Cambridge AS & A Level9701

Organic synthesis (36)

Chemistry 9701 Chapter Notes

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Organic synthesis
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In this chapter

  1. 1Principles of Organic Synthesis
  2. 2Key Functional Group Interconversions
  3. 3Extending the Carbon Chain
  4. 4Chirality in Pharmaceutical Synthesis

1. Principles of Organic Synthesis

Organic synthesis is the art of building complex organic molecules from simpler, more readily available ones. Chemists plan these multi-step reaction sequences, known as synthetic routes. A common strategy is 'retrosynthesis', which involves working backwards from the desired product (the target molecule) to identify suitable starting materials. Each step in the route requires specific reagents and conditions (e.g., temperature, pressure, catalyst, solvent) to ensure the desired transformation occurs with a high yield. Industrial syntheses also consider factors like cost, safety, and atom economy (the efficiency of converting reactants into the final product).

Key term

Retrosynthesis: A strategy for planning a chemical synthesis that involves breaking down a target molecule into simpler precursor structures without assuming knowledge of the starting materials.

Examiner insight

Examiners award marks for each correct step in a synthesis, which includes both the reagent(s) and the specific reaction conditions.

Common pitfall

Forgetting to state the conditions for a reaction, such as 'heat under reflux' or 'UV light', which are just as important as the reagents.

Worked example 14 marks

Devise a two-step synthesis to convert ethene (C₂H₄) into ethanoic acid (CH₃COOH). State all reagents and conditions.

  1. 1

    Step 1: Convert ethene to ethanol. This is an electrophilic addition reaction (hydration). Reagents: Steam (H₂O(g)). Conditions: High temperature (300 °C), high pressure (60-70 atm), and a solid phosphoric(V) acid (H₃PO₄) catalyst.

  2. 2

    Equation for Step 1: CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH

  3. 3

    Step 2: Oxidize ethanol to ethanoic acid. This requires a strong oxidizing agent and heating under reflux to ensure complete oxidation. Reagents: Acidified potassium dichromate(VI) (K₂Cr₂O₇/H₂SO₄). Conditions: Heat under reflux.

  4. 4

    Equation for Step 2: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O

Recap

  • Always plan a synthetic route by considering the functional groups in the starting material and the target molecule.
  • Each reaction step must have correctly stated reagents and conditions.
  • Retrosynthesis involves thinking backwards from the product to the reactant.
  • A good synthesis has a high overall yield and good atom economy.

Quick check

  1. What is the purpose of heating under reflux during the oxidation of a primary alcohol to a carboxylic acid?1 mark

2. Key Functional Group Interconversions

Success in organic synthesis hinges on knowing how to convert one functional group into another. These interconversions are the building blocks of any synthetic route. For example, an alcohol can be oxidized to an aldehyde or a carboxylic acid, while a halogenoalkane can be converted into an alcohol, a nitrile, or an amine. You must memorise the specific reagents and conditions for these key transformations. A reaction map or flowchart is an excellent way to visualise these connections.

Alkene → Alcohol: H₂O(g), H₃PO₄ catalyst, high T & P

Alcohol (1°) → Aldehyde: K₂Cr₂O₇/H₂SO₄, distil

Alcohol (1°/2°) → Carboxylic Acid/Ketone: K₂Cr₂O₇/H₂SO₄, reflux

Aldehyde/Ketone → Alcohol (1°/2°): NaBH₄(aq), warm

Halogenoalkane → Alcohol: NaOH(aq), reflux

Halogenoalkane → Nitrile: KCN in ethanol, reflux

Halogenoalkane → Amine: Excess conc. NH₃ in ethanol, heat in sealed tube

Nitrile → Carboxylic Acid: Dilute HCl(aq), reflux

Nitrile → Amine: LiAlH₄ in dry ether, or H₂/Ni catalyst

Key term

Functional Group: An atom or group of atoms within a molecule that is responsible for the characteristic chemical reactions of that molecule.

Examiner insight

Examiners often test the difference in oxidation products from primary and secondary alcohols, so be clear about the conditions (distillation vs. reflux) for primary alcohols.

Common pitfall

Confusing the reagents for nucleophilic substitution of halogenoalkanes. Use aqueous NaOH for hydrolysis to an alcohol, but ethanolic KCN for forming a nitrile.

Worked example 15 marks

Outline a synthetic route to produce propanone from propene. For each step, give the reagents, conditions, and the structure of the intermediate.

  1. 1

    The target is a ketone (propanone), and the start is an alkene (propene). Ketones can be made by oxidizing a secondary alcohol. Propene can be hydrated to form a secondary alcohol (propan-2-ol).

  2. 2

    Step 1: Hydration of propene to propan-2-ol. Reagents: Steam (H₂O(g)). Conditions: Phosphoric(V) acid catalyst (H₃PO₄), high temperature and pressure.

  3. 3

    Intermediate structure: CH₃CH(OH)CH₃ (propan-2-ol).

  4. 4

    Step 2: Oxidation of propan-2-ol to propanone. Reagents: Acidified potassium dichromate(VI) (K₂Cr₂O₇/H₂SO₄). Conditions: Heat under reflux.

  5. 5

    The final product is propanone, CH₃COCH₃.

Recap

  • Primary alcohols oxidize to aldehydes (distil) then to carboxylic acids (reflux).
  • Secondary alcohols oxidize to ketones (reflux).
  • Tertiary alcohols do not readily oxidize under these conditions.
  • Halogenoalkanes are versatile intermediates for making alcohols, nitriles, and amines.
  • Nitriles can be hydrolysed to carboxylic acids or reduced to primary amines.

Quick check

  1. What reagent is used to reduce a ketone back to a secondary alcohol?1 mark
  2. What are the reagents and conditions to convert bromoethane into ethylamine?2 marks

3. Extending the Carbon Chain

Many syntheses require increasing the number of carbon atoms in the molecule's main chain. The most common method you need to know involves the nitrile group (-C≡N). A halogenoalkane reacts with potassium cyanide (KCN) in an ethanolic solution via nucleophilic substitution. This replaces the halogen atom with a -CN group, adding one carbon atom to the chain. The resulting nitrile is a valuable intermediate; it can be hydrolysed by refluxing with dilute acid (e.g., HCl) to form a carboxylic acid, or it can be reduced using a strong reducing agent like LiAlH₄ to form a primary amine.

R-X + KCN → R-CN + KX (Conditions: ethanol, reflux)

R-CN + 2H₂O + H⁺ → R-COOH + NH₄⁺ (Conditions: dilute acid, reflux)

R-CN + 4[H] → R-CH₂NH₂ (Reagent: LiAlH₄ in dry ether)

Key term

Nitrile: An organic compound that contains a carbon-nitrogen triple bond (-C≡N) functional group.

Examiner insight

Questions that show a starting material with 'n' carbons and a product with 'n+1' carbons are a strong hint that you must use a nitrile intermediate.

Common pitfall

Forgetting that the carbon atom in the nitrile group (-CN) is included in the main chain count when naming the final product.

Fun fact

The nitrile group is found in amygdalin, a compound in bitter almonds, which can release toxic hydrogen cyanide upon hydrolysis.

Worked example 16 marks

Devise a three-step synthesis for propylamine (CH₃CH₂CH₂NH₂) starting from ethene (CH₂=CH₂). State all reagents and conditions.

  1. 1

    The route requires increasing the carbon chain from 2 carbons (ethene) to 3 carbons (propylamine). This points to using a nitrile intermediate. The sequence will be: Ethene → Halogenoethane → Propanenitrile → Propylamine.

  2. 2

    Step 1: Ethene to Bromoethane. Reagent: Hydrogen bromide (HBr). Conditions: Room temperature. This is an electrophilic addition reaction.

  3. 3

    Equation: CH₂=CH₂ + HBr → CH₃CH₂Br

  4. 4

    Step 2: Bromoethane to Propanenitrile. This step adds a carbon atom. Reagent: Potassium cyanide (KCN). Conditions: Dissolved in ethanol, heat under reflux. This is nucleophilic substitution.

  5. 5

    Equation: CH₃CH₂Br + KCN → CH₃CH₂CN + KBr

  6. 6

    Step 3: Propanenitrile to Propylamine. This is a reduction reaction. Reagent: Lithium tetrahydridoaluminate (LiAlH₄) in dry ether, followed by addition of water.

  7. 7

    Equation: CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂

Recap

  • To add one carbon atom, convert the starting material to a halogenoalkane, then react with KCN.
  • The reaction of a halogenoalkane with KCN is a nucleophilic substitution.
  • The resulting nitrile can be hydrolysed to a carboxylic acid (R-COOH).
  • The resulting nitrile can be reduced to a primary amine (R-CH₂NH₂).

Quick check

  1. What two products can be formed from propanenitrile, and what reagents are needed for each conversion?4 marks

4. Chirality in Pharmaceutical Synthesis

A carbon atom bonded to four different groups is called a chiral centre. Molecules with a chiral centre are 'chiral' and exist as a pair of non-superimposable mirror images called enantiomers. While enantiomers have identical physical properties (except for their effect on plane-polarised light), they can have vastly different biological effects. This is because enzymes and receptors in the body are also chiral, and only one enantiomer may fit correctly, like a key in a lock. Synthetic routes in the lab often produce a 50:50 mixture of enantiomers, called a racemic mixture. However, for drugs, it's highly desirable to produce only the single, active enantiomer. This is because the other enantiomer might be inactive (requiring a higher dose of the mixture) or, worse, cause harmful side effects, as famously seen with the drug thalidomide.

Key term

Racemic Mixture (or Racemate): An equimolar (50:50) mixture of two enantiomers of a chiral compound, which is optically inactive.

Examiner insight

When explaining the importance of single-enantiomer drugs, give a benefit for the patient (e.g., fewer side effects) and a separate benefit for the pharmaceutical company (e.g., lower dose needed, more efficient).

Common pitfall

Assuming the 'wrong' enantiomer is always toxic. Often it is simply biologically inactive, which is wasteful.

Fun fact

The morning sickness drug thalidomide was sold as a racemic mixture. The (R)-enantiomer was an effective sedative, but the (S)-enantiomer was a teratogen, causing severe birth defects.

Worked example 12 marks

The drug ibuprofen is chiral. Modern preparations produce only the active (S)-enantiomer. State two reasons why producing a single enantiomer is preferable to selling a racemic mixture.

  1. 1

    Reason 1 (Patient benefit): Fewer side effects. The inactive enantiomer may have unwanted biological effects. Using only the active enantiomer eliminates this risk.

  2. 2

    Reason 2 (Efficacy/Company benefit): Lower dosage required. A racemic mixture is only 50% effective. By using the pure active enantiomer, the dose can be halved for the same therapeutic effect, reducing costs.

Worked example 21 mark

Identify the chiral centre in the molecule of 2-hydroxypropanoic acid (lactic acid), CH₃CH(OH)COOH.

  1. 1

    A chiral centre is a carbon atom bonded to four different groups.

  2. 2

    Examine the central carbon atom (C2). It is bonded to: 1. a -CH₃ group, 2. a -H atom, 3. an -OH group, and 4. a -COOH group.

  3. 3

    Since these four groups are all different, the second carbon atom is the chiral centre.

Recap

  • A chiral centre is a carbon atom attached to four different groups.
  • Molecules with a chiral centre exist as non-superimposable mirror images called enantiomers.
  • A 50:50 mixture of enantiomers is a racemic mixture and is optically inactive.
  • Single-enantiomer drugs offer better therapeutic activity and fewer side effects.
  • Enzymes and cell receptors are stereospecific, meaning they interact differently with different enantiomers.

Quick check

  1. What is the term for a molecule that is non-superimposable on its mirror image?1 mark
  2. Why are most drugs extracted from natural sources single enantiomers?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Name the functional groups present in the paracetamol molecule, which has the structure HO-C₆H₄-NHCOCH₃.2 marks
  2. State the reagents and conditions required to convert chloroethane into ethanol.2 marks
  3. Devise a two-step synthesis to prepare propan-1-ol from propanal (CH₃CH₂CHO). For each step, identify the type of reaction and state the necessary reagents.4 marks
  4. A student plans to synthesise butanoic acid from propan-1-ol. Outline the three steps required for this synthesis, giving reagents and conditions for each step.6 marks
  5. The molecule 2-bromobutane is chiral. Draw the 3D structures of the two enantiomers of 2-bromobutane, clearly showing their mirror-image relationship.3 marks
  6. Explain why the reaction of but-1-ene with HBr produces mainly 2-bromobutane rather than 1-bromobutane. Name the intermediate involved.4 marks
  7. Compound X, C₄H₈O, can be oxidised to form Compound Y, C₄H₈O₂. When Compound X is warmed with NaBH₄, it forms butan-1-ol. Deduce the structures of X and Y, and describe the observation you would make when X is oxidised to Y using acidified potassium dichromate(VI).5 marks
  8. Devise a multi-step synthetic route to convert benzene into phenylamine (aniline, C₆H₅NH₂). You should show the structure of any intermediates and state the reagents and conditions for each step.5 marks
  9. Lactic acid (2-hydroxypropanoic acid) can be converted into 2-aminopropanoic acid (alanine) in two steps. Suggest reagents and conditions for this two-step conversion. Write equations for the reactions.6 marks
  10. Explain, with reference to their mechanism of action, why it is often necessary to produce drugs as single enantiomers rather than as racemic mixtures. Use a specific example to support your answer.4 marks

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