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-dimethylbutan-2-ol is treated with H2SO4, 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) ppm1H NMR of U: 1.60 (singlet) ppm

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Asked Dec 29, 2019
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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-dimethylbutan-2-ol is treated with H2SO4, 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

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Expert Answer

Step 1

Structure elucidation of product T:

Potassium tert-butoxide is a strong bulky base. Due to bulky nature it prefers E2 reaction rather than SN2. Therefore, 2-bromo-3, 3-dimethylbutane undergoes E2 reaction on treatment with K+ -OC(CH3)3 to yield product T (molecular formula C6H12) as shown below:

Chemistry homework question answer, step 1, image 1
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Step 2

Structure validation of product T from 1H spectrum data:

The observed chemical shift value at 1.01 (singlet 9 H) ppm, suggests the presence of three methyl groups, bonded to carbon that has no hydrogen.

The observed chemical shift value at 4.82 (doublet of doublets, 1 H, J = 10, 1.7Hz) ppm, suggests the presence of Ha proton  that is surrounded by two nonequivalent protons, cis proton Hc and trans proton Hb. The trans proton Hb splits Ha signal into a doublet, and the cis proton Hc splits this doublet into two doublets, resulting the doublet of doublets.

The observed chemical shift value at 4.93 (doublet of doublets, 1 H, J = 18, 1.7Hz) ppm, suggests the presence of Hb proton  that is surrounded by two nonequivalent protons, trans proton Ha and the geminal proton Hc. The trans proton Ha splits Hb signal into a doublet, and the geminal proton Hc splits this doublet into two doublets, resulting the doublet of doublets.

The observed chemical shift value at 5.83 (doublet of doublets, 1 H, J = 18, 10Hz) ppm, suggests the presence of Hc proton  that is surrounded by two nonequivalent protons, cis proton Ha and geminal proton Hb. The cis proton Ha splits Hc signal into a doublet, and geminal proton Hb splits this doublet into two doublets, resulting the doublet of doublets.

Chemistry homework question answer, step 2, image 1
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Step 3

Structure elucidation of product U:

Sulfuric acid is a strong acid. It converts primary and secondary alcohols into respective alkenes. The mechanism of the reaction proceeds through an E1 reaction.  An E1 reaction involves carbocation rearr...

Chemistry homework question answer, step 3, image 1
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