# How can I know which way to number the carbon atoms? [closed]

Please help me, I don't know which is the "first" one and what rules apply here...

## 1 Answer

There are three rules to follow for assigning locants (numbers) to a compound like this one.

1. Assign locants to generate the longest possible chain of carbon atoms.

Here is an example of mine:

I can find chains of eight carbon atoms a few different ways. Two of them are below. There is at least one more. Can you find it? How is it similar to one that I highlighted?

The longest chain I can make in your structure is six carbon atoms, and it looks like I can make that chain length multiple ways.

1. Assign locants to maximize the number of substituents.

The goal is to have the most possible substituents. For all of the ways to make a six-carbon atom chain, I count the same number of possible substituents.

Back to my example, the first way to make a chain of eight carbon atoms has four substituents: three methyl groups and one ethyl group. The second way has three substituents: two methyl groups and one isopropyl group. We choose the first one.

I think that five is the maximum number of substituents that can be reached in your structure, and again, there are multiple ways to achieve that number.

1. Assign locants to generate the lowest possible combination of locant values for your substituents.

Using my example, we can number the red parent chain two ways:

In the first one, we have a substituent locant set of {2,3,4,5}. In the second one, we have a substituent locant set of {3,5,6,7}. The first set represents the lower combination of locants, so the first set is the way to go.

For your example, since it has a high degree of symmetry, there may be multiple six carbon atom chains with five substituents having the same set of locant values. In that case, choose the one that puts the alphabetically first substituent at the lowest possible value. I suspect that these sets will be equivalent however given the symmetry. If so, then choose any of them as they are all degenerate.

• Nice guidance without giving the answer! – user55119 Oct 9 '19 at 1:58