Question

Question: The eclipsed conformation of $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{C}\mathit{l}$ is role="math" localid="1648620042367" $\mathbf{15}\mathbf{}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$ less stable than the staggered conformation. How much is the H, Cl eclipsing interaction worth in destabilization?

Short answer

Answer

The H,Cl eclipsing interaction worth in destabilization is 7 kJ per mole.

Step-by-step solution

Energy of the conformation

The energy of conformations arises when the substituents are at a minimum angle (eclipsed conformation).

Staggered and eclipsed conformation

• If the substituents are present at $0^0$then the conformation is an eclipsed conformation, and the energy is calculated as.

$$\begin{gathered} H/H=4kJmo{l}^{-1} \\ C{H}_3/H=6kJmo{l}^{-1} \end{gathered}$$

• If the substituents are present at ${60}^0$then the conformation is a staggered conformation, and the conformation will be more stable.

Explanation

The molecule has an eclipsed conformation and a staggered conformation.

The most stable staggered conformation has 0 kJ energy.

Staggered conformation

The eclipsed conformation is shown as:

Eclipsed conformation

The energy of the eclipsed conformation of the given molecule is shown as:

$2H/H=2\times 4kJ/mol=8kJ/mol$

The given energy is 15 kJ per mole. Therefore, the destabilizing energy of $C{H}_{3}C{H}_{2}Cl$is calculated as:

$$\begin{gathered} Destabilizationenergy=15kJmo{l}^{-1}-8kJmo{l}^{-1} \\ =7kJmo{l}^{-1} \end{gathered}$$