E_strain value for eclipsed n-butane (to 3 decimal places) = ____________ kJ/mol

Question

E_strain value for eclipsed n-butane (to 3 decimal places) = ____________ kJ/mol E_strain value for gauche n-butane (to 3 decimal places) = ____________ kJ/mol E_strain value for anti n-butane (to 3 decimal places) = ____________ kJ/mol Questions What does it mean for a conformation to be " more stable" than other conformation in terms of the strain energy? Based on your answer, which of the above conformations of n-butane is the most stable? What factors cause the observed differences in strain energy among the three conformations of n-butane studied above?

Explanation / Answer

1. A molecule experiences strain when its chemical structure undergoes stress which raises its internal energy. Conformations are the different spacial arrangements due to rotations of groups of atoms about a bond. So, each conformation has different strain energy due to different arrangement of atoms. Thus, strain energy is basically the increased internal energy resulting from rotation of the molecule. Staggered conformations have lower strain energy than eclipsed conformation and are thus more stable.

The strain energy in butane is due to both steric interactions between methyl groups and angle strain caused by these interactions. In n-butane, two of the substituents are methyl groups. Methyl groups are much larger than hydrogen atoms. When eclipsed conformation occur in butane, the interactions are unfavorable, since the largest groups are interacting directly with each other. So this increases the strain energy. As the molecule rotates, it adopts the relatively stable gauche conformation, in which the methyl groups are staggered but next to each other. As rotation continues, the molecule comes to anti conformation, in which substituents are staggered and the methyl groups are as far away from each other as possible. As the large substituents are anti to each other, this decreases the energy and the molecule is most stable in this arrangement. Thus anti n-butane is most stable.

2. The strain energy in butane is due to torsional strain, steric interactions between methyl groups and angle strain caused by these interactions.

Torsional strain - It varies with the different placements of the substituents relative to each other. This leads to different conformations like staggered and eclipsed, with different energy levels and hence different levels of stability. Thus, staggered conformations like gauche and anti are more stable than eclipsed.

Steric strain - It is also known as van der waals strain. It is due to repulsion between two bulky groups which are not directly bonded to each other but become too close to each other and hence there isn't enough space for them. This steric strain makes gauche conformation less stable than anti conformation.

Angle strain - Angle strain is caused when the actual bond angles in the atoms of the molecule differ from the ideal bond angles of their geometry.

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