Question

When pyrrole is added to a dilute solution of ${\mathbf{D}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$in${\mathbf{D}}_{\mathbf{2}}\mathbf{O}$ , 2-deuteriopyrrole is formed. Propose a mechanism to account for the formation of this compound.

Short answer

When pyrrole is reacted with ${\mathrm{D}}_2{\mathrm{SO}}_4$in${\mathrm{D}}_2\mathrm{O}$ , the 2-deuteriopyrrole is formed. The product formed is shown below:

Product of the reaction

Step-by-step solution

Step-by-Step SolutionStep 1:Reaction of pyrrole with D2SO4  in D2O

When pyrrole is reacted with ${\mathrm{D}}_2{\mathrm{SO}}_4$in${\mathrm{D}}_2\mathrm{O}$ , the 2-deuteriopyrrole is formed. The product formed is shown below:

Product of the reaction

Step 2: Reaction mechanism

The lone pair of electrons on the nitrogen atom gets shifted and bond movement occurs. Therefore, it forms a positive charge on the nitrogen atom. Thereby the action of ${\mathrm{D}}_2{\mathrm{SO}}_4$in ${\mathrm{D}}_2\mathrm{O}$occurs. At ${\mathrm{C}}_2$, deuterium gets added and when ${\mathrm{D}}_2\mathrm{O}$acts on it, the hydrogen on ${\mathrm{C}}_2$gets removed and that bond shifts forming a $\mathrm{\pi }$bond between ${\mathrm{C}}_2$and ${\mathrm{C}}_3$. Then, the $\mathrm{\pi }$bond between ${\mathrm{C}}_3$and${\mathrm{C}}_4$ moves to form a $\mathrm{\pi }$bond between${\mathrm{C}}_4$ and ${\mathrm{C}}_3$. Moreover, the $\mathrm{C}-\mathrm{N}$$\mathrm{\pi }$bond again shifts to the nitrogen atom regaining the lone pair of electrons on nitrogen atom. Thus, deuterium is placed on the second position and therefore, 2-deuteriopyrrole is formed.

Reaction mechanism