Cambridge International AS & A Level Chemistry 9701 · A Level topic 36

Organic synthesis (A Level)

What this chapter covers36.1

Paracetamol, procaine, serotonin and almost every drug or natural product contain several functional groups in one molecule. Chemists who make such molecules must know how each group reacts, which reagents attack which groups, and in what order the groups can be introduced. Topic 36 brings together all the organic reactions of the AS and A Level syllabus and asks you to use them: to identify the groups in an unfamiliar molecule, to predict its reactions, to design multi-step routes, and to analyse routes that others have proposed.

No new reactions are introduced in this chapter. Instead it organises the reactions you already know into reaction maps and tables, shows how to plan a synthesis by working backwards, and highlights the details — reagent, condition, order of steps, selectivity, by-products — on which examination marks depend.

Identifying functional groups by their reactions36.1.1(a)

Each functional group has characteristic reactions with standard reagents. A positive result shows that the group is present; a negative one, that it is absent. Table 36.1 collects the tests that the syllabus reactions provide. When a molecule contains several groups, each gives its own result, so a set of tests identifies the whole collection.

Predicting properties and reactions36.1.1(b)

To predict what a reagent does to a molecule with several groups, take each group in turn and ask whether the reagent reacts with it. Groups that do not react are carried through unchanged. Four points trip up many answers:

• An excess of reagent reacts with every group it can. Excess ethanoyl chloride acylates every –OH and every –NH₂ or –NH–; excess bromine water substitutes at every free activated ring position; excess H₂ with a Pt catalyst hydrogenates every C=C, including those of a benzene ring.
• Reactions that need a catalyst do not happen without it. Ethanoyl chloride does not acylate a benzene ring unless AlCl₃ is present; a halogen does not substitute into benzene without a halogen carrier.
• Similar-looking groups behave differently. Phenols react with NaOH but alcohols do not; an amide N is not basic; a halogenoarene does not undergo nucleophilic substitution.
• Charges must balance. A sodium salt has O⁻ and Na⁺; an ammonium salt has N⁺ and a counter-ion.

Past-paper practice · Set 36A · Predicting the reactions of a multi-functional molecule

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 36A.1[8]

Chapters 29–35 contain further questions in which the reactions of a molecule with several functional groups must be predicted, for example 32C.5 and 32C.6.

Answer and marking guidance

Na: the phenol –OH becomes –O⁻Na⁺ (the –NH₂ and indole N–H are unchanged) ✔; redox (reduction) ✔. Excess Br₂(aq): Br substitutes into the benzene ring at the two positions next to the –OH ✔; (electrophilic) substitution ✔. Excess CH₃COCl: the phenol –OH is acylated to –OCOCH₃ ✔ AND the –NH₂ is acylated to –NHCOCH₃ ✔; condensation (or addition–elimination) ✔. Excess H₂/Pt: every C=C, including those of the benzene ring, is hydrogenated, giving a saturated bicyclic amine with the –OH and –CH₂CH₂NH₂ groups kept ✔; reduction / hydrogenation / addition ✔. Examiner insight: full credit was seldom awarded. With Na, the ion was drawn with a − or a + charge but not both, and the reaction was called neutralisation. With Br₂(aq), many did not realise that the phenol ring undergoes electrophilic substitution. With CH₃COCl, many did not see that both the phenolic OH and the NH₂ react, or acylated the benzene ring, which does not happen without AlCl₃. With H₂/Pt, many did not hydrogenate both the C=C and the benzene ring.

The aliphatic reaction map36.1.2

Figure 36.1 links the functional groups of aliphatic compounds by the reactions of the syllabus. Each numbered arrow is one reaction, keyed in Table 36.2. A synthetic route is a path through this map; the shortest path is usually the best, because every step loses some product.

Esters, amides and acyl chlorides are also hydrolysed back to carboxylic acids (esters by dilute acid or alkali and heat; amides by aqueous acid or alkali and heat; acyl chlorides by water at room temperature).

The aromatic reaction map36.1.2

Planning a route36.1.2

Changing the carbon skeleton

Most syllabus reactions leave the carbon skeleton unchanged. Only three add carbon atoms: nucleophilic substitution by CN⁻ (halogenoalkane → nitrile), nucleophilic addition of HCN to a carbonyl group (→ hydroxynitrile), and Friedel–Crafts alkylation or acylation of an arene. If the target has one more carbon than the starting material, a nitrile or hydroxynitrile step is almost always needed; the –CN is then hydrolysed to –COOH or reduced to –CH₂NH₂. The side-chain oxidation of an alkylbenzene is the syllabus reaction that removes carbon atoms.

Order of steps on a benzene ring

Substituents already on a ring direct new ones: –CH₃, –OH and –NH₂ to the 2- and 4-positions; –NO₂ and –COOH to the 3-position. The order of steps must use these effects. To make 3-nitrobenzoic acid from methylbenzene, oxidise the CH₃ to COOH first, then nitrate; nitrating first would put NO₂ at the 2- or 4-position. Conversely, to put an alkyl group opposite an amino group, alkylate first (the alkyl group directs nitration to position 4), then nitrate and reduce.

Choosing a selective reagent

A reagent must convert the target group without spoiling the others. NaBH₄ reduces aldehydes and ketones but not carboxylic acids, whereas LiAlH₄ reduces both. An acyl chloride must not meet water before it meets the intended nucleophile. Br₂(aq) would add to a C=C bond elsewhere in the molecule as well as substituting into a phenol ring.

Past-paper practice · Set 36B · Devising multi-step syntheses

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 36B.1[6]

Answer and marking guidance

(i) G = HOCH₂CH₂CH₂CH₂OH ✔; H = NCCH₂CH₂CH₂CH₂CN ✔. (ii) step 1: NaOH(aq), heat ✔; step 2: acidified KMnO₄ or acidified K₂Cr₂O₇, heat ✔; step 3: KCN or NaCN (CN⁻), heat ✔; step 4: LiAlH₄ (or Na in ethanol, or H₂ with Ni, Pd or Pt) ✔. Examiner insight: (i) was answered well; some gave a nitrated product for H, which would not work. Answers to (ii) were disappointing — the reagents were not well done and the conditions even less so, and some confused steps 2 and 3.

Question 36B.2[5]

Answer and marking guidance

(i) M = benzoic acid ✔; N = 3-nitrobenzoic acid ✔. (ii) step 1: hot KMnO₄ (MnO₄⁻) ✔; step 2: concentrated H₂SO₄ and concentrated HNO₃ ✔; step 3: Sn and concentrated HCl, heat ✔. Examiner insight: (i) was answered well, though some nitrated the ring before oxidising the side chain — that would not work, because methylbenzene nitrates at the 2- or 4-position, not the 3-position. In (ii) most knew the reagents but the conditions were less well known.

Question 36B.3[6]

Answer and marking guidance

(i) 4-bromo-2-nitrobenzoic acid ✔. (ii) E = 2-nitromethylbenzene (1-methyl-2-nitrobenzene); F = 4-bromo-2-nitromethylbenzene ✔✔ — the CH₃ group directs both substitutions, and is oxidised last. (iii) step 1: concentrated H₂SO₄ and concentrated HNO₃ ✔; step 2: Br₂ and AlBr₃ ✔; step 3: hot (alkaline or acidified) KMnO₄ ✔. Examiner insight: many gave an unambiguous name for G. In (ii) a common error was oxidising CH₃ to COOH first. (iii) was answered well; common errors were HNO₂ instead of HNO₃, or omitting "concentrated", for the nitration; Br₂(aq) for the bromination; and omitting heat for the oxidation.

Question 36B.4[6]

Answer and marking guidance

Any three different starting compounds, each with its reagent [1] + [1]: a 1-halobutane, e.g. CH₃CH₂CH₂CH₂Br, with NH₃ under pressure or heated in a sealed tube ✔✔; butanenitrile, CH₃CH₂CH₂CN, with H₂ and Ni or Pt, or LiAlH₄, or Na and ethanol ✔✔; butanamide, CH₃CH₂CH₂CONH₂, with LiAlH₄ or Na and ethanol ✔✔. Examiner insight: some misunderstood the question and wrote a sequence of three reactions ending in butylamine. Other correct starting materials, such as CH₃CH₂CH₂CH₂NO₂ or CH₃CH₂CH₂CONHCH₂CH₃, also scored. Structural or displayed formulae are safer than names, where many slips cost marks.

Question 36B.5[5]

Answer and marking guidance

(i) X = (1-methylethyl)benzene, C₆H₅CH(CH₃)₂ ✔; Y = 1-(1-methylethyl)-4-nitrobenzene ✔. (ii) step 1: (CH₃)₂CHBr and FeBr₃ or AlBr₃ ✔; step 2: concentrated HNO₃ and concentrated H₂SO₄ ✔; step 3: Sn and concentrated HCl ✔. Examiner insight: (i) was answered well; a small number nitrated first (X = nitrobenzene), which does not work because the nitro group is 3-directing and the alkyl group would go to position 3, not 4. Sn is a reactant, not a catalyst, in the reduction. In (ii) some used CH₃CH₂CH₂Cl in place of CH₃CHClCH₃, or did not say the acids must be concentrated.

Naming the type of each reaction36.1.3

To analyse a given route, compare the structures before and after each step and decide what has changed. The change identifies the type of reaction, and the type points to the reagent.

By-products36.1.3

Most reactions give some unwanted products as well as the target. Recognising them is part of analysing a route, and they are often what a question asks for.

• Free-radical substitution continues past the first substitution: bromination of methylbenzene in UV light also gives C₆H₅CHBr₂ and C₆H₅CBr₃, and termination steps join radicals together (for example C₆H₅CH₂CH₂C₆H₅).
• Halogenoalkanes with ammonia give secondary and tertiary amines and a quaternary ammonium salt, because each amine formed is itself a nucleophile.
• Electrophilic substitution of a substituted ring gives isomers: methylbenzene is nitrated at both the 2- and 4-positions; Friedel–Crafts reactions can give isomeric products.
• Substitution and elimination compete when a halogenoalkane reacts with hydroxide: aqueous conditions favour the alcohol, ethanolic conditions the alkene.
• Unsymmetrical alkenes add HX to give two isomeric halogenoalkanes, the major one predicted by Markovnikov's rule.
• Inorganic by-products follow from the equation: HCl from acyl chlorides, H₂O from condensations, SO₂ and HCl from SOCl₂, POCl₃ and HCl from PCl₅.

Past-paper practice · Set 36C · Analysing given routes

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 36C.1[5]

Answer and marking guidance

(i) Br₂ and UV light ✔. (ii) Q = C₆H₅CH₂CN ✔. (iii) step 2: KCN in ethanol, heat ✔; step 3: HCl(aq), heat under reflux ✔. (iv) a further radical-substitution product such as C₆H₅CHBr₂ or C₆H₅CBr₃ (or any viable product such as C₆H₅CH₂CH₂C₆H₅) ✔. Examiner insight: (i) was usually correct. In (ii) common errors were C₆H₅CH₂OH and C₆H₅CH₂CH₂OH. (iii) was generally answered well; the reagents for the substitution were better recalled than those for the hydrolysis.

Question 36C.2[3]

Answer and marking guidance

(i) step 2: KCN / NaCN / CN⁻ ✔. (ii) step 1: PCl₃ and heat, or PCl₅, or SOCl₂ ✔; step 4: NaBH₄ ✔. Examiner insight: (i) was generally answered well, and the reagent for step 1 usually correct. The reagent for step 4 was rarely credited: many suggested LiAlH₄, but that would reduce the –COOH group as well as the ketone.

Question 36C.3[6]

Answer and marking guidance

(i) Y = CH₃COCO₂CH₃ ✔; Z = CH₃C(OH)(CN)CO₂CH₃ ✔. (ii) step 1: CH₃OH ✔ and concentrated H₂SO₄, heat ✔; step 2: HCN with NaCN catalyst ✔; step 3: heat with Al₂O₃ above 100 °C, or heat with concentrated H₂SO₄ (dehydration) ✔. Examiner insight: stronger candidates identified both Y and Z; Y was more often known than Z. (ii) was not well known: many identified methanol for the esterification but omitted the conditions, and the reagents for steps 2 and 3 were less often credited.

Misconceptions and how the topic is assessed36.1

Self-test36.1

Ten questions on the whole chapter. Each gives its reason once you answer.

Summary

• Functional groups are identified by characteristic reactions: Br₂(aq), Na, NaOH, Na₂CO₃, acidified dichromate, Tollens', Fehling's, 2,4-DNPH, alkaline iodine, AgNO₃, and diazotisation with coupling.
• In a multifunctional molecule each group reacts independently; an excess of reagent reacts with every group it can; catalysed reactions need their catalyst.
• Routes are planned by comparing start and target, working backwards, and checking carbon count, order of steps and selectivity.
• Carbon atoms are added by CN⁻, HCN and Friedel–Crafts reactions; side-chain oxidation of arenes removes them.
• Directing effects fix the order of steps on a ring: –CH₃, –OH, –NH₂ to 2,4; –NO₂, –COOH to 3.
• Analysing a route means naming the type of each step and predicting by-products from further substitution, isomer formation and competing reactions.

Examination checklist

• Can I identify every functional group in an unfamiliar molecule and give a test for each?
• Can I predict the product of each common reagent with a multifunctional molecule?
• Can I plan a route of up to four steps, with reagents and full conditions?
• Can I use directing effects to order the steps of an aromatic synthesis?
• Can I choose a reagent that reacts with only the intended group?
• Can I name the type of each reaction in a route and suggest likely by-products?

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