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Aldehydes & Isomers
2017 Paper Two
Question Four
Looking at past AQA A-level Chemistry questions, I wanted to go beyond the answers and focus on thought processes, ideas and tips that will help in examinations.
The series will help you to spot weaknesses and help with revision, or it can serve as addenda to your notes.If you haven't gone through the paper, please look up the questions at www.aqa.org.uk, or click the direct link below, and have a go...
AQA A-level CHEMISTRY Paper 2 Organic and Physical ChemistryMonday 19 June 2017
The first thing to do is count the carbons in the chain. In this case, there are 6 so we know we have ‘hex’ in the name. The functional group is cyanide, so this will be a nitrile with the carbon at this side carbon 1. The alcohol group is therefore on carbon 2. Put it all together…
2-hydroxyhexanenitrile
Take care - a common error is calling it a ‘hexanitrile’.Traditionally, many students have found naming nitriles difficult.
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A bit of general knowledge!
The question tells us that there are 2 stereoisomers existing in a racemate (50:50 mix). They will have the same properties and they can only be separated by the way they interact with polarised light
Each one will rotate polarised light in different directions – they have optical activity: a chiral compound.
Be precise with language: the enantiomers themselves don't rotate and plane polarised light is not ‘refracted’ and definitely not ‘bent’.
It's easy to lose marks on this question, highlighting the need for precise answers.
There is something about the molecule that will lead to a racemate. The carbonyl group itself is trigonal planar (bond angles = 120°)… the nucleophile can attack either side (the top and bottom looking at the diagram below) with an equal probability. You must indicate that the group is planar and not the molecule.
Read the question carefully in order to deduce the correct product. Examiners noted that many different products were seen for this answer.
If there's no stereoisomer, then the molecule is achiral – there is no chiral centre. This means it doesn’t contain a carbon atom with four different groups attached to it.
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