Question

When 2-bromo-3,3-dimethylbutane is treated with K+ −OC(CH3)3, a single product T having molecular formula C6H12 is formed. When 3,3-dimethyl-2-butanol is treated with H2 SO4, the major product U has the same molecular formula. Given the following 1H NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T.

1H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet of doublets, 1 H, J = 18, 10 Hz) ppm

1H NMR of U: 1.60 (singlet) ppm

Answer 1

Treatment of 2-bromo-3,3-dimethylbutane with potassium tert-butoxide results in the formation of an alkene through the formation of a carbocation intermediate.

The reaction is as follows:

The splitting of compound A is as follows:

The three methyl groups are in the same chemical environment and hence produce a singlet by rule at .

The H’s in the methylene group has two different protons trans to each other. Each Hydrogen couple with the adjacent proton and produce a peak at their respective chemical shifts.

The methane proton couple with its adjacent protons and produce a doublet of doublet by rule.

When 3,3-dimethyl butanol is treated with followed by methyl shift, it produces 2,3-dimethyl-2-butene.

The reaction is as follows:

Since all the methyl groups are in the same chemical environment, they produce a singlet at .

Therefore, the compound A and compound B are as follows: