In this tutorial, I want to talk about the nomenclature of the complex substituents. In the last tutorial of this series, we discussed the fundamentals of the nomenclature and how it works. And now, we are going to add one more layer to it and learn how to name molecules that have complicated branches for substituents.
To start with, let’s talk about the groups that only contain 3 and 4 carbons. When it comes to the 3- and 4-carbon substituents, the IUPAC allows the use of the common names. Those are called the “retained common names” as they were retained when the modern system was accepted
There are only two 3-carbon chains. The first one has the connection at the end carbon, while the second one—at the middle carbon.
We call them the “n-propyl” and “isopropyl” groups. The “n” in n-propyl stands for “normal” which means that we have a straight chain with no branching. The “iso” in isopropyl refers to the “isomeric” propyl. This is a historic or a common name, so don’t try to put too much sense into it. Like many things in nomenclature, this is something that you just have to memorize like the dictionary words when you’re learning a new language.
One thing that I do want to point out though, is that the “i-” in the iso- prefix will count for the alphabetical order when we alphabetize our groups in the name. While the “n-” in the n-propyl doesn’t count. As the matter of fact, we will typically skip the “n” part altogether as it is redundant.
Now, the 4-carbon branch has quite a few variations!
We’ve got a normal butyl group, the sec-butyl, the isobutyl, and tert-butyl. The prefix “sec” stands for secondary, while the prefix “tert” stand for tertiary. Again, those are the traditional names, so you don’t have to remember the logic behind why there’s a “sec” or a “tert” prefix. But you will need to memorize those names and their corresponding structures for the test.
Also, just like in the case with isopropyl, the “i-” in the isobutyl will count for the alphabetical order while other prefixes won’t. Here’s a quick rule of thumb you can remember about which prefixes count and which ones don’t. If the prefix is spelled through the dash, it definitely won’t count. If the prefix modifies the structure and is spelled together, like cyclo- or iso- among the ones we’ve already seen, it will count. Finally, any numeric prefixes like di-, tri- tetra-, etc. won’t count either.
Alright, so those were the simple cases that you just have to memorize. How about something more complicated that we can’t commit to memory?
Here’s the example with 5 carbons. This branch is connected at the very end of the chain and doesn’t have any extra branches sticking off it.
So, as we know from the last tutorial, it’s going to be just a simple pentyl group.
Now, how about the other two? Should we use “sec” or maybe “iso” or something else?
This is where the IUPAC steps in and tells us the rule: we start numbering at the place of attachment, find the longest chain, and name the rest of the branch as if it was a standalone molecule. Then, we’ll change the ending -ane to -yl to signify that this is still a branch or a substituent and not a molecule.
Here’s how it works.
We’ll number at the carbon where it connects to the parent molecule. We then find the longest chain, which in this case is a 4-carbon chain. Then, we say that there’s a methyl group in the 1st position here and put together the rest of the name. Also, to specify that this is a complex substituent, we are going to enclose it in the parentheses. The parentheses are not optional here, so make sure you don’t forget about those on the test!
Ok. Let’s look how it works on the example.
Here we have a 6-membered ring with two branches.
The first rule of the nomenclature is to find the longest continuous chain. In this case, it’s going to be the ring itself.
Thus, our parent is going to be cyclohexane.
Next, the branch on the top is a simple ethyl group.
The bottom group is the one that we have just looked at: 1-methylbutyl.
Now, we’ll need to arrange those groups alphabetically in the name. Here’s something important about what counts for the alphabet in the complex substituents. The very first letter in the parentheses will count for the alphabetical order regardless of what it may be (a prefix or anything like that). So, in this case, we have “e” in ethyl vs “m” in the complex substituent. And since “e” is before “m” in the alphabet, the ethyl group will be first in the name. This also gives it precedence over the complex substituent for the numbering preferences since we don’t have the more substituted atom in this cycle.
So, the name for this compound is 1-ethyl-3-(1-methylbutyl)cyclohexane. Notice how we enclosed the complex substituent in the parentheses. Also, notice the punctuation. We are still using the same principles we’ve learned in the last tutorial: we use commas between numbers and dashes between the numbers and letters. So, there’s no dash, or space, or comma, or anything else between the parentheses of the complex substituent and the rest of the molecule as we treat the parentheses as a “letter” for the purposes of nomenclature.
Let’s look at a few more examples.
The first thing we want to do is to find the longest chain. In this case it’s 8-carbon chain. So, the parent will be an octane. We’ll also number the chain to give the branches the lowest possible number.
Now, we have a complex substituent, which we already know how to call. It’s an isopropyl group. And as I’ve mentioned earlier, the isopropyl is a retained common name. If we wanted to use the strict IUPAC rules, however, we’ll call it 1-methylethyl. Thus, there are two ways that we can go with, when naming this molecule.
Option one will use the common name for the group. Notice that we’re adding the prefix di- to signify that we have two isopropyl groups in this molecule.
And we have option two. In option two we use the strict IUPAC name. Notice that here we changed the numeric prefix di- to bis-.
There’s actually a difference which set of numeric prefixes we use depending on what type of a group we have.
For the simple substituents like ethyl, methyl, etc. we use the di-, tri-, tetra-, etc. We also use the same prefixes for the retained common names like isopropyl or tert-butyl. For the complex substituents, however, we use slightly modified prefixes. For 2 groups it’s bis-, for 3 it’s tris-, tetrakis- for 4, pentakis- for five, etc. It’s a rare case when you’re going to see multiples of the same complex substituent, but it’s still possible. So, it’s a good idea to know about these special prefixes in the case your instructor tries to throw a curved ball at you on the test.
Alright, here’s another example.
First thing first, we need to find the longest continuous chain in this molecule. While it may be tempting to take the two tert-butyl groups and classify them as the substituents, it won’t be correct. Here, our longest chain has 8 carbons. Also, notice the numbering that gives the smallest numbers to the branches near to the end of the molecule.
Now, let’s look at the substituents that we have here. We have a few simple methyl groups here. And we also have one complex substituent.
Just like in the previous case, we can use the retained common name here because it’s a 4-carbon group (tert-butyl). Or we can use the strict IUPAC rules and call is 1,1-dimethylethyl group.
And like in the last case, we can make two names for this compound.
One, using the retained common name. And the other one with the IUPAC name for our complex substituent.
In terms of alphabetization, we have “b” vs “m” in the first case, and “d” vs “m” in the second. Notice, that since the “dimethyl” part belongs to the complex substituent, the letter “d” in this case counts for the alphabet. Remember, that we use the first letter of the complex substituent’s name regardless of what that letter is or what part of the complex substituent it’s coming from.