What this chapter covers21

Organic synthesis adds no new reactions. It asks you to use the ones you already know from topics 14 to 20, in a new way. You may be shown a molecule with three functional groups and asked what one reagent does to each of them. You may be given a starting material and a target and asked for the steps between them. Or you may be given a route and asked to name each step, give its reagents and say what else ends up in the flask. All three depend on the same skill: looking at a structure, seeing its functional groups, and knowing what each group does with each reagent.

Two phrases in these statements set the limits. "Using the reactions in the syllabus" means you are never expected to invent chemistry. Every step of every route you write must be a reaction named somewhere in topics 14 to 20, with the reagents and conditions the syllabus gives for it. An examiner will not credit a step that works in the real world but is not in the syllabus, such as a Grignard reagent. "Several functional groups" means the molecules in these questions are usually larger and less familiar than the ones in the earlier units. The groups themselves are the familiar ones.

The page has three parts, one for each outcome. The first part lists the functional groups and the tests that identify them, then shows how to predict what a reagent does to a molecule that has several groups. The second part puts every AS reaction on one map, sets out the conditions that must be stated exactly, and builds routes by counting carbons and working backwards. The third part names the type of each step and its mechanism, and follows the by-products that each kind of reaction produces. Most of the models on the page use a single reaction engine. It holds every AS reaction as an operation on a molecule's structure. The predictions, routes, maps and analyses are therefore calculated for each case, not stored.

Why a group reacts where it does13.2 · 21.1.1(b)

A reaction needs a reagent that is attracted to the group. Each AS functional group has one feature that decides what can attack it. Knowing that feature lets you predict most reactions without memorising a table.

The table explains the pairings examiners like to test. NaBH₄ reduces a C=O but leaves a C=C untouched. It works by delivering a hydride ion, H⁻, which is a nucleophile. A nucleophile is attracted to the δ+ carbon of a C=O and repelled by the electron-rich π bond of a C=C. Bromine is the opposite case. It is an electrophile, so it attacks the C=C and ignores the C=O. Cyanide ions attack C=O and C–X, which both have a δ+ carbon, but not a C=C.

The AS functional groups21.1.1(a)

These are the only groups you need to recognise for AS synthesis. The first step with any unfamiliar molecule is to find every one of them in it, and the model below does that for a set of molecules with two, three or four groups. For alcohols and halogenoalkanes, also note the class (primary, secondary or tertiary). The class changes what the group does, so an alcohol is not fully identified until you have classified it.

Three pairs are easy to confuse on a skeletal formula. An aldehyde and a ketone both contain C=O. In an aldehyde the carbonyl carbon is at the end of the chain and carries a hydrogen, which a skeletal formula does not draw. In a ketone it sits between two carbons. A carboxylic acid and an ester both contain –COO–. In the acid the single-bonded oxygen carries H; in the ester it carries a second carbon chain. An alcohol –OH next to a C=O is not an alcohol at all: that combination is the –COOH of an acid, and it reacts like an acid.

The tests that identify each group21.1.1(a)

21.1.1(a) says "identify … using the reactions in the syllabus". In practice that means the test-tube reactions below. Each has a reagent, a condition and an observation, and a mark needs all three. It is just as important to know which groups do not respond to each test.

Two tests on the list give the same result for two different groups, and questions often use that. Sodium fizzes with an alcohol and with a carboxylic acid. Carbonate fizzes only with the acid. If a compound fizzes with sodium but not with carbonate, it contains an alcohol –OH and no –COOH. Acidified dichromate(VI) turns green with a primary alcohol, a secondary alcohol and an aldehyde. Tollens' and Fehling's respond to the aldehyde alone. So a green dichromate tube with no silver mirror means an alcohol, not an aldehyde.

Working out an unknown21.1.1(a)

A typical question gives a molecular formula and a set of observations and asks for the structure. Deal with it in three passes.

1. Count what the formula allows. Compare the number of hydrogens with the alkane C_nH_2n+2. Each C=C or C=O removes two hydrogens. C₄H₈O is two short of the saturated C₄H₁₀O, so it has one C=C or one C=O (at AS there are no rings in these questions).
2. Turn each result into a statement about a group. Write one line per test: "no precipitate with 2,4-DNPH, so no aldehyde or ketone; decolourises bromine water, so C=C".
3. Draw every structure that fits, then use the tests to rule them out. A test that gives the same result for every structure left tells you nothing new. A question that expects a single answer will always include a test that splits the last two.

The model picks one isomer at random and hides it. Each test you run is applied to the hidden structure, and every candidate that would have given a different result is greyed out. Try to reach a single candidate in as few tests as possible. Then try the C₄H₈O₂ set, where ethyl ethanoate and methyl propanoate cannot be separated by any test on the list. Their hydrolysis products can. That is the reason for hydrolysing an ester before identifying it.

Predicting reactions, one group at a time21.1.1(b)

To predict what a reagent does to a molecule with several groups, work through the groups one by one. For each group, ask whether this reagent reacts with it under these conditions. If it does, rewrite that group as its product. If it does not, leave it exactly as it was. The product is the original skeleton with only the reacting groups changed. In these questions the reagent is taken to be in excess, so every group that can react does react.

The grid below is the whole of 21.1.1(b) in one table. Read across a row to see what one reagent does. Read down a column to see what can be done to one group.

Choose any molecule and any reagent. For each group, the model reports whether it reacts and what it becomes. If it does not react, the model says why. Then it draws the product. Side products (the minor isomer of an addition, the alkene that competes with substitution) are shown separately, below the main product.

Reagents that pick out one group21.1.1(b)

Most of the marks in multi-group questions come from a few reagent pairs. In each pair, both reagents do something, but they differ in which groups they affect. Learn these pairs as pairs.

How many moles react?21.1.1(b)

A favourite question gives a molecule with several groups and asks how many moles of a reagent react with one mole of it, or what volume of gas is released. Count the groups the reagent reacts with, then apply the stoichiometry of one group.

Every AS reaction on one map21.1.2

21.1.2 says "using the reactions in the syllabus". The map below shows all of them: every arrow is a reaction from topics 14 to 20, and every box is a class of compound. Before the map was drawn, the reaction engine was run on each arrow's example to confirm it gives the product shown. A route is simply a path along the arrows. The routes that make up most exam questions go through the alcohol box, the halogenoalkane box, or both.

The same reactions as a table, grouped by what they start from. The final column gives the syllabus statement, so you can go back to the unit where each reaction is taught in detail.

Conditions that must be exact21.1.2

A route loses marks through its conditions much more often than through its reagents. Many AS reagents work with more than one group, or give a different product under different conditions, so the conditions are what make the reagent specific to one step. For each step, write the reagent, its state or solvent, and the heating. In each pair below, getting the condition wrong gives a different product.

The trainer below gives one conversion at a time. Choose the reagent and conditions from the full list. The engine applies your choice to the starting compound and tells you what that choice would actually make. A wrong answer therefore shows you the product your conditions give, not just "incorrect".

Count the carbons first21.1.2

Before looking for any route, count the carbon atoms in the starting material and in the target. Almost every AS reaction keeps the carbon skeleton exactly as it is. Only a few steps change the number of carbons, so the count tells you at once whether one of them has to be in the route.

The +1 steps are the ones that matter most. If the target has one carbon more than the starting material, the route must contain a cyanide step. There are only two. From a halogenoalkane, KCN puts the new carbon where the halogen was. From an aldehyde or ketone, HCN puts it on the carbonyl carbon, next to an –OH. Hydrolysing the nitrile afterwards turns the new carbon into –COOH. Reducing that acid with LiAlH₄ turns it into –CH₂OH.

Working backwards21.1.2

Work from the target, not from the starting material. Find the target's functional group and ask what makes it. Usually two or three reactions do, and one of their starting materials is close to what you have been given. Keep going back one step at a time until you reach something you recognise.

Position matters as much as the group does. Substitution, oxidation, reduction and hydrolysis all leave a group on the carbon where it started. Two reactions can move a group along the chain, and together they make a useful pair. Elimination creates a C=C that includes the old carbon and its neighbour. A Markovnikov addition across that C=C then puts the new group on whichever of the two carbons carries more alkyl groups. So 1-bromopropane can be converted into 2-bromopropane, and propan-1-ol into propan-2-ol.

Worked routes21.1.2

These six cover the patterns behind most AS route questions. Each gives the reagent and conditions for every step. The equations are balanced, because questions often ask for one of them.

The route finder21.1.2

The model searches breadth-first. From the starting compound it applies every synthetic reagent on this page. Then it applies every reagent to each product, and so on until it reaches the target, so the first routes it finds are the shortest. It follows only the main product of each step. It acidifies after any alkaline hydrolysis, as the syllabus statements say. It lists alternatives with the most standard reagents first. Try the pairs from the worked routes, then try some that need three or four steps.

Set the limit to one or two steps and some pairs have no route. The reason is worth understanding. Ethanol cannot be turned into propan-1-ol in fewer than four steps, because adding a carbon needs the cyanide route: bromoethane, then propanenitrile, then propanoic acid, then LiAlH₄. Going the other way, from propanoic acid to ethanoic acid, needs a carbon removed. The only AS reaction that does this in a preparation is splitting a C=C with hot concentrated manganate(VII). The route therefore has to make an alkene first: reduce the acid to propan-1-ol, dehydrate it to propene, and then split the propene into ethanoic acid and CO₂.

Choosing between routes21.1.2

When two routes both work, a question may ask which is better, or why a particular one was chosen. Four considerations come up.

• Number of steps. Every step loses material, so the overall yield is the product of the step yields. Three steps at 80% each give 51% overall, not 80%.
• Mixtures. A step that gives several products needs a separation, and the unwanted isomers are wasted. Free-radical substitution and addition to an unsymmetrical alkene both do this. Making bromoethane from ethene and HBr gives one product; making it from ethane and bromine in UV light gives a mixture.
• Reversibility. Esterification and acid hydrolysis of an ester reach equilibrium and stop short of completion. Alkaline hydrolysis goes to completion.
• Hazards and conditions. HCN is extremely toxic and KCN is highly toxic. Steam hydration needs high pressure. LiAlH₄ needs dry conditions.

Naming the type of each step13.2 · 21.1.3

21.1.3 asks you to analyse a route "in terms of type of reaction and reagents used for each step". The words for the types come from 13.2, and an examiner expects exactly those words. There are two layers. The type of reaction describes what happens to the molecule overall. The mechanism name adds the kind of species that attacks, and it exists for only four AS reactions.

Some steps can be described correctly in more than one way. Hydrogenation of an alkene is an addition and a reduction. Dehydration of an alcohol is an elimination. Hydrolysis of a halogenoalkane by aqueous NaOH is a nucleophilic substitution. Give the name the question asks for. If it asks for a mechanism, use one of the four mechanism names, in full. "Nucleophilic" or "substitution" on its own is not a mechanism.

Free-radical substitution: a mixture14.1.3 · 21.1.3

"Possible by-products" in 21.1.3 refers to two things. There are the inorganic products that every equation has (HBr, H₂O, NaBr). There are also organic side products, which come from the reaction doing more than one thing. Free-radical substitution gives the most side products of any AS reaction, for three reasons that appear in its mechanism.

1. Every hydrogen can be replaced. A chlorine radical takes whichever hydrogen it collides with. Propane gives 1-chloropropane and 2-chloropropane.
2. Substitution does not stop at one. The chloroalkane still has C–H bonds, and they are attacked in the same way. Dichloro-, trichloro- and more highly substituted products build up, the more so the more chlorine there is.
3. Termination joins radicals. Two ethyl radicals give butane. With propane, the propyl and 1-methylethyl radicals combine in three ways.

The model applies the propagation step to every hydrogen of an alkane and groups identical products. It then does the same again to find the disubstituted products, and joins every pair of radicals to find the termination products. The percentages assume that every C–H bond is equally likely to be attacked. Real mixtures contain more of the product from a secondary or tertiary hydrogen, but the conclusion is the same either way: the product is a mixture.

Addition to an alkene: the minor isomer14.2.4 · 14.2.5 · 21.1.3

When HBr, HCl or steam adds to an unsymmetrical alkene, two products are possible, and both form. The H⁺ adds first, to one carbon of the C=C, and the positive charge goes onto the other carbon. The carbocation with more alkyl groups on its positive carbon is more stable. Each alkyl group pushes electron density towards the charge (the positive inductive effect). The more stable carbocation forms faster and leads to the major product. The less stable one gives the minor product, and that minor product is the by-product.

Counting the alkyl groups is simple, apart from one detail. The carbon that was the other end of the double bond also counts, because once the H⁺ has added to it, it is an ordinary alkyl group attached to the positive carbon. For propene, CH₃CH=CH₂, the H⁺ can add to C1, leaving the charge on C2. C2 then carries two carbon atoms, making a secondary cation. Or the H⁺ can add to C2, leaving the charge on C1, which carries one carbon: a primary cation. Secondary wins, so the major product is 2-bromopropane.

Nucleophilic substitution in a route15.1.3 · 21.1.3

Three of the most useful steps in AS synthesis are nucleophilic substitutions of a halogenoalkane. They have the same mechanism: the nucleophile's lone pair attacks the δ+ carbon and the C–X bond pair leaves with the halogen. Each has its own by-products.

The amine side products follow from the mechanism. The primary amine made in the first substitution also has a lone pair on nitrogen, so it is a nucleophile too. It competes with ammonia for the halogenoalkane that is left. Using ammonia in large excess means a halogenoalkane molecule is far more likely to meet NH₃ than the amine. That is why "excess" is part of the conditions.

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

CH₃CH₂NH₂ + CH₃CH₂Br + NH₃ → (CH₃CH₂)₂NH + NH₄Br

Substitution or elimination?15.1.4 · 21.1.3

Hydroxide ions can do two different things to a halogenoalkane. As a nucleophile, OH⁻ attacks the δ+ carbon and replaces the halogen. As a base, it removes a hydrogen from the next carbon, and the C–X bond pair leaves. The result is a C=C. Both reactions happen together, so each step's product is the other step's by-product. Two factors decide the balance:

• The solvent. In water, substitution is favoured and the alcohol is the main product. In ethanol, elimination is favoured and the alkene is the main product. This is the difference the syllabus tests: NaOH(aq) for 15.1.3(a) and NaOH in ethanol for 15.1.4.
• The structure. The more alkyl groups around the C–X carbon, the more elimination wins. A primary halogenoalkane still gives plenty of substitution product even in ethanol. A tertiary one gives mostly alkene even in water.

Elimination has a by-product of its own. If the C–X carbon has two different neighbouring carbons that carry hydrogen, each can lose one, and two different alkenes form. If either of those alkenes can show cis–trans isomerism, it forms as both isomers. 2-Bromobutane gives but-1-ene and but-2-ene. But-2-ene forms as its cis and trans isomers, so three compounds come from one elimination.

What else is in the flask21.1.3

The remaining AS reactions each have their own by-product or complication. A question on "possible by-products" expects the one that belongs to the step it names.

CH₃CH₂OH + PCl₅ → CH₃CH₂Cl + POCl₃ + HCl

CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl

CH₃COOCH₂CH₃ + NaOH → CH₃COONa + CH₃CH₂OH

CH₃CH₂CN + NaOH + H₂O → CH₃CH₂COONa + NH₃

Analysing a whole route21.1.3

An analysis question gives a route, often with some intermediates or reagents missing, and asks about each step. Deal with one step at a time.

1. Compare the two structures. Which group has gone, which has appeared, and has the carbon count changed?
2. Name the reaction. The change of group decides the type (section 16). Give the mechanism name as well where one exists.
3. Give the reagent and conditions that make exactly that change (section 10).
4. Name the by-products: the inorganic one from the equation, and any organic side product that the type of reaction produces (sections 17 to 21).

For each step of each route below, the model tries every reagent on the page and keeps the ones that make exactly that change. It then reads the type of reaction, the inorganic by-products and the side reactions from the engine.

Ten things that lose marks in this topic21

Each of these appears somewhere above. Together they make a checklist to run through before handing in a synthesis answer.

Self-test21.1

Thirty questions across the three outcomes of 21.1. Each answer comes with its reasoning, so a wrong choice teaches as much as a right one.

Definitions to learn21.1

Data used on this page21.1

These are the values and assumptions that the tables and models on this page use. The values are typical literature values, not the official data booklet. Where an examination question gives data, use the data it gives.