last updated Friday, October 10, 2014

Meeting 3 (Condensed Key)

1.a. In order to determine the equilibrium constant for the reactions, the following three equations have to be considered (assuming that the reactants are reacted in a 1:1 ratio).

(i) The first reaction is used as is and the second reaction is reversed leading to an equilibrium constant of K_a1r=10^6.70. After adding the two equations, one obtains the equilibrium constant for the deprotonation reaction of K_R=K_DA*K_a1r = 10^-9.94*10^6.70=10-^3.24= 5.75*10^-4. Thus, only a very small fraction of the mono-protonated salt would be deprotonated by the bicarbonate ion.

(ii) The first reaction is used as is and the third reaction is reversed leading to an equilibrium constant of K_a2r=10^10.33. After adding the two equations, one obtains the equilibrium constant of K_R=K_DA*K_a2r = 10^-9.94*10^10.33=10^0.39= 2.45. Much more of the mono-protonated salt would be deprotonated by the carbonate ion (about 4266 times more to be exact) than in the bicarbonate case. A stronger base (i.e., hydroxide) would probably be better for this step but any excess of the base would also lead to a deprotonation of the phenol group of the aldehyde, which has a similar pK_a-value compared to the mono-protonated ammonium salt..

b. The reaction in a. is carried out in water because both reactants are ionic. Afterwards, 95 % ethanol has to be added to lower the polarity of the solution, which increases the solubility of salicylic aldehyde in the solvent mixture.

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.

d. 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.

e. The treatment of the organic layer with saturated sodium chloride solution removes the bulk of the water from the organic layer. Thus, less anydrous drying agent (Na₂SO₄) is needed afterwards to dry the organic layer.

f. The sharp peak at n=1630 cm^-1 is due to the imine function (n(C=N)) and the broad peak at n=2300-3200 cm^-1 is due to the phenolic OH group, which is is shifted to lower wavenumbers due to the strong hydrogen-bond to the imine nitrogen atom,, are the most important peaks to observe.

g. The signal at d=~8.35 ppm is due to the imine hydrogen atom, which is located at higher d-values compared to alkenes and lower d-values compared to aldehydes. The signal at d= ~13.6 ppm is due to the phenolic hydrogen atom that is shifted downfield due to the strong intramolecular hydrogen-bond with the imine nitrogen.

h. The concentration of the solution is 0.50 % (=0.0300/6.00 mL). Thus, the specific optical rotation for the sample is [a]= -306.0^o (=-1.53^o/(0.005 * 1)). This results in an optical purity of 97.1 % (= -306.0^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.