Electrophilic substitution (A-level only) — AQA A-Level Chemistry
Test yourself on Electrophilic substitution (A-level only) with AQA A-Level practice questions.
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Electrophilic substitution (A-level only) explained
The delocalised ring is a region of high electron density above and below the plane, so it attracts species that are short of electrons.
Read the full explanation
The attack happens in two steps. A pair of electrons leaves the delocalised system and forms a bond to the electrophile; that leaves an unstable positively charged intermediate in which the delocalisation is broken and only part of the ring, drawn as a horseshoe, is still delocalised. In the second step the C-H bond at the attacked carbon breaks heterolytically and the pair of electrons returns to the ring, restoring the full delocalisation and releasing H⁺. The net change is that a hydrogen atom is replaced, so only one substituent is introduced at A-level.
Nitration is an important step in synthesis, including the manufacture of explosives and formation of amines.
Nitration puts an -NO₂ group onto the ring using a mixture of concentrated nitric acid and concentrated sulfuric acid, warmed to about 50 °C. The sulfuric acid is the catalyst: it protonates the nitric acid so that the nitronium ion NO₂⁺ can form, and it is regenerated at the end. Temperature control matters, because above about 50 °C a second nitro group is substituted and dinitrobenzene forms. The product is the gateway to two important families. Reducing the nitro group, with tin and concentrated hydrochloric acid and then excess sodium hydroxide, gives an aromatic amine such as phenylamine, the starting point of azo dyes. Repeated nitration of methylbenzene gives TNT, an explosive.
Friedel–Crafts acylation reactions are also important steps in synthesis.
Friedel-Crafts acylation attaches an acyl group to the ring, forming a new carbon-carbon bond, which is exactly what a synthesis usually needs and what nitration cannot do. The reagents are an acyl chloride and anhydrous aluminium chloride, AlCl₃, acting as a halogen carrier. AlCl₃ pulls the chloride off the acyl chloride to give the acylium ion RCO⁺ and AlCl₄⁻; the acylium ion is the electrophile that attacks the ring. At the end AlCl₄⁻ takes the H⁺ released from the ring, giving HCl and regenerating AlCl₃, so it is a catalyst and not a reactant. The product is an aromatic ketone, which can then be reduced or built on further. Conditions must be dry, since water destroys the AlCl₃.
Students should be able to outline the electrophilic substitution mechanisms of: nitration, including the generation of the nitronium ion acylation using AlCl as a catalyst.
Both mechanisms begin by making the electrophile. For nitration the two concentrated acids react, HNO₃ + H₂SO₄ giving NO₂⁺ + HSO₄⁻ + H₂O, and the positive charge in the nitronium ion sits on the nitrogen atom. For acylation, RCOCl + AlCl₃ gives RCO⁺ + AlCl₄⁻. Then the ring attacks: a curly arrow runs from inside the hexagon to the nitrogen or to the acylium carbon, giving the horseshoe intermediate with its positive charge. A second arrow runs from the C-H bond at that carbon back into the hexagon, releasing H⁺. The negative ion mops up that proton, regenerating H₂SO₄ in the nitration and AlCl₃ with HCl in the acylation.
Your focus
- Draw the general two-step mechanism for electrophilic substitution on benzene using curly arrows.
- Draw the positively charged intermediate correctly, showing the horseshoe, the charge and the hydrogen atom.
- Explain why the ring loses H⁺ rather than adding the electrophile permanently.
Show all 12 objectives
- State the reagents and conditions for the nitration of benzene and explain why the temperature is limited.
- Write a balanced equation for the reduction of nitrobenzene to phenylamine using [H].
- Outline a two-stage synthesis that converts benzene into an aromatic amine, naming the reagents at each stage.
- Write equations for the formation of the acylium ion and for the regeneration of aluminium chloride.
- Predict and name the product of a named acyl chloride reacting with benzene in the presence of AlCl₃.
- Point to the new carbon-carbon bond formed in a Friedel-Crafts acylation and state what it puts onto the ring that nitration cannot.
- Write the equations that generate the nitronium ion and the acylium ion from the reagents given.
- Draw the complete electrophilic substitution mechanism for the nitration of benzene, with all charges and arrows.
- Draw the mechanism for the acylation of benzene and write the equation regenerating the aluminium chloride.
Electrophilic substitution (A-level only) exam tips
Marking Points
- one mark for a curly arrow from inside the hexagon to the electrophile
- one mark for the intermediate drawn with a horseshoe and a positive charge inside the ring
- one mark for a curly arrow from the C-H bond back into the hexagon
- one mark for naming the mechanism as electrophilic substitution and identifying H⁺ as the species lost
- one mark for concentrated nitric acid and concentrated sulfuric acid as the reagents
- one mark for a temperature of about 50 °C, kept low to prevent further substitution
- one mark for reduction of the nitro group using tin and concentrated hydrochloric acid
- one mark for adding excess sodium hydroxide to release the free amine from its salt
- one mark for naming a use, such as dye manufacture from the amine or an explosive from repeated nitration
- one mark for aluminium chloride generating the acylium ion from the acyl chloride
- one mark for an equation showing AlCl₄⁻ formed alongside the acylium ion
- one mark for the aromatic ketone as the organic product
- one mark for AlCl₃ being regenerated, so it is a catalyst rather than a reagent
- one mark for the equation generating the nitronium ion from the two concentrated acids
- one mark for the electrophile drawn with the positive charge on the nitrogen atom of NO₂⁺
- one mark for the curly arrow from inside the hexagon to the nitrogen or to the positive charge
- one mark for the curly arrow from the C-H bond into the hexagon
- one mark for the equation that regenerates the catalyst using the H⁺ released
Examiner Tips
- 💡Draw the intermediate large, with the attacked carbon at the top carrying both the H and the new group side by side.
- 💡The horseshoe must not pass through the attacked carbon; that carbon is now saturated.
- 💡State the temperature and say why: 'about 50 °C, to avoid further substitution' is worth more than '50 °C'.
- 💡Use [H] in reduction equations and balance the hydrogens; nitrobenzene to phenylamine takes 6[H] and gives 2H₂O.
- 💡Write both halogen-carrier equations, the one that makes the electrophile and the one that regenerates the catalyst; they are usually a mark each.
- 💡The acylium ion carries its positive charge on the carbon of the C=O; put it there when you draw it.
- 💡Every curly arrow starts from a lone pair or a bond; an arrow drawn from a charge symbol is penalised each time.
- 💡If you use skeletal formulae in a mechanism and a C-H bond breaks, you must draw in both that bond and the H to earn the mark.
Common Mistakes
- drawing the first arrow from a localised C=C bond rather than from inside the ring
- leaving the hydrogen off the attacked carbon of the intermediate, so the second arrow has no bond to start from
- showing the electrophile adding across the ring to give a non-aromatic product
- omitting the positive charge from the intermediate
- using dilute nitric or sulfuric acid
- writing sulfuric acid into the overall equation as a reactant that is consumed
- stopping the reduction at the ammonium salt and omitting the sodium hydroxide stage
- giving phenylamine as the direct product of nitration
- treating aluminium chloride as a reactant that appears in the overall equation
- drawing the product as an ester or an aldehyde rather than a ketone
- forgetting that the conditions must be anhydrous
- giving HCl as the only inorganic product and omitting the regeneration of the catalyst
- putting the positive charge of the nitronium ion on an oxygen atom
- drawing the attacking arrow from a Kekulé double bond instead of from inside the ring
- omitting the hydrogen from the attacked carbon of the intermediate, leaving no C-H bond for the second arrow
- regenerating the catalyst without accounting for the H⁺ that was released