How many isomers of C4H11N are there?

One of the most popular queries that lands people at this blog is about the isomers of C4H11N - which I suspect may be some kind of organic chemistry question on student homework. In any case, this post will describe how to find all members of a small space like this by hand rather than using software.

Firstly, lets connect all the hydrogens to the heavy atoms (C and N, in this case). For example:


Now eleven hydrogens can be distributed among these five heavy atoms in various ways. In fact this is the problem of partitioning a number into a list of other numbers which I've talked about before. These partitions and (possible) fragment lists are shown here:


One thing to notice is that all partitions have to have 5 parts - even if one of those parts is 0. That's not strictly a partition anymore, but never mind. The other important point is that some of the partitions lead to multiple fragment lists - [3, 3, 2, 2, 1] could have a CH+NH2 or an NH+CH2.

The final step is to connect up these fragments in all possible ways. Easy to write, but hard to do - one way is shown below for [3, 3, 2, 2, 1]:


The two possible fragment lists are shown at the top, and below is one way to connect each list. It so happens that there are two ways to connect each list to give different molecules. The molecules are on the right of the figure.

The rules to follow here are : 1) divide the fragments into two lists as close as possible to the same number of 'spare bonds' (clear circles), 2) connect between the lists (dark circles + lines), 3) leave spare circles unless it's the final step.

Using this procedure should lead to all molecules with the same formula, with some trial and error. Hope this helps! Oh, and as a poster on reddit/r/chemistry pointed out, remember to count stereoisomers...

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