= Phosphate diester
A \b[phosphate diester] is a <phosphoric ester> with two organic groups linked through <oxygen> to the same <phosphorus>. Its neutral-acid representation is $\mathrm{RO-P(=O)(OH)-OR'}$; deprotonating the remaining O–H group gives $\mathrm{RO-P(=O)(O^{-})-OR'}$ without breaking either ester linkage.<Nomenclature of Phosphorus-Containing Compounds of Biochemical Importance by IUPAC and IUBMB>[§4, phosphodiesters and the bridging phosphoric group]{ref}
For an original connectivity example, let the two organic groups be methyl groups. The fragment $\mathrm{CH_3O-P(=O)(OH)-OCH_3}$ contains one phosphorus center but two organic O–P connections. Two separate molecules of $\mathrm{CH_3O-P(=O)(OH)_2}$ instead contain two phosphorus centers, each monoesterified. Counting phosphorus atoms alone therefore cannot determine the ester class.
A diester can bridge two different organic units or join two positions within one unit. In a cyclic nucleotide, the two attachments close a ring; they need not belong to two separate molecules.<Nomenclature of Phosphorus-Containing Compounds of Biochemical Importance by IUPAC and IUBMB>[§4 comments(iii–iv) and Table3, polynucleotide names and cyclic phosphates]{ref} In <DNA> and <RNA>, successive sugar units are joined through <phosphodiester bonds>. Cleaving one ester linkage changes chain connectivity, which is why the same local chemistry can govern the integrity of a much longer <nucleic acid>.
The word “diester” specifies connectivity at the phosphoric group. It does not mean a P–O–P chain of two phosphate units, and it does not by itself specify a reaction rate or the conditions required for <hydrolysis>.
Back to article page
