humptydumpty5
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For b, count the longest chain (7). Both the branches leading off are methyl groups, so you’ll end up with 3,5-dimethyl-heptane
For c, prioritise the triple bond that makes it an alkyne, count the longest chain from there, and add the substituent. Note that there will be a carbon at the beginning and end of the triple bond (before it connects to where you wrote 5). In this case, it will be called 3-butyl-1-octyne
For d, well, this is one I think you just have to know and be able to recognise; it’s benzene
I’ve attached screenshots where I have drawn around the longest chain for your reference. Hope this helps
They can test aromatic compounds to some extent. Look at the data sheet for the C13 NMR it has the aromatic ring in the 120 ppm area. All you need to know for this is the simple benzene one, they won't ask further than that thoughI'm not sure if this is right or not (someone can confirm with me), but in the HSC chemistry syllabus, they focus on what's called 'aliphatic chemistry', which is essentially compounds with an open chain structure, whereas 'aromatic chemistry' focuses on closed chain structures, like benzene. If I recall correctly, apart from like cyclohexane vs cyclohexene experiment in Module 7, they wouldn't test you on aromatic compounds.