Aromatic Compounds and Reactions Questions and Answers

Answer: C

The major product is 5-nitrobenzene-1,3-dicarboxylic acid (5-nitroisophthalic acid), structure (C).

Both carboxyl groups withdraw electrons and deactivate the ring, and each directs an incoming electrophile to the positions meta to itself. In isophthalic acid the only position meta to both carboxyl groups is C5, so the nitronium ion attacks there.

Structure (D) puts the nitro group at C2, ortho to both carboxyl groups, which is electronically disfavoured and sterically crowded. Structure (B) puts it at C4, ortho to one carboxyl and para to the other. Structure (A) is a dinitro product; a strongly deactivated ring does not take a second nitro group under these mild conditions (30 °C), and its positions are wrong as well.

Answer: C

The product is (C), indane-1-carbonitrile.

Sodium amide in liquid ammonia removes HCl from the aryl chloride through a benzyne intermediate. The strong base also removes a proton from the carbon next to the nitrile (the acidic CH₂ that bears CN). That carbanion sits three carbons from the ring and attacks the benzyne carbon next to the chain, closing a five-membered ring. Protonation then gives the bicyclic nitrile with CN on the carbon of the new ring.

(A) and (B) are the products of amide ion adding to the benzyne, which loses to the faster intramolecular carbanion attack. (D) is a ketone, and no step in the sequence converts a nitrile into a ketone.

Answer: C

The product is (C), 3-chloropyridine.

Potassium pyrrolide with chloroform and a base is the Ciamician-Dennstedt ring expansion. The base generates dichlorocarbene from CHCl₃; the electron-rich pyrrole carbon adds to it, giving a dichlorocyclopropane that opens by loss of chloride and expands the five-membered ring to a six-membered one. The new ring carbon carries one chlorine and sits at the 3-position of the pyridine.

The 2-substituted pyridines in (A) and (B) would need the carbene carbon to end up next to nitrogen, which it does not. The ethoxy compounds (B) and (D) would need ethoxide to substitute the chlorine, which does not happen under these conditions.

Answer: B

The sequence is (i) Friedel-Crafts acylation; (ii) Friedel-Crafts alkylation; (iii) Reduction.

The two groups in the product are meta to each other, so the first group introduced must direct the second to the meta position. An acyl group does that; an alkyl group directs ortho/para.

1. Friedel-Crafts acylation with propanoyl chloride gives propiophenone; the acyl group also avoids the rearrangement that direct n-propylation would suffer.
2. Friedel-Crafts isopropylation then places the isopropyl group meta to the acyl group.
3. Clemmensen or Wolff-Kishner reduction turns C=O into CH₂, giving the n-propyl group.

The key rests on this directing-group logic; in the laboratory, alkylating an acyl-deactivated ring is sluggish. Reducing before the second step (option C), or starting with alkylation (A, D), gives an ortho/para product.

Answer: B

The major product W is the 1-methoxycyclohexa-1,4-diene carrying a hydrogen and a deuterium on each of its two CH₂ carbons, which is option (B).

Na in liquid NH₃ with an alcohol is the Birch reduction. The methoxy group is electron donating, so the radical anion is protonated at the ortho and meta positions, leaving the methoxy carbon on a double bond in the product. EtOD is the proton source, so the two new hydrogens are deuterium. Two carbons become sp³: the one ortho to OMe and the one meta to it (para to the first), and each ends up as CHD, because one H was already present and one D is added.

Option (A) puts D on the methoxy carbon and reduces the para positions, which is the pattern for an electron-withdrawing substituent. Options (C) and (D) put the new sp³ carbons on the wrong positions, so they are not the 1,4-diene.

Answer: C

X is 3-chloropyridine and Y is 4-aminopyridine, which is option (C).

Pyrrole with CHCl₃ and KOH reacts with dichlorocarbene (the Ciamician-Dennstedt ring expansion). The carbene adds across the 2,3 C=C bond to give a dichlorocyclopropane that opens by loss of chloride and expands the ring, so the product is 3-chloropyridine, not 2-chloropyridine. With NaNH₂ the aryl chloride loses HCl by elimination to give 3,4-pyridyne, and amide adds to the aryne mainly at C-4, giving 4-aminopyridine (the 3-amino isomer is the minor product).

Options (A) and (D) have 2-chloropyridine, which carbene ring expansion does not give. Option (B) has the right X but a 3-amino product, which is the minor isomer from the pyridyne.

Answer: B

The correct match is P → 1, Q → 2, R → 4, S → 5.

(P) The aldoxime has its OH on the side nearest the ring bromine. Aqueous NaOH forms the oximate, which displaces the activated bromide (it is ortho to the oxime carbon and para to the nitro group) to give a 1,2-benzisoxazole with no substituent at C-3. With a hydrogen at C-3, base opens it (Kemp elimination) to the 2-cyanophenol: (1).

(Q) Acetic anhydride acetylates the oxime OH, and Na₂CO₃ then eliminates acetic acid from the aldoxime acetate to give the nitrile with the bromine unchanged: (2).

(R) The ketoxime closes in the same way, but the methyl group at C-3 prevents ring opening, so the benzisoxazole stays: (4).

(S) This oxime acetate has the O-acetyl group on the side away from the ring, so it cannot cyclize. Aqueous carbonate simply hydrolyses the acetate back to the oxime: (5). The anilide (3) would need an acid-catalysed Beckmann rearrangement.

Answer: B

The product is a mixture of o- and p-chlorotoluene.

With a Lewis acid catalyst (Fe, which forms FeCl₃) and in the dark, chlorine substitutes on the ring by electrophilic aromatic substitution. The methyl group is ortho-para directing, so 2-chlorotoluene and 4-chlorotoluene are formed.

Benzyl chloride and benzal dichloride are side-chain products, formed by free radical chlorination in sunlight or on heating without a catalyst. Di- and trichlorotoluenes need excess chlorine, not the simple monochlorination asked about.

Answer: A

This is the Fittig reaction.

Two molecules of an aryl halide coupled by sodium in dry ether give a biaryl: 2C₆H₅Cl + 2Na → C₆H₅–C₆H₅ + 2NaCl.

The Wurtz-Fittig reaction couples an aryl halide with an alkyl halide to give an alkylbenzene. The Swarts reaction makes alkyl fluorides, Williamson's synthesis makes ethers, and Kolbe's reaction makes salicylic acid from sodium phenoxide and CO₂.

Answer: E

The heterocyclic aromatic compound is Furan.

A heterocyclic compound has at least one atom other than carbon in its ring. Furan is a five-membered ring of four carbon atoms and one oxygen atom. It is aromatic: the four carbon p electrons and one lone pair from oxygen give a planar ring of six delocalised pi electrons, satisfying Huckel's (4n + 2) rule with n = 1.

Phenol, aniline and toluene are aromatic but homocyclic: their ring is benzene, made of carbon only, and the O, N or CH₃ is a substituent outside the ring. Naphthalene has two fused benzene rings, again all carbon.