[Nora]

Hi everyone!

I have a question about structural isomers.

Sometimes I get a question like draw all isomers for C₄H₁₀, is there any way for me to know how many isomers of C₄H₁₀ exist, before I start drawing them? Like a general formula showing how many isomers a molecule has?

I also wonder about another thing. If I were to name this compound:
CH₃CH₂CH₂COOH, then why is it named "butanoic acid", and not "methylbutanoic acid", I mean, is there not a methyl group in the beging of the condensed structural formula?

I would really appreciate help, thank you =D

[fledarmus]

Unfortunately, I am not aware of any way to calculate the number of possible isomers that a given formula might represent. You just have to work your way through the different possibilities systematically, and be able to recognize which structures are unique. For your C_nH_2n+2 formula, it isn't that difficult because there are no rings or double bonds, but as you increase the hydrogen deficiency, the number of possible structures is vastly increased.

As for your second question, the philosophy of the IUPAC nomenclature is that you base the names on the longest carbon chains. Your molecule has four carbon atoms in a line, so it is named as a "butan-". The methyl group you refer to is not a substituent, it is part of the longest chain. If it was attached to the next carbon in, that would be a different story -

CH₃CHCOOH
     /
   CH₃

...would be 2-methylpropanoic acid.

If you look at the way you tried to name your butanoic acid as a methylbutanoic acid, you will note a couple of things that should pop up as red flags. First, butanoic acid is four carbons (butan- means four) and methyl is one carbon, so the name you propose is actually claiming five carbons in a molecule whose structure only has four. That should be your first warning that something is wrong with your name. Second, if you put the appropriate number for your substituent, even if you get the carbon number right (as a methylpropanoic acid rather than a methylbutanoic acid), you will get 1-methylpropanoic acid. In general, if you are claiming a substituent in the 1 position, you should double-check to make sure that it isn't simply an extension of the chain rather than a substituent. That is a common mistake, and unnecessary if you just stay aware of 1- substituents.

[Nora]

Thank you so much for your wonderful reply! It all makes sense now!