Q43E - Organic Chemistry (Mcmurry)

Question

What are the relative energies of the three possible staggered conformation around the C-2, C-3 bond in 2,3-dimethyl butane?

Step-by-Step Solution

Answer

Step 1: Conformation

An infinite number of spatial arrangement of group or atoms possible, in a molecule due to the free rotation of constituent group or atoms around carbon-carbon single bond is called conformation. The conformation can be represented both by Sawhorse and Newman’s Projections

Conformations can be classifies into two types- Eclipsed and staggered,

In case of Newman’s projection of 2,3-dimethyl butane, eclipsed is when the groups  present on  both the singly bonded carbon atoms are as near as possible or positioned at a dihedral angle of ${\text{0}}^{\text{o}}$.

Staggered conformation is when groups attached to both the singly bonded carbon atoms (C-2 and C-3) are as far apart as possible or positioned at at the dihedral angle of atleast ${60}^{\circ }$. Staggard conformations are more stable because, due to large separation between the electron clouds of C-H bonds, the repulsion is minimum and hence energy is minimum.

Step 2: Staggared conformations of 2,3-dimethyl butane

   Staggered conformations, by rotating the back Carbon atom each time by  ${\text{120}}^{\circ }$angle.

Step 3: Relative energies of Staggared conformations of 2,3-dimethyl butane

In the first staggered conformation, we see that the bulky methyl groups are anti to each other but there are 2 gauche interactions or steric strain (strain due to two bulky methyl groups on C-1 and C-2 respectively)

The energies because of  two gauche interactions  $\text{=\hspace{0.17em}2×3.8\hspace{0.17em}=\hspace{0.17em}7.6\hspace{0.17em}KJ/Mol}$

In the second staggered conformation we see that there are 3 gauche interations, so

The energies because of three gauche interactions $\text{=\hspace{0.17em}}3\text{×3.8\hspace{0.17em}=\hspace{0.17em}}11.4\text{\hspace{0.17em}KJ/Mol}$

While if you see the third conformation , you will still find gauche interactions because of which,

 The energies because of three gauche interactions $\text{=\hspace{0.17em}}3\text{×3.8\hspace{0.17em}=\hspace{0.17em}}11.4\text{\hspace{0.17em}KJ/Mol}$

Therefore, The energy of the I staggered conformation (7.6KJ/Mol) will be the least of all the conformers, hence conformation I will be most stable.