My book says to use carboxamide when it is a carboxylic acid derivative (acyl). Ethanamide $\ce{CH3CONH2}$ is a carboxylic acid derivative, so shouldn’t it be named ethan-1-carboxamide rather than ethanamide?

enter image description here

I am confused with this nomenclature and use of amide, carboxamide and imide. When should I use carboxamide and ethanamide?

  • $\begingroup$ Preferred IUPAC name: Acetamide // Systematic IUPAC name: Ethanamide// Other names Acetic acid amide Acetylamine en.wikipedia.org/wiki/Acetamide $\endgroup$
    – Poutnik
    Mar 27 at 9:18

1 Answer 1


The first thing to do when constructing a systematic name is to find the parent structure - be it a cycle, or in this case an acyclic chain.

You choose the parent structure with the most skeletal carbon atoms - that is the longest chain. Here the longest chain in 2 carbon atoms so we start with a parent chain of ethane.

Ethane been substituted with -$\ce{NH2}$ and $\ce{O}$ which describes an amide group. So combining the amide suffix and the ethane parent, we get ethanamide.

NB: carboxyamide group

A carboxyamide group includes the carbon atom, so if you incorrectlydeclared the $\ce{CH3}$ group as the parent methane, you could say it is substituted with a carboxyamide group and would be called methanecarboxamide.

An example of the correct use of carboxyamide is cyclohexanecarboxyamide (pictured below). Here the cyclohexane ring system is the parent and the can be described as carboyamide substituted.


The minutiae of systematic naming is described in the IUPAC Blue Book. Here the relevant rule is P-44.3.2

  • 1
    $\begingroup$ This answer does not explain why the preferred IUPAC name is acetamide (not ethanamide) and how such names are formed. Also, the parent structure of an amide is not always the longest chain. The name of the second example is cyclohexanecarboxamide (not cyclohexanecarboxyamide). Rule P-44.3.2 is not relevant in this case; the relevant rules are in Subsection P-66.1. $\endgroup$
    – Loong
    Apr 26 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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