Questions on electrophilic substitution: nitration and Friedel–Crafts acylation

Question 1

Draw the mechanism for the nitration of benzene, including an equation for the formation of the electrophile, and explain why the reaction is a substitution and not an addition.

Accepted Answer

Indicative content

an equation forming NO₂⁺: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻; accept HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O or the two steps
a curly arrow that starts inside the hexagon, on the delocalised electrons, and ends on the nitrogen atom of NO₂⁺; reject an arrow to an oxygen
the intermediate drawn with H and NO₂ on the same carbon, and a horseshoe with + inside it over the other five carbons only
a curly arrow from the middle of the C–H bond on that carbon back into the ring, with H⁺ shown leaving and nitrobenzene formed
H⁺ + HSO₄⁻ → H₂SO₄, showing that the sulfuric acid is regenerated and acts as a catalyst
losing H⁺ restores the delocalised ring and keeps its stability, which an addition product with no delocalised ring would lose

Levels

Level 3 (5–6 marks) the equation for NO₂⁺ formation, every curly arrow and the intermediate with its horseshoe and charge are correct, the sulfuric acid is shown to be regenerated, and the loss of H⁺ is explained as restoring the delocalised ring's stability
Level 2 (3–4 marks) the electrophile is formed correctly, with the arrow into the ring and the intermediate drawn; the arrow from the C–H bond back into the ring, the regenerated catalyst, or the stability argument for substitution is absent
Level 1 (1–2 marks) NO₂⁺ is named or one arrow drawn; the intermediate commonly keeps a full circle, or nitric acid itself is shown attacking the ring

Question 2

Explain, with equations, how the nitronium ion forms in the nitrating mixture, and why concentrated sulfuric acid is described as a catalyst.

Accepted Answer

B1 HNO₃ + H₂SO₄ → H₂NO₃⁺ + HSO₄⁻, sulfuric acid protonating nitric acid B1 H₂NO₃⁺ → NO₂⁺ + H₂O; accept the single equation HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ for both marks B1 the H⁺ lost from the ring at the end of the mechanism combines with HSO₄⁻ B1 H⁺ + HSO₄⁻ → H₂SO₄, so the sulfuric acid is regenerated and not used up overall

Question 3

Identify the electrophile and the role of the aluminium chloride when a student heats benzene under reflux with ethanoyl chloride and anhydrous aluminium chloride, and give the equations that show the electrophile forming and the catalyst re-forming.

Accepted Answer

B1 the electrophile is the acylium ion, CH₃CO⁺, with the positive charge on the carbonyl carbon B1 aluminium chloride is a catalyst acting as a halogen carrier, accepting a lone pair from chlorine B1 CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻ B1 H⁺ + AlCl₄⁻ → AlCl₃ + HCl, re-forming the catalyst and giving off hydrogen chloride

Question 4

Outline the mechanism for the reaction of benzene with ethanoyl chloride and aluminium chloride, including an equation for the formation of the electrophile.

Accepted Answer

B1 CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄⁻ B1 a curly arrow from within the hexagon to the carbonyl carbon of CH₃CO⁺, the carbon carrying + B1 the intermediate with H and COCH₃ on the same carbon and a horseshoe with + inside it across the other five carbons B1 a curly arrow from the C–H bond into the ring, with H⁺ lost

Question 5

Show that the overall equation for the nitration of benzene is C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O, by combining the equations for the formation of the electrophile, the substitution step and the regeneration of the catalyst, and state what the result shows about the sulfuric acid.

Accepted Answer

B1 HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻; accept HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O, with the regeneration then written as one step B1 C₆H₆ + NO₂⁺ → C₆H₅NO₂ + H⁺, the substitution step summarised from the mechanism B1 H⁺ + HSO₄⁻ → H₂SO₄ and H₃O⁺ + HSO₄⁻ → H₂SO₄ + H₂O, the two hydrogensulfate ions taking back the two protons B1 adding the equations cancels NO₂⁺, H⁺, H₃O⁺, 2HSO₄⁻ and 2H₂SO₄, leaving C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O; the sulfuric acid is regenerated unchanged, so it is a catalyst

Question 6

Compare the reaction of ethene with hydrogen bromide and the nitration of benzene, in terms of the electrophile, the intermediate and the type of reaction.

Accepted Answer

B1 ethene is attacked by a neutral polar molecule, HBr, whereas benzene needs a positive ion, NO₂⁺, made with a catalyst B1 both form a positively charged intermediate after the electrophile accepts a pair of π electrons B1 ethene's carbocation is attacked by Br⁻, giving electrophilic addition B1 benzene's intermediate loses H⁺ and regains its delocalised ring, giving electrophilic substitution

Question 7

Suggest a two-step synthesis of phenylamine from benzene, giving the reagents and conditions for each step and naming the intermediate.

Accepted Answer

B1 step 1: concentrated nitric acid and concentrated sulfuric acid at 50 °C B1 the intermediate is nitrobenzene, C₆H₅NO₂ B1 step 2: tin and concentrated hydrochloric acid, heated under reflux, reduce the nitro group; allow iron for tin B1 sodium hydroxide is then added to release phenylamine from the phenylammonium salt

Question 8

Outline the mechanism for the bromination of benzene with bromine and iron(III) bromide, including equations for the formation of the electrophile and the regeneration of the catalyst.

Accepted Answer

B1 Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻, the carrier accepting a lone pair from bromine; accept the polarised Br–Br–FeBr₃ complex B1 a curly arrow from within the hexagon to Br⁺, or to the δ+ bromine of the complex B1 the intermediate with H and Br on one carbon and a horseshoe with + over the other five, then an arrow from the C–H bond into the ring B1 H⁺ + FeBr₄⁻ → FeBr₃ + HBr, re-forming iron(III) bromide

Question 9

State the reagents and conditions for the nitration of benzene.

Accepted Answer

B1 concentrated nitric acid; reject dilute nitric acid B1 concentrated sulfuric acid, which acts as a catalyst B1 a temperature of about 50 °C, kept below about 55 °C to prevent further nitration; 'heat' alone is not enough

Question 10

Explain why benzene reacts with electrophiles by substitution rather than by addition.

Accepted Answer

B1 the delocalised ring gives benzene extra stability, about 152 kJ mol⁻¹ B1 addition would leave a product with no delocalised ring, so that stability would be lost B1 in substitution H⁺ is lost and the delocalised ring is restored, giving a more stable product

Question 11

Calculate the percentage yield of nitrobenzene when a batch of benzene is nitrated, using the masses and Mr values given in the table.

Accepted Answer

M1 amount of benzene = 31.2 ÷ 78.0 = 0.400 mol M1 maximum mass of nitrobenzene = 0.400 × 123.0 = 49.2 g A1 percentage yield = 37.6 ÷ 49.2 × 100 = 76.4%, to three significant figures

Question 12

Calculate the percentage atom economy for making phenylethanone by the reaction C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl. Mr: C₆H₆ 78.0, CH₃COCl 78.5, C₆H₅COCH₃ 120.0.

Accepted Answer

M1 total Mr of the reactants = 78.0 + 78.5 = 156.5 M1 atom economy = 120.0 ÷ 156.5 × 100 A1 atom economy = 76.7%, to three significant figures

Question 13

Suggest why little phenylethanone forms when a student uses aluminium chloride from a bottle that had been left open, and what the student would have seen when the bottle was opened.

Accepted Answer

B1 water from the air hydrolyses aluminium chloride, giving steamy fumes of hydrogen chloride B1 the hydrated aluminium compound can no longer accept a lone pair from the chlorine of the acyl chloride B1 so little acylium ion forms, there is little electrophile, and the catalyst is lost

Question 14

Suggest the electrophile, an equation for its formation, and the organic product when benzene is heated with benzoyl chloride, C₆H₅COCl, and aluminium chloride.

Accepted Answer

B1 the electrophile is the acylium ion C₆H₅CO⁺ B1 C₆H₅COCl + AlCl₃ → C₆H₅CO⁺ + AlCl₄⁻ B1 the product is the ketone C₆H₅COC₆H₅, diphenylmethanone, with HCl as the other product

Question 15

Explain why benzene needs a positive ion such as NO₂⁺ as an electrophile, when ethene reacts with neutral bromine.

Accepted Answer

B1 the π electrons in benzene are delocalised over six carbons, so the electron density at any one bond is lower than in a C=C bond B1 benzene's electron density is too low to polarise a neutral molecule such as Br₂ B1 a positive ion is attracted strongly enough to the ring to accept a pair of the delocalised electrons

Question 16

Explain why phenol reacts with bromine water at room temperature, whereas benzene reacts with bromine only when a halogen carrier is present.

Accepted Answer

B1 a lone pair on the oxygen of phenol overlaps with the ring and is partly delocalised into it B1 this increases the electron density of the ring above benzene's B1 phenol's ring can therefore polarise a bromine molecule and attack it directly, whereas benzene needs a halogen carrier to make Br⁺

Question 17

Explain why Friedel–Crafts alkylation of benzene gives some disubstituted product, whereas acylation stops after one group, and how the alkylation is run to limit it.

Accepted Answer

B1 an alkyl group releases electron density into the ring, so the alkylbenzene reacts faster than benzene and is alkylated again B1 an acyl group withdraws electron density from the ring through its C=O group, so the ketone is less reactive than benzene and is not acylated again B1 benzene is used in large excess, so the electrophile is far more likely to meet benzene than the alkylbenzene

Question 18

Suggest the organic product when methylbenzene reacts with chlorine in ultraviolet light, with chlorine and aluminium chloride, and with hot alkaline potassium manganate(VII) followed by acid.

Accepted Answer

B1 in ultraviolet light the methyl group is substituted by free radicals, giving (chloromethyl)benzene, C₆H₅CH₂Cl B1 with aluminium chloride the ring is substituted at positions 2 and 4, giving 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene B1 with alkaline manganate(VII) and then acid the side chain is oxidised to COOH, giving benzoic acid, C₆H₅COOH

Question 19

Suggest the main organic product when nitrobenzene reacts with bromine and iron(III) bromide, and explain the position of the bromine.

Accepted Answer

B1 the product is 1-bromo-3-nitrobenzene B1 the nitro group withdraws electron density from the ring and directs the incoming electrophile to position 3

Substance	Mass / g	Mr
benzene, C₆H₆	31.2	78.0
nitrobenzene, C₆H₅NO₂	37.6	123.0