I want to esterify with a carboxylic acid using 4-hydroxy pyridine. However, since 4-hydroxy pyridine is in equilibrium with 4-pyridone, does the carboxylic acid react with the N atom to form amide? How do I prevent this?
The chances of more towards the ester than amide bcoz of stability. you have to see the effect of high-temperature covert which chemoselective groups on the product. at high-temperature aromaticity of 4-hydroxy pyridine does not lose therefore only chances to hydroxy grp react and form an ester.
Only N-acyl-pyridones are isolable from the acylation of 4-pyridone with aliphatic carboxylic anhydrides, and chlorides or free acids in the presence of dicyclohexylcarbodiimide. Similarly the reaction of 4-pyridone with ortho substituted derivatives of benzoic acid leads only to N-acylation products , while benzoyl chloride as well as benzoic acid derivatives 7i-s (meta or para substituted), 3,5-dinitrobenzoyl chloride, and the 2.4-dimethyl substituted benzoyl chlorides 7v - z give exclusively 4-(acyloxy)pyridines 9. Reaction of phthaloyl chloride (10) with 1 yields both N- and 0-acylation products. In solution even at room temperature the N-acylation compounds 2 and 8 are in equilibrium with the (acyloxy)pyridines 3 and 9, respectively. The position of the equilibrium depends mainly on the structure of the acyl residue, it is, however, also effected by temperature and polarity of the solvent.
https://elib.uni-stuttgart.de/bitstream/11682/1064/1/eff123.pdf
if and when you've formed the product, you may find that it's not very stable, and it equilibrates to the N-acylated stuff. If you look in chemical suppliers' websites, you can't find any 4-acyloxypyridine with unsubstituted nitrogen (even the simplest ones), which seems to suggest that these products are unsuitable for long-term storage.
Synthesis of ester by using acid chlorides, the formation of ester is more. As you mention since 4-hydroxy pyridine is in equilibrium with 4-pyridone , the formation of ester is faster than amide so these two are equilibrium SMs, only 4-hydroxy pyridine will react faster. Once the formation of ester starts immediately 4-pyridone will convert to 4-hydroxy pyridine to maintain equilibrium.
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