allylbenzene + HCl -> unknown product; four options

Shouldn't it be 4th option since since the carbon attached to the ring is secondary as well as resonance stabilized?

  • $\begingroup$ What you say about d is mostly true, but you have to remember to think about the mechanism a little to see which two carbocations you are picking between $\endgroup$ – Dennis Cao Jul 17 '18 at 1:36
  • $\begingroup$ I'm confused since I don't know the exact mechanism. Could you please explain me instead of me making absurd guesses? $\endgroup$ – Dante Jul 17 '18 at 1:44
  • $\begingroup$ Ah so it sounds like you should first figure out the mechanism for HCl adding to a double bond, and go from there to solve the problem $\endgroup$ – Dennis Cao Jul 17 '18 at 2:01

I've searched SciFinder for this and related reactions (allyl benzene + H-X). In the results, there are no examples of the halide at the benzylic position in the product, and there are many examples of the halide at the secondary postion arising from the more substituted end of the alkene. One such example is below.

enter image description here

It is certainly warranted to consider both the direct addition and the product from rearrangement (since a resonance stabilized, benzylic carbocation would be formed as an intermediate), and if one were performing the experiment in the lab, it would be a good idea to look for both products. It seems that in practice, the addition of halide to the secondary carbocation is faster than the hydride shift. Predictions about kinetics like this are very difficult, and exam questions that require them could be considered unfair unless there is an overwhelming and predictable rationale for one product over another.

|improve this answer|||||

I believe the fourth option would require a removal of a hydrogen.

As well, the double bond is electron donating so it acts as a nucleophile to pick up the proton.

Double bonds like protons/hydrogen more than a Carbon with four single bonds to other carbon and hydrogen. Its happy with a full octet and (nearly) equally electronegative buddies, while double bonds could sure go for a few more protons to fill up.

|improve this answer|||||
  • 2
    $\begingroup$ The fourth would require a hydride shift of the intermediate which is not uncommon. $\endgroup$ – A.K. Jul 17 '18 at 2:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.