Road map 5

[hoyas19]

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On the right about 3 lines down, (CH3)3CO-K+ removes Br and a H to make a double one on carbon 1 instead of carbon 2, which I thought would be the more stable product. Anyone know why?

Also, I know this is probably a dumb question but is there an easy way to figuring out which is the most acidic proton? Not just in this problem, but all of them..?

 

[Albuterol]

big bulky base, because of steric hinderance will cause the reactant to form the least substituted product.

 

[gentile1225]

the only reasoning i could make of it is a little trick that i remember from orgo. A big bulky base such as (CH3)3CO-K+/(CH3)3COH will always dig out and form the double bond, not necessarily the more stable product. also, it is a secondary halide which can undergo either E1 or E2, and in the E2 elimination then the more stable product would be formed if a cutting tool such as KOH/EtOH reflux or C2H5O-Na+/C2H5OH were used.

hope this helps.