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It's the last one, where 1,4-dimethylcyclohexane breaks apart into... something.

enter image description here

Here's the workup I have. I want to put out there that I'm not concerned about getting the optimal yield yet. Just making the product.

Free radical addition of $\ce{Br}$ to the 2 position.

Removal of that $\ce{Br}$ and the formation of a double bond via E2 using tBuO-

Breaking apart the structure via ozonolysis.

However, no matter where I place the double bond, I can't get the correct configuration of double bonded Oxygens (ketones and aldehydes, right?). So, I must be doing something wrong. Thoughts?

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Everything you said is correct except for the first step. Free radical bromination is reasonably selective, tertiary positions are about 20 time more reactive than secondary carbons, so the 1-bromo, 1,4-dimethyl compound is formed. Everything's a go from there!

enter image description here

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Radical bromination will occur at the 1-position (tertiary carbon). Your strategy from there is correct. Elimination using a small base such as hydroxide will give 1,4-dimethylcyclohex-1-ene. Ozonolysis with reductive workup gives the product.

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  • $\begingroup$ Ah, yep! Forming the more stable radical does indeed give the right product. Thanks! $\endgroup$ – user9810 Nov 16 '14 at 2:34

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