Question 5 of 5: Electrophilic Aromatic Substitution Mechanisms & Combustion

Question 5: Electrophilic Aromatic Substitution Mechanisms & Combustion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Both reactions are electrophilic aromatic substitutions (EAS): a strong Lewis or Brønsted acid first generates a potent electrophile from the reagent, the ring's π-system attacks that electrophile to form a resonance-stabilised, non-aromatic arenium (Wheland) intermediate, and losing H⁺ from the sp³ ipso carbon restores full aromaticity in the product. The two reactions differ only in which electrophile is generated.

(i) Nitration: benzene + HNO₃/H₂SO₄.

1. Generate the electrophile. H₂SO₄ protonates HNO₃, which then loses water to form the nitronium ion, NO₂⁺ — a strong electrophile. $$\mathrm{HNO_3 + 2H_2SO_4 \longrightarrow NO_2^+ + H_3O^+ + 2HSO_4^-}$$
2. Ring attack → arenium intermediate. A pair of ring π-electrons attacks NO₂⁺, forming a new C–N bond; the positive charge is now delocalised over the ring's ortho/para positions relative to the new substituent, while the ipso carbon becomes sp³ (bearing both H and NO₂), breaking aromaticity temporarily.
   Q5a(i) — sp3 ipso carbon, + delocalised over 3 ring positions
3. Deprotonation restores aromaticity. A base (HSO₄⁻) removes the ipso H⁺, and the ring's six π-electrons redelocalise into the full aromatic sextet. $$\mathrm{C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O}$$

Product: nitrobenzene.

(ii) Friedel–Crafts acylation: benzene + acetyl chloride/AlCl₃.

1. Generate the electrophile. AlCl₃ (a strong Lewis acid) coordinates to the chlorine of CH₃C(=O)Cl, weakening and then breaking the C–Cl bond to form a resonance-stabilised acylium ion, CH₃C≡O⁺ ↔ CH₃C⁺=O, plus AlCl₄⁻. The acylium ion is far less prone to rearrangement than a simple alkyl carbocation, since the positive charge is already resonance-stabilised on oxygen. $$\mathrm{CH_3COCl + AlCl_3 \longrightarrow CH_3CO^+ + AlCl_4^-}$$
2. Ring attack → arenium intermediate. The ring's π-electrons attack the electrophilic acylium carbon, forming the new C–C(=O)CH₃ bond and the same kind of sp³-ipso, charge-delocalised arenium cation as in part (i).
   Q5a(ii) — arenium intermediate of the acylation
3. Deprotonation and catalyst regeneration. AlCl₄⁻ removes the ipso H⁺, restoring the aromatic ring and regenerating HCl + AlCl₃ (a true catalyst, recovered unchanged). $$\mathrm{[\text{arenium}]^+ + AlCl_4^- \longrightarrow C_6H_5COCH_3 + HCl + AlCl_3}$$

Product: acetophenone.

b) Combustion of butane in pure oxygen. Combustion of any hydrocarbon in excess/pure O₂ goes to completion: every carbon becomes CO₂ and every hydrogen becomes H₂O. Balance carbon first, then hydrogen, then oxygen last (oxygen often needs a common-denominator doubling to clear a half-integer coefficient):

1. Balance C and H. Butane, C₄H₁₀, has 4 C and 10 H, so one butane needs 4 CO₂ and 5 H₂O. $$\mathrm{C_4H_{10} + O_2 \longrightarrow 4CO_2 + 5H_2O}$$
2. Balance O, doubling to clear the half-integer. The right side now needs 4(2)+5(1)=13 oxygen atoms, i.e. 6.5 O₂ — doubling every coefficient clears the fraction: $$\boxed{\mathrm{2C_4H_{10} + 13O_2 \longrightarrow 8CO_2 + 10H_2O}}$$