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

(2​E)-3-bromo-3-iodo-2-(trihydroxymethyl)prop-2-enoic acid

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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

enter image description here

is treated as

enter image description here

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

whereas

enter image description here

is treated as

enter image description here

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.

(2E)-3-bromo-3-iodo-2-(trihydroxymethyl)prop-2-enoic acid

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    $\begingroup$ 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. $\endgroup$ – Jan Oct 15 '15 at 12:00
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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.

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  • $\begingroup$ 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 $\endgroup$ – WooTangClan Oct 15 '15 at 3:23
  • $\begingroup$ 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. $\endgroup$ – Martin - マーチン Oct 15 '15 at 4:06
  • $\begingroup$ @WooTangClan You should edit your answer instead of putting edits in the comments. $\endgroup$ – bon Oct 15 '15 at 7:20
  • $\begingroup$ Whether or not the orthoacid exists is irrelevant! $\endgroup$ – user55119 Jun 10 at 18:36
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Simple. The rule is that at the tie..(carbon)...there are three ligands of equivalent priority...however, since their not both acids, the rule is that sp2 hybridization beats sp3.

Also, you'll never isolate that geminal (or triplet) alcohol, someone called it an acid hydrate..(even hydrates of aldehydes are difficult to isolate). Only as an orthoester can such structures be isolated.

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