Question

Draw the major organic product formed when the structure shown below undergoes a reaction with t-BuONa.

Answer 1

The reactant given in the question is as follows: Br (A) Convert the given structure into chair form. Yo-I / 58 Br (B) The hydrogen atom attached to carbon-9 in structure (A) is above the plane, therefore, in chair form, it will be at axial position at carbon-9 in structure (B). The bromine atom attached to carbon-10 in structure (A) is below the plane, therefore, in chair form, it will be at axial position at carbon-10 in structure (B). The alkyl halides on reaction with a base give elimination reaction. The carbon bonded to the halide is the alpha carbon and the carbons bonded to the alpha-carbon are beta-carbons. The elimination of beta-hydrogen takes place (hydrogen bonded to the beta-carbon). The elimination takes place by E2 mechanism in case of tertiary halides with strong base. E2 mechanism is a 1 step mechanism, the base removes a proton from the beta-carbon and consequently the elimination of halide ion takes place. Potassium t-butoxide is a bulky base, therefore, it removes a proton from the less hindered carbon atom. The bonds that break in the elimination reaction should be anti-periplanar. There are 3 alpha-hydrogens which are anti-periplanar to the bromine atom, present in the reactant which are shown below:

Hydrogen atom at C-9 and bromine atom at C-10 are anti-periplanar, but the elimination of this hydrogen atom leads to the formation of double bond at the bridge-head position, which is very unstable, therefore the elimination of hydrogen at C-9 will not take place. Thus, the elimination of two possible alpha-hydrogens leads to the two products, which are as follows: Digibo- propter - -