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Double Bond Equivalent/Degree of Unsaturation

25. Degree of unsaturation in
Naphthalene
is

(a) 6; (b) 7; (c) 8.

There are two rings and six double bonds in structure so degree of unsaturation should be 8, but it is not. Why?

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    $\begingroup$ Note that it is generally not acceptable to use curves in two adjacent fused rings, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the two rings. $\endgroup$ – Loong Mar 7 at 6:53
  • $\begingroup$ Could you edit your question to tell us why you think there are six double bonds (I don't see any) and how you know that the answer is not 8? $\endgroup$ – Karsten Theis Mar 7 at 15:40
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You determined the number of double bonds incorrectly, probably due to structure representation with aromatic rings, which, as Loong pointed out, is not entirely correct. You can redraw the structure of naphthalene as such:

Naphtalene

and make sure that there are five double bonds and two rings giving degree of unsaturation

$$\text{DU} = n_\text{rings} + n_\text{π-bonds} = 2 + 5 = 7$$

You can also compare formula of naphthalene $\ce{C10H8}$ with fully saturated alternative $\ce{C10H22}$ and proceed from there:

$$\text{DU} = (H_\mathrm{sat} - H)/2 = (22 - 8)/2 = 7$$

You again arrive at $\text{DU} = 7$ using a simplified formula:

$$\text{DU} = C - H/2 + 1 = 10 - 8/2 + 1 = 7$$

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