Question

When 2-bromo-3,3-dimethylbutane is treated with \({{\bf{K}}^{\bf{ + }}}{\bf{ - }}{}^{\bf{ - }}{\bf{OC}}{\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}} \right)_{\bf{3}}}\), a single product T having molecular formula \({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}\) is formed. When 3,3-dimethylbutan-2-ol is treated with \({{\bf{H}}_{\bf{2}}}{\bf{S}}{{\bf{O}}_{\bf{4}}}\), the major product U has the same molecular formula. Given the following \({}^{\bf{1}}{\bf{H}}\)-NMR data, what are the structures of T and U? Explain in detail the splitting patterns observed for the three split signals in T. 1 H 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 1 H NMR of U: 1.60 (singlet) ppm.

Short answer

The organic reaction involving the removal of a molecule of hydrogen halide where a proton and a halide ion from adjacent carbon atoms are removed to form a pi-bond is known as a dehalogenation reaction.

2-bromo-3,3-dimethylbutane reacts with the strong base potassium tert-butoxide to form the Hoffman product undergoing dehalogenation.

Step-by-step solution

Dehalogenation

The organic reaction involving the removal of a molecule of hydrogen halide where a proton and a halide ion from adjacent carbon atoms are removed to form a pi-bond is known as a dehalogenation reaction.

2-bromo-3,3-dimethylbutane reacts with the strong base potassium tert-butoxide to form the Hoffman product undergoing dehalogenation.

Dehydration

3,3-dimethylbutan-2-ol reacts with sulfuric acid to remove a water molecule during dehydration to form the corresponding Zaitsev product.

Analysing data

1. Given data:

Chemical formula =\({{\rm{C}}_6}{{\rm{H}}_{{\rm{12}}}}\)

The degree of unsaturation (IHD) \({\rm{ = }}\frac{{\left( {\left( {{\rm{2n + 2}}} \right){\rm{ - x}}} \right)}}{{\rm{2}}}\),

where

n= number of carbon atoms

x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting the values,

the degree of unsaturation (IHD) \(\begin{aligned}{c} = \frac{{\left( {\left( {{\rm{2}} \times 6{\rm{ + 2}}} \right){\rm{ - 12}}} \right)}}{{\rm{2}}}\\ = 1\end{aligned}\).

The value of IHD suggests the presence of a pi-bond.

From the NMR spectra,

1. the singlet at 1.01 ppm attributes to the 9H with adjacent carbon with no hydrogen atoms;
2. the doublet of doublets at 4.82 ppm attributes to the 1H ppm, which attributes to the protons having J=10 at 1.7 Hz;
3. the doublet of doublets at 4.93 ppm attributes to the 1H ppm, which attributes to the protons having J=18 at 1.7 Hz;
4. the doublet of doublets at 5.83 ppm attributes to the 1H ppm, which attributes to the protons having J=18 at 10 Hz.

2. Given data:

Chemical formula =\({{\rm{C}}_6}{{\rm{H}}_{{\rm{12}}}}\)

The degree of unsaturation (IHD) \({\rm{ = }}\frac{{\left( {\left( {{\rm{2n + 2}}} \right){\rm{ - x}}} \right)}}{{\rm{2}}}\),

where

n= number of carbon atoms

x= (Number of hydrogen atoms) + (Number of halogen atoms) - (Number of nitrogen atoms)

On substituting the values,

the degree of unsaturation (IHD) \(\begin{aligned}{c} = \frac{{\left( {\left( {{\rm{2}} \times 6{\rm{ + 2}}} \right){\rm{ - 12}}} \right)}}{{\rm{2}}}\\ = 1\end{aligned}\).

The value of IHD suggests the presence of a pi-bond.

From the NMR spectra, the singlet at 1.60 ppm attributes to 12H of the four methyl groups that are equivalent in nature.

Proposing structure

The required structures of T and U are as follows: