MediumChemistryHigh School (11–12)HydrocarbonsConceptualNumerical

The picture shows four Newman projections of butane, seen along the bond between carbon 2 and…

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The picture shows four Newman projections of butane, seen along the bond between carbon 2 and carbon 3.

Four Newman projections of butane, looking along the bond between carbon 2 and carbon 3. In each, the front carbon has one methyl group at 12 o'clock and two hydrogen atoms, and the back carbon has one methyl group and two hydrogen atoms. P: the back methyl group is at 10 o'clock, so the angle between the methyl groups is 60 degrees, staggered. Q: the back methyl group is at 6 o'clock, directly opposite the front one, an angle of 180 degrees, staggered. R: the back methyl group is behind the front one, drawn a little to its left, an angle of 0 degrees, eclipsed. S: the back methyl group is at 8 o'clock, an angle of 120 degrees, eclipsed, with the methyl group of the back carbon behind a hydrogen atom of the front carbon.

Put P, Q, R and S in order of increasing energy.

Using the energies of the conformations of butane, find how much more energy R has than P, in kJ/mol.

Find the energy of S above Q, in kJ/mol.

Need a reminder? The key ideaShow

A single bond turns, so a molecule has conformations. The relative number of molecules in a form of energy $E$ above the lowest is $e^{-E/RT}$ . For butane about the C2-C3 bond: anti $0$ , gauche $3.8$ , eclipsed $16$ , fully eclipsed $19\ \text{kJ/mol}$ . Eclipsing costs: H with H $4$ , H with $\text{CH}_{3}$ $6$ , $\text{CH}_{3}$ with $\text{CH}_{3}$ $11\ \text{kJ/mol}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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