Organic synthesis

In a nutshell

A level organic synthesis has two jobs: make and purify an organic solid, and plan multi-stage routes that use every reaction in the course.

The practical side adds three techniques to the AS liquid skills: filtration under reduced pressure, recrystallisation and measuring melting points. The routes side asks you to spot every functional group in a complicated molecule, predict how it reacts, and chain reactions together, sometimes lengthening the carbon chain on the way.

Assumed knowledge: Organic synthesis (AS), Carbon-carbon bond formation, Amount of substance (percentage yield).

Core content

Practical skills: organic preparation with Quickfit apparatus

This is the same Quickfit kit as AS. What changes at A level is that the product is often a solid.

Heating under reflux is for slow reactions that need prolonged heating. The vapours condense and return to the flask, so no volatile reactant or product is lost.

Distillation is for removing a volatile liquid from a mixture, collecting it close to its boiling point.

The drawing details that still earn marks:

• Reflux condenser vertical and open at the top; water in at the bottom, out at the top.
• Distillation thermometer bulb level with the side-arm of the still head.
• Anti-bumping granules in the flask; heat electrically, because organic liquids are flammable.
• No lines drawn across the joints (a line looks like a seal).

The full AS method for an organic liquid (separating funnel, drying with an anhydrous salt, redistillation) is in Organic synthesis (AS).

Practical skills: filtration under reduced pressure

Filtration under reduced pressure separates a solid from a liquid using a Büchner funnel on a Büchner (side-arm) flask connected to a vacuum pump (often a water pump).

Why reduce the pressure:

• The low pressure in the flask sucks the liquid through quickly, much faster than gravity filtration.
• It pulls most of the solvent out of the solid, leaving drier crystals.

Points of technique:

• The filter paper lies flat and covers all the holes; wet it with a little solvent so it seals before you pour.
• The solid (residue) stays on the paper. The liquid (filtrate) collects in the flask.

Filtration only separates a solid from a liquid. It does not purify the solid on its own; that is what recrystallisation is for.

Practical skills: recrystallisation

Recrystallisation purifies an impure solid. It relies on the product being much more soluble in a hot solvent than in the same solvent cold.

1. Dissolve the impure solid in the minimum volume of hot solvent, so the solution is saturated with the product.
2. If there are insoluble impurities, filter the hot solution quickly to remove them.
3. Cool the solution, then stand it in ice. The product crystallises out.
4. Soluble impurities stay in solution, because there is only a small amount of each, so the cold solution is not saturated with them.
5. Filter under reduced pressure to collect the crystals.
6. Wash the crystals with a small volume of ice-cold solvent to rinse off the impurity-containing solution.
7. Dry the crystals, then weigh them.

What makes a suitable solvent:

• The product dissolves well when hot but poorly when cold.
• Soluble impurities stay dissolved when the solution is cold.
• It does not react with the product.

Every step loses a little product (some always stays dissolved in the cold solvent), so the yield after recrystallisation is lower, but the product is purer.

Still don't get it? · why the impurities stay behind

Think of stirring sugar into a mug of very hot water until no more will dissolve. Now there is a pinch of salt in there too. As the mug cools, cold water cannot hold as much sugar, so sugar crystals start to appear at the bottom. The pinch of salt never comes out: there is so little of it that even cold water can easily keep it dissolved.

Recrystallisation uses exactly that:

1. Using the minimum hot solvent means the solvent is holding as much product as it possibly can.
2. When it cools, the solvent can hold much less product, so the extra product has to come out as crystals.
3. Each impurity is only a small amount. The cold solvent is nowhere near full of it, so it stays dissolved.
4. Filtering now separates the crystals (product) from the solution (impurities).
5. Too much solvent at step 1 ruins this: the cold solvent could then hold most of the product too, and very little crystallises.

In exam wording: dissolve in the minimum volume of hot solvent; cool so the product crystallises; soluble impurities remain in solution; filter under reduced pressure; wash with cold solvent; dry.

Practical skills: measuring melting points

A melting point tells you how pure a solid is, and can help identify it.

• A pure solid melts sharply, over a narrow range (typically 1 to 2 °C), at the data-book value.
• An impure solid melts at a lower temperature and over a wider range.

How to measure it:

1. Pack a small amount of the dry, powdered solid into a capillary tube sealed at one end.
2. Heat it in melting point apparatus (electrical, or an oil bath with the tube beside the thermometer bulb).
3. Heat slowly near the expected melting point, so the thermometer keeps up with the sample.
4. Record the temperature when the solid starts to melt and when it has completely melted. That is the melting range.
5. Compare with the data-book value.

The same idea identifies an unknown carbonyl compound from the melting point of its 2,4-DNP derivative (see Carbonyl compounds).

PAG6: synthesis of an organic solid

PAG6 is OCR's practical activity group for making and purifying an organic solid. It covers heating under reflux, filtration, recrystallisation, using melting point apparatus, thin layer or paper chromatography and a risk assessment. The spec's example is the synthesis of aspirin.

Aspirin is made by esterifying the phenol -OH group of 2-hydroxybenzoic acid with ethanoic anhydride. A carboxylic acid would not do this: phenols are not readily esterified by carboxylic acids.

C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH

2-hydroxybenzoic acid + ethanoic anhydride → aspirin + ethanoic acid

Risk assessment: concentrated acid and ethanoic anhydride are corrosive, so wear eye protection and gloves and handle them in a fume cupboard; heat with a water bath or electric heater, not a naked flame. TLC itself is taught in Chromatography and qualitative analysis.

Synthetic routes: identifying functional groups

Functional group: a group of atoms responsible for the characteristic reactions of a compound.

At A level you must spot every functional group from Module 4 and Module 6, even when several sit in one molecule. The Module 4 ones (alkene, alcohol, haloalkane, aldehyde, ketone, carboxylic acid) are in the AS note. These are the Module 6 additions:

Two-group families to recognise on sight: an amino acid is an amine plus a carboxylic acid; a hydroxynitrile is an alcohol plus a nitrile.

The quickest way to sort the C=O groups is to ask what the carbonyl carbon is bonded to:

And for the -OH and -N groups:

• -OH on a ring carbon is a phenol; -OH on a chain carbon is an alcohol. They react differently.
• N next to C=O is an amide; N with no C=O neighbour is an amine.

Synthetic routes: predicting properties and reactions

Treat each functional group as reacting the way it always does, with the reagents you know for it. A group with no reaction to a reagent is left unchanged.

For 4-hydroxybenzaldehyde, HOC₆H₄CHO:

Equations for two of these rows:

HOC₆H₄CHO + 2[H] → HOC₆H₄CH₂OH

HOC₆H₄CHO + CH₃COCl → CH₃COOC₆H₄CHO + HCl

Properties follow from the groups too. The phenol -OH can hydrogen bond, so this compound is more soluble in water and has a higher melting point than benzaldehyde.

Synthetic routes: the A level reaction toolkit

These are the Module 6 transformations to add to the Module 4 toolkit from the AS note. Learn the reagents and the conditions for every arrow: route questions mark both.

Carbonyls, acids and their derivatives

Nitrogen compounds and chain lengthening

Aromatic compounds

Exam papers may give extra information about a reaction you have not met. Treat it as one more arrow in the toolkit.

Synthetic routes: changing the carbon skeleton

Before planning anything, count the carbon atoms in the start and the target. Most reactions leave the skeleton alone. Only these make a new C-C bond:

Two things to notice:

• The nitrile carbon stays in the chain whatever you do next: H₂/Ni turns -CN into -CH₂NH₂, and acid hydrolysis turns it into -COOH.
• Esterification and amide formation join two molecules through an O or N atom, not a C-C bond. Hydrolysis splits them apart again.

For example, ethanal to 2-hydroxypropanoic acid (two carbons to three):

CH₃CHO + HCN → CH₃CH(OH)CN (NaCN(aq)/H⁺(aq))

CH₃CH(OH)CN + 2H₂O + HCl → CH₃CH(OH)COOH + NH₄Cl (HCl(aq), heat under reflux)

Synthetic routes: aromatic routes and the order of steps

On a benzene ring, the order of the steps decides where the next group goes. A group already on the ring directs the incoming group:

So when a route ends in a 3-substituted product, put the -NO₂ on first and use it to steer, then convert it to -NH₂ last if you need an amine. Worked example 2 shows this.

Activation matters too. Phenol reacts with bromine water and dilute nitric acid with no catalyst, while benzene needs a halogen carrier or concentrated acids. Choose conditions to match the ring you actually have at that stage.

Synthetic routes: devising a multi-stage route

1. Identify the functional groups in the start and the target.
2. Count the carbons. One more in the target means a CN step (or, on a ring, Friedel-Crafts).
3. Work backwards from the target: which group can become the target in one step?
4. Keep working back until you reach something the starting compound can become in one step.
5. On a ring, check the order against the directing effects.
6. For each stage, write the reagents, the conditions and the intermediate. Add equations if asked.

Still don't get it? · planning a route that changes the chain

Imagine building a Lego model of a dog, and you are handed a model of a cat. Most of your moves just swap a piece for another of the same size: change the ears, change the tail. But if the dog needs one more brick in its body, only a couple of special moves let you add a brick, and they only work on certain parts of the model.

A synthesis route works the same way:

1. Swapping one functional group for another (alcohol to haloalkane, aldehyde to acid) is a normal move. The number of carbons does not change.
2. Adding a carbon is a special move. There are only two on the chain: CN^- on a haloalkane, or HCN on an aldehyde or ketone.
3. So if the target has one more carbon, the route must pass through a haloalkane or a carbonyl, then a nitrile.
4. Once the nitrile is made, the new carbon is part of the chain for good. You can turn the -CN into -CH₂NH₂ or -COOH.
5. Plan backwards: target, then the nitrile it came from, then the haloalkane or carbonyl that made the nitrile, then how to make that from your start.

In exam wording, state for each stage the reagent, the conditions and the intermediate, for example "Stage 2: NaCN in ethanol, heat under reflux, forms butanenitrile".

Worked examples

1. Three-stage route with chain lengthening: propan-1-ol to butylamine

Plan. Propan-1-ol has 3 carbons; butylamine (butan-1-amine), CH₃CH₂CH₂CH₂NH₂, has 4. A carbon must be added, so go through a nitrile. Working back: amine ← nitrile (H₂/Ni) ← haloalkane (CN^-) ← alcohol (NaBr/H₂SO₄).

Stage 1: NaBr and H₂SO₄ (which make HBr), heat. Intermediate: 1-bromopropane.

CH₃CH₂CH₂OH + HBr → CH₃CH₂CH₂Br + H₂O

Stage 2: NaCN in ethanol, heat under reflux. Intermediate: butanenitrile.

CH₃CH₂CH₂Br + CN^- → CH₃CH₂CH₂CN + Br^-

Stage 3: H₂ with a nickel catalyst. Product: butylamine.

CH₃CH₂CH₂CN + 2H₂ → CH₃CH₂CH₂CH₂NH₂

Where marks go:

• The nitrile has four carbons, so it is butanenitrile, not propanenitrile: the CN carbon counts.
• "Ethanol" is a condition, not decoration. In water, the haloalkane would be hydrolysed to an alcohol instead.
• Heating 1-bromopropane with ammonia gives propylamine (3 carbons), not butylamine.

2. Aromatic route where order matters: benzene to 3-bromophenylamine

Plan. The target has -NH₂ and -Br 3-positions apart. -NH₂ directs to the 2- and 4-positions, so it cannot be on the ring when the Br goes on. -NO₂ directs to the 3-position, so: nitrate, then brominate, then reduce.

Stage 1: concentrated HNO₃ and concentrated H₂SO₄, about 50 °C. Intermediate: nitrobenzene.

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

Stage 2: Br₂ with a halogen carrier such as FeBr₃. The -NO₂ group directs Br to the 3-position. Intermediate: 3-bromonitrobenzene.

C₆H₅NO₂ + Br₂ → C₆H₄BrNO₂ + HBr

Stage 3: Sn and concentrated HCl, heat under reflux, then NaOH(aq). Product: 3-bromophenylamine.

C₆H₄BrNO₂ + 6[H] → C₆H₄BrNH₂ + 2H₂O

Where marks go: reducing first gives phenylamine, and -NH₂ would send the bromine to the 2- and 4-positions. The order of stages 2 and 3 is the whole question.

3. Percentage yield of a recrystallised solid

Task: 2.76 g of 2-hydroxybenzoic acid is reacted with excess ethanoic anhydride. After recrystallisation and drying, 2.43 g of aspirin is obtained. Calculate the percentage yield.

C₇H₆O₃ + (CH₃CO)₂O → C₉H₈O₄ + CH₃COOH (1 : 1 ratio)

Step 1: molar masses (A_r values from the data sheet).

$M(\text{C}_7\text{H}_6\text{O}_3) = (7 \times 12.0) + (6 \times 1.0) + (3 \times 16.0) = 138.0 \text{ g mol}^{-1}$

$M(\text{C}_9\text{H}_8\text{O}_4) = (9 \times 12.0) + (8 \times 1.0) + (4 \times 16.0) = 180.0 \text{ g mol}^{-1}$

Step 2: amount of the limiting reagent. Ethanoic anhydride is in excess, so 2-hydroxybenzoic acid sets the maximum yield.

$n(\text{C}_7\text{H}_6\text{O}_3) = \frac{2.76 \text{ g}}{138.0 \text{ g mol}^{-1}} = 0.0200 \text{ mol}$

Step 3: theoretical mass of aspirin (1 : 1, so 0.0200 mol).

$m_{\text{theoretical}} = 0.0200 \text{ mol} \times 180.0 \text{ g mol}^{-1} = 3.60 \text{ g}$

Step 4: percentage yield.

$\text{percentage yield} = \frac{2.43 \text{ g}}{3.60 \text{ g}} \times 100 = 67.5\%$

The data are given to 3 significant figures, so the answer is 67.5%.

Where marks are dropped:

• Using the moles of ethanoic anhydride. It is in excess, so it does not limit the yield.
• Dividing the wrong way up: 3.60 ÷ 2.43 × 100 gives 148%, which is impossible.
• Explaining a low yield as "the reaction did not finish" only. Recrystallisation itself loses product: some stays dissolved in the cold solvent and some is lost on filtering and washing.
• A yield over 100% in a real prep usually means the crystals were not fully dry.

4. Overall yield of a multi-stage route

Task: In Worked example 1, the three stages have yields of 80.0%, 65.0% and 90.0%. Calculate the overall percentage yield, and the mass of butylamine made from 0.150 mol of propan-1-ol.

Step 1: multiply the stage yields as decimals. Each stage only works on what the stage before it produced.

$\text{overall yield} = 0.800 \times 0.650 \times 0.900 = 0.468 = 46.8\%$

Step 2: amount of product.

$n(\text{C}_4\text{H}_{11}\text{N}) = 0.150 \text{ mol} \times 0.468 = 0.0702 \text{ mol}$

Step 3: mass of product.

$M(\text{C}_4\text{H}_{11}\text{N}) = (4 \times 12.0) + (11 \times 1.0) + 14.0 = 73.0 \text{ g mol}^{-1}$

$m = 0.0702 \text{ mol} \times 73.0 \text{ g mol}^{-1} = 5.12 \text{ g}$

Where marks are dropped: adding or averaging the yields (giving 235% or 78.3%) instead of multiplying them. Every extra stage lowers the overall yield, which is why chemists prefer routes with fewer stages.

Common exam mistakes

• Dissolving in "plenty of" hot solvent to recrystallise. It must be the minimum volume of hot solvent; extra solvent keeps product dissolved when cold and cuts the yield.
• Saying a good solvent "dissolves the impurities but not the product". The product must dissolve when hot and be much less soluble when cold; soluble impurities stay in the cold solution.
• Washing the crystals with warm solvent, or lots of it. Use a small volume of ice-cold solvent, or the product redissolves.
• Saying impurities raise the melting point. Impurities lower the melting point and make it melt over a wider range.
• Describing a pure solid as melting "over a range". A pure solid melts sharply (narrow range) at the data-book value.
• Saying filtration under reduced pressure "purifies" the solid. It separates the solid from the liquid quickly and leaves it drier; recrystallisation does the purifying.
• Weighing crystals that are still wet, then explaining a yield above 100% as "experimental error". The mass includes solvent: dry to constant mass.
• Esterifying a phenol with a carboxylic acid. Phenols are not readily esterified by carboxylic acids; use an acyl chloride (or an acid anhydride, as for aspirin).
• Expecting phenol to fizz with Na₂CO₃. Phenol reacts with NaOH but not with carbonates; only a -COOH gives CO₂.
• Calling the C=O of an ester or amide a "ketone", or a phenol -OH an "alcohol". Look at what the carbonyl carbon (or the -OH) is attached to.
• Using NaBH₄ to reduce a nitrile to an amine. The reagent is H₂/Ni. NaBH₄ reduces only aldehydes and ketones.
• Writing "KCN(aq)" for haloalkane to nitrile. The CN^- must be in ethanol; in water the haloalkane is hydrolysed to an alcohol.
• Losing the extra carbon when naming. Bromoethane + CN^- gives propanenitrile; reducing it gives propylamine.
• Getting the order of aromatic steps wrong. Reduce -NO₂ to -NH₂ before substituting and the new group goes to the 2- and 4-positions, not the 3-position.
• Mixing up conditions for benzene and phenol: benzene needs concentrated HNO₃/H₂SO₄ or a halogen carrier; phenol reacts with dilute HNO₃ and bromine water without one.
• Giving reagents without conditions or intermediates. Each stage of a route needs reagent, conditions and the intermediate.

Key definitions

• Functional group: a group of atoms responsible for the characteristic reactions of a compound.
• Heating under reflux: heating a mixture with a vertical condenser so that vapours condense and return to the flask, allowing prolonged heating without loss of volatile substances.
• Filtration under reduced pressure: filtering through a Büchner funnel into a Büchner flask connected to a vacuum, so the liquid is drawn through quickly and the solid is left drier.
• Recrystallisation: purifying a solid by dissolving it in the minimum volume of hot solvent and cooling, so the product crystallises while soluble impurities remain in solution.
• Melting point as a test of purity: a pure solid melts sharply at its data-book value; impurities lower the melting point and widen the melting range.
• Multi-stage synthetic route: a sequence of reactions that converts a starting compound into a target compound through one or more intermediates.

Specification

• I can describe organic preparations using Quickfit apparatus, including heating under reflux and distillation, and draw both set-ups correctly. (6.2.5 a i)
• I can describe filtration under reduced pressure and explain why it is used. (6.2.5 a ii)
• I can describe how to purify an organic solid by recrystallisation and explain each step, including the choice of solvent. (6.2.5 a ii)
• I can describe how to measure a melting point and use it to judge purity and identify a compound. (6.2.5 a ii)
• I can outline the preparation and purification of an organic solid such as aspirin, and calculate its percentage yield. (6.2.5 a, PAG6)
• I can identify every functional group from Modules 4 and 6 in a molecule that contains several of them. (6.2.5 b i)
• I can predict the properties and reactions of a molecule with several functional groups, including which group reacts with a given reagent. (6.2.5 b ii)
• I can devise multi-stage synthetic routes between all the functional groups in the specification, giving reagents, conditions and intermediates. (6.2.5 c)
• I can use C-C bond-forming reactions to lengthen a carbon chain, and directing effects to choose the order of steps on a benzene ring, within a route. (6.2.5 c)
• I can use extra reactions given in a question as part of a route. (6.2.5 c)

Related notes

• Organic synthesis (AS): liquid preparations, the Module 4 toolkit and two-stage routes
• Amount of substance: percentage yield calculations
• Aromatic compounds: electrophilic substitution, phenol and directing effects
• Carbonyl compounds: oxidation, NaBH₄, HCN and the 2,4-DNP and Tollens' tests
• Carboxylic acids and esters: esterification, hydrolysis and acyl chlorides
• Amines: preparing aliphatic and aromatic amines
• Amino acids, amides and chirality: amides and multi-group molecules
• Polyesters and polyamides: condensation polymers and their hydrolysis
• Carbon-carbon bond formation: nitriles and Friedel-Crafts reactions
• Chromatography and qualitative analysis: TLC and functional group tests to check a product
• Spectroscopy: confirming the structure of what you have made