Draw The Structure Of 3 Methyl 1 Pentene

2. a) Determine the heat of formation of 3-methyl-1-pentene by using the heat of formation tables to determine typical heats of hydrogenation for monosubstituted alkenes. Show work. The hydrogenation product of 3-methyl-1-pentene is 3-methylpentane, ΔH_f ^o = -41.0 kcal/mol. Looking at the ΔH_f ^o of typical 1-alkenes, one can see: 1-butene (ΔH_f ^o= 0 kcal/mol), 1-pentene (ΔH_f ^o = -5.0 kcal/mol) and 1-hexene (ΔH_f ^o = -10.2 kcal/mol). Notice each one differs by ~5 kcal/mol. Their respective alkanes are: n-butane (ΔH_f ^o = -30.0 kcal/mol), n-pentane (ΔH_f ^o = -35.1 kcal/mol) and 1-hexene (ΔH_f ^o = -39.9 kcal/mol). The heats of hydrogenation of each of these is, respectively, the difference in the heats of formation: -30.0 kcal/mol, -30.1 kcal/mol and -29.7 kcal/mol. The average is -29.9 kcal/mol. Therefore, the ΔH_f ^o of 3-methyl-1-pentene is x = -41.0 -(-29.9) = -11.1 kcal/mol. The reported ΔH_f ^o of 2-methyl-1-butene is -6.1 kcal/mol. Add a methylene to this structure (-5.0 kcal/mol) and you obtain ΔH_f ^o = -11.1 kcal/mol for 3-methyl-1-pentene.

b) Calculate the heat of hydrogenation of (E)-and (Z)-3-methyl-2-pentene. Show work. From the Heats of Formation Table, (E)-3-methyl-2-pentene is ΔH_f ^o = -15.2 kcal/mol and (Z)-3-methyl-2-pentene is ΔH_f ^o = -14.8 kcal/mol. The heat of formation of 3-methylpentane is ΔH_f ^o = -41.0 kcal/mol. The heat of hydrogenation of the (E)-isomer is -41.0 - (-15.2) = -25.8 kcal/mol. The heat of hydrogenation of the (Z)-isomer is -41.0 - (-14.8) = -26.2 kcal/mol .

c) Use a diagram to illustrate that the difference in the heat of hydrogenation of the two geometrical isomers in 2b is equal to the difference in their heats of formation. Which isomer is more stable based upon the heats of formation? Why?

From the diagram on the right, ΔH_f ^o(E) + H_H ^o(E) = ΔH_f ^o(Z) + H_H ^o(Z). Therefore, ΔH_f ^o(E) - H_f ^o(Z) = ΔH_H ^o(Z) - H_H ^o(E). The (E)-isomer is more stable. more negative heat of formation; lower heat of hydrogenation.
d) There is only one monosubstituted alkene having the carbon skeleton of the (E)-and (Z)- isomers in 2b. What is its structure? Assuming that ΔG^o = ΔH^o, which of the three isomeric alkenes would dominate in an equilibrium mixture? How much heat is liberated in the isomerization of the monosubstituted alkene to the (E)-isomer? Show work. Add the monosubstituted alkene to your diagram in 2c and illustrate the heat of isomerization. The isomer is 3-methyl-1-pentene, the answer to part a). The (E)-isomer would dominate in an equilibrium mixture because it is the most stable. The heat of isomerization equals -15.2 - (-11.1) = -4.1 kcal/mol.