last updated Wednesday, April 09, 2008

Problems Meeting #3 (Condensed Key)

1. a. Neither DMSO nor HBr are very healthy, which means that work under the hood is highly advisable here. The reaction affords Me₂S, which has a foul odor to it and is fairly volatile (b.p.: 37 ^oC).

b. The alcohol (reactant), aldehyde (product) and the reaction mixture should be spotted here. The samples should be dissolved in a low boiling, low polarity solvent i.e. dichloromethane.

c. The aldehyde is less polar than the phenol and the alcohol because of the intramolecular hydrogen formed between the carbonyl function and the phenolic hydrogen atom. This makes the hydroxy function less accessible for binding to the polar stationary phase, resulting in a higher R_f-value.

d. The solution is poured into water in order to accomplish a phase separation. The organic compounds (aldehyde, unreacted alcohol, etc) precipitate as an oil or solid, while the majority of the DMSO and the hydrobromic acid dissolve in the aqueous layer.

e. If a significant amount of precipitate (fine needles) was observed at this point, the reaction did not go well. These needles are formed due to the presence of significant amounts of the alcohol (reactant).

f. The main impurities that have to be removed from the aldehyde are the unreacted alcohol and some other oxidation products. These compounds are more polar than the aldehyde, thus a polar solvent is used. The temperature of the solvent (~50 ^oC) has to be kept below the melting point of the compound (57-59 ^oC). Otherwise there is a significant danger of the formation of an oil.

2. a. The pH-value (=how many of the amine functions are protonated) and the temperature play a very important role in the precipitation. In the beginning the solution is fairly acidic, which means that both amine groups are protonated. Thus, the desired product precipitates. When more of the diamine is added, s a partial deprotonation of the ammonium functions is observed. The product dissolves better under those conditions. Upon addition of glacial acid acid, the pH-value decreases (without adding more solvent!), which in turn causes the diammonium salt to precipitate because both amine functions are protonated then.

b.Based on the quantities used in the reader, about 5.16 g of the tartrate salt should be isolated. The diamine (65.2 mmol=(8 mL*0.930 g/mL)/114.2 g/mol) is the limiting reagent here! Keep in mind that only 30% of the total amount of amine are used here (=19.5 mmol). 20.0 mmol (=3.0 g/150 g/mol) of the acid are used here.

c. The optical purity of the isolated salt is 89.6 % (=+11.2/+12.5*100%). The lower value for the optical rotation is indicative of a lower concentration of the optical active species. Either the resolution was not 100% successful, which means that some of the other isomers were co-precipitated, or the sample was still wet when preparing the solution. Another common problem is that the salt is not completely dissolved. The problems that would arise from a wet sample are that a lower yield for the ligand is observed. If the resolution itself was not clean, the product distribution for the epoxide would change. The reaction would appear less stereoselective.

d. In this project, the (R,R) enantiomer of the diamine is isolated because the (R,R) form of the tartaric acid is used. The (R,R)-form of the diammoniumcyclohexane forms the strongest bonds with the hydroxy functions of the tartrate ion, which causes the corresponding salt to precipitate first because it is less polar overall.

e.