Video Transcript: Naming Aldehydes Using IUPAC Nomenclature

Below is the written transcript of my YouTube tutorial video Naming Aldehydes Using IUPAC Nomenclature.

If you prefer to watch it, see Video HERE, or catch the entire Naming Organic Compounds Series.

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Leah here from Leah4Sci.com. And in this video, I will show you how to name aldehydes. An aldehyde is a molecule that can be represented by R bound to a C, double bound to an oxygen, single bound to a hydrogen where R represents the rest of the molecule and the C double bound to a single bound H represents the aldehyde. Be careful not to confuse the functional group and an aldehyde with that of a carboxylic acid which has an extra oxygen between the carbon and hydrogen and the ketone which has another R group attached rather than the hydrogen.

To name an aldehyde, we use the rules explained in previous naming videos and apply the suffix “al”. An interesting thing about the aldehyde is because it ends in an H, it’s a terminal functional group. And when the aldehyde is the highest priority on your molecule, it will always be considered number 1 and therefore you do not have to include the number. When drawing an aldehyde in line structure, you can represent it with simply the carbonyl showing the carbon double bound to the oxygen. Some people will add a hydrogen. I prefer to draw it this way because it helps me recognize that it is indeed an aldehyde. We identify and highlight the parent chain and number to give the aldehyde number 1. In this case, I have two carbons for a first name of eth. Only single bonds on the actual carbon chain giving me a last name of ane. The aldehyde appears on carbon 1 which would give me 1-al but remember that 1 is understood and thus not have to be included. Since the suffix end in a vowel, we have to drop the e from ane for a final name of ethanol.

When you have a substituted aldehyde, you treat it the same way. We start by identifying and highlighting the parent chain and numbering to give the aldehyde number 1. On this molecule, I have four carbons for a first name of but. Only single bonds on the chain giving me a last name of ane. I have an aldehyde to give me the ending al and the methyl group on carbon 3 to give me the prefix 3,methyl. Putting the name together, we start with the prefix, first name, last name, and al remembering to drop the e for a final name of 3-methylbutanal. When you have another group on your molecule that can potentially get the same number as your aldehyde, you have a choice of giving the aldehyde a 1 or the pi bond the 1. However, the aldehyde takes higher priority and so we number from the right. Having five carbons on my parent chain gives me a first name of pent. A double bond on carbon 4 gives me a last name of 4-ene. The aldehyde gives me a suffix of al for a final name of 4-pentenal remembering that we drop the e before placing the al suffix.

When you have the aldehyde coming off of a cyclic compound, you name the cyclic compound and give it the ending carbaldehyde showing this is a carbon aldehyde. Since the cyclic compound has six carbons and no substituents, we simply call it cyclohexane. The aldehyde coming off the chain gives me an ending of carbaldehyde for a final name of cyclohexanecarbaldehyde.

When you have a ridiculously substituted aldehyde, you treat it the same way by first highlighting and identifying your parent chain. Since only groups like a carboxylic acid will have higher priority than aldehyde, unless you see a carboxyl, aldehyde will be number 1 and everything else gets numbered accordingly. In this molecule, I have six carbons on my parent chain for a first name of hex. Only single bonds gives me a last name of ane. The aldehyde will give me a suffix of al and I will carefully mark every substituents to insure I don’t miss one. Be careful that you don’t confuse the isopropyl with the 2-methyl substituents and the way you can recognize the difference is to see that the isopropyl has a carbon coming off the parent chain whereas the methyls come directly off the parent chain.

An isopropyl on the third carbon gives me 3-isopropyl. Two methyls off the fifth carbon gives me 5,5-dimethyl. And a benzene-looking ring off the sixth carbon gives me 6-phenyl, not 6-benzyl. When ordering the substituents in alphabetical order, remember to look at I for isopropyl, the M for dimethyl since di simply means two, and P for phenyl. Since I comes before M which comes before P, the substituents are already in order. Remember to drop the e and insure that you include each part of the puzzle for a final name of 3-isopropyl-5,5-dimethyl-6-phenylhexanal.

Here, we have an interesting example. Given that the alcohol is not your highest priority and therefore gets demoted from a functional group that ends in ol to a substituent with a prefix hydroxene. I start by identifying and highlighting my longest carbon chain and numbering to give the aldehyde number 1. Five carbons gives me a first name of pent. Only single bonds gives me a last name of ane. My OH substituent on carbon 2 gives me the prefix 2-hydroxy. And the methyl substituent on carbon 3 gives me the prefix 3-methyl. And let’s not forget the aldehyde giving me the suffix al. Since H comes before M, we get a final name of 2-hydroxy-3-methylpentanal remembering to drop e before the al.

We’ll end off with an interesting example where a single carbonyl carbon has a hydrogen attached to either side. You may recognize this molecule as having a common name of formaldehyde. And let’s see how to name it. While this may look like a di-al given that it’s an aldehyde from either direction, we still start by identifying the one carbon in the parent chain for a first name of meth. Since there are no other bonds in the chain, this is by default an ane and the suffix is still al for the aldehyde. Drop the e for a final name of methanol.

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