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azulene bromination

I found this question and the answer as shown in my book. I didn't understand why the bromine particularly entered the position shown in the answer.

First of all, if a free radical was generated there it would be 2°, and also will have resonance. Is any other product possible or this is the only major product?

Also, is this allylic bromination? Every allylic position is also a vinylic position — I am not getting it.

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    $\begingroup$ Are you sure that this a radical bromination? :) $\endgroup$
    – Zhe
    Commented Sep 26, 2018 at 18:07
  • $\begingroup$ I have a question on that too , I really confused , like there is no allylic position and every allylic position is like a vinylic position. I am not getting this $\endgroup$
    – Chloritone_360
    Commented Sep 26, 2018 at 18:11
  • $\begingroup$ This is aromatic azulene and it's simple reaction with Br2 $\endgroup$
    – Mithoron
    Commented Sep 26, 2018 at 18:40
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    $\begingroup$ Usually, these types of problems can be explained by looking at the intermediate that has resonance structures that don't disrupt the aromaticity of the other ring to where you've attacked. $\endgroup$
    – Zhe
    Commented Sep 26, 2018 at 18:42
  • $\begingroup$ Can anyone tell me where to learn how to write chemical formulas and reaction is stack exchange ? $\endgroup$
    – Chloritone_360
    Commented Oct 8, 2018 at 4:56

1 Answer 1

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As a follow-up to the excellent comments provided thus far, the following thoughts are added. Electrostatic Potential Map of azulene (1a) shows that there is high electron density in the 5-membered ring (red color signifies high electron density). The 5-membered ring is reminiscent of the aromaticity of cyclopentadienide anion and the 7-membered ring bears a positive charge and has aromaticity similar to cycloheptatrienyl cation. This phenomenon is represented in structure 1b. The reaction does not involve bromine atom (a radical) because a bromine atom would be expected to attack the electropositive ring, not the electron-rich ring. The reaction is ionic and likely involves bromine rather than N-bromosuccinimide (NBS, 4) directly. Bromine is liberated from NBS by trace amounts of HBr formed by the slow decomposition of NBS. In the event, bromination of azulene (1c) produces a resonance stabilized cation 2a. Its canonical structure is 2b that is identical to 2b'. Loss of a proton from 2b (or 2b') provides bromoazulene 3.

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