Ochem destroyer 2011 191

[sagarkanodiya]

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Here is my input Cl replaces OH in first step.
Second step E2 elimination occurs and most substituted Alkene forms.
In third step H and Br adds in anti-markovnikov fashion but here neither side of Alkene is more substituted so why answer is B not C?

 

[helpme123]

In the second step (E2), the least substituted alkene forms because of the use of a bulky base. Even though Chad says this is questionable, most texts use bulky base = always least subs.

 

[ahcten]

The second step: You have a secondary halide and a strong, big, bulky base Potassium t-butoxide (it should be (CH3)3CO-K+ ). Do E2, but show the Hofmann product which is the least substituted alkene. Here the product is butene.
The third step: H and Br add in anti- Markovnikov fashion so the answer is B.
Hope this helps.

 

[sagarkanodiya]

Thanks ! I thought Tertiary halide and bulky base is always hoffman. But I guess this will remain questionable as always. So I guess most people go by "bulky base always gives hoffman product."