1. a. The reaction of an aldehyde with an amine affords an imine (see ligand synthesis).

(A) (B) (P)

b. The aldehyde should be used in excess here because it is easier to separate from the product by recrystallization. Compound B and compound P are too similar in polarity, which means that they would have to be separated using column chromatography.

c. The higher temperature increases the solubility of the reactants and the rate of the reaction as well. In addition, the higher temperature favors the elimination of water from the intermediate.

d. The addition of hydrochloric acid would terminate the reaction above because the amine function on compound B would be protonated. The formed ammonium salt is not a nucleophile anymore.

e.

Compound A is the least polar of the three compounds in the mixture. Thus, it travels the farthest up the TLC plate. On the other side, compound B is the most polar because it has two polar groups (nitro and amine) that interact strongly with the stationary phase. The product P is in between in terms of the polarity because the imine function is less polar than the amine function.

f. In order to determine the yield, the limiting reagent has to be identified first.

n_A = 2.15 g/214.05 g/mol = 10.0 mmol

n_B = 1.60 g/138.13 g/mol = 11.6 mmol

Thus, compound A is the limiting reagent in this reaction.

n_P = 2.80 g/334.16 g/mol = 8.38 mmol

Therefore, the yield is

Yield = n_A/n_P * 100% = 8.38 mmol/10.0 mmol *100 % = 83.8 %

g. Extra Credit:



2. a. Since amino acids have an amine function and a carboxylic acid function, it is possible to separate the enantiomers by addition of a chiral base or chiral acid. After fractionated crystallization, the amino acid can be recovered by adjusting the pH-value accordingly. Alternatively, enzymatic processes could be used i.e. lipase to esterify one enantiomer but not the other. The resulting product mixture can be separated by extractions.

b. The concentration of the solution is 7% (=0.56 g/8 mL). Based on that, the specific optical rotation of the sample can be determined

_D²⁰ = (-2.4^o)/(0.07 * 1 dm) = (-34.3^o)

Thus the optical purity is

X = (-34.3^o)/(-35.2^o) * 100% = 97.4 %

c. The addition of the hydrochloric acid was a bad idea because the species measured now is not the free amino acid, but much more a protonated form. As a result, his optical rotation measurement will be useless except the appropriate reference data is available for this solvent system.


3. a. The ortho isomer has a lower melting point and a lower boiling point because this compound exhibits relatively strong intramolecular hydrogen bond between the amino function and the nitro group. This makes both groups less accessible for intermolecular hydrogen bonds.



b. The 1% triethylamine solution was applied to neutralize the remaining acidic functions on the silica. This step is necessary to minimize the rearrangement of the epoxide during the subsequent column chromatography step.

c. Which drying agent to pick depends on the solvent that constitutes the solution, the type of compounds dissolve in the solvent (acidic or basic) and how much water has to be removed.

d. The proper solvent should be as immiscible as possible with the aqueous layer and dissolve the compound is question well (=high partition or distribution coefficient). In addition, the solvent should also have a lower boiling point or the melting point of the compound in order to allow for easy removal later on.

e. Deuteroacetone exhibits two signals in the ¹³C-NMR spectrum. The signal at 30 ppm consists of 7 lines (=2*I*n+1=2*3*1+1), while the signal at 207 ppm is single line.

f. The coating of the TLC plate contains a fluorescence indicator that ‘glows’ under UV-light. If the compound is UV-active, it appears as a spot on the TLC plate. For instance, the alkenes used in the epoxidation experiment are UV-active, and can be visualized, while the epoxides are not.

g. The cluster of m/z=200, 202 and 204 in a ratio of 10:6:1 is indicative of two chlorine atoms in the molecule/fragment. Chlorine has two main isotopes, ³⁵Cl and ³⁷Cl in an abundance ration of 3:1.


4. a. Esterification reactions are equilibrium reactions, which usually exhibit relatively low Keq. Water is one of the products in this reaction, which means that if it was present in the beginning, the yield of the ester would decrease (Le Chatelier Principle). It also competes with the oxygen on the carbonyl group for the proton of the sulfuric acid. It is imperative that the carboxylic acid is dried thoroughly to obtain a high yield for the ester.

b. Sulfuric acid is the catalyst in the reaction. The carbonyl group of the carboxylic acid is protonated and becomes a better electrophile becomes the second resonance structure below has a higher contribution.



c. The addition of water causes a phase separation. A significant part of the unreacted alcohol and the catalyst dissolve in the aqueous layer while the ester, the carboxylic acid and the remaining alcohol form the organic layer.

d. The extraction of the organic layer with NaHCO₃ removes the unreacted carboxylic acid and the acid catalyst from this layer.



Once all the acids are extracted from the organic layer, the formation of carbon dioxide ceases.

e. Since the ester is a liquid, the student should use AgCl plates to acquire the IR spectrum for the pure compound. The most important changes would be the disappearance of the hydroxyl functions of the alcohol and the carboxylic acid. In addition, the carbonyl group of the ester is about 20 cm^-1 higher than the one found in the carboxylic acid. In addition, the ester exhibits two very strong peaks in the range of 1100-1300 cm^-1 which are due to the COC-function in the ester.


5. a. The reaction of the diammonium tartrate with the potassium carbonate in water is an acid-base reaction, which results in the formation of the free diamine.



The diamine subsequently serves as nucleophile in the reaction.

b. The excess of the aldehyde is necessary in order to ensure the complete conversion of the diamine into the diimine. The mono-imine exhibits similar solubility behavior than the diimine (Jacobsen’s ligand).

c. The student did not isolate a pure compound, but probably the compound that he was planning to synthesize. A melting point difference of 5 ^oC is indicative of a significant amount of impurities e.g. unreacted aldehyde, monosubstitution product or a wet sample.

d. The maximum concentration at a given wavelength is determined using Beer’s Law. The maximum absorbance is A_max =1, the path length l=10 mm=1 cm and the -value is 32000 for =258 nm and 7200 for =320 nm.

c_max= A_max/*l = 1/32000 L/mol*cm * 1 = 3.13*10^-5 mol/L

The peak at =320 nm would show an absorbance of

A= *l* c_max = 7200 L/mol*cm * 1 cm * 3.13*10^-5 mol/L = 0.225 A.U.

This absorbance also lies with the desired range of A=0.1-1 A.U. only one solution has to be prepared.

Since the measurements are done in the UV-range, a quartz cuvette should be used here. A good solvent would be dichloromethane since the ligand dissolves very well.


6. a.Mono and 1,1-disubstituted alkenes do not produce cis and trans epoxides, which makes the analysis of the acquired spectra (NMR, GC/MS) easier. 1,2-disubstitued alkenes produce four different compounds as shown in the reader.

b. Bleach has the advantage that it is cheap, but also the disadvantage that the reaction has to be run at room temperature. Iodosobenzene and MCPBA can be used at -78 ^oC, which makes the reaction more selective because the reaction is much slower. Another advantage of these reagents is that they are usually used in anhydrous conditions, which means that the hydrolysis of the catalyst and the epoxide is suppressed.

c. A two-phase system is used because bleach is aqueous and the alkene does not dissolve in aqueous solutions. The epoxide remains in the organic layer, while the byproducts are transported into the aqueous layer. After completion, the two layers are separated and the epoxide isolated from the organic layer.

d. The five peaks can be assigned as follows:

1.30 min (solvent), 2.55 min (alkene), 5.50 min (R-epoxide), 5.65 min (S-epoxide), 5.90 min (aldehyde)

Therefore, the third and fourth peak are used to determine the e.e.-value

e.e. = (3000-800)/(3000+800) * 100% = 57.9%

7. a. The stannous chloride (SnCl₂*2 H₂O) would not dissolve completely. As a result, only part of it is available for reduction of the nitro compound. In addition, the precipitate would contain basic tin salts that are causing problems in the work-up, because they do not dissolve in water, only in very strongly bases.

b. Glacial acetic acid is used as solvent in this reaction, because the nitro compound dissolves fairly well in it. In addition, it provides a strongly acidic medium for the reaction.

c. Potassium hydroxide is used to deprotonate the ammonium function of the xylidium salt. The high concentration of the base makes the aqueous layer more polar, which in turn makes it easier to extract the amine afterwards.



d. Anhydrous magnesium sulfate is a very bad choice since it is slightly acidic and would absorb the basic amine much stronger. As a result, the yield of the product would be lower.


8. a. The best way to control highly exothermic reactions is to cool the reactants. In addition, the reactant should be combined slowly in diluted form in an inert solvent.

b. The reaction container used in such a reaction should be a little larger than normal to avoid explosions and spillage of solvent/solution onto the heat source.

c. A polar solvent like an alcohol e.g. ethanol would work best here. The polar reactants would dissolve and the non-polar product precipitates in the course of the reaction. In addition, the high boiling point would allow increasing of the rate of reaction by refluxing the mixture.


9. a. In the dominant resonance structure of the nitronium ion, the nitrogen atom carries the positive charge. In the acylium ion, there are two resonance structures of which the dominant form has the positive charge on the oxygen atom. The other (minor) form has the positive charge on the carbon atom which only exhibits a sextet.



b. The problem in the Friedel-Crafts-Alkylation is that the product formed after the first alkylation is more reactive than the starting material because the alkyl group activates the benzene ring. In order to obtain mono-alkylated rings, obe has to use Friedel-Crafts-acylation followed by reduction.

c. The amine function is a Lewis base and reacts with many catalysts that are usually Lewis acids e.g. AlCl₃, BF₃, etc. As a result, the directing effect of the amine group changes from ortho/para to meta directing because the nitrogen atom now carries a positive charge. The protection can be done using an acetyl group which forms an amide which us still mildly ortho/para directing.

10. a. The degree of unsaturation is D.U.=5 (=(10*2+2-13+1)/2)

b. The most important peaks in the IR spectrum are: 3286 (NH, sec. amide), 3132-3193 (CH, sp2), 2928-2982 (CH, sp3), 1661 (C=O, amine), 1511, 1605 (C=C, aromatic), 1370, 1441 (

c. The signal at 1.38 ppm (t, 3H) is due to methyl group next to a methylene group, the signal at 2.09 ppm is due to an isolated methyl group either next to a carbonyl function or a benzene ring. The signal at 3.98 ppm (q, 2H) is a result of a methylene function next to a methyl group and connected to an oxygen atom. The signals at 6.80 and 7.36 ppm (d+d, 2+2) are due to a para-substituted benzene ring. Finally, the broad signal at 7.94 ppm is due to the secondary amide function.

d. The carbon spectrum exhibits two CH3 group (15 and 24 ppm), one CH2 function (64 ppm), two CH functions (122 and 115 ppm) and two quaternary carbon atoms (156 and 131 ppm).

e. Based on the discussion above, the compound is p-acetophenetidine.