Part 6: THERMODYNAMICS

a. Structures i and ii are constitutional isomers. (0.5 marks)

b. Structures i and iii are constitutional isomers. (0.5 marks)

c. 5-methylhex-1-ene (1 mark)

d. C₇H₁₄ + 10.5 O₂ --> 7 CO₂ + 7 H₂O (1 mark)

e. DHc elements = (7 x -93.9) + (7 x -68.4) = -1136.1 kcal/mol

DHf ii= -1136.1 - (- 1095.44) = -40.66 kcal/mol (1 mark)

f & g

g.Given the thermodynamic data in the table provided in the question, and as shown in the figure above, ii is the most stable isomer (i.e. lowest energy) and iii is the least stable isomer (1 mark)

h. ii, the most stable isomer, is a cycloalkane. The primary contribution to its stability is that ii has all the CC bonds are sigma bonds while i and iii each contain a double bond and hence a pi bond. CC sigma bonds are stronger than pi bonds. For i versus iii, the issues are increased branching or the degree of substitution of the alkene unit increases the stability of the system. Increased branching has the net effect of increasing the number of stronger bonds, for example, the number of primary C atoms and hence the number of primary C-H bonds. Since, in terms of bond strength, primary C-H bonds > secondary C-H > tertiary C-H, a molecule with more primary C-H (more branching) is more stable (1.5 marks)

Common errors:

• incorrect coefficients in the balanced equation
• missing signs in calculated values
• incorrect or no labelling of the most and least stable isomers based on the thermodynamic numbers. ii is the most stable because it is at the lowest energy (see the figure).
• incomplete or vague explanation based on bonding of why a more branched isomer is more stable. Typically students did not talk about the strengths of the bonds involved which is the key part.
• incorrect terms being used (e.g. "ii has the most tertiary C atoms" when it has no tertiary C atoms!