last updated Tuesday, January 12, 2016

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. Thus, two equivalents of potassium carbonate are needed to deprotonate the diammonium salt.

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

c. The addition of 95 % ethanol lowers the polarity of the solution and allows for the salicylic aldehyde to dissolve.

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

e. Dichloromethane is used as solvent for the extraction step in the literature. However, due to safety concerns this solvent cannot be used in Chem 30CL. In the lab, the extraction is performed using a solvent mixture (ethyl acetate:hexane=1:1). The dry ligand dissolves well 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 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. Due to the presence of the hydroxyl groups, the ligand displays some affinity towards the polar (ionic) drying agent. Thus, the drying agent should be minimized to reduce the loss due to absorption on the drying agent.

g. The signal at δ=8.35 ppm is a singlet, which is due to the imine hydrogen. The chemical shift is a result of the sp2-hydridization of the carbon atom and the inductive effect of the nitrogen atom. The chemical shift is between alkenes and aldehydes.

h. The concentration of the solution is 0.50 % (=0.0350/7.00 mL). Thus, the specific optical rotation for the sample is [α]= -302.0^o (= -1.51^o/(0.005 * 1)). This results in an optical purity of 95.9 % (= -302.0^o/ -315^o * 100 %). Since the value is 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.