Why is the degree of unsaturation 9 for C9H6N4?

[phoenixXL]

Question
Find out the degree of unsaturation in a compound having the molecular formulae C₉H₆N₄.

Attempt
The point lies in making the possible structure(s).
The structure that I felt possible is as follows

In the compound above there are 8-π electrons,
Hence the degree of unsaturation is 8(Ans)

Problem
The problem comes from the topic of structural isomerism and the book says the answer to be 9

Please help me out. Thanks for your time.

 

[Borek]

Hint: you are aware of the fact degree of unsaturation for cyclohexane - despite the compound being saturated - is 1?

In other words, degree of unsaturation is not only about pi electrons. There are formulas to calculate degree of unsaturation based on the molecular formula alone (without any structural information).

 

[phoenixXL]

Okay, I didn't knew about that.
I looked Wikipedia and found that rings are counted as a degree of unsaturation, therefore I get the missing "1" from my answer. Also I was unaware of the formula(curious to see its derivation).
Thank You.

 

[Borek]

Also I was unaware of the formula(curious to see its derivation).

There is no derivation - it is more like a definition.

 

[PFuser1232]

Okay, I didn't knew about that.
I looked Wikipedia and found that rings are counted as a degree of unsaturation, therefore I get the missing "1" from my answer. Also I was unaware of the formula(curious to see its derivation).
Thank You.

$DU = 1 + \frac{1}{2}∑[n_i (v_i - 2)]$ where n_i is the number of atoms of an element and v_i is the valency of that element. Note that this gives you the total number of rings and π bonds.