IUPAC Naming Question

[dentalwannabe]

Test in < 2 weeks. Eek.

Question about naming prioritization:

1) Cyclohexane with OH and Br. Do I give naming priority to the OH or the Br? Which is in the "1" spot?

2) Is the prioritization different for OH and Br when determining R and S?

I feel like in Case1 I end with -ol, so the alc is most important, but for Case2, R and S determination, I go with Br as "1".

Sounds good?

---

Members don't see this ad.

 

[pdog]

I think it's bromocyclohexanol, with the hexanol in the #1 spot.

There is a whole priority system for determining which functional group gets top billing (http://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry#Order_of_precedence_of_groups). If you can't remember the whole list, you can generally get away with giving the priority to the group with the highest state of oxidation.

 

[pdog]

And priority for the R/S and E/Z determinations is just based on atomic number IIRC. So yeah, each determination is based on a different set of priority rules.

 

[dentalwannabe]

Thank you!

 

[dentalwannabe]

And priority for the R/S and E/Z determinations is just based on atomic number IIRC. So yeah, each determination is based on a different set of priority rules.

Actually, one more question. So for R/S and E/Z determinations is it the mass of the single atom that forms a successive atom in a substituent group that matters, as in Br or O, OR is it the mass of the entire chain, if there is a lot hanging off of an atom? What if there is a Br and an ether that's (in total) way more massive than the Br?

 

[pdog]

IIRC it's just based on the first atom, unless there is a tie. So Br trumps an ether. But if you have 2 C's, you have look at what the next 3 things are bonded to those C's. And double bonds count as two of that thing bonded to that atom. So if a chiral carbon has four things attached to it: a methyl (CH3), a methanol (CH2OH), an H, and an aldehyde (C(=O)H), the H is obviously lowest priority (4). The others are all C, so you have to look at the next stuff attached to them. The methyl is 3, because it just has H's, the methanol is 2, b/c it has one O, and the aldehyde is 1, b/c it has the equivalent of 2 O's due to the double bond.

I'm not an expert on this, but this is how I remember it from o-chem lecture way back when.

 

[dentalwannabe]

Thanks you two! And thank you for the last explanation pdog, and congrats on your absolutely incredible DAT performance!