The correct IUPAC name of the following compound is:


(A) 1,1-diformyl propanal
(B) 3-formyl butanedial
(C) 2-formyl butanedial
(D) 1,1,3-ethane tricarbaldehyde

The structure of 1,1-diformyl propanal has been given. And the IUPAC name has been written. But I can't get the same IUPAC name because I'm a beginner. Also I don't know how to get structure here in the question, so I didn't do it. Help me with the IUPAC naming of the compound with step-by-step explanation . Please.

  • 1
    $\begingroup$ @JamilAhmed The problem is that the first carbon atom in propanal is already almost full, so it cannot have two more functional groups. $\endgroup$
    – DHMO
    Sep 2, 2016 at 15:09
  • 2
    $\begingroup$ @JamilAhmed Does your exercise look like Exercise I, Q.4 on page 7 of this pdf document? tekoclasses.com/IITJEE%20CHEMISTRY%20%20ENGLISH/… $\endgroup$
    – user7951
    Sep 2, 2016 at 15:23
  • 1
    $\begingroup$ @Loong, Q.4 exactly. $\endgroup$ Sep 2, 2016 at 15:32
  • 1
    $\begingroup$ @JamilAhmed Then the answer is not A. It is not 1,1-diformyl propanal. Problem solved? $\endgroup$
    – DHMO
    Sep 2, 2016 at 15:49

1 Answer 1


The proper and straightforward way to solve nomenclature problems is to construct the systematic name for the given structure according to the current IUPAC recommendations. However, for such multiple-choice questions, it may be easier to check the given answers and try to draw a structure for each proposed name as follows – especially when the particular nomenclature rules that are considered in the exercise are not exactly known.

The name 1,1-diformyl propanal (A) can be ruled out since the atom ‘1’ (i.e. the $\ce{C}$ atom of the aldehyde group) of propanal has only one $\ce{H}$ atom that could be substituted; therefore it is impossible to place two formyl groups at atom ‘1’. Furthermore, after substitution of the $\ce{H}$ atom of the aldehyde group, the compound would no longer be an aldehyde.


The name 1,1,3-ethane tricarbaldehyde (D) does not make sense since ethane has only two $\ce{C}$ atoms; therefore it is impossible to place a carbaldehyde group at atom ‘3’ of ethane.


It is possible to draw structures for the names 3-formyl butanedial (B) and 2-formyl butanedial (C) based on the structure of butanedial:


3-formyl butanedial (B)

2-formyl butanedial (C)

Note that the resulting structures for 3-formyl butanedial and 2-formyl butanedial are actually identical, and both structures correspond to the structure that is given in the question. The correct choice is 2-formyl butanedial (C) rather than 3-formyl butanedial (B) since the locant ‘2’ is lower than ‘3’. (Note that the correct spelling of the name is actually 2-formylbutanedial, i.e. without space.)

Anyway, even the name 2-formylbutanedial is actually not in accordance with current IUPAC recommendations. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), aldehydes can be systematically named in various ways:

P-66.6.1 Systematic names of aldehydes

Aldehydes are systematically named in three ways:

(1) substitutively, using the suffixes ‘al’ for $\ce{-(C)HO}$ and ‘carbaldehyde’ for $\ce{-CHO}$;

(2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;

(3) by using the prefixes ‘oxo’, denoting $\ce{=O}$, or ‘formyl-’, denoting the substituent group $\ce{-CHO}$.

In particular,

P- The suffix ‘carbaldehyde’ is used when more than two $\ce{-CHO}$ groups are attached to an alkane.

Therefore, the preferred IUPAC name for the compound given in the question is ethane-1,1,2-tricarbaldehyde.


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.