2010 Destroyer Ochem Question #119

[Mistro4321]

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CH3CH2(OH)CH2CH3 -->

1)SOCl2
2)(CH3)3CO-K+
3)CH3COOOH
4)CH3OH, H+

Answer is CH2(OH)CH(OCH3)CH2CH3


My question is why did the t-BuO cause the elimination product to be anti markovnikov. I remember on Chad's videos that he stated that the Hoffmann product can only form from a TERTIARY halide where the less substitued alkene is favored. The halide formed is clearly a SECONDARY halide.

Please help!

 

[Dotoday]

No, T-butyl is a big base. This will make a less substituted double bond ( Hoffman). This does not have to be a tertiary halide. T- butyl will always make a less substituted alkene, whether it is primary, secondary or tertiary halide. Hope this helps.

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[dental13]

No, T-butyl is a big base. This will make a less substituted double bond ( Hoffman). This does not have to be a tertiary halide. T- butyl will always make a less substituted alkene, whether it is primary, secondary or tertiary halide. Hope this helps.

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Yes!! Whenever you see a reaction with Tbutyl, think of E2 Eliemination where it would form least substituted alkene (Hoffman). But whenever you see small bases, think of E2 reaction with most substituted alkene!! (Zaitsev).

Hope this helps too