The total number of alkene products possible from the dehydrobromination of 3-bromo-3-cyclopentylhexane using alcohol KOH is :

This question was asked here before but I am getting more products than those mentioned there

Dehydrobromination of 3-bromo-3-cyclopentylhexane and number of possible alkene products

My doubt is, why doesn't Hydride shift and ring expansion take place here? I would expect the Elimination reaction to be E1 as Br is a good leaving group and it is 3° halide. So there should be a carbocation intermediate on C3.

There is an alpha hydrogen in the cyclopentyl ring wrt this carbocation. Why won't hydride shift or ring expansion into cyclohexane (more stable ring) take place here?

  • $\begingroup$ The bromide is tertiary. There are no cations involved. There are five alkenes. There are stereoisomers as your link suggests. $\endgroup$ – user55119 Jan 9 at 3:55
  • $\begingroup$ Yes it is a 3° halide(fixed the typo). So are you suggesting that E2 would take place? Why wouldn't E1 take place? $\endgroup$ – thewitness Jan 9 at 4:04
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    $\begingroup$ With a strong base, such as KOH, present E2 predominates over E1. E2 is kinetically controlled and given sufficient time, the alkenes from an E1 will be equilibrated by the liberated HBr to a thermodynamic mixture of alkenes. The E2 mixture of alkenes is usually different from the E1 mixture. $\endgroup$ – user55119 Jan 9 at 19:44

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