Home >
Community >
What is acidity order of o-cyanophenol m-cyanophenol and p-cyanophenol?
Upvote
18
Downvote
+ Acids
+ Chemistry
+ Organic chemistry
Posted by
Mark Pinsker
What is acidity order of o-cyanophenol m-cyanophenol and p-cyanophenol?
Honestly, before this question i had never heard of “anisic acid”, so i googled it.
Now coming to your question, in o-anisic acid there is a phenomenon taking place called “ortho effect” in which there is a strong steric hinderance of 2 bulky OCH3 and COOH groups due to which the COOH group goes out of the plane of the ring, so it its conjugate base cannot take part in resonance with the benzene ring,
so COOH will undergo resonance with itself, so it forms 2 equivalent resonance structure so their contribution to resonance hybrid is 50–50 % and there is proper delocalisation of the negative charge on oxygen atom, so its conjugate base is more stable than the other two as we know that more stable is the conjugate base more is the acidic strength of the acid.
where as in meta and para there is less steric hinderance- no ortho effect-complete resonance with the benzene ring-no equivalent resonance structures( negative charge on carbon atom is less stable than negative charge on oxygen atom as oxygen atom is able to hold negative charge more efficiently).
the contribution to the resonance hybrid of the resonance structure in which there is negative charge on carbon is very less.
So negative charge is not properly delocalised so it is more on the oxygen atom, so less stable than the ortho one.
Comparing meta and para, we know that when a electron donating group like OCH3 present on meta position it cannot give electron through hyperconjugation or +H effect, but it can do withdraw through -I or inductive effect.
But when OCH3 group is present in para position it can donate electron through both +H and -I.
But we know that +H effect has more impact than -I effect. So in para one the +H effect will destabilize the negative charge on oxygen atom where as in meta there will be no such problem.
So meta will be more stable than para.
So the final order of acidic strength is ortho anisic acid > meta anisic acid > para anisic acid
Hope you have understood well what i have explained.
I have tried my level best to give you the best possible explanation.
Honestly, before this question i had never heard of “anisic acid”, so i googled it.
Now coming to your question, in o-anisic acid there is a phenomenon taking place called “ortho effect” in which there is a strong steric hinderance of 2 bulky OCH3 and COOH groups due to which the COOH group goes out of the plane of the ring, so it its conjugate base cannot take part in resonance with the benzene ring,
so COOH will undergo resonance with itself, so it forms 2 equivalent resonance structure so their contribution to resonance hybrid is 50–50 % and there is proper delocalisation of the negative charge on oxygen atom, so its conjugate base is more stable than the other two as we know that more stable is the conjugate base more is the acidic strength of the acid.
where as in meta and para there is less steric hinderance- no ortho effect-complete resonance with the benzene ring-no equivalent resonance structures( negative charge on carbon atom is less stable than negative charge on oxygen atom as oxygen atom is able to hold negative charge more efficiently).
the contribution to the resonance hybrid of the resonance structure in which there is negative charge on carbon is very less.
So negative charge is not properly delocalised so it is more on the oxygen atom, so less stable than the ortho one.
Comparing meta and para, we know that when a electron donating group like OCH3 present on meta position it cannot give electron through hyperconjugation or +H effect, but it can do withdraw through -I or inductive effect.
But when OCH3 group is present in para position it can donate electron through both +H and -I.
But we know that +H effect has more impact than -I effect. So in para one the +H effect will destabilize the negative charge on oxygen atom where as in meta there will be no such problem.
So meta will be more stable than para.
So the final order of acidic strength is ortho anisic acid > meta anisic acid > para anisic acid
Hope you have understood well what i have explained.
I have tried my level best to give you the best possible explanation.
The strength of an acids can be rationalized by the stability of its conjugate base normally. Stable conjugate bases (products of the acid-base reaction) are lower in energy and are favored at equilibrium. The number of conjugate base resonance structures that delocalize formal charges correlates with the stability of the conjugate base. Other factors that affect acidity include: the atom bearing the hydrogen, molecular polarization and orbital overlap with hydrogen. But these are theoretical considerations. Anyhow, practically, it is favoured on the basis of pKa values. Greater the pKa value less is the acidic strength. Drawing resonance structures, one may get the following order of acidic strength of position isomers of cyanophenol
Meta > Ortho = Para
But here is the reality on the basis of pKa values
The strength of an acids can be rationalized by the stability of its conjugate base normally. Stable conjugate bases (products of the acid-base reaction) are lower in energy and are favored at equilibrium. The number of conjugate base resonance structures that delocalize formal charges correlates with the stability of the conjugate base. Other factors that affect acidity include: the atom bearing the hydrogen, molecular polarization and orbital overlap with hydrogen. But these are theoretical considerations. Anyhow, practically, it is favoured on the basis of pKa values. Greater the pKa value less is the acidic strength. Drawing resonance structures, one may get the following order of acidic strength of position isomers of cyanophenol
Meta > Ortho = Para
But here is the reality on the basis of pKa values
Honestly, before this question i had never heard of “anisic acid”, so i googled it.
Now coming to your question, in o-anisic acid there is a phenomenon taking place called “ortho effect” in which there is a strong steric hinderance of 2 bulky OCH3 and COOH groups due to which the COOH group goes out of the plane of the ring, so it its conjugate base cannot take part in resonance with the benzene ring,
so COOH will undergo resonance with itself, so it forms 2 equivalent resonance structure so their contribution to resonance hybrid is 50–50 % and there is proper delocalisation of the negative charge on oxygen atom, so its conjugate base is more stable than the other two as we know that more stable is the conjugate base more is the acidic strength of the acid.
where as in meta and para there is less steric hinderance- no ortho effect-complete resonance with the benzene ring-no equivalent resonance structures( negative charge on carbon atom is less stable than negative charge on oxygen atom as oxygen atom is able to hold negative charge more efficiently).
the contribution to the resonance hybrid of the resonance structure in which there is negative charge on carbon is very less.
So negative charge is not properly delocalised so it is more on the oxygen atom, so less stable than the ortho one.
Comparing meta and para, we know that when a electron donating group like OCH3 present on meta position it cannot give electron through hyperconjugation or +H effect, but it can do withdraw through -I or inductive effect.
But when OCH3 group is present in para position it can donate electron through both +H and -I.
But we know that +H effect has more impact than -I effect. So in para one the +H effect will destabilize the negative charge on oxygen atom where as in meta there will be no such problem.
So meta will be more stable than para.
So the final order of acidic strength is ortho anisic acid > meta anisic acid > para anisic acid
Hope you have understood well what i have explained.
I have tried my level best to give you the best possible explanation.
If you like my answer please upvote !!!
Thanks a lot !!!
Honestly, before this question i had never heard of “anisic acid”, so i googled it.
Now coming to your question, in o-anisic acid there is a phenomenon taking place called “ortho effect” in which there is a strong steric hinderance of 2 bulky OCH3 and COOH groups due to which the COOH group goes out of the plane of the ring, so it its conjugate base cannot take part in resonance with the benzene ring,
so COOH will undergo resonance with itself, so it forms 2 equivalent resonance structure so their contribution to resonance hybrid is 50–50 % and there is proper delocalisation of the negative charge on oxygen atom, so its conjugate base is more stable than the other two as we know that more stable is the conjugate base more is the acidic strength of the acid.
where as in meta and para there is less steric hinderance- no ortho effect-complete resonance with the benzene ring-no equivalent resonance structures( negative charge on carbon atom is less stable than negative charge on oxygen atom as oxygen atom is able to hold negative charge more efficiently).
the contribution to the resonance hybrid of the resonance structure in which there is negative charge on carbon is very less.
So negative charge is not properly delocalised so it is more on the oxygen atom, so less stable than the ortho one.
Comparing meta and para, we know that when a electron donating group like OCH3 present on meta position it cannot give electron through hyperconjugation or +H effect, but it can do withdraw through -I or inductive effect.
But when OCH3 group is present in para position it can donate electron through both +H and -I.
But we know that +H effect has more impact than -I effect. So in para one the +H effect will destabilize the negative charge on oxygen atom where as in meta there will be no such problem.
So meta will be more stable than para.
So the final order of acidic strength is ortho anisic acid > meta anisic acid > para anisic acid
Hope you have understood well what i have explained.
I have tried my level best to give you the best possible explanation.
If you like my answer please upvote !!!
Thanks a lot !!!
More
VOTE
The strength of an acids can be rationalized by the stability of its conjugate base normally. Stable conjugate bases (products of the acid-base reaction) are lower in energy and are favored at equilibrium. The number of conjugate base resonance structures that delocalize formal charges correlates with the stability of the conjugate base. Other factors that affect acidity include: the atom bearing the hydrogen, molecular polarization and orbital overlap with hydrogen. But these are theoretical considerations. Anyhow, practically, it is favoured on the basis of pKa values. Greater the pKa value less is the acidic strength. Drawing resonance structures, one may get the following order of acidic strength of position isomers of cyanophenol
Meta > Ortho = Para
But here is the reality on the basis of pKa values
Ortho > Para > Meta
pKa of o-cyanophenol = 7.0
pKa of p-cyanophenol = 7.97
pKa of m-cyanophenol = 8.61
The strength of an acids can be rationalized by the stability of its conjugate base normally. Stable conjugate bases (products of the acid-base reaction) are lower in energy and are favored at equilibrium. The number of conjugate base resonance structures that delocalize formal charges correlates with the stability of the conjugate base. Other factors that affect acidity include: the atom bearing the hydrogen, molecular polarization and orbital overlap with hydrogen. But these are theoretical considerations. Anyhow, practically, it is favoured on the basis of pKa values. Greater the pKa value less is the acidic strength. Drawing resonance structures, one may get the following order of acidic strength of position isomers of cyanophenol
Meta > Ortho = Para
But here is the reality on the basis of pKa values
Ortho > Para > Meta
pKa of o-cyanophenol = 7.0
pKa of p-cyanophenol = 7.97
pKa of m-cyanophenol = 8.61
More
VOTE