1.a.

The reaction above is a Claisen Schmidt reaction between ferrocene carboxaldehyde and a substituted acetophenone to an a,b-unsaturated ketone. Note that the carbonyl function is connected to the benzene ring here and not to the ferrocene part (opposite to what was done in the lab). The stereochemistry on the newly formed double bond is trans.

b. The derivative with X=OCH₃ is more reactive in this reaction that the one with X=NO₂, because the methoxy group increases the electron-density in the benzene ring, making the enolate formed from the acetophenone a better nucleophile. The nitro group would reduce the nucleophilicity of the enolate.

c. The reaction of the chalcone formed in (a.) with hydroxylamine affords a mixture of the syn- and the anti-oxime initially. The anti-oxime subsequently forms a dihydro-isoxazoline, while the syn-form remains unchanged.




d. In order to determine the yield, the limiting reagent has to be determined.

Compound A: 2.0 g/(214.1 g/mol)=9.34 mmol
Compound B: 2.0 g/(150.2 g/mol)=13.3 mmol

Therefore, compound A is the limiting reagent in this reaction. The product was obtained

Compound P: 2.8 g/(346.2 g/mol)=8.09 mmol

Thus, the yield is

yield = actual yield/theoretical yield * 100% = 8.09 mmol/9.34 mmol * 100% = 86.6 %

e. The newly formed C=C-bond shows as two doublets in the ¹H-NMR spectrum that exhibit a large coupling constant than the aromatic signals. Usually they are relatively far apart from each other if the double bond is connected to a strong group that causes a lot of resonance.

f. Extra credit: Ferrocene carboxaldehyde is orange-red like acetylferrocene. The color of the compound is most likely going to be dark red or purple.


2. a. The addition of the chiral amine leads to the formation of two diastereomeric ammonium salts. Since these compounds usually exhibit different solubilities in solvents like water or alcohols, they can be separated by fractionated crystallization.



b. The sample shows a positive sign for the optical rotation, which means that the S-enantiomer was isolated in the procedure. In order to determine the optical purity, the specific rotation of the sample has to be determined.

Concentration of solution: 0.35 g/5 mL = 7%

Specific optical rotation of sample: [a] = 10.0^o/(0.07*1) = 142.86^o

Optical purity = specific rotation of sample/specific rotation of the pure sample * 100%
= 142.86^o/150^o *100 % = 95.2%

c. The fact that the observed melting point is two degrees below the literature melting point indicates that there are impurities in the sample. However, the relatively sharp melting point together with the fact that the melting is relatively close to the literature melting point, indicates a relatively high purity and also makes it very likely that the correct compound was isolated.

3. a. 95% ethanol contains aside of ethanol also several other compounds like methanol, water, toluene, pyridine, etc. which makes it useless for alcoholic beverages. Particularly the aromatic compounds cause problems when acquiring UV-Vis spectra in the range below l=300 nm, because these compounds absorb there very strongly themselves (p-p* transitions). Even though these additives are only present in small quantities, the cause significant problems and makes 95% ethanol useless as a solvent here.

b. Since the two absorbances are very different in intensity (as documented by the e-values), two solutions have to be prepared here. Using Beer’s Law and the fact that A_max=1 and l=1 cm, one can obtain the maximum concentrations for each peak.

c_max(390 nm) = 1/(3020*1) = 3.3*10^-4 M
c_max(610 nm) = 1/(20*1) = 0.05 M

c. A rotary evaporator allows removing a large quantity of solvent relatively quickly. The main reason is that the surface area is increased due to the spinning motion. In addition, the process is also more environmentally friendly because the solvent is collected and properly disposed off. Finally, one is also able to evaporate solvents with higher boiling points relatively easily, because most rotary evaporator are usually equipped with a vacuum hook-up as well.

d. The broad signal at d~1.6 ppm is due to water in the NMR sample. The water is either a result of a wet sample, old CDCl₃ (which has absorbed moisture from the air) or a wet NMR tube.

e. The signal shift to higher d-values (~3.33 ppm) due to the stronger interaction of the water with the more polar solvent, DMSO. Water forms hydrogen bonds to the oxygen atom of the DMSO, which deshields the hydrogen atoms further.

f. The best way would be to acquire a DEPT spectrum. The signal for the OH group should not show a signal in the DEPT spectrum since only carbon atoms show there. Alternatively, one could treat the aldehyde with D₂O to convert the OH-function against an OD-function, which would not show up in the ¹H-NMR spectrum.

g. Phenolic OH-groups usually show up in the range from 4-7 ppm. The fact that it shows up significantly higher is an indication of additional hydrogen bonds. In this case, there is a chance for intramolecular hydrogen bonds between the phenolic OH group and the carbonyl group, which causes this shift. The polarity of the solvent plays a very important role here. If the solvent is non-polar, this intramolecular hydrogen bond is very strong, in more polar solvents the OH-group interacts stronger with the solvent.



4. a. Ethyl acetate is not a good choice as an alternative here since the solvent reacts with the bleach as well. Esters usually hydrolyze in the presence of bases, which would use up some of the bleach and also change the pH-value, which leads to the hydrolysis of the formed epoxide.

b. Na₂HPO₄ is used as buffer to maintain the pH-value between 9.5-11.5. Outside this window, the alkene undergoes chlorination (low pH-values) or is hydrolyzed. At higher pH-values, the catalyst also deteriorates more rapidly, which means that the reaction slows down as well. Since NaH₂PO₄ solution has a significantly lower pH-value, this solution cannot be used as buffer as is.

c. Silica (SiO₂) is slightly acidic by nature. Since epoxides are very sensitive towards acids (in presence and absence of water), the column is pre-treated with a dilute triethylamine solution (in hexane) to neutralize these acidic residues. Afterwards, the column is flushed with the solvent mixture to remove the excess amine.

d. The two signals are due to the two enantiomers of the epoxide formed. The first one is due to the R-epoxide and the second is due to the S-epoxide (assuming the R,R-form of the catalyst is used). The e.e.-value can be obtained by

e.e. = (area 1 – area 2)/(area 1 + area 2) * 100 %
= (200-50)/(200+50) * 100%
= 60 %

e. Using dichloromethane to “wet” the column was not a smart move, because it is more polar than the solvent mixture that he uses for the actual chromatography step. As a result, he would observe that the alkene and the epoxide pretty much elute at the same time, because they have no chance to absorb on the column. When performing chromatography on polar stationary phases, one should always go non-polar to a polar solvent, and not vice versa.


5. a. The product is the black spot that moved up the least on the TLC plate. The

R_f=6 mm/36.5 mm=0.164

b. The reaction is incomplete, because both starting materials are still present in the mixture as can be seen on the TLC plate (three spots in the product lane).

c. The reaction is completed. The spot for compound A disappeared in sample lane 3. Compound A has to be the limiting reagent here, and the reaction has to have a relatively high K_eq.

d. The polarity of the solvent is higher in this case because more of the polar solvent is present. As a result, all spots move higher up the plate. The separation of the spots would decrease as well.


6.a. Manganese is a transition metal and undergoes redox-reactions more readily than magnesium. Transition metals often times have a broad variety of oxidation states and the redox potentials are not as far away from each other. On the other hand, magnesium has only two oxidation states, the metal and the Mg²⁺ ion, which are relatively far away from each other in terms of potential (E^o=-2.37 V).

b. The 5-position is the position opposite to the OH-group. The t-butyl groups in this position control the approach from the side. If this group is increased in size, this approach pathway is blocked more efficiently. As a result, the e.e.-value should go up slightly.

c. The (R,S)-form of the catalyst is achiral (has a mirror plane) and as such cannot yield a specific enantiomer for the epoxidation. It can be used as an epoxidation catalyst, but just not in asymmetric synthesis.

d. The use of bleach makes the reaction cheaper, because iodosobenzene is fairly expensive. The reaction can be carried out at room temperature, which makes it easier to handle, but the observed e.e.-values are a little lower as well. In addition, the pH-value has to be controlled using a buffer system in order to minimize the chlorination and hydrolysis of the produced epoxide.


7. a. Grignard reagents are strong nucleophiles and strong bases. Thus, water or other impurities react with it, destroying the reagent. Anhydrous diethylether does not contain these impurities. Another aspect is the polarity of the reagent formed in the reaction. Petroleum ether is a mixture of hydrocarbons and not polar enough to solvate the weakly to medium polar Grignard reagent.

b. Magnesium is a very reactive metal, which means that it also reacts with the oxygen in the air. As a result, the metal is covered with a more or less thick layer of the oxide. This layer acts as an insulation layer in the electron transfer from the metal surface to the aryl halide. Crushing the Mg-turnings up exposes a fresh surface and makes the electron transfer easier.

c. The bromobenzene solution has to be added slowly in order to minimize the formation of biphenyl. Once the Grignard reagent is formed, any addition of bromobenzene can lead to the formation of more Grignard reagent or the formation of biphenyl. At lower temperatures and low PhBr concentrations, the formation of the Grignard reagent is favored. The rate of addition should be regulated in a way that the reaction mixture is gently boiling.

d. After the reaction, the mixture contains benzoic acid, unreacted bromobenzene and biphenyl. Benzoic acid is isolated by extraction of the organic layer with sodium hydroxide, which make the acid water-soluble in form of sodium benzoate. The other two compounds remain in the organic layer. After acidifying the aqueous layer, benzoic acid is precipitated as a white solid.




8. a. Acid chlorides are more reactive than the corresponding carboxylic acids because they have a better leaving group (Cl- vs. OH-). This allows for the reaction to be carried out under milder conditions.

b. After the initially reaction, the amide function is still protonated, which makes the compound ionic. As a result, it dissolves well in a polar solvent (acetic acid). The sodium acetate acts as a base and removes this proton. The resulting compound is now neutral and does not dissolve well in glacial acidic acid anymore.

c. A polar solvent is used to dissolve the starting material. However, it cannot contain any water or alcohols since they would react with the acid chloride as well. In addition, the unreacted amine remains in solution (as ammonium salt) while the amide precipitates.

d. The primary amine function is converted into a secondary amide function. As a result, one of the peaks in the range from 3200-3400 cm^-1 disappears. Secondly, the new molecule contains a carbonyl group that should show as a strong peak around 1660 cm^-1.


9.





The first step of the reaction is the protection of the amine function of one the amino acids with a t-Boc group. In addition, the carboxylic acid group has to be deactivated as well by converting it into an ester. The two fragments are then combined in the presence of DCC, which neutralizes the water formed in the reaction. The resulting protected amide can be deprotected using acidic conditions, which allow for the removal of the ester and the t-Boc group.


10. 10. a. D.U.=(2*12+2-14)/2 = 6

b. Important peaks in the IR are found at 3033-3091 cm^-1 (CH(sp²)), 2914-2973 cm^-1 (CH(sp³)), 1718 cm^-1 (C=O), 1677 cm^-1 (C=C, alkene), 1602, 1492 (C=C, aromatic), 1451, 1381 (d(CH₂, CH₃)), 1271, 1108 cm^-1 (C-O), 711 (oop bending, mono-substitution)

c. The two singlets at d=1.74 and d=1.76 ppm (3H each) are due to two almost identical methyl groups. The signal at d=4.82 ppm (d, 2H) is a result of a methylene group next to a CH-group and an oxygen atom, which the triplet at d=5.46 ppm (1H) is due to a CH-group (alkene?) next to a methylene group. Finally, the signals at d=7.39 ppm (t, 2H), 7.49 ppm (t, 1H) and 8.05 ppm (d, 2H) are indicative for a monosubstituted benzene ring. The fact that the doublet is shifted to values above 8 ppm allows the conclusion that a strongly anisotropic group is attached to the ring e.g. nitro, of carbonyl, etc.

d. The signal at 167 ppm is a quartenary carbon, most likely a carbonyl function that has already been identified in the IR spectrum. The range from 118-139 ppm contains four CH-functions and two quaternary carbons (the two smaller signal that do not show up in the DEPT spectra). The signal at 62 ppm is due to a CH₂ function connected to an oxygen atom, while the signals at 18 ppm and 26 ppm are due to two methyl groups.

e.
Based on the discussion above the molecule is 3-Methyl-2-butenyl-benzoate