DAT destroyer 2009 road map 3

[kcng]

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Hey guys,

I have a question on the O-chem road map. You start with a 1-bromo-1-methylcyclopentane and in one path you use: (CH3)3CO-Na+/(CH3)3COH; in the other pathway you use: C2H5O-K+/C2H5OH. Can someone explain why there are 2 different products? I thought of Zaitev's rule where the most substituted alkene will form in greater yield.

T-butoxide is a strong bulky base, so E2 would occur in the first pathway. In the second pathway, ethoxide will also be E2 since OR is strong base. But I still can't figure out why there are 2 different products? Thanks!

 

[DAT12]

T butoxide is a bulky base so does e2 Hoffman (towards outside)
Ethixide is a strong base but not as big so it can squeeze in easily so it does zaitsev

 

[Dotoday]

Whenever you have T-butoxide it is less substituted alkene, when you have ethoxide or methoxide, it is more substituted alkene. They are both E2, but T butoxide makea a Hoffman product, while other small bases make Zaitsev product. Hope this helps.

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[kcng]

Thanks!