Aromatic Compounds and Reactions Questions and Answers

Answer: A

The product is (A), 2-chloro-1-(2,4-dichlorophenyl)ethanone.

Chloroacetyl chloride with anhydrous AlCl₃ forms the acylium ion ClCH₂C≡O⁺, which acylates the ring (Friedel-Crafts acylation). Chlorine is deactivating but ortho/para directing. In 1,3-dichlorobenzene, position 4 is para to one chlorine and ortho to the other, so both chlorines direct the acyl group there; position 2, between the chlorines, is also activated by both but is too hindered. Substitution at C4 puts the new COCH₂Cl group with chlorines at 2 and 4 of the product ring.

(B), the 3,5-dichloro product, would need attack meta to both chlorines. (D) is attack at the hindered C2. (C) is an arylacetic acid, which acylation does not give.

Answer: A

The product is (A), 1-acetyl-2-naphthol (1-(2-hydroxynaphthalen-1-yl)ethanone).

Heating an aryl ester with anhydrous AlCl₃ is the Fries rearrangement: the Lewis acid complexes the ester oxygen, the acyl group leaves as an acylium ion and re-attacks the ring ortho or para to the oxygen, and aqueous acid work-up frees the phenol (here a naphthol). In 2-naphthyl acetate the most reactive position is C1, the alpha position ortho to the oxygen, since attack there keeps a complete benzene ring in the intermediate. The acetyl group therefore ends up at C1, next to the OH at C2.

(B) puts the acetyl group at C3, the less reactive ortho position. (C) and (D) move it to the other ring, which the oxygen does not activate.

Answer: A

The Birch reduction gives (A), 1-methoxy-5-methylcyclohexa-1,4-diene.

Lithium in liquid ammonia (with an alcohol as proton source) reduces a benzene ring to a non-conjugated cyclohexa-1,4-diene. With electron-donating substituents such as OMe and Me, protonation of the radical anion occurs away from them, so the substituted carbons stay on the double bonds of the product. In 3-methylanisole both substituted carbons therefore remain sp², which is option (A): one double bond carries OMe and the other carries Me, separated by CH₂ groups.

(B) is a conjugated 1,3-diene, which the Birch reduction does not give. In (C) and (D) one substituted carbon has become sp³, the pattern expected for electron-withdrawing groups, not donors.

Answer: A

The product is 5-bromo-2-methoxypyridine, A.

Pyridine itself resists electrophilic substitution, but the 2-methoxy group donates electrons into the ring and activates the positions ortho and para to it, C3 and C5. Bromination under these mild conditions occurs at C5, para to OMe, where attack is also less hindered than at C3.

C4 (option B) and C6 (option C) are not activated by the 2-methoxy group, and C6 is adjacent to the deactivating ring nitrogen. Option D replaces the methoxy group by bromine, which is not an electrophilic substitution at all.

Answer: D

The product is 2-(5-bromopyridin-2-yl)-2-methylpropanenitrile, D.

LDA deprotonates isobutyronitrile at the carbon α to the nitrile, giving a tertiary carbanion. This attacks 2,5-dibromopyridine by nucleophilic aromatic substitution. The C2 position, next to the ring nitrogen, is activated because the negative charge of the addition intermediate can rest on nitrogen; C5 has no such activation. So the bromine at C2 is replaced and the one at C5 stays.

A has lost both bromines and is attached at C3. B and C are ketimines, which would need the carbanion to add to a nitrile carbon instead of the ring, and C also has the wrong ring position.

Answer: D

The answer is (D): X is 1-methoxycyclohexa-1,4-diene and Y is cyclohex-2-enone.

Birch reduction of anisole (Na, liquid NH₃, EtOH) adds hydrogen at C-2 and C-5, leaving the electron-donating OMe on a remaining double bond. The product is the non-conjugated 1-methoxycyclohexa-1,4-diene, the X drawn in (A) and (D).

This product is an enol ether. Mild acid hydrolysis would give the non-conjugated cyclohex-3-enone, but with HCl, water and heat the double bond moves into conjugation with the carbonyl, giving the more stable α,β-unsaturated cyclohex-2-enone. That is the Y of (D); the Y of (A) is the non-conjugated ketone, and (B) and (C) start from the wrong diene.

Answer: B

The major product is (B).

Sodium in liquid ammonia with ethanol does two things. It reduces the internal alkyne by dissolving-metal reduction through a vinyl radical anion, which adopts the less hindered trans geometry, so the alkyne becomes an E-alkene. And it carries out a Birch reduction of the benzoic acid ring: with an electron-withdrawing carboxyl group, hydrogen is added at C-1 and C-4, so the carboxyl-bearing carbon becomes sp³ and the ring is a 1,4-diene.

(B) has both features: the E-alkene side chain and COOH on the sp³ carbon of a cyclohexa-2,5-diene. (A) has the Z-alkene that Lindlar hydrogenation would give, and (C) and (D) put the carboxyl group on a double bond, the pattern seen for rings with electron-donating groups.

Answer: A

The major product is (A), 1-benzyl-3-nitroindole.

Indole is an electron-rich heterocycle that undergoes electrophilic substitution at C-3. Attack there gives a cation in which the positive charge is stabilised by the nitrogen lone pair without disturbing the aromaticity of the benzene ring. Concentrated nitric acid with acetic anhydride at -70 °C generates acetyl nitrate, a mild nitrating agent that avoids the acid-promoted polymerisation of indole and delivers the nitro group to C-3.

Attack at C-2 gives a less stabilised intermediate, and nitration of the benzene ring (C-4 or C-7) happens only under strongly acidic conditions where C-3 is protonated and deactivated, which these conditions avoid.

Answer: B

The major product is (B), 1,6-dimethyl-1,2,3,4-tetrahydroquinoline.

PhLi removes the proton ortho to bromine that lies between Br and the side chain, and loss of LiBr gives a benzyne. The amide nitrogen at the end of the three-carbon chain then attacks the aryne carbon next to the chain, closing a six-membered ring, and the resulting aryl anion is protonated.

The new C-N bond forms at the former aryne carbon between the old Br and chain positions. The ring methyl is para to that carbon, which is C-6 of the tetrahydroquinoline. Options A, C and D place the methyl at C-8, C-7 or C-5, which would need a different aryne or a different point of attack.

Answer: C

The major product is (C), PhCO-C(OEt)=CH₂.

This is a carbonylative Stille coupling. Pd(0) inserts into the C-I bond of iodobenzene, CO inserts into the Pd-Ph bond to give an acyl-palladium species, the 1-ethoxyvinyl group transfers from tin to palladium, and reductive elimination joins the acyl and vinyl groups, giving an aryl vinyl ketone in which the vinyl carbon still carries OEt.

(A) is the direct Stille product without CO insertion, which the CO atmosphere suppresses. (B) has no carbonyl and no vinyl group, and (D), an ester, would need CO inserted next to oxygen.