When salicylic acid deprotonates (losing the proton from the carboxyl group) it forms a hydrogen bond with between the oxygen in the carboxylate anion and the hydrogen in the alcohol group. However, surely there will be a steric clash between these groups which would surely be make more significant by the long hydrogen bond? Could the hydrogen bond still form with the carboxyl group twited out of the plane? Also, why doesn't it hydrogen bond intramolecularly before deprotonation which would disfavor the loss of a proton?


If the COO(-) group on salicylic acid's conjugate base didn't hydrogen bond, then wouldn't the COO(-) therefore be more inclined to pick up a proton and return the entire molecule to salicylic acid? In other words H-bonding is a stabilizing interaction. You can view H-bonding as diluting the negative charge density in the COO(-) group. This in turn makes proton abstraction less of a favorable move.

Structure of salicyclic acid's conjugate base:

enter image description here

| improve this answer | |
  • $\begingroup$ Presumably it's not planar though? $\endgroup$ – RobChem Dec 2 '14 at 18:12
  • $\begingroup$ Why wouldn't everything be planar? Each atom is sp2 hybridized. $\endgroup$ – Dissenter Dec 2 '14 at 18:31
  • $\begingroup$ There might be some steric hindrance to planarity though $\endgroup$ – Dissenter Dec 2 '14 at 18:41
  • $\begingroup$ Conformational isomers are possible with rotation about the single bonds between the carbon in the ring and to the hyroxyl group and the carboxylate group. $\endgroup$ – RobChem Dec 2 '14 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.