Chemistry: Aromatic Chemistry, Carbonyls and Amines
Aromatic chemistry, carbonyls and amines is the A-level Chemistry topic covering three families of Year 13 organic reactions: electrophilic substitution reactions of benzene and its derivatives, nucleophilic addition reactions of aldehydes and ketones (carbonyl compounds), and the basicity, nucleophilicity and preparation of amines. It builds on the naming rules, isomerism and mechanism ideas met earlier in the course, and is frequently examined synoptically alongside analytical techniques such as infrared spectroscopy and mass spectrometry to identify an unknown organic product.
Before you start
Make sure you're comfortable with these topics first:
Method
- Represent benzene using the delocalised model (a ring of six carbon atoms, each contributing one electron to a ring of delocalised electrons above and below the plane of the ring) rather than the Kekule structure of alternating single and double bonds, and cite the evidence for this: benzene's C-C bond lengths are all equal (intermediate between a single and a double bond), and its enthalpy of hydrogenation is less exothermic than three times that of cyclohexene, showing benzene is more stable (has lower energy) than the Kekule structure predicts.
- For nitration of benzene, generate the electrophile NO2+ (the nitronium ion) by reacting concentrated nitric acid with concentrated sulfuric acid (the catalyst), then draw the curly-arrow mechanism: a pair of electrons from the delocalised ring attacks NO2+, forming a positively charged intermediate that then loses H+ to reform the stable delocalised ring.
- For Friedel-Crafts acylation, use an acyl chloride (RCOCl) with an AlCl3 catalyst to generate the electrophile RCO+, which reacts with the benzene ring by the same two-step electrophilic substitution mechanism as nitration, producing a phenyl ketone and regenerating the AlCl3 catalyst.
- For carbonyl compounds, first distinguish an aldehyde (C=O at the end of a carbon chain, general formula RCHO) from a ketone (C=O between two carbon chains, general formula RCOR'), then identify which nucleophilic addition reagent is being used: NaBH4 (a source of hydride ions, H-, which act as the nucleophile) reduces a carbonyl to an alcohol, while HCN (a source of cyanide ions, CN-) adds across the carbonyl to form a hydroxynitrile, extending the carbon chain by one carbon and creating a new chiral centre.
- Distinguish an aldehyde from a ketone using a mild oxidising agent: Tollens' reagent (ammoniacal silver nitrate) produces a silver mirror with an aldehyde (oxidised to a carboxylic acid) but no change with a ketone, because a ketone has no hydrogen atom on the carbonyl carbon left to be removed by oxidation.
- For amines, explain basicity and nucleophilicity in terms of the nitrogen atom's lone pair: a primary aliphatic amine (such as ethylamine) is a stronger base than ammonia because the alkyl group pushes electron density onto the nitrogen (a positive inductive effect), making the lone pair more available to accept a proton, whereas an aromatic amine such as phenylamine is a weaker base than ammonia because the nitrogen's lone pair is partly delocalised into the benzene ring, making it less available.
- For preparing a primary amine, react a halogenoalkane with excess ethanolic ammonia (nucleophilic substitution, with the lone pair on nitrogen acting as the nucleophile), using a large excess of ammonia to favour a single substitution over further substitution to a secondary or tertiary amine.
Worked example
Phenylethanone, C6H5COCH3, can be prepared by the Friedel-Crafts acylation of benzene with ethanoyl chloride, CH3COCl, using an AlCl3 catalyst: C6H6 + CH3COCl -> C6H5COCH3 + HCl. Using Ar: C = 12.0, H = 1.0, O = 16.0, Cl = 35.5, calculate the atom economy of this reaction for the production of phenylethanone.
- Calculate the Mr of each reactant: Mr(C6H6) = (6 x 12.0) + (6 x 1.0) = 78.0. Mr(CH3COCl) = (2 x 12.0) + (3 x 1.0) + 16.0 + 35.5 = 78.5.
- Calculate the Mr of the desired product, phenylethanone (C8H8O): Mr = (8 x 12.0) + (8 x 1.0) + 16.0 = 96.0 + 8.0 + 16.0 = 120.0.
- Calculate the total Mr of all reactants used: 78.0 + 78.5 = 156.5.
- Apply the atom economy formula: atom economy = (Mr of desired product / total Mr of all reactants) x 100 = (120.0 / 156.5) x 100.
- Final answer: atom economy = 76.7% (3 s.f.). The remaining mass (the HCl by-product) is not part of the desired product, so it counts against the atom economy even though the equation is perfectly balanced and, in principle, all of the reactant mass could be converted with a 100% yield.
Practice questions
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Q1State the reagent and catalyst used to generate the electrophile NO2+ in the nitration of benzene.Show answer
Answer: Concentrated nitric acid, with concentrated sulfuric acid as the catalyst.
Q2Explain why benzene, unlike cyclohexene, does not readily undergo addition reactions with bromine water.Show answer
Answer: The delocalised ring of electrons in benzene gives it extra stability (lower energy) compared with a structure containing three localised double bonds, so benzene resists addition reactions that would break up this delocalised system; it undergoes substitution reactions instead, which preserve the ring's stability.
Q3Name the type of reagent (nucleophile or electrophile) that NaBH4 provides in the reduction of a ketone to an alcohol.Show answer
Answer: A nucleophile (the hydride ion, H-).
Q4Butanone reacts with HCN to form a hydroxynitrile product. State why this product is optically inactive even though it contains a chiral centre.Show answer
Answer: The HCN adds to the planar carbonyl carbon with equal probability from either face, producing equal amounts of both optical isomers (a racemic mixture); a racemic mixture has no net effect on plane-polarised light because the effects of the two isomers cancel out.
Q5Describe the observation that distinguishes propanal from propanone when each is warmed separately with Fehling's solution.Show answer
Answer: With propanal (an aldehyde), the blue Fehling's solution is reduced to give a brick-red precipitate of copper(I) oxide; with propanone (a ketone), there is no colour change and the solution remains blue.
Q6State and explain whether phenylamine (C6H5NH2) is a stronger or weaker base than ammonia.Show answer
Answer: Weaker; the lone pair on the nitrogen atom of phenylamine is partly delocalised into the aromatic ring, making it less available to accept a proton than the lone pair on ammonia's nitrogen atom.
Q7State the reagents and conditions used to prepare a primary aliphatic amine from a halogenoalkane.Show answer
Answer: Excess ethanolic ammonia, heated under pressure (in a sealed tube).
Q8Calculate the empirical formula of a carbonyl compound containing, by mass, 66.7% carbon, 11.1% hydrogen and the remainder oxygen. (Ar: C = 12.0, H = 1.0, O = 16.0)Show answer
Answer: Remainder = O = 100 - 66.7 - 11.1 = 22.2%. Moles: C = 66.7/12.0 = 5.56; H = 11.1/1.0 = 11.1; O = 22.2/16.0 = 1.39. Dividing by the smallest value (1.39): C = 4.0, H = 8.0, O = 1.0. Empirical formula = C4H8O.
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid to form nitrobenzene. (a) Write an equation to show how the electrophile is generated from nitric acid and sulfuric acid. (b) Describe, using curly arrows in words, the mechanism by which this electrophile reacts with benzene to form nitrobenzene, and state the role of sulfuric acid once the electrophile has reacted.
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An unknown carbonyl compound, R, has the molecular formula C4H8O. When a small sample of R is added to Brady's reagent (2,4-dinitrophenylhydrazine), an orange precipitate forms immediately. A separate sample of R produces no silver mirror when warmed with Tollens' reagent. (a) State what the positive result with Brady's reagent confirms about R. (b) Use the Tollens' reagent result to deduce whether R is an aldehyde or a ketone, and hence give a structural formula and name for R. (c) The melting point of the crystalline 2,4-dinitrophenylhydrazone derivative of R is measured and compared with a data book value for a known compound. Explain why this comparison, rather than the melting point of R itself, is used to identify the exact identity of R.
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Describe how propylamine, CH3CH2CH2NH2, could be prepared from bromoethane, CH3CH2Br, in a two-stage synthesis via a nitrile intermediate, giving the reagents and conditions for each stage, and explain why this route is used rather than direct substitution with ammonia.
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