**Part
8: THERMODYNAMICS
**

a.

**b. **CH_{2
}

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

**d. **(CH3)_{2}CHCH=CH_{2}

**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)_{2}CHCH_{2}CH_{2}OH

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 H
_{2}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.