Q.16. - Organic Chemistry

Question

Question: The torsional energy in propane is 14 kJ/mol (3.4 kcal/mol). Because each H,H eclipsing interaction is worth 4.0 kJ/mol (1.0 kcal/mol) of destabilization, how much is one H, $\mathit{C}{\mathit{H}}_{\mathbf{3}}$  eclipsing interaction worth in destabilization? (See Section 4.10 for an alternate way to arrive at this value.)

Step-by-Step Solution

Answer

Answer

 

The destabilization energy of one $C{H}_3-H$ eclipsing interaction is 1.4 kcal/mol.

Step 1: Conformation of propane

There are three carbon atoms present in the propane. For the eclipsed conformation of propane, the dihedral angle between H of the first carbon atom and H of the second carbon atom is ${\mathbf{0}}^{\mathbf{0}}$ .

Step 2: Eclipsed conformation of propane

The eclipsed conformation is shown below:

Eclipsed form

Step 3: Calculation of the destabilization energy of one C H 3 - H eclipsing interaction

• The torsional energy in propane is 3.4 kcal/mol.

• H-H eclipsing interaction is 1.0 kcal/mol.

 

There are 2 H-H eclipsing interactions in the eclipsed conformation of propane. The value of the destabilization energy of one$C{H}_3-H$  eclipsing interaction is calculated as follows:

Thus, the destabilization energy of one $C{H}_3-H$eclipsing interaction is 1.4 kcal/mol.