2
$\begingroup$

IUPAC name of the given compound is

enter image description here

(1) 3-Hydroxycyclohex-4-en-1-one
(2) 5-Hydroxycyclohex-3-enone
(3) 5-Ketocyclohex-2-enol
(4) 3-Ketocyclohex-5-enol

By what I understand by assigning the priority the parent is the ketone. If I number carbonyl carbon as 1, then I should move towards the alcohol group because it has higher priority than the alkene and both are at number 3, so the alcohol should be assigned the number 3 and alkene as 4?

So, according to me, the answer should be (1), but in the key it is given as (2).

$\endgroup$
4
$\begingroup$

As for your other question, the simplified criteria for the numbering are:

  1. lower locants for the principal characteristic group that is expressed as suffix
  2. lower locants for multiple bonds
  3. lower locants for prefixes
  4. lower locants for substituents cited first as a prefix in the name

The corresponding actual wording 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:

(…)

(c) principal characteristic groups and free valences (suffixes);

(…)

(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;

(…)

For the compound that is given in the question, you have correctly identified the principal characteristic group as ketone. It is expressed as suffix “one”. Thus, a low locants is assigned first to the ketone according to Rule (c), which yields “…1-one”.

Next, a low locant is given to the double bond according to Rule (e), which gives “3-en-1-one”.

Finally, a low locant is assigned to the remaining $\ce{-OH}$ substituent according to Rule (f).

Thus, the complete name is 5-hydroxycyclohex-3-en-1-one.

| improve this answer | |
$\endgroup$

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.