If both double and triple bonds are at the terminals,so prefreference would be given to? [closed]

Question

I want to know about the preference level in order to givr IUPAC name

Answer 1

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

P-14.4 NUMBERING

When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

(…)

(e) saturation/unsaturation:

  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

(…)

 

P-31.1.1.1 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.
(…)

If a compound contains both double and triple bonds, low locants are given first to all multiple bonds together as a set. However, if the multiple bonds are at the terminals of the chain, there still remains a choice. For example, $\ce{H2C=CH-CH2-C#CH}$ could be named “pent-1-en-4-yne” as well as “pent-4-en-1-yne” since both names correspond to the same locant set “1,4”. In this case, the lowest locant is given to the double bond. Therefore, the correct name is pent-1-en-4-yne.

Answer 2

Double bonds are given the preference.