$\ce{CO2}$ 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).
$\ce{CO2}$ 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).
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.More
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.
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.
$\ce{CO2}$ 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).
$\ce{CO2}$ 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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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.
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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VOTE