Organic synthesis

In a nutshell

Organic synthesis is the skill of joining up the reactions you already know into a route that turns a given starting material into a target molecule, in up to four steps.

Nothing here is a brand-new reaction: this subtopic rewards you for recognising the functional groups at the start and the end, and picking the reagents and conditions that get you between them.

It also asks you to explain why chemists design greener routes: fewer steps, high atom economy, no solvent and safer starting materials.

Assumed knowledge: Halogenoalkanes, Alkenes, Alcohols, Amount of substance (atom economy and percentage yield).

Core content

Designing a greener synthesis: fewer steps and high atom economy

AQA wants the reasons behind good process design, not just the slogans. Lead with the reason that earns the mark.

Fewer steps is better because:

• Yield is lost at every step, and the losses multiply, so a shorter route gives a higher overall yield.
• Each step costs time, energy and money and usually needs its own separation and purification, so fewer steps means less energy, less waste and lower cost.

How badly yield drains away with extra steps, if each step runs at 80%:

High atom economy is better because a larger fraction of the mass of the reactants ends up in the desired product, so less is wasted and the process is more sustainable and economical.

Atom economy is set by the equation, not by how well the reaction is carried out (that is percentage yield). A reaction can have a high yield but poor atom economy if it makes a lot of by-product.

The rule worth remembering: an addition reaction has 100% atom economy (everything goes into one product), while substitution and elimination always lose atoms as a by-product.

Why addition beats substitution: a worked contrast

Compare two routes to the same product, ethanol.

Addition (hydration of ethene):

$CH_2{=}CH_2 + H_2O \rightarrow C_2H_5OH$

$\text{atom economy} = \frac{46.0}{28.0 + 18.0} \times 100 = 100\%$

Substitution (bromoethane with aqueous alkali):

$C_2H_5Br + NaOH \rightarrow C_2H_5OH + NaBr$

$\text{atom economy} = \frac{46.0}{108.9 + 40.0} \times 100 = 30.9\%$

The substitution route throws away the NaBr as waste, which is why its atom economy is so low. (The full atom-economy method is in Amount of substance.)

Greener still: no solvent, safer starting materials

Chemists also aim to design processes that do not need a solvent and that use non-hazardous starting materials.

Avoiding a solvent is better because:

• The solvent would later have to be separated from the product, which costs money and loses yield.
• Many organic solvents are flammable, volatile or toxic, so leaving them out cuts hazard, waste and pollution.

Using non-hazardous starting materials is better because it is safer for workers and the environment and produces less harmful waste.

The aliphatic reaction toolbox

A synthesis is only as good as the reactions you can call on. The tables below are the AQA toolbox: each row is one step you are allowed to use. The roadmap shows the same reactions as a map.

From an alkene (C=C):

From a halogenoalkane (R-X):

Note how the solvent for NaOH decides the product: water gives substitution to an alcohol, ethanol gives elimination to an alkene.

From an alcohol (R-OH), using acidified potassium dichromate(VI), K₂Cr₂O₇ / H₂SO₄ (orange to green):

An alcohol can also go to:

From an aldehyde or ketone (C=O):

From a nitrile (R-CN):

From a carboxylic acid or acyl chloride:

Esterification is reversible: carboxylic acid + alcohol ⇌ ester + water, which is why the acyl chloride routes (not reversible) are used when a good yield of ester or amide is needed.

The only ways to change the carbon chain length

Most steps swap one functional group for another but leave the carbon skeleton the same length. Only a few steps change the number of carbons, and they are the ones exam routes are built around:

• KCN on a halogenoalkane adds one carbon (R-X becomes R-CN).
• KCN (with dilute acid) or HCN on an aldehyde or ketone adds one carbon (a hydroxynitrile).
• On a benzene ring, Friedel-Crafts acylation adds the carbons of the acyl group.

So if your target has more carbons than your starting material, a nitrile step almost always has to appear in the route.

The aromatic reaction toolbox

Benzene chemistry gives a short, high-value route to aromatic amines and ketones.

How to devise a route

A method that works under exam pressure:

1. Draw the start and the target and label the functional group on each.
2. Count the carbons. If the target has more, plan a nitrile step to add them.
3. Work backwards from the target: ask what single reaction in the toolbox makes that functional group, and what it would be made from.
4. Chain the steps so each step's product is the next step's starting material, and check you are within four steps.
5. Write reagents and conditions for every step (and the intermediate structures if asked). You do not need equations or mechanisms unless the question asks for them.

Still don't get it? · devising a multi-step route

Think of it like planning a train journey when there is no direct line. You know the station you start at (your starting material) and the station you must reach (the target), and the map of lines is your reaction toolbox. The job is to find a sequence of connections, with no more than four legs.

Rebuild it one step at a time:

• First look at both ends. A C=C at the start and an NH2 at the target are two different "stations", so you will need connections between them.
• Count the carbons at each end. If the target is bigger, you have to pass through a nitrile, because KCN is the main line that adds a carbon.
• Now work backwards. "How do I make an amine? By reducing a nitrile. How do I make that nitrile? From a halogenoalkane and KCN. How do I make the halogenoalkane? From the alkene and HBr." You have just built the route in reverse.
• Read it forwards and check every product feeds the next step, and that there are four legs or fewer.

Back to the exam wording: each "leg" is one step, and for each step you must give the reagent and the conditions (for example "KCN in ethanol, reflux"), plus the structure of each intermediate if the question asks for it.

Worked examples

Example 1: synthesise propanoic acid, CH₃CH₂COOH, from bromoethane, CH₃CH₂Br.

The target has three carbons; bromoethane has two. You need to add one carbon, so a nitrile step is required.

• Step 1: CH₃CH₂Br + KCN (in ethanol, reflux) → CH₃CH₂CN (propanenitrile). Nucleophilic substitution; this adds the extra carbon.
• Step 2: CH₃CH₂CN + dilute HCl, reflux → CH₃CH₂COOH (propanoic acid). Hydrolysis.

Two steps. The reagent in Step 1 is KCN, not HCN, and there is no acid in that step.

Example 2: synthesise butane-1,4-diamine, H₂NCH₂CH₂CH₂CH₂NH₂, from ethene, CH₂=CH₂.

The target has four carbons with an NH₂ on each end; ethene has two carbons. You must add one carbon at each end, so you need two nitrile groups, which means a dibromide first.

• Step 1: CH₂=CH₂ + Br₂ → BrCH₂CH₂Br (1,2-dibromoethane). Electrophilic addition.
• Step 2: BrCH₂CH₂Br + KCN (in ethanol, reflux) → NCCH₂CH₂CN (butanedinitrile). Nucleophilic substitution; a carbon added at each end.
• Step 3: NCCH₂CH₂CN + H₂ / Ni (or LiAlH₄ in dry ether) → H₂NCH₂CH₂CH₂CH₂NH₂ (butane-1,4-diamine). Reduction.

Three steps. The carbon count is the whole puzzle: a single CN would give the wrong product, and HBr or H₂ would add no carbon at all.

Example 3: synthesise phenylamine, C₆H₅NH₂, from benzene, C₆H₆.

• Step 1: C₆H₆ + conc HNO₃ + conc H₂SO₄, about 50 °C → C₆H₅NO₂ (nitrobenzene). Electrophilic substitution (nitration).
• Step 2: C₆H₅NO₂ + Sn + conc HCl, then NaOH → C₆H₅NH₂ (phenylamine). Reduction.

Two steps. Both nitration acids must be written as concentrated, the reducing agent must be written in full as Sn and HCl (not just "H⁺"), and the NaOH at the end is needed to free the amine from its salt.

Common exam mistakes

• Giving an incomplete reagent: writing "potassium dichromate" without the acid, or "dichromate" without saying acidified. The acid is part of the reagent and the mark is lost without it.
• Confusing the two cyanide reactions: a halogenoalkane needs KCN with no acid (substitution), but an aldehyde or ketone needs KCN with dilute acid, or HCN (addition). Writing HCN for the halogenoalkane, or adding concentrated acid to the carbonyl step, is rejected.
• Using NaBH₄ to reduce a nitrile: it does not work. Nitriles need H₂/Ni, LiAlH₄ (dry ether) or Na in ethanol. Adding acid in the same step as LiAlH₄ also loses the mark.
• Reducing nitrobenzene with just "H⁺": the reagent must be written in full as Sn and HCl (or Fe/HCl, or H₂/Ni).
• Forgetting "concentrated" on both acids in nitration; one or both missing costs a mark.
• Not counting the carbons: picking HBr or H₂ when the target has more carbons than the start, so no carbon is ever added, or adding only one CN to a target that needs one on each end.
• Mixing up the NaOH conditions: aqueous NaOH gives the alcohol (substitution), ethanolic NaOH gives the alkene (elimination).
• Writing full equations and mechanisms when the question asks only for reagents and structures. These earn no extra marks and waste exam time.
• A wrong reagent is penalised as a chemical error and usually loses the product mark too, though a later step can still be marked consequentially from your (wrong) structure.

Key definitions

• Atom economy is a measure of the proportion of the mass of the reactants that ends up as the desired product:

$\%\ \text{atom economy} = \frac{\text{mass of desired product}}{\text{total mass of reactants}} \times 100$

• Percentage yield compares how much product was actually made with the maximum possible from the equation:

$\%\ \text{yield} = \frac{\text{actual amount (mol) of product}}{\text{theoretical amount (mol) of product}} \times 100$

(Both are set out in full in Amount of substance; atom economy depends only on the equation, yield on how the reaction is carried out.)

Specification

• I can explain why chemists aim to design processes that do not require a solvent and that use non-hazardous starting materials.
• I can explain why chemists aim to design production methods with fewer steps that have a high percentage atom economy.
• I can use reactions in the specification to devise a synthesis, of up to four steps, for an organic compound, giving the reagents, conditions and intermediate structures.

Related notes

• Halogenoalkanes (substitution to alcohols, nitriles and amines; elimination)
• Alkenes (electrophilic addition, the usual way into a route)
• Alcohols (oxidation, dehydration, esterification)
• Amount of substance (atom economy and percentage yield)
• Aldehydes and ketones (3.3.8): reduction and the hydroxynitrile step
• Carboxylic acids and derivatives (3.3.9): esters, acyl chlorides and amides
• Aromatic chemistry (3.3.10): nitration and Friedel-Crafts acylation
• Amines (3.3.11): reduction of nitriles and of nitrobenzene