Unit 20: Aromatic Hydrocarbons — Short Questions

C₆H₆ + Cl₂ --AlCl₃--> C₆H₅Cl + HCl

AlCl₃ is a Lewis acid; it accepts a lone pair of electrons from one chlorine atom of Cl₂, polarising and then breaking the Cl-Cl bond heterolytically. This generates the strongly electrophilic Cl⁺ species (with AlCl₄⁻ as the counter-ion), which is reactive enough to attack the stable, electron-rich but delocalised benzene ring. Without the halogen carrier, Cl₂ alone is not a strong enough electrophile to react with benzene.

Step 1 (generation of electrophile) AlCl₃ attacks Cl₂ with a lone pair, shown with a curly arrow from a lone pair on one Cl of Cl⁻Clᵟ⁺ (the Cl nearer AlCl₃ carries δ-, the far Cl carries δ+) to Al; the Cl-Cl bond breaks heterolytically (curly arrow from the bonding pair onto the departing Cl), giving Cl⁺ and AlCl₄⁻. Step 2 (electrophilic attack): a curly arrow from the delocalised π electron cloud of benzene points to Cl⁺, forming a new C-Cl σ bond and a positively charged arenium ion (sigma complex) in which the ring is no longer fully aromatic and the + charge is delocalised over the remaining five carbons. Step 3 (restoration of aromaticity): a curly arrow from the C-H bond on the sp³ carbon points into the ring to re-form the delocalised π system, while another arrow shows the H⁺ being removed by AlCl₄⁻, regenerating AlCl₃ and releasing HCl, giving chlorobenzene.

Electrophilic (aromatic) substitution.

Delocalised π electrons are π electrons that are not fixed between just one pair of atoms (as in a localised, single C=C or C=O double bond) but are spread out (shared) over three or more atoms in a conjugated system. This happens because the unhybridised p-orbitals on the conjugated atoms are all parallel and overlap sideways with one another, so the π electron density forms a continuous cloud above and below the plane of the atoms rather than being confined to one bond.

CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻

This is electrophilic (aromatic) substitution — specifically Friedel-Crafts acylation. The delocalised π electrons of benzene attack the electrophilic carbon of CH₃CO⁺, forming a new C-C σ bond and a positively charged arenium ion (sigma complex) in which aromaticity is temporarily lost. A base (AlCl₄⁻) then removes the H⁺ from the sp³ carbon; the C-H electrons move back into the ring to restore the delocalised aromatic system, the H⁺ combines with Cl⁻ to form HCl, and AlCl₃ is regenerated. The product is phenylethanone, C₆H₅COCH₃.

2-Hydroxybenzoic acid (2-hydroxybenzenecarboxylic acid), common name salicylic acid. COOH, being the highest-priority substituent, is numbered C-1.

1-Iodo-2-nitrobenzene. Since iodo and nitro are of similar (non-priority-list) status, the carbon bearing the substituent earlier in the alphabet (iodo before nitro) is numbered C-1, giving the lower locant set {1,2}.

1-Bromo-3,5-dimethylbenzene.

Each carbon in the benzene ring is sp² hybridised. The three sp² orbitals on each carbon form σ bonds — two to neighbouring ring carbons and one to a hydrogen atom — all lying in the same plane with bond angles of 120°, giving a planar, regular hexagonal σ-framework. [Diagram: hexagon of six sp² carbons, each bonded to one H, with C-C-C and H-C-C angles of 120°.] Each carbon also has one unhybridised p-orbital perpendicular to this plane; these six p-orbitals, all parallel to each other, overlap sideways above and below the ring to form a continuous π-electron cloud in which the six π electrons are shared (delocalised) equally over all six carbons. [Diagram: hexagon with a circle inscribed inside it, the circle representing the delocalised π system, drawn as two doughnut-shaped electron clouds above and below the ring plane.] This delocalisation makes all six C-C bonds identical in length (139 pm), intermediate between a single and a double bond.

In phenol, one of the lone pairs on the oxygen of the -OH group occupies a p-orbital that is aligned parallel to, and overlaps with, the p-orbitals of the ring's π system. This lone pair becomes partly delocalised into the ring, increasing the electron density of the ring (particularly at the ortho and para positions). Because benzene's bromination already requires a Lewis acid (AlBr₃) to polarise Br₂ into a strong enough electrophile, the much more electron-rich ring of phenol can polarise Br₂ itself, without a catalyst, and reacts far more readily with the electrophile than the comparatively electron-poor, unsubstituted benzene ring does.

Iodobenzene is the main product. Bromine is more electronegative than iodine, so in the I-Br bond the iodine atom carries the partial positive charge (Iᵟ⁺-Brᵟ⁻). Since the benzene ring (a nucleophile/electron donor) attacks the more electrophilic, δ+ atom, it is the iodine end that bonds to the ring, with Br⁻ (as HBr after deprotonation) being the leaving group.

C₆H₆ + IBr → C₆H₅I + HBr

Step 1 (generation of electrophile) H₂SO₄ + HNO₃ → NO₂⁺ + HSO₄⁻ + H₂O. Here H₂SO₄ protonates HNO₃, which then loses water to give the nitronium ion, NO₂⁺, the electrophile. Step 2 (electrophilic attack): the -COOH group on the ring is electron-withdrawing and meta-directing, so the delocalised π electrons of the ring attack NO₂⁺ at the position meta to -COOH (C-3), forming a new C-N bond and a positively-charged arenium ion (sigma complex) in which the ring's aromaticity is temporarily lost. Step 3 (restoration of aromaticity/catalyst regeneration): HSO₄⁻ removes the H⁺ from the sp³ carbon; these electrons return to the ring to restore aromaticity, and HSO₄⁻ + H⁺ → H₂SO₄, regenerating the sulphuric acid. Because H₂SO₄ is used up in step 1 and reformed in step 3, it is not consumed overall and acts purely as a catalyst, giving 3-nitrobenzoic acid as the organic product.