1.
a. Water reacts with the PhMgBr to form Benzene and an ether insoluble Mg-salt that coats teh surface. The reaction stops because the electron-transfer is blocked.

PhMgBr + H₂O --- > Ph-H + Mg(OH)Br

b. Sulfuric acid is used as proton source to protonate the Mg-salt.

PhCOOMgBr + H₂SO₄ --- > PhCOOH + 'Mg(HSO₄)Br'

c. If bromobenzene is added too fast, a lot of the Grignard reagent reacts woth the excess Ph-Br to form biphenyl.

PhMgBr + Ph-Br --- > Ph-Ph + MgBr₂

d. Water is one of the products in teh esterification. According to Le Chatelier Principle, the presence of water would lower the yield since the esterification is an equilibrium reaction.


2. a. Reducing reagent: SnCl2, 3 equivalents are required, strongly acidic conditions. For equation see reader.

b. The acid chloride is more reactive, because the Cl- is a better leaving group than the OH-. The reaction requires less rigorous conditions when the acid chloride is used and usually leads to higher yields as well.

c. The first equivalent is used as reactant, the second one to neutralize the formed HCl and the third one to push the the reaction towards the desired product. The higher temperature is necessary to speed up the reaction.

d. For diagram, see reader. It should consist of a heat source, a flask (with the starting solution), a threeway distilling head, a water jacket condenser, a vacuum adapter, and a round bottom flask (to collect the distillate)


3.
a. The NMR machine needs a deuterated solvent to lock in the magnetic field.

b. The triplet is due to CDCl3. Since the nuclear spin of D is I=1, teh number of lines observed is 2*n*I=3

c. H-NMR is much more sensitive due to the higher abundance. It allows also to get information about the neighborhood based on the splitting. The integration permits conclusion about the relative ratios of protons.


4.
a. A capillary spotter is needed to apply the sample to the TLC plate since it is a micro-technique, which only requires very small quantities.

b. The ceric stain method is used. The Mo(VI) (as molybdate, pale yellow) is reduced to Mo(V), which is dark blue in color. This method is used for compounds that are not visible under the UV light and are polar enough to attach the molybdate.

c. Flash chromatography is used because the quantity of material used is biiger than before. As a rule of thumb, ~25 times as much stationary phase as material to separate should be used to obtain a good separation. The air pressure allows to speed up the process.

d. TLC is used to monitor the reaction and to optimize the solvent mixture for the flash chromatography later on.


5.

6.
a. The major product is the cis epoxide since most of the substrates go preferentially through the concerted mechanism. The reason why most of you observe ee~50% is that there is a competion between different mechanisms. Which one is dominant depends on the type of substrate used. Alkyl substituted alkene lead usually high yields of cis epoxides. Conjugated alkenes (e.g. styrene) afford higher percentages in trans product, because the pi system is able to stabilize the radical better (see diagram in reader).

b. GC on a chiral can be used (see experiment), H-NMR employing a NMR shift reagent (see lecture/reader) or a chiral solvent can be used as well.

c. The ligand is yellow because of a p-p*-transition at l=420 nm. The catalyst itself shows d-d-transitions that originate from the transition metal (Mn(III)). These transitions are in the visible range, hence the color changes to dark brown.

d. In order to separate the amine, a chiral acid has to be used e.g. tartaric acid, camphoric acid, etc. Like in the resolution of the 1,2-Diaminocyclohexane in the lab, one enantiomer will precipitate more likely than the other (in form of a diastereomeric salt).

e. The Na₂HPO₄ serves as buffer because the catalyst and the product are sensitive to pH changes.

f. The presence of water (insufficient drying) leads to the formation of acid during the chromatographic step. This acid causes a rearrangement reaction of the epoxide to form aldehydes or ketones due to hydride or alkyl shift.


7.
a. The specific rotation for the sample is a=-22.5^o (=-0.45^o/0.04). Based on this number, it can be concluded that the sample does not contain significant amount sof S-(+)-MSG, but more likely is R-(-)-MSG. The optical purity of the sample is 93.75% (= -22.5^o/-24^o).

b. The optical purity of this sample is 50% (75%-25%). Since the R-(-)-enantiomer is presence in excess, the specific rotation is a= - 12^o.


8. The following information can be extracted from

a. Empiric formula: Degree of unsaturation = 9

b. IR spectrum: presence of =CH (sp2) at 3050 cm^-1, C=C (aromatic) at 1600 and 1500 cm^-1, oop bending for monosubstitution at 695, 745 cm^-1, no peak ~1700 cm^-1 indicates absence of carbonyl group!

c. H-NMR: two "singlets" at 3.8 ppm and 7.4 ppm (ration 1:5)---> symmetry

d. C-NMR: 5 signals total (-> symmetry), one at 83 ppm (O-CH), three in the range from 125-130 pp (CH, aromatic), one at 136 ppm (-C-, aromatic

e. The compound is cis-stilbene oxide.




9 Extra credit problem



In order to explain the product distribution, the Cram chelate model has to be used. The approach of the nucleophile via pathway B is favored over the approach via pathway A due to less steric hindrance (H-atom vs. Ph-group). See also reader.