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How do I find the oxidation numbers for ester?

My main issue is about finding the numbers for ester groups.

         /O\
     -    ||
R -- O -- C -- R'
     -

I want the oxidation numbers of those two O atoms and C atom.

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$\ce{CO2}$ and derivatives (e.g., carbonates and their esters, urea, guanidine, $\ce{CCl4}$) are formally C(+4), fully oxidized. Carboxylates are then C(+3) (but not a chloroformate ester, of course).

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  • $\begingroup$ I added a schema as much as i can can you check it please $\endgroup$ – Ozan Kurt Apr 16 '14 at 19:00
  • $\begingroup$ That is an ester, yes. The paired electrons on the oxygen are usually seen as rabbit ears on the sp^2 hybridized oxygen, normal to the pi orbitals of the double bond. $\endgroup$ – Uncle Al Apr 17 '14 at 2:45
  • $\begingroup$ Rabbit ears is a rather bad description though, @UncleAl $\endgroup$ – Jan Sep 27 '15 at 0:51
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By finding oxidation numbers a systematic way is to go from the element with the highest electronegativity to the lowest. Also always consider neighbouring atoms.

I the presented case of an arbitrary ester $\ce{R\bond{-}O\bond{-}(C\bond{=}O)\bond{-}R'}$, this would be oxygen. Also the $\ce{R}$ moieties are of alkyl or aryl type, so they are bonded to a carbon.

Now assign $-2$ to all Oxygens, assuming they will form a full valence shell (which is a requirement for oxidation states). Now consider that electrons between two carbons will be shared equally. Therefore you can ignore $\ce{R'}$, meaning you assign the total group a value of $0$. There is one Oxygen bonded to the carbon and to $\ce{R}$, it's effect of the oxidation number will be divided in both directions. Therefore the carboxyl carbon will "feel" only $-1$. Add up all the oxidation numbers around the carboxyl carbon ($-3$). Since it is a neutral compound the oxidation state of the carbon has to mirror its surroundings - the value is therefore $+3$.

Always keep in mind, that oxidation states are formal numbers only. They do not represent the actual charge of an atom. They will still give you an idea about the polarity of the molecule.

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