Question

How many electrons does each of the four nitrogen atoms in purine contribute to the aromatic system?

Short answer

A total of five electrons are donated by the nitrogen in the ring.

The NH shown donate 1 lone pair that 2 electrons which take part in delocalization and other three donate 1 electron.

Step-by-step solution

Purine

It’s a heterocyclic aromatic five and six-membered fused compound the N is the heteroatom present in the ring with lone pair and it is a nitrogenous base having four unsaturated double bonds.

Aromaticity

The compound is aromatic only when it follows Huckel’s rule that 4n+2 π electrons are present like 6 pi electrons in which n=2 hence the aromatic nature of the compound.

Like in pure the 4 double bonds are present that is 8 π electron which does not follow the 4n+2 rule because the n value is not a whole number which means the lone pair of nitrogen must take in the delocalization to make the compound aromatic in nature.

$\begin{array}{c}4n+2=8\\ 4n=8-2\\ 4n=6\\ n=\frac{3}{2}\end{array}$

Step 3: π- electrons of nitrogen take part in aromaticity

As clear from the diagram the NH bond is not shared by the double bond so it’s alone pair that 2 electrons take part in delocalization and the rest of the three N is shared by the double bond so that only 1 electron is taken part in the delocalization to make the compound aromatic.