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Reaction with H2SO4 and H2O


Function

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Hello everyone

 

Reaction of this molecule

post-100256-0-88812200-1417963546_thumb.png

With sulfuric acid and water gives us the following molecule.

post-100256-0-65769200-1417963554_thumb.png

Main question: why not this molecule?

post-100256-0-62126600-1417963561_thumb.png

 

I understand that OH binds to the 3rd, and also why not to the 4th molecule (Markovnikov's rule, if I'm not mistaken; correct me if I'm not right). But not why a methyl-group is being replaced by a H-atom on the single-bonded O-atom of the ester, leaving a carbonic acid. Is it because H2O leaves another electrophile H+-atom when dissociating in H and OH, and that perhaps a bond with H would be more stable than a bond with CH3?

 

Secondary question: what's the name of the reacting molecule? I can think of nothing better than methyl-4-cyclopentyl-3-methyl-but-3-enoate.

I hardly doubt, however, that's the real name of the molecule. Can someone give me the correct IUPAC name? (Doesn't necessarily have to be the newest nomenclature that's been introduced few weeks or months ago.)

 

Thanks.

 

Function

 

P.S. don't mind incorrect rotations etc. of my drawings; quickly drew them in Paint

Edited by Function
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It's two reactions: acid catalysed ester hydrolysis and addition across a double bond. The acid protonates the carbonyl oxygen, which helps to promote nucelophilic attack of water at the carbonyl carbon and eventual expulsion of MeOH (note: it is not the methyl group that is leaving, but the whole -OMe component).

 

And example mechanism I stole from http://osxs.ch.liv.ac.uk/java/Nucleophilic%20substitution%20at%20the%20carbonyl%20group%20-%20Acid-catalysed%20ester%20hydrolysis%20and%20transesterification.html

 

post-35291-0-42066800-1417965958_thumb.jpg

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The ester is hydrolyzed by the acid. What do you know about the amide/ester/carboxylic acid/carbonyl halide functional group transformations?

 

The ester hydrolysis proceeds through a tetrahedral intermediate around what was the carbonyl carbon. Basically the acid protonates the C=O oxygen so that water can then act as a nucleophile and attack the (now even more electron deficient through resonance) carbonyl carbon. Push some arrows and eliminate a molecule of MeOH and there is your product.

 

EDIT: cross posted with my esteemed colleague above who never let's me answer an organic question :)

Edited by mississippichem
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Thanks, people!

If you care to know more the mechanism for acid catalyzed ester hydrolysis was actually proven through a classic experiment where one does the the reaction with 18-O labeled reagents IIRC.

 

It turns out the tagged oxygen ends up in the products such that it proves the reaction proceeds through nucleophilic attack by water at the carbonyl carbon rather than direct protonation of the ester oxygen followed by elimination of ROH.

 

This is beyond the scope of your question but chemistry threads here are resolved too quickly (actually a good thing) and I haven't posted in a while so I'll ramble a bit.

 

Read about that experiment if you care to. It's one of those "classic" ones that seem to be instructive.

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You can do some amazing experiments with doubly tagged water (heavy H and O) in vivo rather than in vitro - the ratios of the heavier isotopes after testing in different part of of body, excreta, and waste can give you quantifiable data on all sorts of bodily functions (basal or field metabolic rate especially). IIRC lots of amazing research areas are/were tied up waiting for MRECs and similar to give go ahead as some campaigners waved the "radiation bad" flag because the word isotope reminds them of something

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