Oxygen Charge & Electron Density in Simple Carboxylic Acids

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SUMMARY

The discussion focuses on the electron density and charge distribution between the two oxygen atoms in simple carboxylic acids, specifically during the process of Fischer esterification. It is established that both oxygens are equivalent due to resonance and that the carboxylic group can freely rotate around the carbon bond, making the specific attachment of protons to either oxygen largely irrelevant. However, once protonation occurs, the symmetry is lost, leading to a doubly protonated carboxyl group that significantly influences the reaction mechanism. The key intermediate formed during this process is a carbocation, which is susceptible to nucleophilic attack.

PREREQUISITES
  • Understanding of carboxylic acid structure and properties
  • Knowledge of resonance structures in organic chemistry
  • Familiarity with Fischer esterification mechanism
  • Basic concepts of carbocation stability and nucleophilic attack
NEXT STEPS
  • Study the mechanism of Fischer esterification in detail
  • Learn about resonance structures and their implications in reaction mechanisms
  • Investigate the role of carbocations in organic reactions
  • Explore the effects of protonation on functional groups in acidic environments
USEFUL FOR

Chemistry students, organic chemists, and anyone interested in understanding the mechanisms of esterification and the behavior of carboxylic acids in acidic conditions.

Karan Punjabi
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In any simple carboxylic acid there are two oxygen atoms then i have a confusion that which oxygen has more negative charge on it or which one has the most electron density on it?
 
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They are equivalent due to resonance, plus the carboxylic group freely rotates around the single bond - so it doesn't matter.
 
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Borek said:
They are equivalent due to resonance, plus the carboxylic group freely rotates around the single bond - so it doesn't matter.
But in mechanism of esterification in presence of an acid...the proton attacks tje double bonded oxygen
 
Once the carboxylic group becomes protonated (in other words: is not dissociated) it is no longer symmetrical (no resonance). Which oxygen becomes protonated is in most cases completely random. But it still doesn't matter to which oxygen the H is attached, as the group can freely rotate around "carbon-rest of the molecule" bond, so the final product will be exactly the same.
 
I'm assuming you're talking about acidic Fischer esterification, but it's hard to tell from your question what step is tripping you up. Since the environment is acidic, the carboxyl is going to be overwhelmingly neutral. However, the key intermediate in the reaction is a doubly protonated carboxyl group. Both oxygens become protonated and the two OH groups attached to the carboxyl carbocation draw a lot of electron density away from this carbon. The carbocation therefore has a significant partial positive charge which is readily attacked by a nucleophile (in the case of esterification, the nucleophile is the oxygen on a hydroxyl group). The link below gives a good overview:

http://www.organic-chemistry.org/namedreactions/fischer-esterification.shtm

The carbocation intermediate that I referred to is given by the resonance structure in the first set of brackets.
 

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