destroyer ochem #28 and #31

[jlee1986]

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hey guys,

as i was reviewing through some of the destroyer ochem q's i came across a couple of questions that were confusing to me:

#28 after the addition of Br to the methyl cyclohexane, why would potassium tertbutoxide deprotonate the carbon right next to the carbon attached to the methyl group to make a methyl cyclohexene? i thought we were supposed to deprotonate the carbon that makes the most stable alkene...

#31 i don't know if i'm missing something but i don't think i've ever seen this reaction before...why does deuterium just replace the methyl hydrogens...?

thanks in advance for your help! 🙂

 

[jlee1986]

anyone pleeeeeeeeease?😀

 

[Flipper405]

Sorry, I would love to help, but I don't have destroyer so I don't know what's going on in the question...

 

[RyanF]

Edit: This is just a repeat of what was said above. I was too slow with quick-reply 🙁

#28 - The second reaction adds a strong base (an E2 reaction) but the substrate and base are sterically hindered, so the predominate elimination product will be the anti-zaitsev form (less substituted alkene).

#31 - This is an alpha-carbon question. This is acid catalyzed so the carbonyl oxygen is deuteriated and then pi electrons migrate to the oxygen as a relatively acidic alpha proton leaves the compound, an enol is formed. From there the D from D2O adds to the enol (on that alpha carbon) and the carbonyl group reforms. This process is repeated two more times

I also was confused by this problem and at first and thought it was a hydration reaction forming a gem-diol. After referring to my O-chem book, I believe the terms that imply an alpha-carbon reaction are DCl (catalytic)-it will be regenerated and leave the carbonyl oxygen-and the excess D2O because a hydration reaction would only need 1 equivalent.

 

[D.D.S.]

What about destroyer ochem # 29? is CH3CH2O-K+ not a bulky base, so is that why it forms the more substituted alkene?


What do you consider bulky???