Elimination reaction for C2H5O-K+/C2H5OH is E2 since it uses the strong base and gives more substituted alkene. Is that right? How about (CH3)3CO-Na+/(CH3)3COH?
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- Thread starter em07
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I think it has something to do with hm... hoff man and zets ball
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yea the first base is strong. it does an E2 reaction and gives the zaitseu product (more substituted) for secondary and tertiary halides.
the second other one is big and bulky so it gives the less substituted product which is known as the hoffman product.
the second other one is big and bulky so it gives the less substituted product which is known as the hoffman product.
