# Cahn–Ingold–Prelog priority of carboxylic group versus ortho carboxylic acid

Should the C=C double bond below be labelled as (E) or (Z)?

When using the Cahn–Ingold–Prelog (CIP) priority system and the ‘Sequence Rules’, double bonds are split into two bonds and the doubly bonded atoms are duplicated.

Thus

is treated as

where (O) and (C) are duplicate representations of the atoms at the other end of the double bond;

whereas

is treated as

Therefore, the name of the structure that is given in the question is (2​E)-3-bromo-3-iodo-2-(trihydroxymethyl)prop-2-enoic acid.

• You should note that in the first step (turning $\ce{C(=O)OH}$ into $\ce{C(-O-(C))(-(O))(-O)}$ and comparing that to $\ce{C(-O)(-O)(-O)}$ the two are still equivalent. You need to step down one step further to see that the highest priority O of the carboxyl group is bonded to a dummy C whose priority is higher than that of the true H on the carboxylic hydrate. That might help clear remaining confusion.
– Jan
Oct 15 '15 at 12:00

The compound doesn't exist as drawn, it would dehydrate to the symmetric diacid. That being said, I'd guess Z as the extra hydrogen atoms on the imaginery acid hydrate increase it's priority over the acid.

• no changed my mind - according to this The acid is higer priority. looks like your old professor was right.. The old dog. vanderbilt.edu/AnS/Chemistry/Rizzo/chem220a/priority.pdf Oct 15 '15 at 3:23
• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. Oct 15 '15 at 4:06