Naming Alcohols - Online Tutor, Practice Problems & Exam Prep

How to name alcohols

Similar to double bonds and triple bonds, alcohols are also going to get a modifier. But that modifier is going to be instead of "ene" or "yne", it's going to be "ol". Okay? So, actually, you can remember that the modifier is "ol" just by looking at the word alcohol. Okay? It tells you that the modifier is going to be "ol". And it turns out that a lot of the same rules are going to apply for the alcohol as the double bond. For example, alcohols are actually going to receive highest priority when numbering alkanes. What that means is that alcohols get more priority than anything else. So that means if I have an alcohol and an alkane, I'm going to try to give the alcohol the lowest number. But even if I have an alcohol and a double bond, guess what? If they're competing against each other, I'm still going to go with the alcohol. The way that I've always said it in organic chemistry 1 is that alcohol beats all. Alright? It means that it's going to beat all the other substituents that you could have. Alright? So alcohols are going to beat all.

Old school vs. new school naming

The thing is that there's actually this is I'm just going to star this. Whenever we get into the idea of modifiers, now we have differences in where we can put the location. Okay? So basically, in the previous example, I showed you that you could put the location before the root. And if you do that, that's fine. If you put your location before the root, so for example, 1-pentene, that would be considered like the old-school method. Okay? And that's fine. But if you have more than one modifier present, then many times it's going to be more advantageous to use it within the root. If it's within the root, that's what we call the new school. So I want to show you an example of what that looks like. So 1-pentene would be the old-school method. The new school method would be, and by the way, new school and old school isn't really like a scientific term, it just has to do with as names start getting more complicated, there's starting to become more of a need to use it within. And it's pent-1-ene. Alright. So I know that that doesn't look quite as nice. It's like yuck. There's numbers inside of my word. But sometimes that becomes essential, especially when you have more than one modifier in the word. So for example, let's say that I had a double bond and an alcohol. So let's say that I had pentanol. Notice that the e-n describes that I have a double bond and the o-l describes that I have an -OH somewhere. Okay? Let's say that they are in positions 1-4. Okay? If I just say 1,4-pentenol, that's really confusing because I don't know which one is where. I don't know is the ene on the 1 or is the ene on the 4. So that's why I'm going to have to put at least one of these inside of the root. So then this would be a lot more clear. For example, if I did 1-pent-4-en-ol. That would be a lot more descriptive because now I would know, okay, the 4 directly describes the alkene. Okay? So I'm just trying to show you that that would be more like a new school method to put at least one of the numbers inside of there. Conversely, on the other hand, you also could have written it, you would have been able to write it as 4-pent-1-enol. That would have also been fine. Okay? It's just about making sure that each modifier has its own number. Okay? So for this compound, notice that I was kind of giving you all this this whole peptide because you're going to need it for. So for this compound, I want you to try to solve it on your own, realize that it has a double bond and an alcohol, so it's going to provide a special challenge and go ahead and try to get the right answer.

Name the following compound

So the first thing to notice here is that we have a ring, so this is going to be a cycloalkane. Remember when I told you guys that if you had a cycloalkane and it had one substituent, then you could omit the number? In this case, we're not going to want to do that because we have more than one thing coming off of it. This is a little bit more complex. We want to say the location of everything. Alright?

So the first thing is, what's the root going to be? Let's go ahead and just box this off because I know that came from the last example. What is going to be the name of this root? Well, the root is going to be a combination of a lot of stuff. It's going to be the number of carbons that are in this ring, plus it's going to have the word cyclo in it, plus it's going to have the two modifiers in it. Right? So I'm just going to start off with the easiest one. The root would just be cyclohexane. But after adding my two modifiers, this is actually going to turn into cyclohexenol. Does that make sense? Because I have a double bond and I have an alcohol.

Now what I have to do is figure out the locations of everything. So because this is a ring, I can start wherever I want. So where am I going to want to put my number one carbon, my number 1? Obviously, the alcohol. Why? Because alcohol beats all. So that means this is my 1, and now I have two options. I can either go counterclockwise or clockwise, red or green, depending on our rules of priority. What would be the next step? Which one would come next? And the answer is that double bonds receive more priority than halogens. So what that means is that the iodine might be really close, but the double bond has more priority, so the double bond is going to get my 2 here. And then we're just going to keep numbering around, so then I get 3, 4, 5, 6. So that means that if I were to name my substituents, what kind of substituents do I have? Well, all I have in terms of substituents is the iodine, so that would be a 6-iodo. Remember that double bonds and alcohols aren't called substituents. They're called modifiers. So that's my only substituent.

So now the only thing I have to do is figure out where the locations of these are going to go. What are the locations? Well, it seems like I have a 1 and I have a 2. So is it okay for me just to say that this is 1,2-cyclohexanol? No, that's not okay. We're going to have to put at least one of these numbers inside of the word so it's more specific what this is. Okay? So the one that I'm going to choose to put inside, you could pick whichever number you want. I'll just pick the 1. Okay? So what that means is that the final name here would be 6-iodo-2-cyclohexene-1-ol. Okay? But like I said, if you've made it 1-cyclohex-2-en-1-ol. That would have also been fine. Also, just so you know, some people put all the numbers inside. So another option would have been to say, this is getting annoying at this point, but cyclohex-2-ene-1-ol. That would also be completely fine. Alright? So any of these, any combinations of these you want, go ahead and use them. I'm going to stop telling you all the different combinations after this lesson because I think it's just a waste of time to do this every single permutation for every single name. But I did want you to see one example of me showing you everything. Okay? The one that you can't do is this one up here. That one just sucks because it's not specific enough. Alright? So very good. Let's go ahead and move on.

Here’s what students ask on this topic:

What is the IUPAC naming convention for alcohols?

The IUPAC naming convention for alcohols involves identifying the longest carbon chain containing the hydroxyl group (-OH) and replacing the 'e' at the end of the corresponding alkane name with 'ol'. The position of the hydroxyl group is indicated by a number placed before the name. For example, CH₃CH₂OH is named ethanol. If other functional groups are present, the hydroxyl group takes priority in numbering. For instance, CH₃CH(OH)CH₃ is named 2-propanol.

How do you prioritize functional groups when naming compounds with alcohols?

When naming compounds with multiple functional groups, the hydroxyl group (-OH) takes precedence over double and triple bonds in numbering. This means you should number the carbon chain to give the hydroxyl group the lowest possible number. For example, in a compound with both a double bond and an alcohol, the alcohol gets the lower number. If you have CH₃CH=CHCH₂OH, it is named but-3-en-1-ol, not but-1-en-4-ol.

What is the difference between old school and new school naming methods for alcohols?

The old school method places the position number of the functional group before the root name, such as 1-pentene. The new school method incorporates the position number within the root name, like pent-1-ene. This becomes particularly useful when multiple modifiers are present, as it clarifies the positions of each functional group. For example, 4-pentene-1-ol (old school) can be written as pent-4-en-1-ol (new school) to clearly indicate the positions of the double bond and the hydroxyl group.

How do you name a compound with both a double bond and an alcohol?

When naming a compound with both a double bond and an alcohol, the hydroxyl group takes priority in numbering. The positions of both the double bond and the hydroxyl group must be indicated. For example, if you have a compound with a double bond at position 1 and an alcohol at position 4, it can be named as pent-1-en-4-ol or 4-pentene-1-ol. The new school method, pent-1-en-4-ol, is often preferred for clarity.

Why is it important to include the position of the hydroxyl group in the name of an alcohol?

Including the position of the hydroxyl group in the name of an alcohol is crucial for accurately identifying the structure of the compound. The position number ensures that there is no ambiguity about where the hydroxyl group is located on the carbon chain. For example, 1-butanol and 2-butanol are different compounds with distinct properties, and the position number helps differentiate them clearly.