Due: November 9, 2009

Friedrich Wöhler (1800-1884) (Wöhler possessed a wry sense of humor) 1 2

Connections Aluminum was once a precious metal although it was plentiful. The problem was how to remove it from its ore. Friedrich Wöhler, of urea synthesis fame, was able to accomplish this feat but by an impractical method. He was to meet a young chemist, Frank Jewett, recently arrived in Göttingen from Yale. Aware of the difficulty Wöhler had had and probably encouraged by Wöhler, Jewett, as a professor at Oberlin College, passed the problem onto Charles Martin Hall, a young student at the college. Hall solved the problem in his family garage. Thus was born Alcoa. At the same time in Spray, North Carolina, Thomas Willson, a Canadian, and American James Moorhead were unsuccessfully trying to refine aluminum using an electric arc. Unsuccessful in purifying aluminum, they sought calcium metal. Heating coal tar and lime in an electric furnace they obtained a brittle material that produced a combustible gas upon exposure to water. The material was not calcium nor was the gas hydrogen. The pair was calcium carbide and acetylene, the basis for Union Carbide (RIP).

Charles Martin Hall (1863-1914)

1. Determine the structures A-K. Explain your reasoning. Structure A seems to be the center of action. From its reactions it appears to be an alkyne. How do we learn about its structure? Hydrogenation over a nickel catalyst affords a straight chain alkane B (normal chain). The heat of hydrogenation is too much for a double bond. Using 3-hexyne and 1-hexyne as models for internal and terminal alkynes in the heats of formation table, the respective heats of the respective heats of hydrogenation are -65.5 and -69.2 kcal/mol for the two alkynes. The heat of formation of n-hexane is -39.9 kcal/mol, the product of hydrogenation. Clearly, A looks like an internal alkyne. The heat of combustion of this n-alkane gives a clue to the chain length. Compute the heat of combustion of n-hexane (C6H14) from the elements: 6 C (-94.05 kcal/mol) + 7H2 (-68.3 kcal/mol) = -1042 kcal/mol. The heat of combustion of n-hexane is -1042 - (-40 (ΔHfo)) = -1002 kcal/mol. This value is less negative than -1317 kcal/mol by -315 kcal/mol. The heat of combustion for an unstrained -CH2- group is -157 kcal/mol. The value -315 kcal/mol accounts for two more -CH2- groups. Thus, B is n-octane. A is either 2-, 3- or 4-octyne. Na/NH3 reduction of A gives E, which contains an (E)-double bond. Ozonolysis of E gives two products. This eliminate a symmetrical alkyne for A and symmetrical alkene for E. K is acetaldehyde since it is the only product of ozonolysis of 2-butene. Therefore, J is hexanal (C6 straight chain aldehyde). A is 2-octyne; E is (E)-2-octene. Syn, 1,2-addition of two hydroxyl groups to E gives F [(2R*, 3R*)-2,3-octanediol] as a racemate]. Lindlar reduction of A affords C [(Z)-2-octene]. C upon anti addition of Br2 gives D, (±)-(2R*, 3S*)-2,3-dibromooctane. NaNH2 isomerization of A forms I (1-octyne). Hydration of the terminal alkyne I gives only H, 2-octanone, which also forms along with G (3-octanone) upon hydration of A.

Note: There is another way to find the number of carbons and hydrogens in B: ΔHo(comb) = ΔHo(comb Cn) + ΔHo(comb Hn+1) -ΔHfo ΔHo(comb) = nC + (n+1)H - 5n (The term 5n is -5 kcal/mol/CH2 where n = # carbons.) Now, -1317 = -94.05n + [(n+1)(-68.3)] - (-5n) -1249 = -157n n = 7.9 or C = 8.

1. Determine the structures A-K. Explain your reasoning.

2. Provide reagents for the following reactions. Explain your reasoning.

3. Design a synthesis of muscalure [(Z)-tricos-9-ene], the sex attractant of the common housefly, Musca domestica. As a source of carbon you have available 1-butyne, 1-pentyne and acetylene. You may use 1-pentyne and acetylene only once, i.e, only seven of the carbons may be provided by these two alkynes. All reagents are available.

 

4. Estimate the heat of formation of 1-,2-,3- and 4-octyne. Equilibration of any one of these isomers with KOH at 200oC produces about as much 2-octyne as 3-octyne both of which individually exceed the amount of 1-octyne. However, the amount of 4-octyne is less than the amount of 2- or 3-octyne. Explain. [Hint: 2- and 3-octyne have an entropic advantage over 4-octyne.]

5. Two bottles are found on a laboratory shelf labeled "alkyne A" and "alkyne B". Hydrogenation of A or B over a platinum catalyst gives the same alkane C. Compound A reacts with H2 in the presence of Lindlar's catalyst to form D. Compound D reacts with O3 to form a single compound E, C3H6O. On the other hand, compound B reacts with aq. H2SO4 in the presence of HgSO4 to give two ketones J and K. Under the same conditions, A gives only J. Compound B also reacts with Na/NH3 to give F, which itself reacts with Br2/H2O to give a pair of constitutional isomers, G and H.. Treatment of either G or H with aqueous NaOH gives the same compound I, C6H12O, that is also formed by the reaction of F with peracid. What are the structures of A-K? Explain and illustrate. [Note: G and H are not distinguished from one another. Pay attention to stereochemistry.]

6. The reaction on the right, which was conducted on three different cycloalkynes, was reported in 1985 by Suzanne Abrams and Angela Shaw of the National Research Council of Canada. Rather than use NaNH2 as your text suggests, they used the lithium salt of tetradeutero-1,3-diaminopropane in tetradeutero-1,3-diaminopropane as a solvent at room temperature. a) Name the alkynes used in these experiments. [Note: Not surprisingly, cyclooctyne was found to be unstable to the reaction conditions.] b) The base in this experiment is formed by adding n-butyllithium to the solvent, tetradeutero-1,3-diaminopropane. Use the pKa table to explain why this is a sound way to prepare this base. c) Provide an explanation (mechanism) as to how each methylene group becomes deuterated. [Hint: Such reactions are often called "zipper" reactions. Why?]