Both double and triple bonds are present in the hydrocarbon shown above, which has the name hexa-1,3-dien-5-yne. Why is the numbering chosen such that the alkene has lower locants than the alkyne? Is there a rule which stipulates that the alkene is a more 'senior' functional group?


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The appropriate passage in the 2013 Blue Book is as follows:

P- The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore, the preferred IUPAC name for the above compound is hexa-1,3-dien-5-yne, as the double bonds receive the lower locants (1,3) than in the alternative name hexa-3,5-dien-1-yne.

Note that this preference for double bonds over triple bonds is only relevant when there is no difference between the locant set for the multiple bonds. In the following example, the locant set (1,4) is lower than (2,5) and hence hex-4-en-1-yne is the preferred name, even though it has a higher locant for the double bond:

PIN of hex-4-en-1-yne

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