Identify the structure and the labeled nitrogens

The question refers to a fused heterocyclic system with nitrogens labeled $N_1, N_3, N_7,$ and $N_9$ (as in purine-type numbering).

We must find which nitrogen does not contribute to basicity.

Principle of basicity in heterocycles

A nitrogen contributes to basicity if its lone pair is available for protonation.

• Pyridine-like nitrogen: lone pair is outside the aromatic sextet, so it is available for protonation  basic.
• Pyrrole-like nitrogen: lone pair is part of the aromatic sextet, so it is not available for protonation  does not contribute to basicity.

Examine the labeled nitrogens

In the purine framework:

• $N_1$ and $N_3$ are pyridine-like nitrogens. Their lone pairs are not part of the aromatic sextet, so they can accept a proton.
• $N_7$ is also pyridine-like, so its lone pair is available for protonation.
• $N_9$ is pyrrole-like. Its lone pair participates in maintaining aromaticity of the imidazole-like ring.

Therefore, protonating $N_9$ would disrupt the aromatic electron delocalization, so this lone pair is not available for basic behavior.

Conclusion

The nitrogen that does not contribute to basicity is:

$N_9$

Option check

• A: $N_7$  contributes to basicity
• B: $N_9$  does not contribute to basicity
• C: $N_1$  contributes to basicity
• D: $N_3$  contributes to basicity

Hence, the correct option is B.