Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B2 d Solution 2026-10-20
copper(I) converts the methyl grignard reagent into a softer organocopper nucleophile. Methyl undergoes 1,4 addition to the beta carbon of the cycloheptenone, giving a metal enolate rather than direct 1,2 addition to the ketone carbonyl.
The carbon nucleophilic site of this enolate, at the alpha carbon, attacks the aldehyde carbonyl in the second reagent. Its C=O electrons move to oxygen; protonation of the resulting alkoxide gives the beta-hydroxy ketone side chain. The cyclic acetal is a protected ketone and survives these nucleophilic conditions.
The beta methyl group shields one face of the cyclic enolate. Aldol bond formation from the less crowded opposite face places the new alpha side chain trans to the beta methyl group. Thus the two starred ring centers have an anti relationship; the two mirror-related products are possible from achiral starting materials and reagents. This argument does not assign the separate, unstarred alcohol center, for which the question gives no specified configuration.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B3 c Solution 2026-10-20
Number the nitrogen N1, the phenyl-bearing carbon C2, and then continue through C3–C6 around the piperidine. Call the solid-wedge epoxide face up; the phenyl group is down in both starting structures. The bulky phenyl group favors a pseudoequatorial orientation.
epoxide opening follows the Fürst-Plattner preference for a chair-like, trans-diaxial opening trajectory. Attack at the two epoxide carbons is not equivalent: the favored route evolves toward an ordinary chair, whereas the alternative demands a less favorable twist/boat-like distortion. Such conformational control of epoxy-2-arylpiperidine opening is supported by synthetic studies.[ref]
1,3-Allylic Strain as a Strategic Diversification Element for Constructing Libraries of Substituted 2-Arylpiperidines by Thomas C. Coombs and others. Scheme 1, chair-like trans-diaxial epoxide opening; applied here with the free-NH conformer
For the first epoxide, azide attacks C5 from down, breaking the C5–O bond. The residual C4–O bond is retained, so work-up gives C5–N3 down and C4–OH up, with the original C2–Ph down. For the second, down-face epoxide, azide attacks C4 from up, leaving C4–N3 up and C5–OH down. These assignments are shown in the middle row of the original stereochemical map.
In each mechanism an azide terminal nitrogen attacks the epoxide carbon, the C–O bonding pair goes to oxygen, and protonation gives the alcohol. Each attacked center inverts, and the azide and OH are trans. The phenyl-bearing center is unchanged. Reversing the epoxide face changes which carbon has the favored axial attack trajectory; a steric rule based only on distance from phenyl misses that conformational effect.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B3 d i Solution 2026-10-20
Number the cyclohexane consecutively, putting the top methylene group at C1 and the two original epoxides at C2/C3 and C5/C6. Both epoxide oxygens are up.
A is a diazidocyclohexanediol, with each epoxide replaced by one azide and one alcohol. A standard successive chair-favored opening gives N3 down at C2 and C5, OH up at C3 and C6. This is the 1,4-diazido/1,4-diol connectivity drawn in the bottom row of the original map.
For each opening, azide attacks from the face opposite oxygen, breaking one C–O bond with inversion at the attacked carbon. protonation of the two resulting alkoxides gives the two OH groups. After the first opening, the preferred conformation of the opened portion must be used for the second trans-diaxial opening; simply imposing one unchanged chair on both steps can select the wrong regioisomer.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 1 B4 c Solution 2026-10-20
The first step is a vinylogous aldol condensation at C5 of the unsaturated lactam. Number nitrogen N1, the lactam carbonyl C2, and then the methoxy-substituted alkene C3=C4 and the methylene group C5.
hydroxide reversibly removes a C5 proton. The resulting vinylogous enolate is delocalized through C5–C4–C3–C2–O: it is not necessary to create an unstabilized isolated carbanion beside nitrogen. Its C5 nucleophilic contributor attacks acetone. protonation gives a C5-linked tertiary alcohol, . Further removal of the remaining C5 hydrogen and elimination through an aldol-dehydration sequence forms C5=C(Me)2, leaving the original methoxy enol ether in place.
In the second step, aqueous acid hydrolyses that enol ether to a ketone at C4. protonation at C3 gives an oxygen-stabilized cation at C4. water attacks C4, proton transfers form a hemiacetal-like intermediate, and loss of methanol followed by deprotonation produces C4=O and the C3 methylene group.
Past exam of the chemistry B course of the University of Cambridge 2021 Paper 2 B10 a ii Solution 2026-10-20
For the three elementary transformations just proposed, a qualitative free-energy profile has three transition states and two intermediates. The sugar substituents and indole C3 substituent stay unchanged, so the structural shorthand below suppresses them while retaining the reacting atoms.
Qualitative three-step free-energy profile with intermediate and transition-state connectivity
. Details. The first transition state has a partially broken anomeric C1–O(leaving group) bond and developing oxocarbenium character, . Its following minimum is the glycosyloxocarbenium ion plus counterion. The second transition state has a forming C1–C2 bond, , and developing positive charge in the indole. Its following minimum is the covalently joined arenium ion with an intact C2–H bond. The third transition state has simultaneous C2–H cleavage, B–H formation and restoration of the indole pi bond, . It leads to the aromatic C-linked product and protonated base.
The vertical separations are illustrative because no activation free energies or equilibrium constants are supplied. In particular, the highest barrier cannot be identified experimentally from this drawing alone. protonation of the leaving group is treated as an initial rapid equilibrium in the chosen mechanism; explicitly resolving it would add the corresponding proton-transfer state and intermediate.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B2 b Solution 2026-10-20
Number the cyclopentenone carbonyl carbon , its alkene carbons and with OR at , and its saturated carbons and . The ester alkene has alpha carbon , bearing Me and CO2Et, and beta carbon , bearing the protected carbon substituent. lithium diisopropylamide generates the ordinary ketone enolate at the saturated alpha carbon . The acetonide is retained.
The carbon end of this enolate attacks ester beta carbon Y. Its pi system shifts through X onto the ester oxygen, forming the first new bond and an ester enolate at X. That enolate now attacks the original cyclopentenone beta carbon intramolecularly, giving the second new bond . The original enone electrons move onto its ketone oxygen, and protonation of the final ketone enolate completes the product. The OR substituent is retained at the new bridgehead; it need not leave in either Michael addition.
The atom map provides a useful structural check. Between bridgeheads and , the three paths are , , and . Their numbers of internal atoms are 2, 1 and 2, establishing the norbornane skeleton. Both original carbonyls survive and both initial alkenes are consumed. The two successive conjugate additions give this atom map without requiring a concerted cycloaddition.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B2 c Solution 2026-10-20
The new bridge is made by an intramolecular Michael addition. On the cyclohexanone side, the methyl-bearing fused alpha carbon is quaternary and cannot be deprotonated there. triethylamine can instead generate a small equilibrium population of the enolate at the other, methylene alpha carbon. Its carbon attacks the beta carbon of the enone in the seven-membered ring. The enone pi electrons move towards its carbonyl and then onto oxygen, creating the new bridge and a second enolate.
protonation of that enolate at the former methyl-substituted enone alpha carbon restores the left-hand ketone. The initial enone C=C has therefore become a saturated C–C bond while one new transannular C–C bond forms. Both ketones are retained, and the initial two-ring framework becomes a three-ring framework without changing the total index of hydrogen deficiency. The methyl originally at the fused quaternary centre does not migrate; the other methyl ends up on a protonated, formerly unsaturated carbon. A strong-base concentration large enough for quantitative enolate generation is unnecessary for a favourable intramolecular trapping reaction, and neither an aldol dehydration nor a methyl shift is needed.
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 B3 b Solution 2026-10-20
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.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B4 a Solution 2026-10-20
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]
Felkin-Anh and Cram Chelate by University of Windsor. pp. 1–2, section 1, nonpolar alpha substituents and the small-group trajectory
Original Newman construction of the preferred Felkin–Anh approach and the local stereochemistry corresponding to product A
. Details. Past exam of the chemistry B course of the University of Cambridge 2022 Paper 1 B4 c Solution 2026-10-20
protonation activates the oxygen of the anomeric acetal bridge. Cleavage of its bond to the anomeric carbon releases that tether oxygen as the terminal primary alcohol, while the other ring oxygen stabilizes the developing anomeric electrophile by lone pair donation. The resulting oxocarbenium description has a nearly planar reactive anomeric centre; it need not be drawn as a long-lived free carbocation.
methanol attacks that centre, giving a protonated methyl acetal, and deprotonation regenerates acid. The former bridge becomes CH2OH; the anomeric substituent becomes OMe. The other C–O bonds and existing carbon configurations are retained. The ring can adopt a different chair once the bridge constraint is removed, so an apparent movement of substituents in the projection does not imply inversion at all the other carbons.
The depicted anomeric OMe is axial in the product chair. This geometry allows an endocyclic oxygen lone pair to align antiperiplanar to the exocyclic C–OMe bond, providing stabilization associated with the anomeric effect. Acidic conditions also permit cleavage and recapture at the anomeric centre, so selection is not simply backside displacement at an unchanged tetrahedral starting carbon. The anomeric effect rationalizes the specified axial product; the scheme alone gives neither its numerical equilibrium population nor evidence that every other anomer is absent.
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 c ii Solution 2026-10-20
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 . 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.
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 b ii Solution 2026-10-20
Number the cyclopentanone ring at its carbonyl, then at the methyl-bearing carbon, at the mesylate-bearing carbon, at the carbon joined to cyclohexane , and at the remaining methylene. B has a enone and the enamine. triethylamine removes the hydrogen at while the –O bond breaks to release mesylate. An E2 elimination, with approximately antiperiplanar geometry in the puckered ring, gives the conjugated cyclopentenone. Elimination towards instead would give a nonconjugated alkene.
The enamine carbon now attacks enone beta carbon intramolecularly. nitrogen donates into 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 ; the first was . These two connections account for the bridged skeleton without a concerted cycloaddition or an aldol step. The formula of B is : relative to A, elimination removes overall, namely methanesulfonic acid after proton transfers.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B3 c Solution 2026-10-20
E, with the axial anomeric acetate, hydrolyses faster. The tert-butyl substituent strongly favours the chair in which it is equatorial, so the two substrates cannot freely interchange axial and equatorial acetate simply by an inexpensive ring inversion. protonation activates the anomeric acetoxy leaving group. As its C–O bond cleaves to release acetic acid, a lone pair on the endocyclic oxygen donates into the developing anomeric orbital, producing an oxocarbenium ion. water attacks that carbon, and deprotonation gives the cyclic hemiacetal. Attack from either face accounts for the unspecified anomeric configuration of OH in the product.
The axial C–OAc bond in E can adopt antiperiplanar geometry relative to an endocyclic oxygen lone pair. This favourable interaction assists departure in the ionization transition state. The corresponding bond in F is poorly aligned in its preferred chair; achieving comparable overlap requires an unfavourable chair change, placing tert-butyl axial, or substantial ring distortion. The distinction is stereoelectronic control of acetal hydrolysis, rather than merely counting steric crowding in the products.
An axial acetal can itself receive ground-state stabilization from the anomeric effect, so “axial means unstable” is not an adequate rate argument. What matters is the relative Gibbs energy of activation for the departing bond and the geometry of lone-pair assistance. Conformationally locked model acetals experimentally demonstrate faster acid-catalysed cleavage of the axially oriented isomer, although their numerical rate ratios should not be transferred to these different substrates.[ref]
Stereoelectronic control of the hydrolysis of a conformationally locked acetal by Anthony J. Kirby and Robert J. Martin. 1983, abstract and pp. 1627–1632, acid-catalysed axial/equatorial comparison
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B3 d Solution 2026-10-20
With triethylamine present, proton transfer from thiophenol produces a more nucleophilic thiolate population. sulfur attacks the anomeric epoxide carbon C1 from the face opposite its epoxide C–O bond. Displacement by backside attack opens the epoxide, breaks C1–O, and leaves the former epoxide oxygen attached to C2. protonation of that alkoxide by thiophenol or triethylammonium gives G: SPh at C1 on the front face and OH at C2 on the back face, a trans relationship. This is inversion at C1 with retention of the C2 C–O bond; there is no need to invert every stereocentre. Rapid protonation also suppresses competing silyl transfer by the alkoxide.
Without added base, neutral thiophenol is less nucleophilic. acid-assisted cleavage at the anomeric epoxide bond can instead be accompanied by donation from the ring oxygen, giving an oxocarbenium ion with charge shared with the ring oxygen. thiophenol capture from the face that puts SPh back on the original epoxide face, followed by deprotonation, gives H with SPh and C2–OH both on the back face, the net-retention pathway. An ion-pair or highly asynchronous opening can provide this outcome; the drawing does not establish a long-lived free carbocation.
The distinction is configuration, not a mandatory fixed chair. For the closely matching trisilylated substrate, the primary study obtained both anomers with thiophenol alone, and established an unusual inverted chair for the alpha thioglycoside by NMR and crystallography. Thus H should be explained as an accessible product, not asserted to be the sole product or necessarily to have axial SPh in its preferred conformation.[ref]
Reaction of thiophenol with glucal epoxides: X-ray structure of 3,4,6-tri-O-tert-butyldimethylsilyl-1-S-phenyl-1-thio-α-D-glucopyranoside by Caroline Walford and others. 1997, Schemes 1–2 and Figure 1; neutral reaction mixture and configuration/conformation assignment
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B4 a ii Solution 2026-10-20
Both permitted alternatives follow the same condensation–activation–migration sequence. sodium acetate liberates a useful concentration of hydroxylamine from its hydrochloride while providing a buffered medium. nitrogen of NH2OH attacks the ketone carbonyl; proton transfer gives a carbinolamine. protonation of its carbon-derived OH and loss of water, with nitrogen lone-pair donation, then gives the oxime after deprotonation. The two C=N geometries can be separated.
For the B-to-D route, oxime O-tosylation begins when oxime oxygen attacks sulfur of tosyl chloride and chloride leaves; triethylamine accepts the released proton. The resulting O-tosyl oxime has an excellent leaving group. The bond, anti to N–OTs, migrates to nitrogen while N–O cleaves, giving a ring-expanded nitrilium ion. Its local connectivity is , with the existing scaffold still joining the two paths. water attacks the nitrilium carbon . proton transfers and imidic-acid/amide tautomerization produce D, with nitrogen inserted on the aromatic side of the old carbonyl.
For the C-to-E route, the same O-tosylation activates the other oxime geometry. Now the anti bond is ; its migration gives , and aqueous trapping produces E, with nitrogen inserted on the spiro-junction side. Each Beckmann rearrangement expands the ketone's six-membered ring to a seven-membered lactam. The original carbonyl carbon remains the amide carbonyl carbon; nitrogen is the inserted atom. In this mechanism the ketone oxygen is lost during oxime condensation, the oxime oxygen leaves with the sulfonate group, and water supplies the new amide oxygen.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 1 B5 a iii Solution 2026-10-20
Transfer is favourable when new water–solute interactions compensate the cavity and reorganization costs of accommodating the solute. The Gibbs energy also includes the loss of solute–alkane interactions; it is not simply a count of hydrogen bonds. isobutane cannot compensate those costs with strong directional interactions, explaining its positive transfer Gibbs energy. acetone and both amines can accept hydrogen bonds from water and have favourable electrostatic solvation, making their transfer moderately favourable. isopropylamine can additionally donate to water, consistent with a somewhat greater preference for water than trimethylamine.
acetamide combines a strong carbonyl acceptor with N–H donors and has the most favourable hydration in the group, accounting for its much more negative transfer Gibbs energy. Its pure solid lattice is not an extra sublimation penalty to insert here: the stated process starts from a dilute alkane solution and ends in dilute water, rather than dissolving the pure crystal. The nonmonotonic relationship between hydrogen-bond acceptor parameter and transfer Gibbs energy is expected because donor strength, nonpolar surface, solvent cavity formation, and all other solvation contributions also enter.
For matched standard states, , with the water-to-alkane activity ratio. As an illustrative 298 K calculation, the hydrocarbon value implies , whereas the amide value implies approximately . These are standard-state partition preferences, not absolute aqueous solubilities. amine protonation could introduce an additional pH-dependent contribution to an experimentally measured distribution coefficient, but no such speciation information is supplied and it is not needed for the molecular comparison.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 1 B2 a Solution 2026-10-20
In the strongly acidic nitrating mixture, the dimethylamino nitrogen is predominantly protonated. The substituent is therefore during aromatic substitution, rather than the neutral electron-donating group. Its lone pair is unavailable for conjugation. Ortho or para attack would generate a sigma-complex contributor with positive ring charge at the carbon attached to an already positive nitrogen; meta attack avoids that especially unfavorable contributor.
Sulfuric acid protonates nitric acid, and loss of water produces the nitronium ion. Aromatic attack at a meta position forms a Wheland intermediate; deprotonation restores aromaticity. The bicarbonate workup then returns the amino group to its neutral form. is 3-nitro-N,N-dimethylaniline.
There are six inequivalent aromatic carbons in this unsymmetrical meta-disubstituted ring and one carbon environment for the two equivalent N-methyl groups, giving seven carbon NMR signals. The para isomer would have only four aromatic environments plus the methyl environment. An ortho isomer could also give seven signals, so signal counting supports but does not alone establish the meta assignment; protonation-dependent directing effects are essential.




