Organic synthesis

1. Overview

Organic synthesis is making a target molecule from a starting material in planned steps. This section pulls together the AS reactions of 14.1 to 20.1: you use them to identify the groups in a molecule, predict how it reacts, and plan or analyse a route, with the reagents, conditions and by-products of each step. Mechanisms are on their own pages.

Key Definitions

• Functional group: the atom or group of atoms that gives an organic molecule its characteristic reactions, such as C=C in alkenes or -OH in alcohols.
• Synthetic route: a planned sequence of reactions that turns a starting material into a target molecule, giving the reagents and conditions for each step and the structure of each intermediate.
• Reagent: a substance that reacts with the organic compound in a step and is used up; a catalyst speeds up the step but is not used up.
• Intermediate (in a synthesis): a compound made in one step of a route and used as the starting material of the next; unlike an intermediate in a mechanism, it can usually be isolated.
• By-product: any substance formed in a step besides the wanted product: another product of the main reaction (such as water in esterification) or the product of a competing reaction (such as a minor isomer).
• Heating under reflux: heating a mixture with a vertical condenser fitted, so that vapours condense and run back into the flask, allowing long heating without losing volatile substances.
• Distillation (during a reaction): boiling off a volatile product as soon as it forms and condensing it, for example to remove an aldehyde before it is oxidised further.

Content

The AS reaction toolkit

AS routes are built from these reactions, grouped by starting class. Each line reads: reagents and conditions → product (type of reaction). R is an alkyl group and X a halogen atom.

Alkanes (14.1)

• Cl₂ or Br₂, ultraviolet light → halogenoalkane + HX (free-radical substitution)
• heat, Al₂O₃ catalyst → shorter alkane + alkene (cracking)

Alkenes (14.2)

• H₂(g), Pt or Ni catalyst, heat → alkane (addition: hydrogenation)
• steam, H₂O(g), H₃PO₄ catalyst → alcohol (electrophilic addition)
• HX(g), room temperature → halogenoalkane (electrophilic addition)
• X₂, room temperature → 1,2-dihalogenoalkane (electrophilic addition)
• cold, dilute, acidified KMnO₄ → diol (oxidation)
• hot, concentrated, acidified KMnO₄ → the C=C splits: a =CH₂ end gives CO₂ and H₂O, a =CHR end gives RCOOH and a =CRR′ end gives the ketone RCOR′ (oxidation)
• many monomer molecules join → poly(alkene) (addition polymerisation)

With an unsymmetrical alkene, HX or H₂O adds mainly with the X or OH on the carbon that has fewer H atoms (Markovnikov addition), because that route passes through the more stable carbocation.

Halogenoalkanes (15.1)

• NaOH(aq), heat → alcohol (nucleophilic substitution)
• KCN in ethanol, heat → nitrile with one more carbon (nucleophilic substitution)
• excess NH₃ in ethanol, heat in a sealed tube (under pressure) → primary amine (nucleophilic substitution)
• NaOH in ethanol, heat → alkene (elimination of HX)

Alcohols (16.1)

• HX(g), heat; KCl with concentrated H₂SO₄ or H₃PO₄, heat; PCl₃, heat; PCl₅ or SOCl₂, room temperature → halogenoalkane (substitution)
• Na(s) → sodium alkoxide + H₂ (redox)
• acidified K₂Cr₂O₇ or KMnO₄ (oxidation): primary alcohol → aldehyde if it is distilled off as it forms, or → carboxylic acid if heated under reflux; secondary → ketone, which is not oxidised further; tertiary → no reaction
• heated Al₂O₃ catalyst, or heat with concentrated H₂SO₄ → alkene (elimination: dehydration)
• carboxylic acid, concentrated H₂SO₄ catalyst, heat → ester + H₂O (condensation: esterification)

Aldehydes and ketones (17.1)

• NaBH₄, or LiAlH₄ in dry ether → primary alcohol from an aldehyde, secondary alcohol from a ketone (reduction)
• HCN with KCN catalyst, heat → hydroxynitrile with one more carbon (nucleophilic addition)
• aldehydes only: acidified K₂Cr₂O₇ or KMnO₄, heat under reflux → carboxylic acid (oxidation)

Carboxylic acids (18.1)

• reactive metal → salt + H₂; alkali → salt + H₂O; carbonate → salt + H₂O + CO₂
• alcohol, concentrated H₂SO₄ catalyst, heat → ester + H₂O (condensation)
• LiAlH₄ in dry ether (not NaBH₄) → primary alcohol (reduction)

Esters (18.2)

• dilute acid, heat under reflux → carboxylic acid + alcohol (hydrolysis; reversible)
• dilute alkali, heat under reflux → carboxylate salt + alcohol (hydrolysis; complete), then acidify to get the acid

Nitriles and hydroxynitriles (19.2)

• dilute acid, heat under reflux; or dilute alkali, heat under reflux, then acidify → carboxylic acid (hydrolysis)

Amines (19.1): at AS you only need to make them, from halogenoalkanes.

Reactions that change the carbon chain

• One carbon more: a cyanide step (KCN with a halogenoalkane, or HCN with an aldehyde or ketone). Hydrolysis then turns -C≡N into -COOH.
• Fewer carbons: hot, concentrated, acidified KMnO₄ splits a C=C; cracking breaks a long alkane; alkaline I₂(aq) removes a CH₃ group as CHI₃.
• Pieces joined or split: esterification and ester hydrolysis; addition polymerisation.
• Every other AS reaction keeps the carbon skeleton.

Identifying functional groups by tests

Test	Positive result	Group shown
Br₂(aq)	orange to colourless	C=C
2,4-DNPH	orange precipitate	C=O in an aldehyde or ketone
Tollens' reagent, warm	silver mirror	aldehyde, -CHO
Fehling's solution, warm	blue to orange-red precipitate	aldehyde, -CHO
acidified K₂Cr₂O₇, warm	orange to green	primary or secondary alcohol, or aldehyde
alkaline I₂(aq), warm	pale yellow precipitate of CHI₃	CH₃CO- or CH₃CH(OH)-
Na(s), dry sample	fizzing (H₂)	-OH, in an alcohol or an acid
Na₂CO₃(aq)	fizzing (CO₂)	-COOH
AgNO₃(aq) in ethanol, warm	white, cream or pale yellow precipitate	C-Cl, C-Br or C-I

• Negative results count: a ketone gives the 2,4-DNPH precipitate but no silver mirror; a tertiary alcohol leaves acidified K₂Cr₂O₇ orange.
• -COOH is one group, carboxyl, not a ketone plus an alcohol: no 2,4-DNPH precipitate and no oxidation by acidified K₂Cr₂O₇. Methanoic acid, HCOOH, is the exception: it is oxidised and gives a silver mirror.
• Acidified KMnO₄ (purple to colourless) is a poor test here: it also attacks C=C.
• Esters and nitriles have no direct test in this table. Hydrolyse them and test the products: a nitrile heated with NaOH(aq) gives NH₃, which turns damp red litmus blue.

Predicting properties and reactions

In a molecule with several groups, each group keeps its own reactions, so check every group against the reagent:

• NaBH₄ reduces aldehydes and ketones only; LiAlH₄ also reduces carboxylic acids. Neither reduces C=C.
• Acidified K₂Cr₂O₇ oxidises primary and secondary alcohols and aldehydes but leaves C=C alone; acidified KMnO₄ oxidises C=C too.
• H₂ (Pt or Ni) and Br₂ add across C=C; HX(g) adds across C=C and also substitutes -OH.
• NaOH(aq) neutralises -COOH, and on heating hydrolyses esters and nitriles and substitutes C-X.

Physical properties follow from the same groups:

• Hydrogen bonding: -OH, -COOH and -NH₂ groups form hydrogen bonds, so the compound boils at a higher temperature than a hydrocarbon of similar $M_{\text{r}}$, and small molecules dissolve in water.
• Isomerism: a C=C where each carbon carries two different groups gives cis/trans isomers; a carbon with four different groups is a chiral centre, giving optical isomers (13.4).

Worked example 1: a molecule with three functional groups

Question: Compound K, C₆H₁₀O₂, gives these results: (1) Br₂(aq) turns from orange to colourless; (2) 2,4-DNPH gives an orange precipitate; (3) warm Tollens' reagent gives a silver mirror; (4) warm alkaline I₂(aq) gives a pale yellow precipitate; (5) Na₂CO₃(aq) gives no fizzing; (6) Na(s) fizzes with dry K.

(a) Deduce the functional groups in K. (b) K is CH₃CH(OH)CH=CHCH₂CHO. Predict the organic product with (i) NaBH₄, (ii) an excess of acidified K₂Cr₂O₇, heated under reflux, (iii) HCN with a KCN catalyst. (c) Predict the isomerism K shows.

Step 1: read each result. Result 1: C=C. Results 2 and 3: an aldehyde, -CHO. Results 5 and 6: an -OH group but no -COOH, so an alcohol. The -CHO and the -OH use both O atoms of C₆H₁₀O₂, so there is no other C=O group. Result 4 shows CH₃CO- or CH₃CH(OH)-. The only aldehyde with CH₃CO- is ethanal, so K must contain CH₃CH(OH)-, a secondary alcohol.

Step 2: predict each reaction, group by group.

• (i) NaBH₄ reduces only the -CHO: CH₃CH(OH)CH=CHCH₂CH₂OH.
• (ii) Acidified K₂Cr₂O₇ attacks two groups. The -CHO becomes -COOH and the secondary alcohol becomes a ketone: CH₃COCH=CHCH₂COOH. The C=C is unchanged.
• (iii) HCN adds to the C=O of the aldehyde only, adding one carbon: CH₃CH(OH)CH=CHCH₂CH(OH)CN.

Answer: (a) C=C, an aldehyde and a secondary alcohol with a CH₃CH(OH)- group. (b) The three products in Step 2; LiAlH₄ would give the same product as NaBH₄, because K has no -COOH. (c) Cis/trans isomers, because each C=C carbon carries H and a different second group; and optical isomers, because the CH(OH) carbon has four different groups (CH₃, H, OH and the rest of the chain).

Devising a synthetic route

1. Compare the starting material and the target: the number of carbon atoms, the functional groups and where they sit on the chain.
2. Count carbons. One more carbon needs a cyanide step; the same number means no new C-C bond.
3. Work backwards from the target: which toolkit reactions make its functional group, and can you make one of their starting compounds?
4. Write each step as reagents and conditions, then the structure of the product.
5. Check the other groups, so that no reagent attacks a group you need to keep. Prefer the shortest route, because every step loses some product (overall yield = the step yields multiplied together).

Worked example 2: an alcohol one carbon longer

Question: Suggest a four-step synthesis of pentan-1-ol, CH₃(CH₂)₄OH, from butan-1-ol, CH₃(CH₂)₃OH. Give the reagents and conditions for each step and the structure of each intermediate.

Plan: count carbons and work backwards. Butan-1-ol has 4 carbon atoms and pentan-1-ol has 5, so one step must be a cyanide step. No AS reaction turns -C≡N straight into -CH₂OH. But LiAlH₄ turns -COOH into -CH₂OH, and hydrolysis turns -C≡N into -COOH. So the route is alcohol → halogenoalkane → nitrile → acid → alcohol. The HCN route would not work: it would leave an extra -OH group on the carbon next to the new one.

The route:

1. SOCl₂, room temperature (substitution): CH₃(CH₂)₃OH + SOCl₂ → CH₃(CH₂)₃Cl + SO₂ + HCl
2. KCN in ethanol, heat under reflux (nucleophilic substitution): CH₃(CH₂)₃Cl + KCN → CH₃(CH₂)₃CN + KCl
3. Dilute HCl, heat under reflux (hydrolysis): CH₃(CH₂)₃CN + 2H₂O + HCl → CH₃(CH₂)₃COOH + NH₄Cl
4. LiAlH₄ in dry ether (reduction): CH₃(CH₂)₃COOH + 4[H] → CH₃(CH₂)₄OH + H₂O

Answer: butan-1-ol → 1-chlorobutane → pentanenitrile → pentanoic acid → pentan-1-ol, with the reagents, conditions and structures above.

Worked example 3: an ester from a single starting material

Question: Butyl butanoate, CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃, is to be made with butan-1-ol as the only organic starting material. Suggest a two-step route.

Plan: split the target at the ester link. The C=O side came from butanoic acid, CH₃CH₂CH₂COOH, and the other side from butan-1-ol. Both have a straight chain of four carbons with the group on the end carbon, so the acid can be made from the alcohol.

Step 1: oxidise part of the alcohol. Heat some butan-1-ol under reflux with an excess of acidified K₂Cr₂O₇ (orange to green). Distilling instead would remove butanal before it could be oxidised to the acid.

CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O (oxidation)

Step 2: esterify. Heat the butanoic acid with more butan-1-ol and a few drops of concentrated H₂SO₄ as catalyst.

CH₃(CH₂)₂COOH + CH₃(CH₂)₃OH ⇌ CH₃(CH₂)₂COO(CH₂)₃CH₃ + H₂O (condensation)

Answer: reflux some butan-1-ol with excess acidified K₂Cr₂O₇ to make butanoic acid, then esterify it with more butan-1-ol. By-products: water in both steps, butanal if oxidation is incomplete, and leftover acid and alcohol, as esterification is reversible.

Analysing a given route

For each step, compare the structures before and after the arrow and name the type of reaction from the change:

What happens in the step	Type of reaction
one atom or group replaces another	substitution
atoms add across C=C or C=O	addition
HX or H₂O is removed, forming C=C	elimination
reaction with water breaks a bond	hydrolysis
two molecules join, losing a small molecule such as water	condensation
O gained or H lost	oxidation
H gained or O lost	reduction

Add the mechanism name when there is one (13.2), check the reagents and conditions against the toolkit, and list the possible by-products.

Small molecules from the main equation: HX (free-radical substitution); H₂O (dehydration, oxidation, esterification); NaX or KX (NaOH or KCN with RX); POCl₃ and HCl (PCl₅); SO₂ and HCl (SOCl₂); NH₄⁺ or NH₃ (nitrile hydrolysis in acid or alkali); CO₂ (hot KMnO₄ on a =CH₂ end).

Organic side products from competing reactions:

• a minor positional isomer when HX or H₂O adds to an unsymmetrical alkene
• a mixture of alkenes from an elimination: 2-bromobutane gives but-1-ene and cis- and trans-but-2-ene
• some alcohol with NaOH in ethanol, and some alkene with NaOH(aq)
• further reaction: more halogen atoms (and longer alkanes from termination) in free-radical substitution; the amine reacting with more halogenoalkane (excess NH₃ reduces this); an aldehyde oxidised on to the acid

Worked example 4: analysing a three-step route

Question: This route converts 1-bromobutane, CH₃CH₂CH₂CH₂Br, into butanone, CH₃COCH₂CH₃.

• Step 1: NaOH in ethanol, heat → A
• Step 2: H₂O(g), H₃PO₄ catalyst → B
• Step 3: acidified K₂Cr₂O₇, heat, distil → butanone

(a) Identify A and B and the type of each step. (b) Suggest possible by-products. (c) Why not simply hydrolyse 1-bromobutane and oxidise the alcohol?

Step 1. HBr is removed, so A is but-1-ene, CH₂=CHCH₂CH₃, the only alkene possible with Br on C1. By-products: NaBr, H₂O and some butan-1-ol from competing substitution.

Step 2. Water adds across the C=C. By Markovnikov addition the OH goes to C2, through the more stable secondary carbocation, so B is butan-2-ol, CH₃CH(OH)CH₂CH₃. By-product: a little butan-1-ol, through the less stable primary carbocation.

Step 3. CH₃CH(OH)CH₂CH₃ + [O] → CH₃COCH₂CH₃ + H₂O. By-products: water, and butanal or butanoic acid from any butan-1-ol carried through.

Part (c). Hydrolysis would leave the OH on C1. Butan-1-ol is a primary alcohol, so it oxidises to butanal or butanoic acid, never to a ketone. Steps 1 and 2 move the functional group from C1 to C2.

Answer: (a) A is but-1-ene (elimination); B is butan-2-ol (electrophilic addition); step 3 is oxidation. (b) The by-products listed under each step. (c) A primary alcohol cannot give a ketone.

Key Equations

R and R′ are alkyl groups; X is a halogen; [O] is an oxygen atom from the oxidising agent and [H] a hydrogen atom from the reducing agent.

Oxidation of alcohols (acidified K₂Cr₂O₇ or KMnO₄):

• primary to aldehyde (distil): RCH₂OH + [O] → RCHO + H₂O
• primary to acid (reflux): RCH₂OH + 2[O] → RCOOH + H₂O
• secondary to ketone: RCH(OH)R′ + [O] → RCOR′ + H₂O

Reduction (NaBH₄, or LiAlH₄ in dry ether; only LiAlH₄ for the acid):

• RCHO + 2[H] → RCH₂OH and RCOR′ + 2[H] → RCH(OH)R′
• RCOOH + 4[H] → RCH₂OH + H₂O

Adding one carbon atom:

• RX + CN⁻ → RCN + X⁻ (KCN in ethanol, heat)
• RCOR′ + HCN → RR′C(OH)CN (HCN with KCN catalyst, heat; R′ = H for an aldehyde)

Hydrolysis of a nitrile (dilute acid, heat under reflux): RCN + 2H₂O + H⁺ → RCOOH + NH₄⁺

Esterification (concentrated H₂SO₄ catalyst, heat): RCOOH + R′OH ⇌ RCOOR′ + H₂O

Common Mistakes to Avoid

• Mistake: writing "heat under reflux" for every step. Fix: reflux is for long heating. HBr(g) adds to an alkene at room temperature, and an aldehyde is distilled off as it forms; refluxing would oxidise it to the acid.
• Mistake: mixing up the two cyanide reactions. Fix: a halogenoalkane needs KCN in ethanol and heat (nucleophilic substitution); an aldehyde or ketone needs HCN with a KCN catalyst and heat (nucleophilic addition). Compare step 2 of Worked example 2 with part (b)(iii) of Worked example 1.
• Mistake: confusing NaOH(aq) with NaOH in ethanol, or writing "ethanoic" for "ethanolic". Fix: heated with NaOH(aq), 1-bromopentane gives pentan-1-ol (substitution); heated with NaOH in ethanol, it gives pent-1-ene (elimination). Ethanoic acid would just neutralise the NaOH.
• Mistake: using NaBH₄ to reduce a carboxylic acid. Fix: NaBH₄ reduces only aldehydes and ketones. In Worked example 2, pentanoic acid needs LiAlH₄ in dry ether to become pentan-1-ol.
• Mistake: forgetting a second group the reagent also attacks. Fix: check every group. In Worked example 1, excess acidified K₂Cr₂O₇ oxidises both the -CHO and the secondary -OH of K.
• Mistake: giving the catalyst as a by-product, or "catalyst" as the type of reaction. Fix: in esterification, concentrated H₂SO₄ is the catalyst, water is the by-product and the type is condensation.

Exam Tips

• Suggest a synthesis: for every step give the reagents, the conditions and the structure of the product, for example "Step 1: KCN in ethanol, heat. Step 2: dilute HCl, heat under reflux." Name the reagent fully, with its acid: "acidified K₂Cr₂O₇", not just "Cr₂O₇²⁻" or "dichromate".
• State the type of reaction with the 13.2 terms and the mechanism name where there is one: "nucleophilic substitution", not just "substitution".
• Identify a group from a test: give the reagent, the colour change from start to finish and the group, and make sure each conclusion agrees with every result, including the negative ones.
• Suggest by-products: give formulae for small molecules from the equation (H₂O, HBr, NaBr) and name organic side products, such as a positional isomer.