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A-Level Structural isomerism and stereoisomerism

What the A-Level syllabus expects for Structural isomerism and stereoisomerism, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to explain, sketch, compare, name. About 3% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (3 marks)

Explain how the use of chiral auxiliaries assists in the preparation of drugs that are obtained as a single pure enantiomer.

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A single, enantiomerically pure chiral auxiliary is covalently attached to the starting material. This places a fixed chiral centre next to the reacting site, so the subsequent reaction now produces two diastereomers rather than two enantiomers, and one diastereomer is strongly favoured (the reaction is diastereoselective). Because diastereomers have different physical properties, the desired one can be separated by ordinary means such as crystallisation or chromatography. The auxiliary is then cleaved off to release the required single pure enantiomer (and can often be recovered and reused).

Example 2 (3 marks)

Naloxone carries a terminal alkene in the form of an N-allyl group (-CH2-CH=CH2). Representing naloxone as RCH2CH=CH2, draw the structural formulae of the optical isomers produced when naloxone reacts with bromine in an inert solvent. Label the stereochemistry of each optical isomer with its R or S configuration and justify your assignments.

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Br2 adds across the terminal C=C of RCH2-CH=CH2 to give the 1,2-dibromide RCH2-CHBr-CH2Br. The internal carbon (the CHBr) now carries four different groups: -CH2R, -Br, -H and -CH2Br, so it is a single stereocentre and the product exists as a pair of optical isomers (enantiomers). The terminal -CH2Br carbon is not a stereocentre (two H). CIP priorities at the stereocentre: Br (Z=35) is highest; then compare -CH2Br vs -CH2R at the first atom - both are C, so go to the next atoms: -CH2Br gives (Br,H,H) while -CH2R gives (C,H,H); Br beats C, so -CH2Br > -CH2R; H is lowest. Priority order: Br > CH2Br > CH2R > H. Draw the two mirror-image tetrahedra: with H pointing away, if Br->CH2Br->CH2R is clockwise the centre is R, and the enantiomer (anticlockwise) is S.

Example 3 (2 marks)

Sketch two possible structures for the complex anion present in Reinecke's salt, and name the kind of isomerism that this anion displays.

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Reinecke's salt contains the octahedral anion [Cr(NCS)4(NH3)2]^-, with four thiocyanate and two ammine ligands about Cr(III). Two arrangements are possible: - trans: the two NH3 ligands on opposite vertices (180 deg apart), the four NCS in the equatorial plane. - cis: the two NH3 ligands on adjacent vertices (90 deg apart). Because the two forms differ only in the relative positions of identical ligands, the anion shows cis-trans (geometric) isomerism.

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More A-Level H2 Chemistry topics

The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry