Question

Energy between staggered and eclipsed ethane is?
A.$0.6kcal{\text{ mo}}{{\text{l}}^{ - 1}}$
B.$2.9kcal{\text{ mo}}{{\text{l}}^{ - 1}}$
C.$12kcal{\text{ mo}}{{\text{l}}^{ - 1}}$
D.$14kcal{\text{ mo}}{{\text{l}}^{ - 1}}$

Answer 1

To solve this question, our approach would be to understand the energy levels of the two forms of ethane i.e. the staggered form and the eclipsed form. And we should know which state is most stable and is favoured by the ethane molecule.

Complete answer:
As we know that the compound ethane is the small molecule and it can rotate freely around its carbon bonds
The sigma bond can overlap while the ends of the molecule rotate, the two methyl groups are not held in fixed positions relative to another.
So, let’s first also discuss the definition of the conformers. The conformers can be defined as the spatial arrangement formed by the rotation around the single bonds are said to be conformers and the ‘conformations’.
Ethane molecules can adopt two conformations – staggered and the eclipsed.
The molecule ethane can be rotated around
So, considering all the discussion above and keeping in mind, the energy levels of both the forms of the ethane molecule:
So, the correct energy barrier between the staggered and the eclipsed form is the $2.9kcal{\text{ mo}}{{\text{l}}^{ - 1}}$
Therefore, the correct answer is the option B i.e. $2.9kcal{\text{ mo}}{{\text{l}}^{ - 1}}$ and this energy barrier arises because of the rotation of the atoms around the carbon-carbon bond, which is due to their differential geometrical aspect.

Option B is the correct answer.

Note:
Also, out of the two conformations of the ethane molecule i.e. out of staggered and eclipsed form, the staggered form is more favourable and is the stable form than the eclipsed conformation because the angle between the C-H bonds on the front and rear carbon is maximized, which in turn reduces the energy, thus, the staggered conformation is the stable and most favourable conformation of the ethane molecule.