Aromatic Compounds and Reactions Questions and Answers

Answer: C

The major product is (C), indole-3-carbaldehyde.

DMF and POCl₃ form the Vilsmeier reagent, a chloroiminium ion. Indole is attacked by electrophiles at C-3, the most nucleophilic position, because the cation formed keeps the benzene ring intact and is stabilised by the nitrogen lone pair. The iminium intermediate is hydrolysed by aqueous NaOH to the aldehyde.

(A) and (B) are intermediates or side forms before hydrolysis, not the product after aqueous base. (D) puts the formyl group on the benzene ring at C-4, which is far less reactive than C-3.

Answer: A

P is (A), acetophenone phenylhydrazone.

This is the Fischer indole synthesis. With ZnCl₂ the phenylhydrazone tautomerises to an ene-hydrazine, which undergoes a [3,3]-sigmatropic rearrangement forming a new C-C bond to the ortho position of the aryl ring. Cyclisation and loss of NH₃ then give the aromatic indole. The methyl of acetophenone becomes C-3 (as CH) and the phenyl-bearing carbon becomes C-2, so the product is 2-phenylindole.

Propiophenone phenylhydrazone (D) would give 3-methyl-2-phenylindole. An azo compound (B) and an anil without the N-N bond (C) cannot lose NH₃ in this way.

Answer: A

(A), 1-bromo-2,4-dinitrobenzene, reacts fastest.

Methoxide replaces bromide by nucleophilic aromatic substitution: OMe⁻ adds to the carbon bearing Br, forming a negatively charged Meisenheimer complex, and Br⁻ is then lost. The addition is rate-determining, so anything that stabilizes the negative charge speeds the reaction. Electron-withdrawing groups ortho and para to the leaving group do this by taking the charge onto themselves.

(A) has two nitro groups, ortho and para, the strongest such combination. (B) has a nitro group ortho but a cyano group para; cyano withdraws less strongly than nitro, so it is slower. (C) has only one nitro group, and the para methyl donates electrons and destabilizes the anion. (D), with a single cyano group and a para methyl, is the least activated.

Answer: B

The product is (B), 3-(2-cyanoethyl)-1-methylindole: the silyl group is replaced.

Indoles react with electrophiles at C3, the most nucleophilic position. Acrylonitrile is a Michael acceptor, and here C3 attacks its terminal carbon. C3 already carries the SiMe₃ group, so attack occurs at that carbon (ipso attack). The resulting cation is stabilized by the C-Si bond next to it (the β-silicon effect), and it restores aromaticity by losing the trimethylsilyl group instead of a proton. The net result is substitution of SiMe₃ by CH₂CH₂CN.

(A) puts the new group at C2 while keeping the silyl group, but C2 is less nucleophilic and the silicon directs reaction to its own carbon. (C) and (D) would need the much less nucleophilic benzene ring to react while the activated C3 position is left untouched.

Answer: B

The product is (B), with NHR meta to OMe.

Diazotisation of an anthranilic acid gives the benzenediazonium-2-carboxylate, which loses N₂ and CO₂ to form benzyne. Here the methoxy group ends up next to one end of the triple bond: 3-methoxybenzyne.

The amine adds to the benzyne carbon farther from OMe. That puts the developing aryl anion on the carbon next to OMe, where the electronegative oxygen stabilises it inductively; protonation then gives the 3-methoxy (meta) product. Addition to the near carbon would place the anion away from OMe and give the ortho product (A), the minor pathway. (C), the para product, cannot come from a 2,3-benzyne at all, and (D) would need two additions.

Answer: D

The answer is (D): M is 1,2-dimethyl-1,4-cyclohexadiene and N is 1,2-dimethylcyclohexene.

Birch reduction (Na in liquid NH₃ with an alcohol) reduces a benzene ring to a non-conjugated 1,4-cyclohexadiene. With electron-donating alkyl groups, protonation occurs at the unsubstituted positions, so the methyl-bearing carbons stay on a double bond: M is 1,2-dimethyl-1,4-cyclohexadiene.

Wilkinson's catalyst is very sensitive to steric hindrance and hydrogenates the less substituted double bond first. In M, one C=C is disubstituted (CH=CH) and the other tetrasubstituted, carrying both methyls. Only the CH=CH is reduced, leaving N = 1,2-dimethylcyclohexene.

(B) has the right M but reduces the wrong double bond. (A) and (C) give conjugated dienes, which the Birch reduction does not give.

Answer: B

The antiaromatic species is (B), the cycloheptatrienyl anion.

A planar, fully conjugated ring is aromatic with 4n + 2 π electrons and antiaromatic with 4n. The cycloheptatrienyl anion has three C=C (6 π electrons) plus the lone pair on the carbanion carbon: 8 π electrons, a 4n count, so if planar it would be antiaromatic. It is accordingly very hard to form.

The others are aromatic. The cycloheptatrienyl (tropylium) cation (A) has 6 π electrons. Azulene (C) has 10 π electrons around its perimeter. Indole (D) has 10 π electrons, including the nitrogen lone pair.

Answer: A

The products are 5-nitroquinoline and 8-nitroquinoline.

In concentrated acid quinoline is protonated on nitrogen. The quinolinium ion's pyridine ring is strongly deactivated, so the nitronium ion attacks the benzene ring instead. Positions 5 and 8, the carbons next to the ring fusion, give the most stable cationic intermediates (each keeps the pyridinium ring intact and allows more resonance forms), so nitration gives a roughly equal mixture of 5- and 8-nitroquinoline.

Positions 3 and 4 are on the deactivated heterocyclic ring and are not attacked under these conditions, which rules out options C and D. Position 6 gives a less stabilised intermediate than 5, which rules out option B.

Answer: A

The keyed product is 9,10-dibromo-9,10-dihydrophenanthrene, option A.

The 9,10 bond of phenanthrene has much more double-bond character than any other bond, because both its carbons belong to only one of the three rings and breaking it leaves two intact benzene rings. Br₂ in CCl₄ therefore adds across it like an alkene (via a bromonium ion) rather than substituting on a benzene ring; no Lewis acid such as Fe or AlCl₃ is present, so ring substitution (B and D, at C-3 and C-2) is not expected.

9-Bromophenanthrene (C) is what the dibromide gives on heating, by loss of HBr, and is the product when the reaction is run hot; the stem names no heating, so the addition product is the answer the key takes.

Answer: A

The product is the oxygen-bridged adduct in A, a methyl-substituted 1,4-dihydro-1,4-epoxynaphthalene.

Sodium amide in liquid ammonia removes H-Br from the p-bromotoluene (the proton ortho to Br goes with Br), making a benzyne, 4-methylbenzyne, by elimination. Benzyne is a very strained, reactive dienophile; furan, a diene, traps it in a Diels-Alder reaction. The two new bonds join the benzyne carbons to C-2 and C-5 of furan, so the oxygen becomes a one-atom bridge and the former C-3=C-4 of furan remains as a C=C in the new six-membered ring, exactly as in A.

B has no C=C in the bridged ring, which would need two extra hydrogens. C and D alter the benzene ring (it is no longer aromatic), which the reaction cannot do.