Part 8: THERMODYNAMICS

a. A = 2-methyl-2-butene (or 2-methylbut-2-ene and B = 2-methyl-1-butene (or 2-methylbut-1-ene)

b. CH₂

c. A has a molecular formula = C5H10 so a MW = 70.13 g/mol. Therefore 0.5 moles = 35.65g.

d. (CH3)₂CHCH=CH₂

e. For C, the heat of formation = 5x(-93.9) + 5x(-68.4) - (-796) = -15.5 kcal/mol

f. Since A is the more stable product compared to B, A has a less exothermic heat of combustion than B. Therefore A = -792 kcal/mol (-3329 kJ/mol) and B = -794 kcal/mol (-3336 kJ/mol). Obviously C = -796kcal/mol (-3345 kJ/mol).

g. By comparing the structures of A, B and C we can see that if there are more alkyl groups on the bond it is more stable : A has 3, B has 2 and C has only 1.

h. (CH3)₂CHCH₂CH₂OH

Common errors:

• Common mistakes were no locant for double bond "2-methylbutene" or wrong locant "3-methyl-2-butene"
• The empirical formula : less than half the students got this correct, common wrong answers were the molecular formula, CH5, the molecular weight, the "text" formula...
• Mistakes were due to "outragous" calculator errors. THINK!
• Wrong answers were redrawing structures A or B, or pentene derivatives.
• Calculation errors due to using 10 H₂ instead of 5 (balance the equation)
• Common mistake with lowest energy = most stable = most exothermic heat of formation = least exothermic heat of combustion. Last year's ENDOTHERMIC example seems to have caused some confusion (saw a few "BIGGEST NUMBER + ..." with NO consideration whether it is exo- or endo-thermic.
• Most common mistakes in part h were alcohols that could dehydrate to form more than one product. Also H-OH, phenol, NaOH and all small alcohols made an appearance.