1. a. Conjugated alkenes (R=alkene, aromatic) tend to stabilize radical intermediates better. Therefore, they lead to a higher percentage of trans product than alkyl substituted alkenes. Hence, the ee value is lower. For full credit show mechanism (see reader page 23, figure 3).

b. There are several ways to improve the stereoselectivity for this reaction.

- use a catalyst that has different R-groups on the phenyl rings e.g. OMe

- lower the temperature of the reaction (e.g. use m-Chlorobenzoic acid at -78 C)

- use additives e.g. pyridine oxides

2. a. area of peak A = 2.4 cm², area of of peak B=1.2 cm², area of peak C=1.2 cm²

--> Area A=50%, Area B=Area C=25%

b. ee=(Area A - Area (B))/(Area A+Area B) * 100%=(50-25)/(50+25)*100%=33%

c. The third peak is most likely due to the formation of an aldehyde or ketone which occured during the flash chromatography. Show mechanism for full credit.

d. A chiral column was used (b-Cyclodextrin specifically) to separate the enantionmers. Usually the cis isomer leaves the column first (major product=bigger peak!)

3. a. A. Sodium/Benzophenone

2 Na + 2 H₂O ---> 2 NaOH + H₂

B. Calcium hydride

CaH₂ + 2 H₂O ---> Ca(OH)₂ + H₂

C. Calcium hydride or Phosphorous pentoxide

P₄O₁₀ + 2 H₂O (or 6 H₂O) ---> 3 HPO₃ (or 3 H₃PO₄)

b. Polar compounds are absorbed on the drying agent as well, which would lower the overall yield.

c. Na₂SO₄ absorbs 10 water molecules

4. a. Since he wants to isolate the (S,S)-isomer, he should use D-(-)-tartaric acid (the one that you isolated in the lab was the (R,R)-form and you used L-(+)-tartaric acid!). He uses the fact that this compound is less soluble in water than the other diastereomeric salts.

b. There are various techniques mentioned in the reader e.g.

- Convert the enantiomers into diastereomers and obtain a H-NMR for them

- Use chiral shift reagents e.g. Eu(hfac)3

- use polarimetry (optical rotation)

- Use chiral GC or HPLC column

c.

5.7. g = 0.05 mol of diamine, 30% (50% of 60% of trans) of that is (S,S)-form (=0.015 mol), MW of the product=264 g/mol, leading to a yield of 3.96 g

5.

6. a. To avoid formation of biphenyl

Ph-Br + PhMgBr ---> Ph-Ph + MgBr₂

b. The bromides have a stronger C-X bond and therefore are more stable (longer shelf life). Iodides are more sensitive towards air and light and also much more expensive.

c. Magnesium is a not a noble metal, hence possesses an oxide layer on the surface. Scratching or crushing reduces the size of this surface layer and makes it easier to initiate the reaction. Keep in mind that the reaction can be understood as electron transfer reaction from the surface (see reader).

d. 2 RMgBr <==> R₂Mg + MgBr₂

You can use ²⁵Mg-NMR as a tool to detect the different species

Most Grignard undergo oligomerization (dimers, tetramers, etc) which are observed as well in the NMR spectrum. The degree of association depends on the concentration, solvent and the temperature.

7. a. Degree of unsaturation = 5

b.

c.

d.

e. The fragment at m/z=120 amu is due to a fragment (H₂C=CH-C₆C₅-OH) which was formed by the elimination of butene from the t-Bu-O-group.

f. The molecule is p-tert-butoxy-styrene (t-Bu-O-C₆H₅-CH=CH₂).

8.

a. Concentration of the solution = 0.8 g/10 mL=- 0.08 g/mL

[a]=+0.45^o/(0.08*0.5)=11.25^o

Optical purity = 11.25/12.4 = 90.7 % ~ 91%

b. The sample contains the L-(+)-form of tartaric acid (the same that you used in the lab for the resolution). For structure see reader.