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The question is asking for the product and the answer is C. My expected answer was E. Here's how I tried to solve it: Since Br (the leaving group) is at the equatorial position, the structure would 'ring flip'.

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Since $\ce{CH3}$ is pointing downwards, I assumed it would have methyl group point outwards (dashed), however, the correct answer is a wedge.

After elimination, how would we know whether the substituent would be pointing inwards, outwards, or even right on the plane? (no wedge or dash)

  • $\begingroup$ Related: E2 Elimination; Anti-periplanar beta hydrogen $\endgroup$ – user7951 Dec 13 '16 at 23:52
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    $\begingroup$ You forgot to rotate the molecule over so that the double bond is on the correct side. $\endgroup$ – Zhe Dec 14 '16 at 0:09

You diaxial intermediate is good and you can clearly see that there is no way to achieve a Zaytsev product in an $\mathrm{E2}$ elimination since that would require eliminating the methyl group. Carry on with the elimination and then you have a double bond in the top half of the molecule and the methyl group pointing downards — molecule E if inverted by a vertical plane of symmetry traversing the $\ce{C-Me}$ bond.

But the product molecules are drawn with the double bond on the bottom. Thus, you need to rotate the molecule by $180^\circ$. If you do this with a rotation axis in the paper plane (again one parallel to the $\ce{C-Me}$ bond), the double bond is in the correct position but the methyl group is pointing upwards.

You can verify this by determining the absolute configuration of the ipso-methyl carbon: it should be (R) according to your chair drawing and it is (R) in answer C.

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