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Question

The rotational energy barrier in K cal mol1 between the most stable and the least stable conformations of 2,3- dimethylbutane along C2C3 bond is :
(Given : Energies K cal mol1 for H/CH3 eclipsing = 1.8, CH3/CH3 eclipsing = 3.9, CH3/CH3 gauche = 0.9. Torsional strain of H/H eclipsing is neglected.)

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Solution

The most stable form of 2,3-dimethylbutane is

Energy due to torsional strain is
=2 × (CH3/CH3 gauche)=2×0.9=1.8 K cal mol1

The least stable form of 2,3-dimethylbutane is

Energy due to torsional strain is
=2 × (CH3/CH3 eclipsing)=2×3.9=7.8 K cal mol1

The rotational energy barrier between the most stable and least stable conformations of 2,3-dimethylbutane is
=7.81.8=6 K cal mol1

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