1.a.
b.
In order to prevent this reaction from occurring the amino function has to be protected.
c.
ketone: 1.40 g/150.18 g/mol = 9.3 *10-3 mol
aldehyde: 1.51 g/151.0 g/mol = 1*10-2 mol
KOH: catalyst
product: 2.20 g/283.1 g/mol = 7.78 *10-3 mol
Hence, the ketone is the limiting reagent and 9.3 *10-3 mol of the product are expected here. The molecular mass of the product is 283.1 g/mol.
Yield = actual yield/theoretical yield * 100 %
= 7.78 *10-3 mol/9.3 *10-3 mol * 100%
= 83.6 %
d. The carbonyl peak is shifted to lower wavenumbers due to increased conjugation. In addition, a peak at 1620 cm-1 is going to appear due to the newly formed C=C bond as well as a peak at 960-970 cm-1 for the oop for a trans-substituted alkene.
e. The reactants are relatively polar and so is the catalyst. Therefore, a polar solvent is needed e.g. absolute ethanol.
f. The product is less polar than the aldehyde.
2. a. The acid has to be dry because water is detrimental to yield in the esterification experiment. The esterification is an equilibrium reaction and water is one of the products there. Additional water in the reaction mixture would drive the reaction to the left (Le Chatelier Principle).
b. After the Grignard solution is reacted with dry ice, the Mg-salt of benzoic acid is formed. The addition of the sulfuric acid converts this salt into benzoic acid.
PhCOOMgBr + H2SO4 ---- > PhCOOH + Mg2+ + HSO4- + Br-
c. The nitro-group is not compatible with the Grignard reagent (see reader). It would have to be converted to a nucleophilic group that does not react with the Grignard reagent. However, this also means that the directing effect changes from meta to ortho/para.
3. a.
b. The addition of water serves the purpose to lower the solubility of the ligand in the solvent. The amount of the ligand precipitating increases this way.
c. No. Acetone and water are miscible which means that no phase separation is observed in the washing processes.
d. Here was a mistake in the numbers. It should read 25 mg of the ligand are dissolved in 5.0 mL of solvent and the polarimeter reading is –1.51o. Thus,
c= 0.025 g/5 mL = 0.005 g/mL, l= 1 dm
[a] = a/c*l = -1.51o/(0.005 * 1)= -302o
Optical purity = [a]/[a]ref * 100% = -302o/-315o *100 % = 95.9 %
e.
The infrared spectrum does not show the expected strong OH absorption in the range between 3200 and 3500 cm-1. The shift to lower wavenumbers and the significant broadening of the peak (2500-3100 cm-1) is a result of the strong intramolecular hydrogen bonding of the OH-group with the neighboring nitrogen atom of the imine function. The same effect can be observed in the 1H-NMR spectrum where the phenolic hydrogen atom appears around 13 ppm, which is about 3-5 ppm higher than normal for phenolic protons.
f.
The proton on the imine function shows up around d=8.3 ppm, more shifted than an alkene proton (5-7 ppm), but less than an aldehyde proton (9-11 ppm). The imine carbon can be found around 160-165 ppm following the same argument (alkene: d~130 ppm, aldehyde: d~190 ppm).
4.a. Advantages: Most reactive of all acid derivatives, reaction can be carried out under relatively mild conditions.
Disadvantages: moisture sensitive, deteriorates in quality quickly if not handled properly.
b. The rotation around the C-N bond is relatively slow due to a higher rotational barrier. The two groups on the nitrogen atom are not equivalent on the NMR time scale, and show up in different places in the spectrum. At higher temperature, the rotation is fast enough that the NMR only observes the average value.
c. At this point the lidocaine and the unreacted chloride are still in the organic layer. The addition of the hydrochloric acid causes a protonation of the lidocaine, which makes it water-soluble. The unreacted chloride remains in the organic layer.
5. a.
The shown plate above assumes that the compound A is less polar than compound B, but that both are polar. Compound C is non-polar. A polar stationary phase is used and a non-polar solvent (hexane):
1. Compound A and compound B only moves moderately up the TLC plate due to their strong interaction with the stationary phase. Compound C moves relatively far up the plate due to the lack of interaction with the SiO2.
2. Both, compound A and compound B, are still visible in the sample lane (3) because the reaction mixture still contains reactant in the mixture in the equilibrium. The given equilibrium constant is relatively small (Keq=10).
b.
The same stationary phase is used like above, but a more polar solvent mixture (10% ethyl acetate/90% petroleum ether):
1. the non-polar compound pretty much moves with the solvent front
2. the more polar compounds move higher up the plate due to the higher eluting power of the solvent
6. a. Bleach is used as oxygen source in the epoxidation. Alternatively, iodosobenzene and m-chloroperbenzoic acid can be used (m-CPBA). In some cases, even hydrogen peroxide is used.
b. At lower pH-values (below 9.5), there is a danger that HOCl is formed in the reaction mixture. This compound chlorinates the alkene. If the pH-value is too high, the epoxide is hydrolyzed. The phosphate solution serves as a buffer here.
c. Ceric staining was used for the visualization of the epoxide since most epoxide are not UV-active anymore. The staining solution contains molybdate (“Mo42-“), which is pale yellow in color. It attaches to polar compounds and oxidizes them upon heating. The resulting Mo(V) species is dark blue in color.
d. The deuterated solvent is necessary for the NMR instrument to lock. The lock serves as a reference point for a homogeneous magnetic field.
e. He probably was not very efficient to remove the acidic residues from the silica. The remaining acid rearranged the epoxide on the column to form an aldehyde or ketone.
f. In order to determine the e.e.-value, the relative composition has to be determined.
Peak 1: 200/300 *100% = 66.7%
Peak 2: 80/300 * 100% = 26.7%
Peak 3: 20/300 * 100% = 6.7 %
Peak 1 and peak 2 are due to the cis and trans-epoxide, while peak 3 is most likely due to the rearrangement product. The e.e. can be calculated by
e.e.= (66.7% - 26.7%)/(66.7%+26.7%) * 100 % = 42.8 %
g. A chiral column is needed to separate enantiomers from each other. In the lab, a b-cyclodextrin column was used.
7.a.
b. Acetic acid anhydride is used as solvent here, and not dichloromethane.
c. The peak at n=1660 cm-1 is due to the carbonyl function. The peak is so low due to the conjugation with the electron-rich ferrocene ring. The peak around 500 cm-1 can be attributed to the Fe-C stretching modes.
d.
R=Fc (left: syn-form, right: anti-form)
8. a. A polystyrene cuvette is a very poor choice here due to two reasons. First, polystyrene tends to dissolve in hydrocarbon and chlorinated solvents. Secondly, the cuvette itself absorbs at wavelengths below 340 nm. A silica or quartz cuvette would be more appropriate here.
b. The peak to look at is the one at 440 nm, since it is the largest peak in the range to be measured.
Amax=1, l=5 mm=0.5 cm, e=95 cm-1/M
c=A/(l*e) = 1/(95*0.05)=0.021 M
c. The compound is orange, hence absorbs in the blue-violet range.
9. In order to remove trace amounts of water (aside of other impurities like peroxides) from THF, either sodium metal/benzophenone or lithium aluminum hydride is used.
2 Na + 2 H2O ---- > 2 NaOH + H2 or
LiAlH4 + 4 H2O ---- > LiOH + Al(OH)3 + 4 H2
This drying method is more efficient since the water is chemically destroyed.
b. No! Alkali metals and their hydrides are incompatible with halogenated solvents. Most of the time they initiate radical reactions, which leads to explosions.
c. The problem is that the drying process is an equilibrium reaction. Even though the drying agent absorbs most of the water, it does not absorb everything.
d. The absorption of water is favored at lower temperatures (DG<0), but reverses at higher temperature since the entropy increases then and the TDS overpowers the enthalpy term DH. As a result, DG becomes positive and the reaction reverses.
10. a. The degree of unsaturation can be calculated by
DBE = (2*10+2-10)/2 = 6
b. Important peaks in the IR are found at ~3050 cm-1 (CH(sp2)), 2850, 2900 cm-1 (CH(sp3)), 1650 cm-1 (C=C, alkene), 1600, 1500 cm-1 (C=C, aromatic), 1450, 1380 (d(CH2)), 750 cm-1 (oop, cis-alkene and ortho-subst)
c. The signal at d=3.3 ppm (“s”, 4H) is due to two methylene groups. The signal at d=5.8 ppm (t, 2H) is a result of two alkene protons. Finally, the signal at d=7.1 ppm (“s”, 4H) allows the conclusion that there is a disubstitution on the benzene ring.
d. The signal at 134 ppm is a quartenary carbon, the three signals between 125 and 130 ppm are CH functions and the signal at 30 ppm a CH2-function.
e. Based on the discussion above the molecule is 1,4-dihydronaphthalene.
