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?
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
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:
-
1$\begingroup$ Presumably it's not planar though? $\endgroup$– RobChemCommented Dec 2, 2014 at 18:12
-
1$\begingroup$ Why wouldn't everything be planar? Each atom is sp2 hybridized. $\endgroup$ Commented Dec 2, 2014 at 18:31
-
$\begingroup$ There might be some steric hindrance to planarity though $\endgroup$ Commented Dec 2, 2014 at 18:41
-
2$\begingroup$ 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. $\endgroup$– RobChemCommented Dec 2, 2014 at 23:55
-
1$\begingroup$ @Dissenter but H-bonding is present even before deprotonation why does it not make the acid weaker? $\endgroup$– NehaCommented May 29, 2021 at 3:57