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The compound is given in the image

What is the double bond equivalence of the given compound? The given answer is 14 (8 pi electrons and 6 rings = 14 DBE). But why isn't the lone pair on nitrogen considered? The lone pair on nitrogen is involved in resonance with the other 2 bonds making the ring aromatic, so shouldn't we also consider that lone pair?

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    $\begingroup$ If some double bonds are in resonance, that does not make more or fewer of them. It is still the same number of double bonds. Resonance is about as irrelevant as the weather in the street outside. So is the lone pair of N. $\endgroup$ Commented Apr 2, 2021 at 12:00

2 Answers 2

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As Ivan Neretin pointed out in the comments, $\pi$ bonds will remain same whether or not in resonance.

However, if you have trouble counting the double bond equvalent, you can directly use the formula for it given as $$\ce{DU=\dfrac{2C+2-H+N-X}{2}}$$ so here number of carbons are C=23, H=21, N=1 which also gives 14 as the answer.

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6 rings, and 8 double bonds, adds together for 14 DBE. Resonance will not change how many double bonds there are - you can see that the nitrogen has made 3 bonds and so does not contribute to the DBE total

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  • $\begingroup$ For an alternative method, look here. $\endgroup$
    – user55119
    Commented Mar 1 at 16:27

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