Statement 13.4.6

This statement expects you to be able to draw all the possible isomers of a compound if you are told its molecular formula.

In truth, the only way of preparing for this is to do as many questions as you can find. The problem is that if it comes up in a multiple choice paper, all they can really ask for is a total number of isomers. Although you can find that out from the mark scheme, it doesn't help you to identify the ones you have missed, or the extra ones that are wrong.

This statement is new in the 2022 syllabus, and I can only hope that the questions they ask are simple, involving small numbers of atoms.

Try to do it systematically.

• Ask yourself whether chain isomerism is possible, and then find all the possible variations.

• If you have atoms other than just carbons and hydrogens, ask yourself what functional group(s) it might be a part of - and if it is possible to move that functional group along the chain.

• Have any of the isomers that you have come up with so far got a chiral centre - a carbon atom with 4 different groups attached? If so you have optical isomers. Draw them clearly.

• Is there a possibility of a carbon-carbon double bond in the isomer? If so, could there be geometric isomers? Incidentally, if you seem to be short of hydrogens, don't necessarily think that there must be a carbon-carbon double bond. There could be a carbon-oxygen double bond, for example, if oxygen plays a part in the molecular formula.

• When you have finished check that you haven't done anything stupid like having carbon atoms with 3 or 5 bonds. It is easy to do.

This will become easier when you have done a lot more chemistry, because you will get to know the likely possibilities.

So, for example, if you were given a molecular formula like C₄H₈O₂, you would immediately recognise that compounds with two oxygens in are most likely to be carboxylic acids or esters, and so you would find as many of those as you could.

But there are also others isomers that you would need to think of as well. There could be a C=O group somewhere and an OH group somewhere else on the chain. Or there could be two OH groups and a carbon-carbon double bond, in which case you might have geometric isomers as well.

To be honest, this is a nightmare statement!

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