PS6

,  

Chem 220 -Organic Chemistry

Problem Set 6

Chapter 7, Structure and Synthesis of Alkenes

Due: Monday, March 4, 2013

Solution Set 

Notice: For those of you who are using the 6th or older edition of the textbook, you can learn about olefin metathesis by clicking here.

Reading Assignment: Heats of Formation and Heats of Hydrogenation

1. Read Degree (Elements) of Unsaturation. What is the molecular formula of Prozac? Draw its structure? Confirm its degree of unsaturation by computation (show work). Draw an acyclic compound having the same formula as Prozac (there are many). Does it have the same number of degrees of unsaturation? Google Prozac. Its molecular formula is C17H18F3NO.The structures are below. Using the method I presented in class, first drop the oxygen. This leaves, C17H18F3N. Replace three fluorines with three hydrogens. Thus, C17H21N. Substitute C, H for N. The formula is reduced to C18H22. The most saturated C18 alkane has the molecular formula C18H38. Therefore, (38-22)/2 = 8 degrees of unsaturation. Each aromatic ring accounts for 4 degrees of unsaturation; 3 double bonds, one ring.

Prozac
(How to manipulate JSmol structures)

 

Prozac

 

A typical acyclic compound with the same molecular formula as Prozac, AND the same number of degrees of unsaturation, is shown below.

Paul Sabatier
1912 Co-Nobel Prize in Chemistry
Hydrogenation by Metal Catalysis

 

 

 

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, ΔHfo = -41.0 kcal/mol. Looking at the ΔHfo of typical 1-alkenes, one can see: 1-butene (ΔHfo = 0 kcal/mol), 1-pentene (ΔHfo = -5.0 kcal/mol) and 1-hexene (ΔHfo = -10.2 kcal/mol). Notice each one differs by ~5 kcal/mol. Their respective alkanes are: n-butane (ΔHfo = -30.0 kcal/mol), n-pentane (ΔHfo = -35.1 kcal/mol) and 1-hexene (ΔHfo = -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 ΔHfo of 3-methyl-1-pentene is x = -41.0 -(-29.9) = -11.1 kcal/mol. The reported ΔHfo of 2-methyl-1-butene is -6.1 kcal/mol. Add a methylene to this structure (-5.0 kcal/mol) and you obtain ΔHfo = -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 ΔHfo = -15.2 kcal/mol and (Z)-3-methyl-2-pentene is ΔHfo = -14.8 kcal/mol. The heat of formation of 3-methylpentane is ΔHfo = -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, ΔHfo(E) + HHo(E) = ΔHfo(Z) + HHo(Z). Therefore, ΔHfo(E) - Hfo(Z) = ΔHHo(Z) - HHo(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 ΔGo = ΔHo, 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.

3. The major product B from free radical monobromination of alkane A (C6H14) readily reacts with water to form C, C6H14O. Treatment of bromide B with KOH in ethanol produces two, and only two, compounds, D and E. Compound D liberates 1.6 kcal/mol more heat upon hydrogenation to A than does E. What are the structures A-E? What is the heat of isomerization of D to E? Illustrate and explain. [Hint: What are the structures possible for A?]
There are five possible structures, 1-5, for the acyclic, saturated alkane A. Since bromide B reacts readily with water (SN1) to form alcohol C, structures 1 and 5 are eliminated because they do not have any tertiary hydrogens. Structure 3 is also eliminated because E2 elimination of 3-bromo-3-methylpentane could give three alkenes: 2-ethyl-1-butene and (E)- and (Z)-3-methyl-2-pentene. That leaves 6 and 7 from 2 and 8 and 9 from 4 for the structures of D and E. Lookup the heats of formation of 4, 8 and 9 in the Heats of Formation Tables. From these data, one calculates the difference in the heat of hydrogenation as 1.1 kcal/mol, which isn't the value 1.6 kcal/mol. Alternatively, Table 7-1 in Wade gives the difference as 1.4 kcal/mol. (there is some discrepancy here.) Neither 6 nor 7 appear in the heats of formation table but we can look at lower homologs and calculate the heats of formation using the 5.0 kcal/mol/CH2 rule. Thus, 2-methyl-1-butene (ΔHfo = -8.4 kcal/mol) gives ΔHfo = -13.4 kcal/mol for 7 while 2-methyl-2-butene (ΔHfo = -10.0 kcal/mol) affords ΔHfo = -15.0 kcal/mol for 6. The difference is 1.6 kcal/mol. [There is actually no need to include the 5 kcal/mol correction because the difference would be the same.] Using Table 7-1 in Wade, a value of 1.6 kcal/mol is obtained. So, A = 2-methylpentane; B = 2-bromo-2-methylpentane; C = 2-methyl-2-pentanol; D = 2-methyl-1-pentene (7); E = 2-methyl-2-pentene (6). D, with a less negative heat of formation than E, liberates more heat on hydrogenation.

 

4. In 1968 Wiberg and Fenoglio determined the heat of combustion of cyclobutene: -619.2 kcal/mol. Determine the heat of hydrogenation of cyclobutene to cyclobutane. You will need to use the heat of formation tables. Explain and illustrate with a diagram.

K. B. Wiberg

 

 

You are given the heat of combustion of cyclobutene. The heat of formation of cyclobutene can be calculated by first determining the heat of combustion of 4 moles of graphite and 3 moles of hydrogen from the standard state. The Heat of Formation Tables gives the heat of formation of CO2 as (-94.05 kcal/mol) and water (-68.3 kcal/mol) at 25 oC for a total of -581.1 kcal/mol for 4C and 3H2 from the elements. Thus, the heat of formation of cyclobutene is the difference between these numbers (-581.1) - (-619.2) = +38.1 kcal/mol. Clearly, cyclobutene is less stable than the elements from which it is formed. The Tables give the heat of formation of cyclobutane. The heat of hydrogenation of cyclobutene to cyclobutane is (+6.6) - (+38.1)) = -31.5 kcal/mol.

 

5. Two stereoisomers, A and B, absorb one equivalent of hydrogen upon catalytic hydrogenation to form cyclooctane. Compound A, which is capable of resolution, liberates 34.5 kcal/mol of heat while B liberates 24.3 kcal/mol of heat.
a) What are the structures of A and B ? Heat is liberated means that the reactions are exothermic. The numerical values are negative. Since A and B both absorb only one equivalent of hydrogen to form cyclooctane, they must be cis- and trans-cyclooctene. trans-Cyclooctene is more strained than cis-cyclooctene because a chain of six methylene groups is the minimum length to span a double bond that is trans. It gives off more heat on hydrogenation. Therefore, A is trans; B is cis.
b) What are the heats of formation of A and B ? The heat of formation of cyclooctane is from the Heats of Formation Tables. The heats of hydrogenation are given. The differences gives the respective heats of formation.
c) What is the difference in strain energy between A and B ? The difference between the heats of formation: 10.2 kcal/mol.
d) What is the difference in the heat of combustion between A and B? From the diagram on the right it is obvious than the difference in the heats of combustion is the same as the difference in the strain energy of the two.
e) Why is A capable of resolution? The non-superimposable mirror images of trans-cyclooctene do not interconvert. It is difficult for the vinyl hydrogens to pass through the ring to racemize. Link to JSmol structures here.