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How do we plan multi-step syntheses and account for the handedness of molecules?

Optical isomerism and chirality, the synthesis of organic compounds through multi-step routes, choosing reagents and conditions, and the importance of single enantiomers in the pharmaceutical industry.

An Edexcel 9CH0 answer covering optical isomerism and chirality, planning multi-step organic syntheses, reagents and conditions, and single enantiomers in pharmaceuticals.

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What this topic is asking

Edexcel Topic 19 wants you to recognise chirality and optical isomerism, plan multi-step syntheses by choosing the correct reagents and conditions for each functional-group conversion, predict when a reaction gives a racemate, and explain why single enantiomers matter in medicines.

The answer

Chirality and optical isomerism

To identify a chiral centre, look for a carbon with four genuinely different substituents. For example, in 2-hydroxypropanoic acid the central carbon carries CH3\text{CH}_3, OH\text{OH}, COOH\text{COOH} and H\text{H}, four different groups, so it is chiral. A carbon bearing two identical groups (such as the central carbon of propan-2-ol, which has two CH3\text{CH}_3) is not a chiral centre.

When reactions give a racemate

Whenever a chiral centre is created by attack on a planar group, both products form equally. The classic case is nucleophilic addition to a carbonyl: the C=O\text{C=O} is planar, so a nucleophile such as CNβˆ’\text{CN}^- can attack from either face with equal probability, producing equal amounts of each enantiomer, that is a racemic mixture. Recognising this is a common exam requirement.

Planning a multi-step synthesis

Multi-step synthesis joins individual functional-group conversions into a route. Useful conversions to combine include:

  • alkene to alcohol (steam, H3PO4\text{H}_3\text{PO}_4 catalyst) or to haloalkane (HX);
  • haloalkane to alcohol (warm aqueous OHβˆ’\text{OH}^-), to nitrile (ethanolic KCN\text{KCN}, adding a carbon), or to amine (excess ammonia);
  • nitrile to amine (LiAlH4\text{LiAlH}_4) or to carboxylic acid (acid hydrolysis);
  • primary alcohol to aldehyde (distil with acidified dichromate) then to acid (reflux);
  • acid to ester (alcohol, conc H2SO4\text{H}_2\text{SO}_4) or to acyl chloride (SOCl2\text{SOCl}_2).

At each step you must state the reagent and the conditions (temperature, solvent, catalyst). Plan by working backwards from the target functional group and checking the carbon count, remembering that the nitrile route is the standard way to lengthen the chain.

Single enantiomers in pharmaceuticals

Examples in context

Example 1. Ibuprofen. Ibuprofen has one chiral centre and is active mainly as a single enantiomer, yet it is sold as a racemate because the body slowly converts the inactive form into the active one. This illustrates the central message of the topic: a single enantiomer is usually preferred, but the economics of synthesis and the body's own chemistry decide whether selling a racemate is acceptable.

Example 2. Limonene and the smell of citrus. The two enantiomers of limonene smell different: one of oranges, the other of lemons. Because our olfactory receptors are chiral, they respond differently to each mirror image. This everyday example shows why enantiomers, identical in most physical properties, can behave completely differently in a chiral biological environment, exactly the principle behind single-enantiomer drugs.

Try this

Q1. State what is meant by a chiral centre. [1 mark]

  • Cue. A carbon atom bonded to four different groups.

Q2. Explain why the pharmaceutical industry prefers to use a single enantiomer of a drug. [2 marks]

  • Cue. Enantiomers can have different biological effects; using one reduces side effects and the required dose.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20205 marks2-hydroxypropanoic acid (lactic acid), CH3CH(OH)COOH\text{CH}_3\text{CH(OH)COOH}, is chiral. (a) Explain why it is chiral and draw the two enantiomers. (b) Lactic acid is made by adding HCN\text{HCN} to ethanal followed by hydrolysis. Explain why this route produces a racemic mixture.
Show worked answer β†’

Identify the chiral centre, then explain the equal attack on a planar carbonyl.

(a) The central carbon is bonded to four different groups (CH3\text{CH}_3, OH\text{OH}, COOH\text{COOH} and H\text{H}), so it is a chiral centre (1). The two enantiomers are non-superimposable mirror images, drawn as 3D wedge-and-dash mirror images about that carbon (1).

(b) Ethanal has a planar C=O\text{C=O} group (1). The CNβˆ’\text{CN}^- nucleophile can attack the carbonyl carbon equally from either face (1), so the two enantiomers form in equal amounts, giving an optically inactive racemic mixture (1).

Edexcel 20184 marksDevise a two-step synthesis of propanenitrile (CH3CH2CN\text{CH}_3\text{CH}_2\text{CN}) starting from ethene. For each step give the reagent and the conditions.
Show worked answer β†’

Build the chain by going via a haloalkane, then substituting with cyanide.

Step 1: react ethene with HBr to give bromoethane (1): CH2=CH2+HBr→CH3CH2Br\text{CH}_2=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}, at room temperature (1).

Step 2: heat bromoethane under reflux with potassium cyanide dissolved in ethanol (1). Nucleophilic substitution by CNβˆ’\text{CN}^- adds a carbon: CH3CH2Br+CNβˆ’β†’CH3CH2CN+Brβˆ’\text{CH}_3\text{CH}_2\text{Br} + \text{CN}^- \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{Br}^- (1).

Markers require the ethanolic solvent for the cyanide step (aqueous would favour substitution to an alcohol or elimination) and credit recognising that the nitrile route adds a carbon to the chain.

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