Advanced organic synthesis

Reading a molecule that contains several functional groups

A polyfunctional molecule contains more than one functional group, and almost every A Level synthesis question hands you one. The skill is not to guess but to work through the molecule in the same order every time, because the groups that are easiest to miss are the ones that change the answer. Start with the carbonyl region: find every C=O and decide what is joined to the carbonyl carbon. A hydrogen on it means an aldehyde; a carbon on each side means a ketone; an -OH on it means a carboxylic acid; an O joined to a second carbon means an ester; a nitrogen on it means an amide. Then hunt for the C=C, the C-X of a halogenoalkane or of a halogenoarene, an -OH on a saturated carbon (alcohol) or straight on a ring (phenol), a nitrile, a nitro group, a primary or secondary amine, and, where an amine and a carboxyl group sit in the same molecule, an amino acid.

The same pair of atoms is a different functional group in a different place, and that is the trap. An -OH on a saturated carbon makes a neutral alcohol that acidified dichromate(VI) oxidises; an -OH joined straight to a benzene ring makes a weak acid that dissolves in sodium hydroxide, activates the ring towards electrophiles and is not oxidised by dichromate(VI) under the usual conditions. An NH2 on a saturated carbon is the basic amine group; the same NH2 attached to a carbonyl carbon is an amide, whose nitrogen lone pair is delocalised into the C=O, which is why an amide is not basic at all. Two compounds can even share a molecular formula and behave completely differently, so it is the structure, never the formula, that decides the chemistry.

Use the tests in the syllabus as a second and independent identification. The formula tells you what could be there and a test result tells you what is there, so work out both what a positive result confirms and what it rules out. One test can be answered by more than one group, and that catches people out: bromine water is decolourised by a C=C, by a phenol and by phenylamine, so the observation alone settles nothing until you know which groups the molecule could contain. Where a compound gives an orange precipitate with 2,4-dinitrophenylhydrazine (2,4-DNPH), it has a carbonyl group; if it also gives a silver mirror with Tollens' reagent it is an aldehyde, and if it does nothing with Tollens' reagent it is a ketone.

Predicting reactions when two groups compete

One rule governs every polyfunctional question: a reagent attacks every group in the molecule that it is able to attack. Predict the product by taking each group in turn and asking whether this reagent does anything to it. If two groups react, both products form, and you will be asked either to notice that or to choose a different reagent that leaves one of them alone.

Reducing agents. Sodium tetrahydridoborate, NaBH4, in water or ethanol reduces an aldehyde or a ketone to an alcohol and attacks nothing else, so it is the selective choice when the molecule also carries a carboxylic acid, an ester or a nitrile. Lithium tetrahydridoaluminate, LiAlH4, in dry ether is much stronger and reduces aldehydes, ketones, carboxylic acids, esters, amides and nitriles, so it is the wrong reagent whenever a carbonyl group elsewhere in the molecule has to survive - and it reacts violently with water, so it must be used in dry ether.

Oxidising agents. Acidified potassium dichromate(VI) or acidified potassium manganate(VII) oxidises a primary alcohol to an aldehyde and then on to a carboxylic acid and a secondary alcohol to a ketone, and a tertiary alcohol is not oxidised at all. Dichromate(VI) leaves a C=C alone, but manganate(VII) does not. Hot alkaline manganate(VII) is the reagent that oxidises the side chain of an alkylbenzene all the way to a benzoate ion, and it also attacks primary alcohols, aldehydes and C=C bonds, so it cannot be used on a molecule where those groups have to be kept. An aldehyde is oxidised by either reagent, which is why an aldehyde is distilled out of the mixture as soon as it is formed.

Acid-base behaviour and the reactivity order. In a molecule holding both an -NH2 and a -COOH, such as 2-aminoethanoic acid, both groups show themselves: the compound is neutralised by sodium hydroxide at the carboxyl group and by hydrochloric acid at the amine group, and in water it exists mainly as a zwitterion, with the proton transferred to the nitrogen. The three acidic groups you have met rank as carboxylic acid, then phenol, then alcohol: a carboxylic acid releases carbon dioxide from hydrogencarbonate, phenol dissolves in sodium hydroxide to give sodium phenoxide, and an alcohol is essentially neutral. On a ring, an -OH, an -NH2 or an alkyl group releases electron density and makes substitution faster, while -NO2, -COOH and -CO- withdraw electron density, slow substitution down and direct any further substitution into the 3-position.

The reaction toolkit: reagent and condition for every step

Every step of a route is a triple: what is in the flask, the reagent with its conditions, and the type of reaction. Learn them as triples and a route becomes a sentence you can say out loud. Conditions are marks - the concentration, the catalyst, the temperature, the solvent, and whether the mixture is heated under reflux or distilled. 'Dichromate' is not an answer; 'acidified potassium dichromate(VI), heated under reflux so that the aldehyde is oxidised on to the acid' is.

The table below is the toolkit you need for route questions. It tells you two things a list of reagent names does not: which steps change the carbon skeleton, and which steps destroy part of the molecule. Reaction with KCN adds exactly one carbon atom; Friedel-Crafts alkylation and acylation build a group on to a ring; and the side-chain oxidation of an alkylbenzene throws away everything on that side chain except the one carbon that becomes a carboxyl group.

Three sets of conditions are worth extra care because they are so specific. Nitration of benzene needs a mixture of concentrated nitric and concentrated sulphuric acids at a temperature between 25 C and 60 C: hot enough to form the nitronium ion, cool enough to stop a second nitro group adding. Diazotisation needs sodium nitrite with dilute acid kept below 10 C in an ice bath, because the diazonium salt decomposes as the mixture warms. And the reduction of nitrobenzene with tin and concentrated hydrochloric acid gives the amine as its salt, so sodium hydroxide solution must be added afterwards to liberate the free amine.

Devising a multi-step route

Design a route backwards from the target. Ask which functional group the last step has to create and what compound lies one reaction away from it: an ester comes from an alcohol with a carboxylic acid or an acyl chloride, a carboxylic acid comes from a primary alcohol, an aldehyde or a nitrile, an amine comes from a nitrile, an amide or a nitro compound, and a phenol comes from a diazonium salt and water. Then step back again and ask how that compound would be made, until you reach the starting material you were given.

The carbon skeleton is the part most routes cannot change, and the exceptions are worth knowing by heart. Reaction of a halogenoalkane with KCN in ethanol adds one carbon as a nitrile, which can then be hydrolysed to the acid or reduced to the amine. Friedel-Crafts alkylation builds an alkyl group on to a ring, and acylation builds a carbonyl group on to it. Oxidation of an alkylbenzene side chain with hot alkaline manganate(VII) cuts the other way, throwing away everything on that side chain except the carbon that becomes the carboxyl group. Nothing else in the course changes the number of carbon atoms, so count the carbons at every arrow: a route that quietly gains or loses one is wrong.

Order the steps so that the groups you need survive. Do an oxidation before a reduction when you need both, because the strong reducing agents will reduce what the oxidising agent has just made. Do not plan a side-chain oxidation in a molecule that also holds a primary alcohol or a C=C unless you want those oxidised too. Add any group that deactivates the ring, a nitro group above all, after the ring substitutions you still need, because a deactivated ring will not react. And where a group would be attacked by a step you cannot avoid, put that step last: a nitrile is a good place to keep a nitrogen until the final reduction, and a nitro group is a good place to keep an amine until the tin and hydrochloric acid reduction.

Then check the route forwards, out loud, one sentence per step. 'Benzene, concentrated nitric acid with concentrated sulphuric acid at 25 to 60 C, gives nitrobenzene; nitrobenzene, hot tin with concentrated hydrochloric acid and then sodium hydroxide solution, gives phenylamine.' If a sentence has no reagent and no condition in it, the step is not finished and the examiner cannot give it a mark.

• Work backwards from the target, then check forwards from the starting material.
• Only three things change the carbon skeleton: KCN, which adds one carbon; Friedel-Crafts alkylation or acylation on a ring; and side-chain oxidation of an alkylbenzene, which cuts a side chain down to -COOH.
• Oxidise before you reduce, and put a reduction last whenever the reducing agent would attack another group as well.
• Add a deactivating group such as -NO2 only after the ring substitutions you still need have been done.
• Give a reagent and a condition for every arrow, and count the carbon atoms on both sides of every arrow.

A worked route: benzene to phenol in four steps

Benzene can be turned into phenol using nothing but the reactions in the syllabus, in four steps. Step 1 is nitration: benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid at 25-60 C gives nitrobenzene and water by electrophilic substitution, because the sulphuric acid protonates the nitric acid, which then loses water to give the electrophile, the nitronium ion NO2+. Step 2 is reduction: nitrobenzene is heated with tin and concentrated hydrochloric acid to reduce the nitro group to an amino group, and sodium hydroxide solution is then added to liberate the free amine, phenylamine.

Step 3 is diazotisation: phenylamine with sodium nitrite and dilute hydrochloric acid, kept below 10 C in an ice bath, gives benzenediazonium chloride. The low temperature matters because the diazonium ion is unstable; warm it and it reacts with water and loses nitrogen. Step 4 turns that instability to good use: warming the diazonium salt with water replaces the diazonium group by an -OH group and gives phenol. The same intermediate is used in the opposite sense when it is added cold to phenol in sodium hydroxide solution, because then it couples to give an azo dye instead of decomposing.

Read the whole route and it hands you the four reaction types and the four sets of conditions you have to know. It also shows you where the material disappears, because every step produces a by-product as well as the compound you want; the overall yield is the product of four step yields, so it is much lower than any single one. That is the reason phenol is manufactured industrially by the cumene process, in which benzene and propene are joined with an acid catalyst, the cumene is oxidised with air, and dilute acid cleaves the hydroperoxide to give phenol and propanone together, so far less is wasted.

Analysing a route: reaction types, reagents and by-products

Route analysis runs the opposite way round from route design. You are given a sequence of steps and asked, for each one, the type of reaction and the reagents and conditions; two marks a step is the usual pattern, so a four-step route is worth eight marks and there is no room for vagueness. Name the species that attacks and say what happens to the skeleton - electrophilic substitution, electrophilic addition, nucleophilic substitution, nucleophilic addition, addition-elimination, elimination, hydrolysis, oxidation, reduction, esterification, diazotisation or coupling.

The by-products are the part candidates forget. Every step in the syllabus has a competing reaction, and naming that reaction and the compound it gives is what scores. There are only three patterns to look for: a second group in the molecule reacts, the same group reacts a second time, or the reagent attacks the same group in a different way. Decide which of the three you are looking at and the answer writes itself.

Two examples show the pattern. Reducing nitrobenzene with tin and concentrated hydrochloric acid is not finished until sodium hydroxide has been added, because the reduction product is the phenylammonium ion and until the mixture is made alkaline you do not have phenylamine at all. Nitrating methylbenzene with the nitration mixture gives a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene, because the methyl group directs substitution into the 2- and 4-positions, so the by-product is the isomer you did not want and separating it is where part of the yield goes. In both cases the extra product comes out of the chemistry of the molecule, not out of carelessness.

Running a route at the bench, and choosing between routes

The practical decision in most steps is between heating under reflux and distilling, and getting it the right way round is worth marks. Reflux means heating in a flask fitted with a vertical condenser so that vapour condenses and runs straight back down: nothing is lost and the reaction can be driven to completion, which is what you want for oxidising an alcohol to a carboxylic acid, hydrolysing a nitrile, an amide or an ester, esterifying an alcohol and nitrating benzene. Distil when the product would react again with the reagent you are using, as when a primary alcohol is oxidised to an aldehyde and the aldehyde would otherwise be carried on to the acid.

Then separate and purify, because a route whose intermediates are impure fails at the next step. An immiscible liquid layer is taken off in a separating funnel, running off the lower layer first; a liquid is dried over a drying agent such as anhydrous sodium sulphate and redistilled, collecting the fraction at its boiling point; a solid is recrystallised and filtered under reduced pressure. Add anti-bumping granules before heating anything, and learn the observations that show a step has worked: orange dichromate(VI) turning green, purple manganate(VII) being decolourised, steamy fumes of hydrogen chloride turning damp blue litmus paper red when an acyl chloride meets water.

When two routes reach the same target, compare them properly rather than by opinion. The overall yield is the product of the step yields, never their average, so a four-step route with 80 per cent in every step gives 0.80 x 0.80 x 0.80 x 0.80 = 0.410, that is 41 per cent, and a longer route has to be far better at each step to compete. Atom economy measures how much of the mass of the starting materials ends up in the product you want: adding methanol to carbon monoxide to make ethanoic acid has an atom economy of 100 per cent, while a substitution wastes the atoms that leave as hydrogen chloride or water. Then weigh the hazard of the reagents - potassium cyanide is extremely toxic, chromium(VI) compounds are carcinogenic, LiAlH4 reacts violently with water - the cost and availability of the starting materials, the energy needed for the temperature and pressure, and how easily the product can be separated from its by-products.

Practical questions want practical answers: where the reaction is done (a fume cupboard for anything that gives off hydrogen chloride or uses phenylamine), how the mixture is heated, what you would see, and how the product is purified and checked. 'It fizzes' with no reagent named scores nothing, whereas 'adding sodium carbonate solution gives effervescence of carbon dioxide, which turns limewater milky' scores.

• Reflux with a vertical condenser to drive a slow reaction to completion; distil to remove a product that would otherwise react again.
• Separate immiscible layers in a separating funnel, dry the organic layer over anhydrous sodium sulphate or a suitable drying agent, then redistil and collect the fraction at the boiling point.
• Follow the reaction by observation: orange dichromate(VI) to green chromium(III), manganate(VII) losing its purple colour, steamy fumes of hydrogen chloride in the fume cupboard.
• Compare routes on overall yield (which multiplies), atom economy, hazard, cost, energy and the work needed to separate and purify.