Question

Nucleosides are stable in water and in dilute base. In dilute acid, however, the glycosidic bond of a nucleoside undergoes hydrolysis to give a pentose and a heterocyclic aromatic amine base. Propose a mechanism for this acid- catalyzed hydrolysis.

Short answer

Answer: The key steps in the acid-catalyzed hydrolysis of a nucleoside are: (1) protonation of the glycosidic oxygen, (2) nucleophilic attack by water, (3) proton transfer and breaking of the glycosidic bond, and (4) deprotonation of the pentose. These steps together lead to the formation of a pentose and a heterocyclic aromatic amine base.

Step-by-step solution

Protonation of the glycosidic oxygen

In the presence of an acid, a proton (H+) will protonate the glycosidic oxygen atom. This protonation makes the oxygen more electrophilic, weakening the glycosidic bond.

Nucleophilic attack by water

The weakened glycosidic bond makes the carbon atom, which was connected to the glycosidic oxygen, more susceptible to nucleophilic attack. A water molecule acts as a nucleophile and attacks the carbon atom. This leads to the formation of a pentavalent carbon intermediate with a positively charged oxygen, which is not stable.

Proton transfer and breaking of the glycosidic bond

A proton from the protonated glycosidic oxygen is transferred to one of the other oxygen atoms in the pentavalent carbon intermediate. The transfer of a proton results in the breaking of the glycosidic bond, yielding a pentose molecule with a positively charged oxygen and a heterocyclic aromatic amine base.

Deprotonation of the pentose

The positively charged oxygen on the pentose molecule attracts a water molecule, which acts as a base and captures the extra proton. This results in the formation of a stable pentose and the regeneration of a water molecule. The overall reaction has produced a pentose and a heterocyclic aromatic amine base, and the mechanism for acid-catalyzed hydrolysis of a nucleoside is complete.