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1-methyl-1-quarternary ammonium cation(with methyl groups) cyclohexane is heated with hydroxyl ion. I want to find the major product.

In the solution, it is given that a hydrogen attached to the methyl group is eliminated.

However,
1. the methyl group and the cation are not placed in an anti-periplanar configuration, so there would be considerable amount of energy needed (just a guess) to distort the ring;
2. the hydrogen atom attached to the beta carbon on the ring is equally acidic;
3. The transition state for OH eliminating the hydrogen attached to the beta carbon is not crowded. I don't think sterics play a role here.

So I feel like the hydrogen attached to the beta carbon should be eliminated. Why am I wrong?

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  • $\begingroup$ The anti-periplanar is favored in an E2 elimination, but it is not necessary if an E1cb mechanism is being followed. Further, in an E1cb mechanism deprotonation of the methyl group produces a more stable carbanion. This earlier answer may prove helpful. $\endgroup$ – ron Aug 3 '17 at 18:46
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    $\begingroup$ If it's a methyl group, there will be a hydrogen anti-periplanar. A figure or specific example would help. $\endgroup$ – jerepierre Aug 3 '17 at 18:53
  • $\begingroup$ @ron but isn't e1cb a limiting case of the e2 spectrum, with simply less electron density in the pi orbital and more on the beta carbon. Shouldn't it resemble the E2 transition state in which case? $\endgroup$ – xasthor Aug 4 '17 at 2:12
  • $\begingroup$ (reference for this claim: Peter Sykes, page 256). $\endgroup$ – xasthor Aug 4 '17 at 2:31
  • $\begingroup$ And besides, why would the stability of the carbanion matter? The rate limiting step is the loss of the leaving group $\endgroup$ – xasthor Aug 4 '17 at 2:55

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