Organic synthesis

Like the AS synthesis topic, this asks you to join up known reactions to build a target molecule. Now you also have the A Level reactions of arenes, phenol, amines, amides and acyl chlorides.

Identifying functional groups

A molecule may carry several functional group 官能团 types. Use the test reactions to spot each one, then predict its behaviour. For example:

• decolourises bromine water with no catalyst, giving a white precipitate → a phenol or a phenylamine (the ring is activated).
• reacts violently with cold water, giving fumes of $\text{HCl}$ → an acyl chloride 酰氯.
• gives a purple colour with neutral $\text{FeCl}_3$ → a phenol 苯酚.

A map of the A Level reactions

Planning and analysing a route

To devise a synthetic route 合成路线, work backwards from the target: which single reaction makes it, and from what? Repeat until you reach the starting material, then write each step with its reagent 试剂 and conditions.

When you analyse a route, state the type of reaction for each step (for example electrophilic substitution, addition–elimination 加成消去, oxidation or reduction) and watch for likely by-products 副产物 — for example, making an amine from a halogenoalkane also gives over-substituted amines, lowering the yield.

Worked example. Devise a route from benzene to phenylamine, $\text{C}_6\text{H}_5\text{NH}_2$. Work backwards: phenylamine comes from reducing nitrobenzene, and nitrobenzene comes from nitrating benzene - so the route is two steps. Step 1: benzene with concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ at $55\ °\text{C}$; electrophilic substitution gives nitrobenzene (keep below $55\ °\text{C}$, or further substitution follows). Step 2: reduce the nitrobenzene with tin and concentrated $\text{HCl}$, then add $\text{NaOH}$ to free the amine from its salt. There is no direct route - you cannot put an $\text{–NH}_2$ straight onto a ring, which is exactly why this nitrate-then-reduce pair is worth knowing by heart.

Common mistakes.

• Writing "add $\text{KMnO}_4$" with no conditions. Every step needs its reagent and conditions: hot, acidified or alkaline $\text{KMnO}_4$, under reflux, for methylbenzene → benzoic acid.
• Forgetting the $\text{NaOH}$ after the $\text{Sn}$/$\text{HCl}$ reduction. In acid the product is the phenylammonium salt; the base frees the amine.
• Using $\text{NaBH}_4$ to reduce an amide or a nitrile. $\text{NaBH}_4$ is too weak; use $\text{LiAlH}_4$ (or $\text{H}_2$/Ni for a nitrile).
• Losing count of carbons. $\text{KCN}$ adds one carbon to the chain; most other steps keep the count the same, so check it at every step.
• Warming the diazonium salt before the coupling step. Above $10\,°\text{C}$ it gives phenol and $\text{N}_2$ instead of the azo dye.
• Reaching for $\text{AlCl}_3$ to brominate a phenol or phenylamine. These rings are activated, so bromine water alone works, and gives substitution at all three of the 2-, 4- and 6-positions.

• Combine AS and A-level steps and watch for benzene-ring reactions and carbon-count changes.
• State every reagent and condition, and note when a step produces a racemic mixture.
• Choose the shortest route that reaches the target functional group.