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In the following reaction, why is the molecule protonated at the indicated carbon and not on one of the other double bonds?

enter image description here

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  • $\begingroup$ protonation of the pi bond indicated in your figure is the only one that can produce a carbocation that is both resonance stabilized (allyl) and tertiary at one end. $\endgroup$ – ron Jun 29 '16 at 14:10
  • $\begingroup$ @ron so another pi bond in the ring can attack $\endgroup$ – Koolman Jun 29 '16 at 14:22
  • $\begingroup$ No, attack by the other pi bonds cannot produce such a stable carbocation. $\endgroup$ – ron Jun 29 '16 at 15:02
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The key is delocalization of the positive charge in carbocations.

  1. Most stable carbocation that you can get from the protonation of the compound above. Reasons for it being the most stable:

    • The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)
    • In one of the resonance forms, the positive charge is at a tertiary carbon atom

    enter image description here

  2. Second most stable carbocation.

    • The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)

    enter image description here

  3. When protonation happens on the third pi bond. No delocalization - not as stable as the carbocations shown above.

    enter image description here

  4. There are other possible ways in which the compound could have been protonated (below), however none of them are as stable as (1) or (2), because there is no delocalization.

    enter image description here enter image description here enter image description here

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