Q.51. - Organic Chemistry

Question

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

Step-by-Step Solution

Answer

Answer

 

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

Step 1: Energy of the conformation

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

Step 2: 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.

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

 

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

Step 3: 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:

 $2H/H = 2\times 4 kJ/mol = 8 kJ/mol$

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

$$\begin{gathered} Destabilization energy = 15 kJ mo{l}^{-1} - 8 kJ mo{l}^{-1} \\ =7 kJ mo{l}^{-1} \end{gathered}$$