Question

The rotational energy barrier $in{\text{K cal mol}}^{-1}$ between the most stable and the least stable conformations of $\text{2,3- dimethylbutane}$ along $C_2-C_3$ bond is :
(Given : Energies ${\text{K cal mol}}^{-1}$ 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.)

• A. 6.0
• B. 6.00
• C. 6

Answer 1

Answer: (A), (B), (C)

The most stable form of $\text{2,3-dimethylbutane}$is

$\therefore$ Energy due to torsional strain is
$=2\times \left(C{H}_3/C{H}_3\text{gauche}\right)=2\times 0.9=1.8{\text{K cal mol}}^{-1}$

The least stable form of $\text{2,3-dimethylbutane}$ is

$\therefore$ Energy due to torsional strain is
$=2\times \left(C{H}_3/C{H}_3\text{eclipsing}\right)=2\times 3.9=7.8{\text{K cal mol}}^{-1}$

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