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Topic: Product of rxn of cis-norborene-5,6-endo-dicarboxylic acid + H2SO4  (Read 9987 times)

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Offline SabertoothTiger

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I hydrolized the following compound: cis-norbornene-5,6-endo-dicarboxylic anhydride
then took the cis-norborene-5,6-endo-dicarboxylic acid and added H2SO4 to it (warmed it up and cooled it) then added drop by drop water (then I reheated the mixture and crystallized it again).

What is the compound? It was positive for carboxylic acid and also positive for an ester.

This is the original structure: http://www.chemblink.com/structures/3813-52-3.gif
of my compound before I added H2SO4 and water

It's a 2 five member ring (one with C=C bond) and it's connected via triangle of c-c bonds, with CO2H popping out from 2nd ring without C=C bond

When I was looking for answer online for what the compound was someone said that it was "add water across the double bond, open the anhydride, and then form a lactone with the carboxylic acid that is closer to the hydroxide" but I'm still not sure what the structure is and if they are actually correct. I was wondering if it's possible to explain it with arrow pushing mechanism please

Offline galoisien

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Re: Product of rxn of cis-norborene-5,6-endo-dicarboxylic acid + H2SO4
« Reply #1 on: June 09, 2010, 12:39:01 PM »
Do you have NMR data...?

Offline SabertoothTiger

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Re: Product of rxn of cis-norborene-5,6-endo-dicarboxylic acid + H2SO4
« Reply #2 on: June 09, 2010, 02:58:51 PM »
Do you have NMR data...?

No I don't - the NMR was acting up when I was taking the scan.
One thing that I found out that the COOH stays which is why I have a (+) for carboxylic acid
the other one forms an ester because of carbocation thats formed due to acid interaction with double bond. One COOH bends over to fix the carbocation forming a 5 member ring. I think I answered my question with structure.

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