The rotational energy barrier $i n {K cal mol}^{- 1}$ between the most stable and the least stable conformations of $2,3- dimethylbutane$ along $C_{2} - C_{3}$ bond is :
(Given : Energies ${K cal mol}^{- 1}$ for $H / C H_{3}$ eclipsing = $1.8$ , $C H_{3} / C H_{3}$ eclipsing = $3.9$ , $C H_{3} / C H_{3}$ 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
$= 2 \times (C H_{3} / C H_{3} gauche) = 2 \times 0.9 = 1.8 {K cal mol}^{- 1}$

The least stable form of $2,3-dimethylbutane$ is

$∴$ Energy due to torsional strain is
$= 2 \times (C H_{3} / C H_{3} eclipsing) = 2 \times 3.9 = 7.8 {K cal mol}^{- 1}$

The rotational energy barrier between the most stable and least stable conformations of $2,3-dimethylbutane$ is
$= 7.8 - 1.8 = 6 {K cal mol}^{- 1}$

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