Since dialdehydes have a carbonyl group on either end of an alkane, is it reasonable to call carbon dioxide "methanedial", or is it that ethanedial (glyoxal) is the first dialdehyde?


1 Answer 1


The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P- Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

pentanal ($\ce{CH3-CH2-CH2-CH2-CHO}$)
pentanedial ($\ce{OHC-CH2-CH2-CH2-CHO}$)

In this context, the term ‘aldehyde’ is unambiguously defined:

P-66.6.0 Introduction.

The class name ‘aldehyde’ traditionally refers to compounds containing the $\ce{-CH=O}$ group attached to a carbon atom. However, nomenclature for aldehydes has been extended to describe a $\ce{-CHO}$ group attached to a heteroatom.

Since carbon dioxide ($\ce{O=C=O}$) does not contain a $\ce{-CH=O}$ group, it is not considered an aldehyde and thus Section P- does not apply to carbon dioxide.

Therefore, the shortest dialdehyde is ethanedial ($\ce{OHC-CHO}$). (Note that the preferred IUPAC name is ‘oxaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.)

  • $\begingroup$ What about a diketone nomenclature? Does the blue book specify two adjacent $\ce{R}$ groups, and thus methanedione would also be invalid? $\endgroup$
    – hBy2Py
    Aug 29, 2015 at 15:43
  • $\begingroup$ @Brian To call a carbonyl group a ketone, the oxygen has to be double bonded to a carbon atom bonded to two other carbons. So the ketone series begins with propanone and the diketone series, with butanedione. $\endgroup$
    – GeeJay
    Aug 30, 2015 at 7:07
  • $\begingroup$ @JayJay Gotcha, that's what I'd assumed. Thanks! $\endgroup$
    – hBy2Py
    Aug 30, 2015 at 11:05

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