Carboxylic acid is the main functional group due to higher priority, but since it's an ether, what's the proper way to name the following compound?

Structural formula of ether with ketone and carboxyl groups

My attempt: 2-(1-methyl oxo propyl ether) ethanoic acid.


1 Answer 1


You have already found out that the carboxylic acid group is the principal characteristic group. Therefore, the ether group has to be expressed as a prefix. The corresponding rule in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows.

P- Substituent prefix names for $\ce{R'-O-{}}$ groups are formed by concatenation, i.e., by adding the prefix ‘oxy’ to the substituent prefix name for the group $\ce{R'}$. These compound prefixes require the numerical multiplying prefixes ‘bis’, ‘tris’, etc.

In this case, the substituent prefix name for the $\ce{R'-{}}$ group is 3-oxobutan-2-yl and thus the substituent prefix name for the $\ce{R'-O-{}}$ group is (3-oxobutan-2-yl)oxy. Therefore, the complete name for the compound that is given in the question is [(3-oxobutan-2-yl)oxy]acetic acid.

[(3-oxobutan-2-yl)oxy]acetic acid

Note that locants are omitted for parent compounds when all substitutable hydrogen atoms have the same locant. Therefore, the preferred IUPAC name is [(3-oxobutan-2-yl)oxy]acetic acid and not 2-[(3-oxobutan-2-yl)oxy]acetic acid.

  • $\begingroup$ Thanks but can you just explain 1) why did you add "2-yl" and 2) is it 3 oxo instead of 2 oxo because you went lower numbering to the location of point where the ether connects $\endgroup$
    – Chuck
    Feb 26, 2023 at 12:27
  • $\begingroup$ @Chuck butan-2-yl is an alkanyl-type substituent group with the free valence at position 2, not to be confused with the alkyl-type substituent group butyl with the free valence at position 1. It's 3-oxobutan-2-yl and not "2-oxobutan-3-yl" because the free valence gets a low locant first. $\endgroup$
    – Loong
    Feb 26, 2023 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.