8
$\begingroup$

I am struggling with naming this structure either (E)-2-bromo-1,4-difluoro-3-iodobut-2-ene or (E)-3-bromo-1,4-difluoro-2-iodobut-2-ene.

(E)-2-bromo-1,4-difluoro-3-iodobut-2-ene or (E)-3-bromo-1,4-difluoro-3-iodobut-2-ene

I am aiming at the question whether there is also a prioritization for IUPAC nomenclature given by the order of atomic numbers as we have it for the CIP nomenclature. Or do we just have a prioritization given by the alphabetic order in the case of halides?

$\endgroup$
1
13
$\begingroup$

As per the official IUPAC guidelines, namely Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

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 (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The directly relevant part here is rule (g), which states that the substituent cited first gets the lowest locant. Since substituents are cited in alphabetical order, this indirectly means that the lowest locant is assigned to the substituent which comes first alphabetically, i.e. 'bromo'.

Note that this is only rule (g), and that in general rules (a)–(f) should be applied first (their full text can likely be found elsewhere on Stack Exchange). In this case, rules (a)–(d) are irrelevant and rules (e)–(f) do not distinguish between the two proposed numbering systems, so (g) is used as the tiebreaker.

In any case, the preferred IUPAC name of the above molecule is (2E)-2-bromo-1,4-difluoro-3-iodobut-2-ene. [Technically, the locant of the double bond must be specified with the stereodescriptor, hence (2E)- instead of just (E)-, although in practice this is often ignored when there is only one double bond.]

(2E)-2-bromo-1,4-difluoro-3-iodobut-2-ene
$\endgroup$
0

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.