , intramolecularly, closing the six-membered circuit . Addition to the ketene dithioacetal and ethylthiolate elimination restore the bond, leaving one SEt group at .
base-assisted loss of the proton at , with C–NO2 cleavage at X, creates and expels nitrite. Enolization at supplies the third aromatic double bond and the phenolic OH. proton transfers to the leaving anions complete the acid–base balance. nitroethane supplies the new ring carbon X and its methyl substituent; its nitro group is eliminated. Aromatization, rather than a separate oxidation of an intact six-membered starting ring, provides the final driving force. This is a proposed stepwise nitroalkane benzannulation; related DBU-promoted tandem Michael addition/cyclization/aromatization has direct primary precedent.[ref]
Original atom maps for the open-chain benzannulation, the two bonds of the norbornanone-forming tandem Michael addition, and the intramolecular Michael closure
. Details. Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B2 d Solution 2026-10-20
Let and number the carbon substituent from oxygen: . sodium hydride removes a proton from C3, alpha to the nitrile, forming hydrogen and a nitrile-stabilized carbon anion. That carbon attacks the nitroaryl ipso carbon bearing O. The tether closes a five-membered cyclic Meisenheimer intermediate with sequence C3–C2–C1–O–aryl ipso carbon. The para nitro group stabilizes the negative charge generated by temporary loss of aromaticity.
Rearomatization breaks the aryl–O bond, releasing the tether oxygen as an alkoxide while retaining the new aryl–C3 bond. protonation gives . This is a Truce–Smiles rearrangement: the aryl group migrates from oxygen to carbon. The nitrile remains attached to its original side-chain carbon; the original ether oxygen becomes the terminal alcohol oxygen. The substrate and product have identical atomic composition. Ordinary substitution at a saturated chain carbon would not produce this aryl–carbon bond or restore the terminal OH.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B4 d Solution 2026-10-20
triethylamine removes an alpha proton from nitromethane to generate a nitronate population. Its carbon attacks the cyclic ketone carbonyl of E, forming a new C–CH2NO2 bond while the C=O pi pair moves onto oxygen. protonation by triethylammonium or another proton donor gives the tertiary alcohol F and regenerates base. This is a Henry reaction, also called a nitroaldol reaction: the former ketone carbon now bears both OH and CH2NO2.
The nitro group remains attached to its original nitromethane carbon; attack through oxygen would give a different connectivity. The methyl acetal and the original alcohol substituents need not be cleaved or displaced. This addition produces the isolated connectivity without requiring dehydration to a nitroalkene. The stated complex mixture is compatible with competing additions or stereoisomers, but the scheme supplies no basis for claiming F is the unique or kinetically dominant product. Only its formation is being explained.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B1 a Solution 2026-10-20
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 . It confirms replacement of one ring hydrogen rather than chlorine, but directing effects, rather than the formula alone, establish the position.
Independently drawn aromatic substitution products and the sites selected by protodebromination
. Details. Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B2 a Solution 2026-10-20
The first reagent supplies a vinyl nucleophile, , rather than an allyl group. Two successive conjugate additions install vinyl at C2 and a 2-formylethyl chain at C1. Number the original nitro-bearing alkene carbon C1. Vinylmagnesium bromide attacks the beta carbon C2 of the nitroalkene, shifting the original C1=C2 electrons into a magnesium nitronate. The new C2–vinyl bond remains intact during the second stage; its terminal double bond is not the electrophilic site in that stage.
The carbon end of the nitronate at C1 then attacks the terminal beta carbon of acrolein. Move the acrolein C=C electrons towards its carbonyl and the C=O electrons onto oxygen. This forms the C1–CH2CH2CHO connectivity through an aldehyde enolate; protonation by methanol and tautomerization restore the aldehyde. The nitronate returns to a neutral nitro group as it makes the second C–C bond. Thus the chain is attached through the carbon formerly beta to the acrolein carbonyl, rather than by a direct addition to its aldehyde carbon.
After the first addition, C2 bears the vinyl group while the nitronate carbon C1 remains approximately planar. Approach of the second electrophile to the less crowded face, opposite the adjacent vinyl substituent, gives the new formylethyl chain trans to vinyl. Consequently nitro and vinyl are cis, as depicted: at C1 the chain and nitro group occupy opposite faces. This is a substrate-controlled diastereoselective tandem Michael addition. The reagents and starting material contain no chiral source, so the depicted product represents a racemic pair rather than an enantioselective synthesis. Steric facial selection explains the relative configuration without requiring a uniquely established magnesium aggregate or transition-state conformation.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B2 c iii Solution 2026-10-20
base removes an alpha-amide proton from G, giving a carbon nucleophile. Its intramolecular attack on the nitroaryl ipso carbon, which currently bears the ether oxygen, forms a five-membered cyclic Meisenheimer complex. Count its atoms along the existing tether: alpha carbon, the original phenyl carbon bearing the side chain, its adjacent phenoxy carbon, ether oxygen, and attacked nitroaryl ipso carbon. The ortho nitro group again stabilizes the temporarily dearomatized anion.
Rearomatization now breaks the nitroaryl–O bond, leaving phenoxide on the original ring. proton transfer gives . This is a Truce-Smiles rearrangement: the nitroaryl group migrates from oxygen to carbon, while the original phenoxy oxygen stays on its original ring and becomes OH. No nitro-group migration or amide cleavage is needed. G and H have identical formulas; only the bond connections and a proton position change. carbonate need not produce a high equilibrium enolate concentration: a small population can be consumed by favourable intramolecular cyclization and rearomatization. The specified transformation alone does not establish its detailed proton-transfer sequence.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B2 c ii Solution 2026-10-20
G is the diaryl ether . Here denotes the nitrogen of the intact morpholine ring; the phenoxy oxygen and phenylacetamide chain are ortho on their original ring. potassium carbonate first generates phenoxide from F. Its oxygen attacks the fluorine-bearing carbon of 1-fluoro-2-nitrobenzene, forming a Meisenheimer complex. Delocalization of its negative charge into the ortho nitro group stabilizes this addition intermediate. Fluoride elimination restores aromaticity and completes nucleophilic aromatic substitution.
This first step forms an O–aryl bond, rather than immediately arylating the less acidic alpha carbon. The neutral product has formula , obtained from the two substrates by net loss of HF. The connectivity is shown beside B in the independently drawn diagram above. The amide-substrate O-to-C migration strategy has direct primary precedent.[ref]
Developing the Scope of O→C Aryl Migrations: Exploring Amide Substrates as Potential Precursors for Asymmetric Reactions by Dana Ameen and Timothy J. Snape. 2014, pp. 1925–1934, amide substrates and five-membered rearrangement pathways
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B2 b ii Solution 2026-10-20
The same base first makes . 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 . protonation during acidic workup yields .
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]
Reaction of organic anions. Part 108. On the mechanism of the vicarious nucleophilic substitution of hydrogen in nitroarenes by Mieczysław Mąkosza and Tomasz Glinka. 1983, pp. 3860–3861, isotope effects, base dependence and hydrogen/halogen competition
The displayed para product is consistent with access to the less crowded activated position. Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B2 b i Solution 2026-10-20
tert-Butoxide removes the acidic hydrogen from . 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 . 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.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B2 b ii Solution 2026-10-20
The qualitative nucleophilicity order is -OMe -Me H -Cl -NO, matching the product-yield order at the stated stopping time. Electron donation raises the availability of the aniline lone pair for attack. A para methoxy group donates by resonance, while methyl is a weaker donor. Chlorine withdraws inductively and gives an aniline less nucleophilic than the unsubstituted compound. A para nitro group strongly withdraws by resonance: electron flow from nitrogen through the ring into the nitro group stabilizes a charge-separated form and greatly reduces nucleophilicity.
The key arrows for the nitroaniline comparison move the nitrogen lone pair into its own aromatic ring, shift the ring bonds toward the para nitro group, and shift a nitro N=O bond toward oxygen. The methoxy donor produces the opposite electronic bias. These are finite-time yields, so they support a kinetic explanation qualitatively; they are not by themselves measured rate constants or equilibrium constants.
Past exam of the chemistry B course of the University of Cambridge 2025 Paper 1 B2 b i Solution 2026-10-20
The aniline nitrogen attacks the ipso carbon bearing fluorine. The aromatic electrons shift to give a Meisenheimer complex, with negative charge delocalized to the positions ortho and para to the attacked carbon. At either nitro substituent, that charge can be delocalized further onto nitro oxygen. Subsequent proton transfer and fluoride elimination restore aromaticity and give the diarylamine.
Both the ortho and para nitro groups stabilize the addition intermediate in 1-fluoro-2,4-dinitrobenzene. The para-mononitro compound lacks the additional ortho stabilization, so addition is less favourable and generally slower under the stated conditions. This comparison concerns stabilization of the addition pathway; the presence of fluorine does not make the reaction an ordinary aliphatic SN2 reaction.


