last updated Friday, October 02, 2015

Meeting 3 (Condensed Key)

1.a. The problem is that the bicarbonate is a weak base, which is not strong enough to deprotonate the mono cation, which has a pK_a= ~10.

b. The reaction in a. is carried out in water because both reactants are ionic. They do not dissolve in 95 % ethanol.

c. The reflux is required to keep the aldehyde in solution and to eliminate the water from the intermediate. At low temperatures, the rate of the ligand formation is very low. The jacketed condenser with propylene glycol is used as cooling medium here to reduce the water usage and the possibility of floods in the hood. However, the heating has to be properly controlled to prevent that the solvent evaporates.

d. Upon completion of the reaction, water is added to the hot reaction mixture to increase the polarity of the solution, which facililtates the precipitation of the ligand.

e. The extraction in the solvent mixture (ethyl acetate:hexane=1:1) is necessary to remove the unreacted diamine and other polar compounds. The dry ligand dissolves in either solvent. Unfortunately, the presence of water in the crude ligand after the precipitation from the reaction mixture causes problems because the water and the ethanol dissolve in ethyl acetate as well. Both of them increase the polarity of the solution and therefore lower the solubility of the weakly polar ligand significantly. In case of the hexane, there seems to be a wetting problem due to the large difference in polarity. The addition of the hexane to the ethyl acetate reduces the solubility of water in the solvent and the polarity of the solvent.

f. The extraction with saturated sodium chloride solution serves as pre-drying step that removes the bulk of the water and ethanol from the organic layer.

g. The signal at δ=13.6 ppm is a broad singlet that is due to the phenolic hydrogen atom. It is located at higher δ-values because of the strong intramolecular hydrogen between phenolic hydrogen and the imine nitrogen atom that is also observed in low polarity solvents like CDCl₃.

h. The concentration of the solution is 0.45 % (=0.0450/10.00 mL). Thus, the specific optical rotation for the sample is [α]= -291.1^o (= -1.31^o/(0.0045 * 1)). This results in an optical purity of 92.4 % (= -291.1^o/ -315^o * 100 %). Since the value is only slightly below the desired value of 99+ %, the ligand is probably not completely dry. However, the student needs to look at this result and compare it with the data for his tartrate salt.