Questions & explanations
1. Why does nitrobenzene undergo electrophilic substitution predominantly at the meta position?
- Because ortho and para attack lead to an arenium ion with positive charge adjacent to the positively charged nitrogen, which is highly destabilising.
- Because the nitro group donates electrons to the ring, activating the meta position.
- Because the nitro group is an ortho/para director but steric hindrance forces meta substitution.
- Because the nitro group withdraws electrons by resonance, making the meta position the most electron-rich.
Answer: Because ortho and para attack lead to an arenium ion with positive charge adjacent to the positively charged nitrogen, which is highly destabilising.
In nitrobenzene, the nitro group is strongly electron-withdrawing. For ortho or para attack, one resonance structure of the arenium ion places the positive charge on the carbon bearing the nitro group. This carbon is adjacent to the positively charged nitrogen, creating severe electrostatic repulsion and destabilising the intermediate. Meta attack avoids this, so it is favoured.
2. For the nitration of toluene, which statement about the arenium ion intermediates is correct?
- The ortho and para arenium ions have no resonance structure with positive charge on the methyl-bearing carbon.
- The meta arenium ion has a resonance structure with positive charge on the carbon bearing the methyl group.
- All three arenium ions (ortho, meta, para) are equally stabilised by the methyl group.
- The ortho and para arenium ions have a resonance structure with positive charge on the carbon bearing the methyl group.
Answer: The ortho and para arenium ions have a resonance structure with positive charge on the carbon bearing the methyl group.
For toluene, the methyl group is an ortho/para director. In the arenium ion formed by ortho or para attack, one resonance structure places the positive charge on the carbon bearing the methyl group. This structure is stabilised by the +I effect and hyperconjugation of the methyl group, lowering the activation energy for ortho/para attack.
3. Which of the following groups is a strong activator for electrophilic aromatic substitution due to resonance donation of a lone pair?
- -CH3
- -OH
- -NO2
- -Cl
Answer: -OH
The -OH group has a lone pair on the oxygen atom attached to the ring. This lone pair is donated into the ring by resonance (+M effect), providing an extra resonance structure in the arenium ion that places the positive charge on the oxygen. This stabilises the intermediate and activates the ring strongly.
4. Which sequence of reactions converts aniline to p-bromoaniline?
- Acetylation, bromination, hydrolysis
- Bromination, acetylation, hydrolysis
- Sulphonation, bromination, desulphonation
- Nitration, reduction, bromination
Answer: Acetylation, bromination, hydrolysis
Aniline is too reactive and gives tribromoaniline on direct bromination. Acetylation protects the -NH2 group as -NHCOCH3, which is less activating but still ortho/para-directing. Bromination then gives p-bromoacetanilide. Hydrolysis with dilute acid removes the acetyl group to yield p-bromoaniline.
5. To prepare m-bromonitrobenzene from benzene, which order of reactions is correct?
- Nitration followed by sulphonation
- Bromination followed by nitration
- Sulphonation followed by nitration
- Nitration followed by bromination
Answer: Nitration followed by bromination
To get m-bromonitrobenzene, the meta-directing group must be installed first. Nitration of benzene gives nitrobenzene, which is meta-directing. Subsequent bromination then gives m-bromonitrobenzene. If bromination is done first, the ortho/para-directing bromine leads to p-bromonitrobenzene.
6. Starting from benzene, which sequence of reactions gives m-nitrobenzoic acid?
- Nitration, then Friedel-Crafts methylation, then oxidation
- Friedel-Crafts methylation, then oxidation, then nitration
- Friedel-Crafts acylation, then nitration, then reduction
- Oxidation, then nitration, then Friedel-Crafts alkylation
Answer: Friedel-Crafts methylation, then oxidation, then nitration
First, Friedel-Crafts methylation of benzene gives toluene. Oxidation of the methyl group with hot KMnO4 yields benzoic acid, which has a meta-directing -COOH group. Finally, nitration with conc. HNO3/H2SO4 gives m-nitrobenzoic acid. This order respects directing effects.
7. Which compound does NOT give benzoic acid on oxidation with hot KMnO4?
- Toluene
- Ethylbenzene
- tert-Butylbenzene
- p-Xylene
Answer: tert-Butylbenzene
Side-chain oxidation requires at least one benzylic hydrogen. tert-Butylbenzene has a quaternary benzylic carbon with no hydrogen, so hot KMnO4 cannot oxidize it. Toluene, ethylbenzene, and p-xylene all have benzylic hydrogens and give benzoic acid or terephthalic acid.
8. Which of the following monocyclic systems is aromatic according to the four-part test?
- Pyridine (C5H5N)
- Cyclobutadiene (C4H4)
- Cyclopentadiene (C5H6)
- Cyclohexane (C6H12)
Answer: Pyridine (C5H5N)
Pyridine is cyclic, planar, every ring atom (5 C and 1 N) has a p-orbital in the pi system, and it has 6 pi-electrons (each double bond contributes 2, and the N lone pair is in an sp2 orbital in the plane, not counted). Thus it satisfies all four criteria (4n+2, n=1).
9. Which compound does NOT undergo Friedel-Crafts alkylation with CH3Cl/AlCl3?
- Chlorobenzene
- Toluene
- Nitrobenzene
- Benzene
Answer: Nitrobenzene
Nitrobenzene has a strong deactivating -NO2 group that makes the ring too electron-deficient to attack the electrophile. Friedel-Crafts alkylation fails on rings with -NO2, -CN, -SO3H, -CHO, -COR, -COOH. Toluene, chlorobenzene, and benzene all undergo the reaction.
10. Nitration of tert-butylbenzene gives which product as the major one?
- o-nitro-tert-butylbenzene
- m-nitro-tert-butylbenzene
- 2,4-dinitro-tert-butylbenzene
- p-nitro-tert-butylbenzene
Answer: p-nitro-tert-butylbenzene
tert-Butyl group is ortho/para-directing due to hyperconjugation and inductive effect. However, its bulk causes severe steric hindrance at ortho positions, making para attack strongly favored. Thus p-nitro-tert-butylbenzene is the major mononitration product.
11. Which compound does NOT give benzoic acid on oxidation with hot KMnO4?
- Toluene
- tert-Butylbenzene
- Ethylbenzene
- Propylbenzene
Answer: tert-Butylbenzene
Side-chain oxidation requires at least one benzylic hydrogen. tert-Butylbenzene has no hydrogen on the benzylic carbon, so it does not undergo oxidation. Toluene, ethylbenzene, and propylbenzene all have benzylic hydrogens and are oxidised to benzoic acid.
12. What is the major product when p-cresol is nitrated with dilute HNO3 at room temperature?
- 2-nitro-4-methylphenol
- 3-nitro-4-methylphenol
- 2,4-dinitro-4-methylphenol
- 4-methyl-2,6-dinitrophenol
Answer: 2-nitro-4-methylphenol
p-Cresol has both -OH (strong activator, ortho/para director) and -CH3 (weak activator, ortho/para director). The -OH dominates. Since the para position is blocked by -CH3, nitration occurs ortho to -OH, giving 2-nitro-4-methylphenol as the major product.