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A level Chemistry OCR HELP

I need help with these types of question :
How many structural isomers of C6H14O are tertiary isomers?
i know the answer now obviously but how do i figure these types of questions where they ask how many isomers there are.
please help
Original post by M.ayEL
I need help with these types of question :
How many structural isomers of C6H14O are tertiary isomers?
i know the answer now obviously but how do i figure these types of questions where they ask how many isomers there are.
please help


Tertiary isomers implies that they must have the -OH on a carbon that is bound to three other carbons.

Each carbon must be bound to four other atoms, so I’d start by drawing out the (C)3COH skeleton for one isomer, draw one possible structure by drawing in the remaining atoms (2 carbons and 13 hydrogens). I’d then repeat the process until I can’t think of any more valid ways to draw the any possible structures and count the number of structures drawn, hoping I haven’t missed any.
Reply 2
This is how I would probably work through it, if this helps?
Firstly, I'd separate the -OH to make checking the H atoms easier (C6H13OH).
Then I'd start with a 6C chain, then 5C, 4C etc. until there are no more possibilities (ie. they start repeating).
We know that a 6-carbon chain won't work as that would only result in primary or secondary alcohols.

I had a look and briefly explained what I did in the attached two images. Hopefully the notes make sense!
Btw, I added the blue H atoms just to help me check I had 13 (without the -OH), and remember to consider cyclic structures (which aren't possible here as that would require one less H atom).

I've just remembered that you could add ethyl groups (-CH2CH3), I think this would be less likely, but if an ethyl group replaces any of the methyl groups (must be the methyl attached to the -COH) and one carbon taken away from the straight chain (of any of the three isomers I drew), then we can see that the ethyl group becomes part of the longest chain (so it's no longer an ethyl group and the compound ends up being one of the already drawn isomers).
Reply 3
Original post by simxne_
This is how I would probably work through it, if this helps?
Firstly, I'd separate the -OH to make checking the H atoms easier (C6H13OH).
Then I'd start with a 6C chain, then 5C, 4C etc. until there are no more possibilities (ie. they start repeating).
We know that a 6-carbon chain won't work as that would only result in primary or secondary alcohols.

I had a look and briefly explained what I did in the attached two images. Hopefully the notes make sense!
Btw, I added the blue H atoms just to help me check I had 13 (without the -OH), and remember to consider cyclic structures (which aren't possible here as that would require one less H atom).

I've just remembered that you could add ethyl groups (-CH2CH3), I think this would be less likely, but if an ethyl group replaces any of the methyl groups (must be the methyl attached to the -COH) and one carbon taken away from the straight chain (of any of the three isomers I drew), then we can see that the ethyl group becomes part of the longest chain (so it's no longer an ethyl group and the compound ends up being one of the already drawn isomers).


Thank you!! this was soo useful :biggrin:

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