VII. AROMATIC COMPOUNDS TEST 1 1 / 20 Which side-chain carbon remains attached to the ring in the benzoic acid product? C₆H₅CH₂CH₂CH₃; hot aqueous KMnO₄, then H₃O⁺ No side-chain carbon The original benzylic carbon The terminal methyl carbon All three as a propyl group A new carbon from KMnO₄ Solution: Benzylic oxidation retains the ring-attached side-chain carbon as CO₂H; more remote carbons are oxidized away. A. Incorrect. The carboxyl carbon derives from the benzylic carbon. B. Correct. Benzylic oxidation retains the ring-attached side-chain carbon as CO₂H; more remote carbons are oxidized away. C. Incorrect. The chain does not reverse its attachment. D. Incorrect. Strong oxidation shortens the side chain. E. Incorrect. Permanganate supplies no carbon. 2 / 20 Which count supports aromaticity in its planar conjugated ring? Cyclopentadienyl anion: five sp2 carbons, two C=C bonds and one lone pair 8 pi electrons 6 pi electrons 5 pi electrons 4 pi electrons 10 pi electrons Solution: Four electrons from two pi bonds plus the anionic lone pair give 4n+2 with n=1. A. Incorrect. The lone pair contributes two, not four. B. Correct. Four electrons from two pi bonds plus the anionic lone pair give 4n+2 with n=1. C. Incorrect. Electrons are not counted one per carbon automatically. D. Incorrect. This omits the participating lone pair. E. Incorrect. Sigma electrons should not be included. 3 / 20 Which classification follows the stated geometry? Hypothetical planar cyclooctatetraene with uninterrupted p overlap Aromatic, 8 pi electrons Nonaromatic solely because carbon is present Aromatic because every even count qualifies Antiaromatic, 8 pi electrons No pi electrons because the bonds are alternating Solution: A planar continuous 4n loop has antiaromatic character; actual tub-shaped COT avoids this premise. A. Incorrect. Eight is not 4n+2. B. Incorrect. Carbon rings can be aromatic. C. Incorrect. The Hückel distinction is more specific. D. Correct. A planar continuous 4n loop has antiaromatic character; actual tub-shaped COT avoids this premise. E. Incorrect. Four C=C bonds contribute eight pi electrons. 4 / 20 Why does counting four pi electrons not establish antiaromaticity of the whole ring? Cyclopenta-1,3-diene; one ring CH₂ carbon The molecule has no double bonds Four pi electrons always mean aromatic Every ring carbon is sp The saturated CH₂ interrupts cyclic conjugation The ring must lose a carbon Solution: The ring lacks an uninterrupted p-orbital loop and is nonaromatic. A. Incorrect. It has two. B. Incorrect. Four is a 4n count, but geometry and conjugation also matter. C. Incorrect. The CH₂ is sp3. D. Correct. The ring lacks an uninterrupted p-orbital loop and is nonaromatic. E. Incorrect. No atom loss is needed for classification. 5 / 20 Which organic product follows ordinary Friedel–Crafts acylation? A. B. C. D. E. Solution: The acylium-derived electrophile installs an acetyl group while aromaticity is restored. A. Incorrect. The acyl chloride does not supply methoxide. B. Incorrect. Hydrolysis of reagent is not the intended ring acylation product. C. Incorrect. The reaction transfers the acyl group. D. Incorrect. That would require carbonyl reduction after acylation. E. Correct. The acylium-derived electrophile installs an acetyl group while aromaticity is restored. 6 / 20 Which statement correctly compares their N lone pairs? Pyrrole: four pi electrons from C=C plus N lone pair; pyridine: three C=C Pyrrole's pair enters the sextet; pyridine's in-plane pair does not Neither lone pair can ever affect aromaticity Pyridine has eight aromatic pi electrons including its lone pair Both lone pairs enter the sextet in the same way Pyrrole has no nitrogen lone pair Solution: The two heterocycles reach aromatic sextets using different nitrogen orbital contributions. A. Correct. The two heterocycles reach aromatic sextets using different nitrogen orbital contributions. B. Incorrect. Pyrrole requires its lone pair. C. Incorrect. That incorrectly adds the in-plane pair. D. Incorrect. Pyridine's lone pair is outside its pi sextet. E. Incorrect. Its neutral N has one participating pair. 7 / 20 Which positional ring-nitration product is favored? A. B. C. D. E. Solution: The electron-withdrawing acyl group destabilizes ortho/para sigma-complex contributors involving the ipso region. A. Incorrect. Acyl groups are ordinarily meta directors. B. Incorrect. These are electrophilic aromatic nitration conditions. C. Correct. The electron-withdrawing acyl group destabilizes ortho/para sigma-complex contributors involving the ipso region. D. Incorrect. Ordinary nitration replaces a ring H. E. Incorrect. This confuses withdrawing carbonyl substitution with a donating alkyl group. 8 / 20 Which substrate has the usual structural requirement for benzylic side-chain oxidation? C₆H₅C(CH₃)₃ versus C₆H₅CH₂CH₃; hot KMnO₄/H₂O Only C₆H₅C(CH₃)₃ Neither because aromatic rings prevent all oxidation C₆H₅CH₂CH₃ Only a substrate with an O-H group Both because total side-chain H count alone controls it Solution: It has benzylic H atoms; the tert-butyl benzylic carbon has none. A. Incorrect. That reverses the benzylic-H criterion. B. Incorrect. Suitable side chains oxidize. C. Correct. It has benzylic H atoms; the tert-butyl benzylic carbon has none. D. Incorrect. An initial O-H bond is not required. E. Incorrect. Hydrogens must be benzylic in this standard rule. 9 / 20 Which step restores aromaticity after electrophile addition? C₆H₆ + Br₂/FeBr₃; no light, controlled monobromination Hydride addition to the ring Removal of H⁺ from the sigma complex Addition of a second Br to every ring carbon Loss of an entire CH unit Homolysis of every C-C bond Solution: Deprotonation regenerates the aromatic pi system and gives bromobenzene. A. Incorrect. That does not restore the substitution product. B. Correct. Deprotonation regenerates the aromatic pi system and gives bromobenzene. C. Incorrect. That would not be the ordinary substitution pathway. D. Incorrect. The ring carbon skeleton is retained. E. Incorrect. No such cleavage occurs. 10 / 20 Which description accommodates fluorine's two effects? C₆H₅F; HNO₃/H₂SO₄; para nitration isomer requested Activating because F is electropositive Activating and meta directing Deactivating overall, yet ortho/para directing Deactivating and meta directing in every case No EAS can occur at any conditions Solution: Inductive withdrawal lowers rate while lone-pair donation stabilizes selected sigma complexes. A. Incorrect. Fluorine is strongly electronegative. B. Incorrect. Both descriptions are wrong here. C. Correct. Inductive withdrawal lowers rate while lone-pair donation stabilizes selected sigma complexes. D. Incorrect. Halogens are the standard exception. E. Incorrect. Deactivation does not mean absolute impossibility. 11 / 20 Why is the ring meta position favored? C₆H₅CO₂CH₃; Br₂/FeBr₃, controlled substitution The directly attached carbonyl group withdraws electron density Bromine must replace the ester methyl group Every oxygen-containing group directs ortho/para The ester adds a new aromatic ring The methoxy group is directly attached to the ring Solution: The ring is bonded to C=O carbon, not to the ester oxygen. A. Correct. The ring is bonded to C=O carbon, not to the ester oxygen. B. Incorrect. This is ring EAS. C. Incorrect. Attachment and resonance direction matter. D. Incorrect. No ring is added. E. Incorrect. The actual attachment is through carbonyl carbon. 12 / 20 Which product carbon becomes stereogenic? C₆H₆ -> C₆H₅COCH₂CH₃ -> alcohol; CH₃CH₂COCl/AlCl₃, then NaBH₄/H₂O The reagent chloride atom Every aromatic carbon The terminal methyl carbon None, because all secondary alcohols are achiral The former carbonyl carbon Solution: Reduction gives C₆H₅CH(OH)CH₂CH₃ with four distinct substituents. A. Incorrect. Chloride leaves and is not a carbon center. B. Incorrect. They remain sp2. C. Incorrect. A CH₃ carbon has three identical H atoms. D. Incorrect. This alcohol has different carbon groups. E. Correct. Reduction gives C₆H₅CH(OH)CH₂CH₃ with four distinct substituents. 13 / 20 Which feature is preserved through the reduction? C₆H₅COCH₃ -> meta-nitro ketone -> alcohol; nitration then mild NaBH₄ The side-chain carbon remains planar The nitro group migrates para The molecule loses the nitro group necessarily The meta relationship of NO₂ to the side chain The C=O bond Solution: Reducing C=O changes the side-chain group but does not move its ring attachment. A. Incorrect. It becomes tetrahedral. B. Incorrect. No ring rearrangement is involved. C. Incorrect. Mild NaBH₄ is used to reduce the ketone selectively. D. Correct. Reducing C=O changes the side-chain group but does not move its ring attachment. E. Incorrect. It is reduced to C-OH. 14 / 20 Why does oxidation before nitration change the directing effect? C₆H₅CH₂CH₃ -> C₆H₅CO₂H -> mononitro acid An alkyl donor becomes a carbonyl-containing withdrawing group Oxidation makes the ring saturated The ring gains a nitrogen before nitration Oxidation removes the ring entirely Both substituents must direct identically because carbon is attached Solution: The benzoic acid substituent is meta directing, unlike the starting alkyl group. A. Correct. The benzoic acid substituent is meta directing, unlike the starting alkyl group. B. Incorrect. The aromatic ring remains. C. Incorrect. Nitrogen is added only in nitration. D. Incorrect. The product retains benzene. E. Incorrect. Their electronic effects differ. 15 / 20 Where is deuterium introduced? C₆H₆ -> aryl ketone -> secondary alcohol; acylation then NaBD₄/H₂O In the departing chloride At all aromatic carbons At the former carbonyl carbon In a new methyl carbon Only at the OH proton Solution: Deuteride delivery forms C-D, while aqueous workup supplies the O-H proton. A. Incorrect. Chloride is not part of the alcohol. B. Incorrect. Hydride reduction does not deuterate the ring generally. C. Correct. Deuteride delivery forms C-D, while aqueous workup supplies the O-H proton. D. Incorrect. Deuterium is hydrogen, not carbon. E. Incorrect. That proton comes from workup and is exchangeable. 16 / 20 Which factor can favor para among the ortho/para products? C₆H₅C(CH₃)₃; controlled nitration; compare ortho and para access Nitronium is a radical Para has a different molecular formula from ortho A meta-directing carbonyl on the starting ring Less steric crowding near the bulky substituent Loss of all methyl groups before every nitration Solution: The alkyl group directs ortho/para, and para attack avoids close crowding at an ortho site. A. Incorrect. The ordinary electrophile is NO₂⁺. B. Incorrect. Regioisomers share the formula. C. Incorrect. No carbonyl is present. D. Correct. The alkyl group directs ortho/para, and para attack avoids close crowding at an ortho site. E. Incorrect. That is not required. 17 / 20 Which electronic feature favors ortho/para attack? C₆H₅OCH₃; one equivalent Br₂, AcOH, controlled monobromination The methyl carbon must become a carbocation first The ring has no pi electrons Oxygen lone-pair donation into the ring A benzylic C-H bond is required Oxygen withdraws only by resonance Solution: Methoxy donation stabilizes ortho/para sigma-complex contributors. A. Incorrect. That is not the EAS activation step. B. Incorrect. Benzene has a pi sextet. C. Correct. Methoxy donation stabilizes ortho/para sigma-complex contributors. D. Incorrect. Methoxy bromination occurs on the ring. E. Incorrect. Methoxy can donate by resonance. 18 / 20 Which product property follows in an achiral medium? meta-NO₂-C₆H₄-COCH₂CH₃; NaBH₄/MeOH, mild A racemic benzylic alcohol with the meta nitro group retained A deoxygenated alkylbenzene A nitrile at the side chain Only one enantiomer by definition A para-nitro alcohol by migration Solution: Reduction creates one stereocenter at the carbonyl carbon without a chiral source. A. Correct. Reduction creates one stereocenter at the carbonyl carbon without a chiral source. B. Incorrect. NaBH₄ retains O as OH. C. Incorrect. No nitrile-forming reagent is present. D. Incorrect. No enantioselective reagent is specified. E. Incorrect. Reduction does not relocate NO₂. 19 / 20 Why can this route avoid an alkyl-cation rearrangement problem? C₆H₆ -> C₆H₅COCH₂CH₃ -> C₆H₅CH₂CH₂CH₃; acylation then Zn(Hg)/HCl, heat Friedel-Crafts acylation uses a freely rearranging propyl cation The ring is completely hydrogenated Acylation installs a nonrearranging acyl framework before carbonyl reduction Zinc removes one chain carbon The carbonyl reduction adds a fourth carbon Solution: The acyl carbon becomes CH₂, giving the straight side chain. A. Incorrect. The electrophile is acyl-derived. B. Incorrect. Clemmensen reduction targets the carbonyl. C. Correct. The acyl carbon becomes CH₂, giving the straight side chain. D. Incorrect. The carbonyl carbon is retained. E. Incorrect. It adds no carbon. 20 / 20 Which combination identifies the retained side-chain carbon and preferred nitration family? C₆H₅CH₂CH₃ -> benzoic acid -> nitration Terminal carbon retained; para nitration Benzylic carbon retained; meta nitration Both side-chain carbons retained; ortho only A new oxidant carbon retained; meta No side-chain carbon retained; ipso only Solution: Oxidation converts the attached carbon to CO₂H, which then directs meta. A. Incorrect. Both atom origin and direction are wrong. B. Correct. Oxidation converts the attached carbon to CO₂H, which then directs meta. C. Incorrect. Oxidation shortens the side chain. D. Incorrect. The oxidant contains no carbon. E. Incorrect. The carboxyl carbon remains attached. Your score isThe average score is 0% 0% Restart quiz VII. AROMATIC COMPOUNDS TEST 2 1 / 20 What changes between the aryl-ketone intermediate and the final alcohol? C₆H₆ + CH₃COCl/AlCl₃, then NaBH₄ Oxygen leaves as CO₂ Ring carbon count increases The side chain becomes a carboxylate Carbonyl carbon changes from sp2 to sp3 The aromatic ring must become cyclohexane Solution: Hydride adds to C and protonation gives C-OH while the aryl-acyl C-C bond remains. A. Incorrect. Oxygen remains in OH. B. Incorrect. Reduction adds no carbon. C. Incorrect. That would be a different oxidation state. D. Correct. Hydride adds to C and protonation gives C-OH while the aryl-acyl C-C bond remains. E. Incorrect. The ring remains aromatic. 2 / 20 Which order-sensitive prediction is justified? C₆H₅COCH₂CH₃; nitrate first, then reduce ketone Nitration deletes the carbonyl carbon Reduction must occur spontaneously before nitration NO₂ always appears para regardless of substituent Nitration occurs under the ketone's meta-directing influence The final alcohol's directing effect determines the earlier nitration Solution: The directing group present during nitration controls the position; later reduction does not retroactively change it. A. Incorrect. Ordinary EAS retains the side chain. B. Incorrect. No reducing reagent is present in the first stage. C. Incorrect. Directing effects matter. D. Correct. The directing group present during nitration controls the position; later reduction does not retroactively change it. E. Incorrect. Later chemistry does not set an earlier regioisomer. 3 / 20 Which positions are jointly favored by the two directing effects? p-Cl-C₆H₄-CHO; CHO at C1, Cl at C4; mononitration All six positions inequivalent C2 and C6 C1 only C4 only C3 and C5, equivalent by symmetry Solution: They are meta to CHO and ortho to Cl; symmetry makes them one constitutional product family. A. Incorrect. The para-disubstituted framework retains symmetry. B. Incorrect. These are ortho to CHO and meta to Cl. C. Incorrect. C1 already bears CHO and ordinary nitration replaces H. D. Incorrect. C4 already bears Cl. E. Correct. They are meta to CHO and ortho to Cl; symmetry makes them one constitutional product family. 4 / 20 Which earlier molecular feature is no longer relevant after complete side-chain oxidation? C₆H₅CH₂CH₂CH₃; oxidize to CO₂H, then nitrate The carboxyl group's meta direction Original remote alkyl-chain length for the acid's constitution The six ring carbons Presence of the aromatic ring The location of the ring-substituent bond Solution: Different oxidizable alkyl lengths converge to the same ring-bound CO₂H group. A. Incorrect. That remains important. B. Correct. Different oxidizable alkyl lengths converge to the same ring-bound CO₂H group. C. Incorrect. They remain in the product. D. Incorrect. The ring still defines the substrate. E. Incorrect. It is retained through oxidation. 5 / 20 Which product reflects the favored benzylic C–H substitution? A. B. C. D. E. Solution: Abstraction at the benzylic carbon gives a resonance-stabilized radical; bromination replaces a benzylic hydrogen. A. Incorrect. Terminal-chain bromination lacks the resonance stabilization available at the benzylic position. B. Incorrect. This is ring bromination rather than the favored benzylic radical substitution. C. Correct. Abstraction at the benzylic carbon gives a resonance-stabilized radical; bromination replaces a benzylic hydrogen. D. Incorrect. Meta ring bromination is not the favored NBS/light pathway. E. Incorrect. Para ring bromination is characteristic of a different aromatic-substitution setting, not the requested benzylic pathway. 6 / 20 Why is the resulting alcohol carbon not stereogenic? C₆H₆ + C₆H₅COCl/AlCl₃; then NaBH₄/H₂O It has no hydrogen It is still planar sp2 Oxygen prevents all stereogenicity It bears two identical phenyl groups Both rings are lost during reduction Solution: Reduction gives (C₆H₅)₂CHOH with duplicate substituents. A. Incorrect. Benzhydrol has a C-H bond there. B. Incorrect. The alcohol carbon is tetrahedral. C. Incorrect. O-bearing carbons can be stereogenic. D. Correct. Reduction gives (C₆H₅)₂CHOH with duplicate substituents. E. Incorrect. The phenyl groups are retained. 7 / 20 Which sequence of functional roles is correct? C₆H₆ -> C₆H₅COCH₃ -> meta-NO₂ derivative -> alcohol Three carbon-carbon cleavage steps Nitrile formation followed by hydrolysis Three consecutive radical substitutions Hydride electrophile followed by nitro nucleophile Acyl electrophile, nitronium electrophile, hydride nucleophile Solution: Acylation and nitration are EAS; carbonyl reduction is nucleophilic hydride addition. A. Incorrect. The carbon framework is largely retained. B. Incorrect. No nitrile is involved. C. Incorrect. These are not the stated mechanisms. D. Incorrect. The reagent roles are reversed. E. Correct. Acylation and nitration are EAS; carbonyl reduction is nucleophilic hydride addition. 8 / 20 Which ring substituent controls nitration in the last stage? C₆H₅CH₂CH₃ -> benzoic acid -> ethyl ester -> nitration A new ring-bound chromium group -C(=O)OCH₂CH₃ The original ethyl group No substituent because esterification removes CO₂H carbon -OCH₂CH₃ directly attached to the ring Solution: The ring remains bonded to carbonyl carbon; esterification does not turn it into a directly attached ethoxy donor. A. Incorrect. Chromium is not part of the isolated ester. B. Correct. The ring remains bonded to carbonyl carbon; esterification does not turn it into a directly attached ethoxy donor. C. Incorrect. It has already been oxidized. D. Incorrect. The acyl carbon stays on the ring. E. Incorrect. The actual connection is through C=O. 9 / 20 Which carbon accounting is correct for the final organic molecule? C₆H₆ -> C₆H₅COCH₂CH₃ -> meta-NO₂ ketone -> alcohol Eight carbons Twelve carbons Nine carbons Six carbons Ten carbons Solution: Six ring carbons plus three acyl-derived carbons remain; nitration and reduction add none. A. Incorrect. One acyl carbon has been omitted. B. Incorrect. A second aromatic ring is not supplied. C. Correct. Six ring carbons plus three acyl-derived carbons remain; nitration and reduction add none. D. Incorrect. The side chain is retained. E. Incorrect. No extra carbon enters during nitration or reduction. 10 / 20 Which stage adds two new carbons to the isolated tracked organic molecule? C₆H₅CH₂CH₃ -> acid -> ethyl ester -> nitro ester Every stage adds two carbons Nitration Acid workup Esterification with EtOH Permanganate oxidation Solution: The two ethanol carbons become the ethoxy group; nitration adds nitrogen and oxygen. A. Incorrect. Only ethanol incorporation does. B. Incorrect. NO₂ contains no carbon. C. Incorrect. H₃O⁺ supplies no carbon. D. Correct. The two ethanol carbons become the ethoxy group; nitration adds nitrogen and oxygen. E. Incorrect. It removes remote side-chain carbons rather than adding carbon. 11 / 20 Which final signal combination is compatible with the intended product? C₆H₆ -> acetophenone -> meta-nitroacetophenone -> alcohol Ester OCH₃ plus acid OH Nitrile band plus terminal alkyne H Aldehydic H plus no aromatic protons Only alkane signals and no heteroatoms Aromatic protons plus CH(OH)-CH₃ fragment Solution: Ketone reduction gives a benzylic secondary alcohol while retaining aromaticity and nitro substitution. A. Incorrect. No esterification occurs. B. Incorrect. Neither group is formed. C. Incorrect. Neither feature matches the final alcohol. D. Incorrect. O and N remain. E. Correct. Ketone reduction gives a benzylic secondary alcohol while retaining aromaticity and nitro substitution. 12 / 20 Where is ¹³C in the final ester? C₆H₅-¹³CH₂-CH₃ -> acid -> ethyl ester -> nitration At the ester carbonyl carbon In the nitro group At the ester ethoxy CH₂ Distributed among all ring carbons At the ethoxy CH₃ Solution: Benzylic oxidation converts the labeled carbon to CO₂H; later stages preserve its identity. A. Correct. Benzylic oxidation converts the labeled carbon to CO₂H; later stages preserve its identity. B. Incorrect. The nitro group contains no carbon. C. Incorrect. Those carbons come from ethanol. D. Incorrect. No scrambling is specified. E. Incorrect. That carbon also comes from ethanol. 13 / 20 Which statement separates constitutional and stereochemical predictions? C₆H₆ -> C₆H₅COCH₃; nitrate, then NaBH₄ Every racemate contains two different carbon skeletons Nitration alone creates the benzylic alcohol center Meta and para are enantiomers R/S formation moves NO₂ from meta to para Meta ring substitution is retained, and reduction can create an R/S pair Solution: Ring regiochemistry and carbonyl-face selection are separate decisions. A. Incorrect. Enantiomers share connectivity. B. Incorrect. The center forms on reduction. C. Incorrect. They are constitutional regioisomers. D. Incorrect. Stereochemistry does not relocate substituents. E. Correct. Ring regiochemistry and carbonyl-face selection are separate decisions. 14 / 20 Which workup prevents carrying a benzoate salt forward as though it were neutral acid? C₆H₅CH₂CH₃; hot KMnO₄ then H₃O⁺; nitrate Acidify the oxidation mixture before isolating CO₂H Remove the ring with ozone Evaporate water and assume all salts are neutral Add NaBH₄ instead of acid Add more NaOH only Solution: The oxidation medium can contain carboxylate; acid workup protonates it. A. Correct. The oxidation medium can contain carboxylate; acid workup protonates it. B. Incorrect. That destroys the intended substrate. C. Incorrect. Drying does not remove ionic charge. D. Incorrect. That is not the needed protonation workup. E. Incorrect. That favors carboxylate. 15 / 20 Which product follows the standard activated nucleophilic aromatic substitution? A. B. C. D. E. Solution: Methoxide displaces fluoride at its original ring carbon, leaving nitro para to the new methoxy group. A. Incorrect. The starting aryl fluoride has not undergone substitution. B. Incorrect. Methoxide attacks through oxygen; this ring-methylated fluoride does not replace the activated C–F bond. C. Incorrect. Substitution preserves the relative position of the nitro group; it does not move it to ortho. D. Incorrect. The nitro group does not migrate to meta during displacement. E. Correct. Methoxide displaces fluoride at its original ring carbon, leaving nitro para to the new methoxy group. 16 / 20 What distinguishes the benzyne proposal for the first reaction? Chlorobenzene + NaNH₂/liquid NH₃; compare para-nitrofluorobenzene + methoxide Elimination of HX requires an ortho H before nucleophilic addition It cannot form a new C-N bond It consumes the aromatic carbons as CO₂ It requires a para nitro group in every case It is ordinary backside SN2 at sp2 carbon Solution: Benzyne formation differs from nitro-activated addition-elimination SNAr. A. Correct. Benzyne formation differs from nitro-activated addition-elimination SNAr. B. Incorrect. Amide addition can give aniline. C. Incorrect. The ring skeleton is retained. D. Incorrect. Benzyne does not require that activator. E. Incorrect. Aryl SN2 is not the standard pathway. 17 / 20 In an ideal no-exchange reduction, where is 18O? C₆H₆ -> aryl ketone -> alcohol; carbonyl oxygen is ¹⁸O before NaBH₄/H₂O On a ring carbon In the alcohol OH Necessarily all in water In the chloride leaving group In the newly added C-H bond Solution: Hydride addition retains carbonyl oxygen; protonation gives the alcohol. A. Incorrect. Oxygen does not become carbon. B. Correct. Hydride addition retains carbonyl oxygen; protonation gives the alcohol. C. Incorrect. Exchange is excluded in the stated model. D. Incorrect. The chloride is not the carbonyl oxygen. E. Incorrect. The isotope is oxygen, not hydrogen. 18 / 20 Which correction is needed? C₆H₅COCH₃ -> meta-nitro ketone -> alcohol; creator proposes para-NO₂ final drawing Add one carbon to the methyl group Keep NO₂ meta to the side-chain attachment Add a second NO₂ para Delete the alcohol oxygen Convert the phenyl ring to cyclopropane Solution: The ketone directed the prior nitration; reduction does not migrate NO₂. A. Incorrect. NaBH₄ adds no carbon. B. Correct. The ketone directed the prior nitration; reduction does not migrate NO₂. C. Incorrect. The stated mononitration does not justify this. D. Incorrect. Reduction retains O. E. Incorrect. No such skeletal change occurs. 19 / 20 Which pair of errors would both misrepresent the sequence? C₆H₅CH₂CH₃ -> C₆H₅CO₂H -> nitro acid Acidifying benzoate and retaining the carboxyl carbon Using CO₂H as the directing group during nitration Retaining six aromatic carbons and predicting meta Removing remote side-chain carbon and retaining the ring Retaining the ethyl chain and predicting para solely from that original group Solution: Oxidation changes the side chain to CO₂H before nitration and changes the directing influence. A. Incorrect. Both are appropriate. B. Incorrect. That is correct for the intermediate. C. Incorrect. Both are correct. D. Incorrect. That matches benzylic oxidation. E. Correct. Oxidation changes the side chain to CO₂H before nitration and changes the directing influence. 20 / 20 Which relationship is retained between the two acid groups? m-C₆H₄(CH₂CH₃)₂; hot KMnO₄/H₂O, then H₃O⁺ Meta Ortho Para No acid groups because two side chains block oxidation Geminal on a single ring carbon Solution: Both oxidizable side chains become CO₂H at their existing ring attachment positions. A. Correct. Both oxidizable side chains become CO₂H at their existing ring attachment positions. B. Incorrect. Oxidation does not move attachment sites. C. Incorrect. No ring substitution migration occurs. D. Incorrect. Both have benzylic H atoms. E. Incorrect. The groups began on different carbons. Your score isThe average score is 0% 0% Restart quiz GO TO THE NEXT CHAPTER
