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In this question, I believe there is a mistake for iii). I think that there the tertiary carbocation would be formed (more stable), which would lead to the tetra substituted alkene (even more stable). Am I wrong?

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The answer above only shows one of the two possible isomers that could be formed by elimination from the tertiary carbocation intermediate.

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I suspect that both isomeric cyclohexenes with tri-substituted double bonds would probably be produced in roughly equal amounts, along with a smaller amount of the exo-methylene compound.

Edit: Thanks to Bon and Yolo123 for their comments

Elimination towards the isopropyl group is also a possibility, but two factors argue against this being the major elimination pathway. First there is a statistical factor; there is only proton on the central isopropyl carbon available for elimination, whereas the are 4 (2+2) protons available alpha to the carbocation center in the 6-membered ring. Second, if you look at a model and consider that the proton to be eliminated must be lined up with the carbocation p-orbital you see that each methylene group adjacent to the carbocation has one hydrogen well lined up for elimination, whereas to properly line up the isopropyl proton results in some adverse steric interactions between the isopropyl methyl hydrogens and the ring methylene hydrogens.

My guess is that the gem-dimethyl group in the cyclohexane ring will not favor one in-ring elimination pathway over the other - my model doesn't show them exerting any adverse steric effect one way or the other.

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  • $\begingroup$ Hi ron, imagine that the base pulls off an H from the isopropyl group. That forms tetrasubstituted $\endgroup$
    – yolo123
    Commented Dec 18, 2014 at 17:11
  • $\begingroup$ @yolo123 There is a fair amount of steric hindrance on the central carbon of the isopropyl group so removal of a proton there is much less likely than at the other two positions $\endgroup$
    – bon
    Commented Dec 18, 2014 at 17:14
  • $\begingroup$ @bon, I doubt so. The "ortho-like" positions would also be very much sterically hindered, if there were so much steric hindrance at the central C of the i-Pr group. $\endgroup$
    – yolo123
    Commented Dec 18, 2014 at 17:15
  • $\begingroup$ @ron For the second isomer you have given in your answer, is it possible that there would be some steric hindrance due to the adjacent methyl groups and so the first isomer would predominate. $\endgroup$
    – bon
    Commented Dec 18, 2014 at 17:16
  • $\begingroup$ @bon, let's see what ron has to say about the tetra-substitution. $\endgroup$
    – yolo123
    Commented Dec 18, 2014 at 17:16

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