Condensed Key

1.a.

b. In order to assess the final yield, the yields have to be determine in both steps.

Step 1:

Compound A: m=3.4 g/136 g/mol = 0.025 mol

Thionyl chloride: m = 2 mL * 1.631 g/mL/119 g/mol = 0.0274 mol

Thus compound is limiting reagent here, and 0.025 mol * 0.9 = 0.0225 mol of the acid chloride are obtained in step 1.

Step 2:

Acid chloride: m=0.0225 mol

Diethylamine: m= 5mL * 0.707 g/mol / 73 g/mol = 0.0484 mol

Since two equivalents of amine are needed to react one more of acid chloride, 0.045 mol of amine would be required. Since there is more than thi spresent, the acid chloride is the limiting reagent here. Thus,

0.0225 mol * 0.8 = 0.018 mol of the amide (mw=191.1 g/mol) would be isolated, which equals 3.44 g.

c.

d. The student would only isolate an ammonium salt of the acid, because only an acid-base reaction would take place at room temperature. In order to obtain the amide, the student would have to heat the salt to 180-200 ^oC.

e. As mentioned earlier, a minimum of two equivalents of the amine are needed for the reaction, one which beomes part of the product and one to neutralize the acid formed in the reaction, which precipitate as diethylammonium chloride from ether.

f. Extra Credit: The common name for the product is "OFF" or "DEET" . It is used as a bug repellent.

2. a. Phenylglycine is an amino acid, hence possesses both an amino and an carboxylic acid function. Thus, either a chiral acid e.g. lactic or mandelic acid can be used, or a chiral base like brucine in order to perform the resolution step. In either case, the enantiomers have to be converted into diastereomers in order to be able to separate them based on different solubilities in a given solvent. The target compound can then be obtained by adding an acid or base (opposite as in the first step).

b. A melting is usually not a good tool to access the optical purity of a sample, because most enantiomers has identical melting points. Even if the melting point is different, nothing really can be said about the optical purity, since the differences can originate from all type of impurities.

c. In order to assess the optical purity by polarimetry, first the concentration of the solution and the optical rotation for the sample have to be determined.

c=1 g/10 mL= 0.1 g/mL

[a]= a/(c*l) = +29^o/(0.1*2) = +145^o

The optical purity is then given by

X=a(observed)/a(pure enantiomer) * 100% = +145^o / +155^o * 100% = 93.5 %

3. a. Since the compound is non-polar, a non-polar or weakly polar solvent should be used here e.g. hexane or dichloromethane. The plastic cuvette however is not a good choice here because it will dissolve in these solvent and also interfere with the measurements since it absorbs at l=280 nm as well. Realistically speaking, the student can only measure the range from 300-700 nm with the plastic cuvette.

b. In order to find a good concentration, the student has to look at the peak with the largest e-value, which is at l=280 nm. The concentration can then be calculated using Beer's Law:

c1 = A/(e*l) = 1/(30000 *1) = 3.33 * 10-5 M

However, the expected absorbance for the peak with the lowest e-value (at l=485 nm) would be A=0.083, which is outside the recommended window of A=0.1-1.0. Hence, she would have to prepare a second solution with a slightly higher concentration (cmax=4*10-4 M)

c. She is probably looking at a n-> p* transition here which could be caused by a conjugated carbonyl

d. Extra Credit: The compound is dark-red in color since it only absorbs in l=485 nm in the visible range.

4. a. Ethers are medium polar and therefore absorb moisture from the air. The second, more important problem is that ethers tend to form peroxide when exposed to sun-light and air. These peroxides are explosive and usually have higher boiling points than the ethers themselves. Thus, their concentration increases during a distillation process.

b. In order to remove these peroxides, water and other impurities e.g. BHT (stabilizer in THF), there are two different methods. Both of them destroy the peroxides and the water chemically via a redox reaction or combination of "H⁺" and "H^-".

- Na/Benzophenone: 2 Na + 2 H₂O ---> 2 NaOH + H₂

The mixture turns dark blue after the water, peroxides and the dissolved oxygen is consumed. The dark blue color is due to the presence of a ketyl radical of the benzophenone.

- Lithium aluminum hydride: LiAlH₄ + 4 H₂O ---> LiOH + Al(OH)₃ + 4 H₂

After an extended reflux the purified solvents are usually distilled and then stored under inert gas.

c. Ethers are polar enough to dissolve Grignard reagents, but they do not react with them.

5. a. Air is used as oxidant in the formation of the catalyst. Alternately, a dilute dsolution of hydrogen peroxide could be used, but the yield would be lower.

b. In case of the (R,R) and the (S,S)-form, the asymmetric bridge induces the stereochemistry into the epoxide. The bridge controls the approach of the alkene. In the case of (R,S) and (S,R), the approach is not controlled, and as a result both enantiomers are forms in the same quantity.

c. For diagram see reader. The additive increases the rate of teh reaction and the stereoselectivity in many cases, due to a later transition state.

d. Both peaks are due to the C=N-bond. The shift is due to the coordination of the lone-pair to the Mn(III)-ion which pulls electron-density out of the C=N bond in the catalyst. Thus, a shift to lower wavenumbers is observed.

6. a.

b. Even with adding the concentrated sulfuric acid, the reaction is rather slow. Heating the mixture under reflux increases the rate of the reaction dramatically.

c. Esterification reactions are true equilibrium reactions, with K_eq values around 1-10. In order to increase the yield for the reaction (=shift the equilibrium to the right), an excess of a starting material should be used according to the Le Chatelier Principle. In this reaction, methanol was used in excess because it is easier to separate from the product since it dissolves in water.

d. There are two main causes for this observation:

- The benzoic acid was not properly dried or the used methanol was old, which means that it also absorbed a lot of water (alcohols are hygroscopic!). In either way, the additional water present in the mixture reverses teh reaction and ultimately leads to a lower yield of the ester.

- A low yield is observed as well if the reaction mixture is not sufficiently heated. The reaction would slow down, which means the conversion at a given point in time is less. As a result, less of the ester would be isolated.

7.a. FeCl₂*4 H₂O is light blue in color, while FeCl₃*6H₂O is orange-brown. FeCl₃ oxidizes the cyclopentadienyl ligand to form dimers instead of forming ferrocene.

b. Cylcopentadiene easily undergoes Diels-Alder reactions. As a result, it is not commercially available. However, the dimer can be purchased and be cracked to yield the monomer in a retro-Diels-Alder reaction.

c. The diacylation product has one acetyl group on each Cp-ring, and not both on the same ring, because the first one deactivates the Cp-ring already and makes it a much weaker nucleophile. The monoacylation product is used in higher yield with teh method used in the laboratory (acetic anhydride and phosphoric acid).

The two compounds are separated by column chromatography on silica using petroleum ether first to elute the unreacted ferrocene and then a mixture of PE and ethyl acetate (90:10) to elute the monoacylation product. Under these conditions, the diacylation product will remain on the column because it is more polar.

8. a. Peak A is the unreacted styrene, peak B is due to the cis-epoxide and peak C is most likely the trans-epoxide, which has a sligtly higher retention time than the cis-epoxide.

b. In order to determine the relative percentages, first the individual areas of teh triangles have to be determined (A=b*h/2). The best is just to measure the areas with a ruler.

Peak A: A1 = 22 mm * 6 mm/2 = 66 mm²

Peak B: A2 = 31 mm * 16.5 mm/2 = 256 mm²

Peak C: A3 = 11 mm * 12.5 mm/2 = 69 mm²

Total Area: A = a! + A2 + A3 = 391 mm²

Percentages:

Compound A: 66/391 * 100% = 17%

Compound B: 256/391 * 100% = 65%

Compound C: 69/391 * 100% = 18%

c. In order to determine the ee value, the peaks B and C have to be used

ee= (Compound B - Compound C)/(Compound B + Compound C)*100% = (256 - 69)/(256 + 69) * 100%= 58%

The ee value is a little higher than the one reported in teh Hanson paper. Factors like temperature and choice of oxidant influence the ee-value in some cases dramatically. A reaction run at lower temperature is usually slower, but often also an increased ee-value is observed.

d. A chiral GC-column was used for the separation of the two enantiomers. In the lab, a modified version of b-cyclodextrin was used. The two different enantiomers have different retention times.

9. a. There are various parameter that have to be considered in chosing a solvent for extraction:

- miscibility: The two solvents should not have a significant miscibility in each other. This is only possible for solvents which have very different polarities.

- partition coefficient: How much better does the compound dissolve in the solvent that is used for extraction? The larger the difference is, the less solvent is needed and the fewer extractions to extract the majority of the sample.

- quantity to be extracted: This is crucial when it comes to determine how much solvent is used for the actual extraction. The least amount of solvent should be used based on the solubility of the compound in the organic phase and the miscibility. This saves resources (cost of the solvent) and time (less solvent to evaporate), and also reduces waste.

b. In theory 1 mL (=(200 mL * 0.1 g/mL)/20 g/mL) are sufficient for the extraction. However, since diethyl ether dissolves up to 5% in aqueous solutions, 10 mL are needed to saturate the aqueous layer in the first extraction. For practical purposes, three extractions with 20 mL each would probably do an excellent job here.

c. Low boiling solvents evaporate during the extraction process. If the extraction container e.g. separatory funnel is not vented, the pressure build-up will lead to the uncontrolled removal of the stopper or even an explosion/breakage at some point.

d. An emulsion formed due to vigorous shaking. This often happens in multi-solvent systems or if the polarities or densities of the two phases are similar. The mixture can either be agitated or just left alone in order to separate on its own.

10. a. The degree of unsaturation is given by

DBE = (2*C + 2 - H - X + N)/2 = (2*11 + 2 - 17 + 1)/2 = 4 ==> probably a Phenyl ring present

b.

c.

d.

Since there are nine signals in the C-NMR spectrum for eleven carbons in the empirical formula, there is some kind of symmetry in the molecule.

The compound is

e. The peak at m/z=91 is due to the tropylium ion [C₇C₇]⁺.The fragment at m/z=134 is due to a fragment [C₆H₅CH₂N(CH₃)CH₂]⁺.