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Why doesn't the carbocation form on the carbon that is attached to the ethyl group in this reaction?

Subsequently, why doesn't the $\ce{Cl-}$ attack the carbon that is attached to the ethyl group? Is a tertiary carbocation less stable than a secondary allylic carbocation?

Reaction:

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1,2 -hydride shifting takes place after the formation of the tertiary carbocation. After the hydride shifting, the carbocation becomes resonance stabilised. And the chlorine adds up there(the first product that you have shown). Also, as I said the carbocation was resonance stabilised. So you can draw resonance structure of the carbocation intermediate and the positive charge goes to 4 position and then chlorine adds up there (the second product that you have shown) and thus we get the second product. Hope this helps.enter image description here

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  • $\begingroup$ Thanks. This helps a lot. I never thought to look for potential hydride shifts when I already have a tertiary carbocation. So this suggests that an allylic carbocation is more stable than a tertiary carbocation, yes? $\endgroup$ – nickb Jan 9 '18 at 19:38
  • $\begingroup$ Yes resonance stabilisation is preferred over +I stabilisation. So, I hope it's clear. Do tick mark my answer as this question's answer in Chemistry.SE if my answer helped you. $\endgroup$ – DJ Koustav Jan 9 '18 at 19:43
  • $\begingroup$ Have labeling studies been done that confirm your proposed mechanism? If, as you say, an allyl carbocation is more stable than a tertiary carbocation, then why not protonate at the ethyl end of the double bond to directly produce the allyl carbocation which would then directly lead to the observed products? I'm a bit surprised at the reported products. I would have thought that protonation would occur at the other double bond to give a substituted allyl carbocation which would then lead to 2 different products. $\endgroup$ – ron Jan 10 '18 at 3:59
  • $\begingroup$ @ron First, we should protonate at the second position following Markovnikov's rule then think of carbocation rearrangement. Also, we should add the H+ ar the pi bond at the 1 position because there the electron density increases due to +I of ethyl group and hence the electrophile attacks there. Please reply if you agree and feel free to point out any shortcoming that my answer might contain. $\endgroup$ – DJ Koustav Jan 10 '18 at 5:44

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