$$\ce{(R)-2-Bromooctane + alcoholic~KOH -> E2~reaction}$$

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My answer is that this reaction will have a minor and a major product. In relation to the regiochemistry, the major product is the one which follows the Zaitsev rule, but this major product will still have two forms the trans- and the cis-, of course the major is the trans-. so there will be 3 different products.

Was I correct?

  • $\begingroup$ The cis product would be formed through a higher energy transition state(gauche confirmation) and will hence be a minor product $\endgroup$ – xasthor Aug 6 '17 at 5:48

I am not an expert but I think you are correct. The alkylhalide is secondary, the nucleophile is a strong base and the solvent is polar protic. That leads to the conclusion that the most likely reaction to happen is E2. According to Zaitsev rule the major product should be the most substituted alkene because its transition state is lower in energy. That means that the main product would be 2-octene. In reference to the cis and trans structure, the trans one would be more stable. Taking all of the above under consideration, the main product of the reaction would be trans-2-octene and the minor products: cis-2-octene and 1-octene.

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