Chapter 3: Conformations of Alkanes and Cycloalkanes

Conformational Analysis Answers
 

Ans 1:

First, as in any question you encounter with a compound name, draw out the compound (hence understanding nomenclature is vital). Using your model kit will also help solving this problem. It is convienient to describe the C3-C4 bond system as having a C3- and C4-methyl groups and a C3-t-butyl. In general terms, all the energy maxima will be eclipsed conformations, and the minima will be staggered. The highest energy conformation will be with the C3-t-butyl and C4-methyl groups eclipsed, B. Rotation will move to the least stable of the staggered conformers with the C4-methyl group gauche to both the C3-t-butyl and methyl groups, A. Further rotation gives the eclipsed conformer with the C3- and C4- methyls eclipsed, E, then onto the lowest energy conformation will be with these same two groups at 180o to each other, in the staggered conformation, D. Now to the most favourable eclipsed conformer with each of the R groups eclipsed only by H atoms, AB. The next staggered conformation has the C3-t-butyl group gauche to the C4-methyl group, C, and finally back to B.

 

Ans 2:

The four structures are shown below along with their respective ΔHf values. The most stable conformation will have the most exothermic ΔHf and this will correspond to the situation with both alkyl groups in the more favourable equatorial position. Equatorial substitutents are prefered over axial substituents due to the presence of the destabilising 1,3-diaxial interactions (a steric effect) when the substituent is axial due to its proximity to the other axial positions on the same face of the cyclohexane. The larger the substituent, the more destabilising this steric effect is. So the most stable isomer will be the one with both Me and tBu equatorial, then the one with the larger tBu equatorial and Me axial, then equatorial Me, axial tBu, and the both axial isomer will be the least stable.  ΔHf = -46.84 ΔHf = -44.41 ΔHf = -40.53 ΔHf = -38.16 kcal/mol ΔHf = -195.98 ΔHf = -185.81 ΔHf = -169.58 ΔHf = -159.66 kJ/mol most least stability   The ΔHc of the most stable isomer can be calculated using the Hess's Law relationship for the balanced reactions shown to the left.  From the diagram we get the relationship that:  ΔHc  (elements) = ΔHf (isomer) +  ΔHc(isomer) So, based on the molecular formula, C11H22, we can insert the numbers to get : -1782 = -46.84 + ΔHc,  therefore ΔHc = -1735.16 kcal/mol.  To calculate the equilibrium constant, we need to use the equation ΔG = -RT ln K and make the approximation that ΔG = ΔH, thus ΔG = -46.84 --38.16 = -8.68 kcal/mol  (for the two cis isomers).  Inserting the values,  -8680 cal/mol = -(1.987 cal/molK) (298 K) ln K, so K = 2.3 x 106

 

Ans 3:

Here is a diagram showing the reactions and the data provided in the question: The heat of a reaction (in this case the heat of hydrogenation) can be calculated from the heats of formation of the products and starting materials (also see the figure below): With this data, we can create the required energy diagram. If a C=C is 146 kcal/mol and a C-C is 83 kcal/mol, then we can estimate the strength of a π bond as 146 - 83 = 63 kcal/mol. This means a σ bond is 20 kcal/mol stonger than a π bond.  We need to compare the reactions of cyclopropane and propene to propane. The reaction of cyclopropane breaks a σ bond and that of propene a π bond. Remember that breaking a bond requires an INPUT of energy and therefore, 20 kcal/mol more was required for cyclopropane compared to propene. But the ring strain of cyclopropane will be RELEASED on going to propane. If we correct for the differing bonds, then the heat of hydrogenation of cyclopropane becomes -57.6 kcal/mol. (ie. more exothermic by 20 kcal/mol) Now we can get an estimate of the energy released due to ring strain by -57.6 - - 29.7 = -27.9 kcal/mol This compares very favourably with 28 kcal/mol found by combustion analysis.

© Dr. Ian Hunt, Department of Chemistry