I am trying the hypothetical synthesis below, but I didn't get the product I was expecting. Instead, the reaction formed mostly styrene, in >80% yield (elimination product of the bromide)

enter image description here

The synthesis of an analogous product with one less carbon in the bromide (benzylic bromide) is described in the literature and I thought it would work with this one since they only differ by a carbon. Any thoughts on how I could overcome this selectivity problem?

I thought about using a milder base, but I don't think the problem here is actually the base I use, cause they will all form the enolate, the problem is the enolate abstracting the benzylic proton from the bromide and generating styrene + butyrolactone.

  • 1
    $\begingroup$ What was the procedure? Preform the enolate then add the bromide? Or add the base to the mixture of lactone and bromide? $\endgroup$ – Waylander Apr 30 '19 at 19:35
  • $\begingroup$ Sorry, I forgot to mention... The enolate was formed first and then the bromide was added. $\endgroup$ – Raul Luciano Apr 30 '19 at 20:02
  • $\begingroup$ It is a bit too easy to form the styrene this way. Could you use phenacyl bromide instead then do a Clemmensen to get the desired product? Or react the enolate with styrene oxide and hydrogenate. $\endgroup$ – Waylander Apr 30 '19 at 20:14
  • $\begingroup$ Clemmensen would not reduce the lactone, right? $\endgroup$ – Raul Luciano Apr 30 '19 at 20:22
  • 1
    $\begingroup$ Should not do so. Could always reduce it stepwise with sodium borohydride the ionic hydrogenation. $\endgroup$ – Waylander Apr 30 '19 at 20:35

Your Answer

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

Browse other questions tagged or ask your own question.