Reaction 2 is a reduction, which converts the nitro group into an ammonium function. This ammonium function is subsequently deprotonated by a base and reacts with the aldehyde to form an imine.

b. In order to determine the yield, the amount of compound A has to be determined first.

nA = 3.0 g/150.2 g/mol = 0.020 mol

Since there was no statement about the amount of nitric acid being used, one has to assume that compound A is the limiting reagent. This means that 0.020 mol of the product is formed in theory. However, since the yield is only 85%, the true yield is

nB = 0.020 mol * 0.85 = 0.017 mol
mB = 0.017 mol * 195.2 g/mol = 3.32 g

c. The yield is reduced due to increased amount of dinitration that occurs if the temperature is not properly controlled during the addition of the nitric acid. The dinitration product is thermodynamically more stable, but the activation energy for the dinitration is higher than the one for the mono-nitration. In addition, the dinitro product is more polar, so it remains in solution during the recrystallization process.

d. The reduction requires a high concentration of protons to tie up the oxygen atoms and provide protons to form the ammonium function. If the solution is not acidic enough, the reaction yields different product i.e. hydroxylamines, etc.

e. The potassium carbonate acts as a base that deprotonates the ammonium function of compound B. The amine function subsequently acts as a nucleophile.

f. The solvent in reaction 3 has to be fairly polar in order to dissolve the ammonium salt and the potassium carbonate. It would probably be best to start with water for the formation of the amine, and then add ethanol to increase the solubility of the aldehyde in the reaction mixture (see ligand synthesis).

g. The white crystals are a result of the oxidation of benzaldehyde to benzoic acid by air. The problem is that the acid protonates the amine function, thus makes it inactive for the formation of the imine. In order to remove the acid, the aldehyde should be treated with sodium bicarbonate solution, which removes the acid as benzoate, which is water soluble.

PhCOOH + NaHCO₃ ----- > PhCOO^-Na⁺ + CO₂ + H₂O

h. Extra Credit: Compound B exhibits three signals in the ¹H-NMR spectrum and six signals in the ¹³C-NMR spectrum.


2. a. He has to reheat the glassware because moisture and oxygen are adsorbed onto the glassware when it cools down. Both are not visible but the glass surface attracts water very much. The drying tube is meant to keep this water out as much as possible.

b. The student will observe that the reaction mixture is boiling much more vigorously than the solutions of his fellow students. In addition, the mixture is much darker as well in the end. Finally, the yield of benzoic acid is going to be fairly low, because he consumed the majority of the Grignard reagent already by reacting it with bromobenzene.



c. The addition of sulfuric acid serves the purpose to quench the unreacted Grignard reagent and to convert the Mg-benzoate into benzoic acid. The ice is added to control the temperature, more specifically to keep it low to prevent the diethyl ether from evaporating.



d. The extraction with sodium hydroxide moves the benzoic acid into the aqueous layer, while the unreacted bromobenzene, biphenyl and benzene remain in the organic layer.




3. a. If the other enantiomer of tartaric acid would be used, there are a lot of changes. First off, the (S,S)-form of the 1,2-diaminocyclohexane would be isolated in the resolution step. Consequently, the (S,S)-form of the ligand and the catalyst is formed. The optical rotation of the ammonium salt and the ligand will have the opposite sign. Ultimately, the other enantiomer of the epoxide will be formed in higher quantity.

b. The solution is fairly acidic in the beginning, but the pH-value rises as more and more of the diamine is added. In the beginning all ammonium functions are protonated, and a precipitate forms. Upon raising the pH-value, there is only a partial protonation observed, which makes the compounds more soluble. Thus, the precipitate disappears and only is formed upon addition of glacial acetic acid.

c. In order to assess the purity of the sample, the specific optical rotation of the sample has to be found.

c_S = 0.1 g/5 mL = 0.02 = 2%

[]S = /(cS * l) = 0.24/(0.02*1) = +12.0^o

Optical purity = []S /[]Lit * 100 = +12.0^o / +12.5^o * 100 = 96 %

The optical purity of the sample is fairly high. Ideally, it should be 99%+, but 96% will be sufficient for most reactions as well.

d. The index “D” stands for the wavelength the optical rotation was acquired at, in this case at = 589.6 nm.


4. a. In order to find the ‘proper’ concentration, the molar extinction coefficient for the largest peak is used. Since the range from = 300-600 nm is measured, only the peak at = 320 nm and = 500 nm have to be considered. The first peak is the larger one, thus the maximum concentration is given by

c_max = A_max/( * l) = 1/(1000 * 1) = 10^-3 M

For this concentration, the second peak would show an absorbance of

A₂ = * l * c_max = 120 * 1 * 10^-3 = 0.12

which is in the desired window of 0.1 < A < 1.0.

b. The statement is entirely wrong. There is not direct correlation between the absolute purity of a sample and the melting point ratios (here: 95/100). How much the melting point is influenced depends on the individual compound and as well as the nature of the impurity.

c. Since his compound is very sensitive towards acids and bases, he needs to ensure that the stationary phase does not contain either one. Silica is slightly acidic, which means that it should be pre-treated with a tertiary amine to remove these impurities. Afterwards, the excess amine should be removed as well by washing the column with the solvent used for elution later on.

d. The drying agent is supposed to absorb the water, but not any other polar compounds i.e. alcohols, etc. The removal of the drying agent is necessary to ensure that the drying process does not reverse at higher temperatures (when the solvent is boiled off).

e. After the development, the TLC plate is dried and the place under the UV light. The alkene used in the lab appear there as a pink spot (usually relatively far up the TLC plate). After marking, the plate is dipped in the “ceric staining” solution, whipped clean in the back and then gently heated. The epoxide appears as a blue spot (Mo(V)) about halfway up the TLC plate.

f. Ethers like THF are usually contaminated with water and peroxides (and stabilizers like BHT as well). The most efficient way to remove these compounds is to reflux the solvent over sodium metal and benzophenone until a dark-blue solution is observed.

2 Na + 2 H2O ---- > 2 NaOH + H₂

Afterwards, the dry THF should be distilled under inert gas.

g. The fumes are HCl, which formed in the hydrolysis of the acyl chloride.

ClCH₂COCl + H₂O ---- > ClCH₂COOH + HCl

Most acyl chlorides are very reactive and react more or less readily with water. The bottle should be kept closed if the compound is not used.

h. Extra credit:

Based on the cluster pattern for the parent peak, the compound must contain three halogen atoms. Since the mass is too large for even one bromine atom to be present, the best guess here is that there are three chlorine atoms in the molecule. With this assumption, the remaining fragments have a total mass of 14, which relates to a carbon atom and a deuterium atom. The compound is deuterochloroform (CDCl₃).


5. a. Diethyl ether has a relatively low polarity, which means that many organic compounds dissolve relatively well in it. Also, it does not dissolve significantly in water (~5%), which makes it easy to form two layer. In addition, it is less dense compared to most aqueous layers, which means that it usually is on the top. This is an advantage because fewer transfers have to be conducted during the extraction itself. Finally, the low boiling point makes it to remove it. However, it is highly flammable, so great caution is advisable here.

b. Diethyl ether is less polar than tetrahydrofuran, which means that it is hygroscopic than THF. The lower polarity also makes it less soluble in water, which makes the phase separation easier later on. However, the lower polarity also results in a lower boiling point, which can be a problem for halides that do not react well, and in lower donor ability as compared to THF.

c. The main difference is that diethyl ether forms stronger hydrogen bonds to the stationary phase than dichloromethane. As a result, diethyl ether is more powerful as eluent than dichloromethane when polar stationary phases are used.

d. The low boiling point of diethyl ether is the critical point here. The solvent elutes very early off the column which means that it does not interfere with any other compounds that are of interest.


6. a. The electrophile used in the Friedel-Crafts-Acylation is the acylium ion. It is obtained from the reaction of acetic acid anhydride with phosphoric acid.



The advantage of this method is that it does not require strictly anhydrous conditions (i.e. CH₃COCl/AlCl₃) and produces relatively low concentrations of the electrophile. This slows down the reaction and also makes it more selective. This means that the reaction predominantly yields the mono-acylation product.

b. If the reaction is carried out under the wrong conditions, the amount of the di-acylation product increases. The second acetyl group is attached at the other ring and not at the same ring where the first one was added. The unsubstituted benzene ring is a better nucleophile.



c. The mono-acylation product is less symmetric than the diacylation product, which means that it will generate more signals in both spectra. CpFe(CpAc) shows four signals in the ¹H-NMR spectrum and six signals in the ¹³C-NMR spectrum, while Fe(CpAc)₂ exhibits only three signals in the ¹H-NMR and five signals in the ¹³C-NMR spectrum. In both spectra, the intense signal for the unsubstituted ring disappears in the disubstituted compound.

d. The first peak (tR=3.0 min) is the ferrocene, the second (tR=5.3 min) is due to CpFe(CpAc) and the third peak (tR=6.2 min) is due to Fe(CpAc)₂. The different retention times are a result of the different polarities, which are also used in TLC or column chromatography to separate them.


7. a. Epoxides are not very stable under acidic or basic condition, which means that they can easily be hydrolyzed to form diols.



Therefore, extended exposure to aqueous solutions should be avoided since it would lower the overall yield.

b. Cinnamates are starting materials in the synthesis of diltiazem and taxol. Studies have shown that donor groups increases while acceptor groups decrease (in some cases even invert) the selectivity to form cis epoxides. The reason can be seen in the possibility to stabilize the radical intermediate. Donor groups destabilize the radical intermediate, thus disfavoring this intermediate and therefore leading predominantly to the cis epoxide. Electron-withdrawing groups do the opposite, which means that the radical has time to rearrange and form the thermodynamically more stable trans epoxide.

c. The phosphate buffer is added to stabilize the pH-value during the reaction. Both the catalyst and the epoxide are sensitive towards hydrolysis. If the pH-value is too low, the alkene is chlorinated due to the presence of hypochlorous acid (HOCl) in the system.

d. Pyridine N-oxide is added as a promoter. As a strong Lewis base it adds to the sixth coordination position of the catalyst and therefore stabilizes the catalyst. The reaction is a little slower initially, but this allows for better stereochemical control during the reaction (late transition state). Also, the catalyst is less prone to hydrolysis and therefore retains it activity longer.

e. First off, the solvent mixture is way too polar for this system, which causes a relatively poor separation of the compounds. Secondly, the compounds can be identified as a-methylstyrene, a-methylstyrene oxide (2-methyl-2-phenyloxirane) and 2-phenylpropanal. Based on the polarities, styrene moves the highest up the TLC plate and the aldehyde the least.

Rf=0.95 Rf=0.85 Rf=0.82

f. The addition of copious amounts of drying agent is not advisable because the product (epoxide) is fairly polar. This means any excess of drying agent will adsorb the compound as well, leading to a lower final yield.

g. Just spotting the solution onto the TLC plate is not enough to identify the fractions containing the epoxide. The problem is that not only the epoxide produces a blue spot, but also the aldehyde/ketone since the molybdate attaches to those as well. As a conclusion, these fractions do not necessarily contain the epoxide. The student has to run a normal TLC to evaluate the presence of the epoxide.


8. a. The product (oxime) is more polar than the reactant (benzophenone). Thus, a medium polar solvent like ethyl acetate would be appropriate here since benzophenone dissolves fairly well in it.

b. If the pH-value is too high, the reaction does not proceed quickly because the hydronium concentration is too low to produce water as leaving group in the molecule below.



c. The reactant is of medium polarity, which means that the predominant forces are dipole-dipole interactions here. The product has a hydroxyl function, which means that there are hydrogen bonds between the molecules. Thus, the melting point of the product is higher than the one of the reactant.

d. Extra credit

The addition of a strong acid causes the rearrangement of the oxime to form an amide (Beckman rearrangement).



9.a. The degree of unsaturation is

D.B.U. = (2*C + 2 – H)/2 = (2*12+2-16)/2 = 5

Indicating the presence of a phenyl ring and one additional double bond.

b. The IR spectrum shows peaks at 3035-3066 cm^-1 (sp²-CH), 2872-2973 cm^-1 (sp³-CH), 1721 cm^-1 (C=O, ester), 1603, 1493 cm^-1 (C=C, aromatic), 1276, 1120 cm^-1 (C-O, ester), 712 cm^-1 (oop, monosubstituted arene).

c. The ¹H-NMR spectrum exhibits signals at =1.18 ppm (d, 6H), 3.65 ppm (m, 1H), 3.75 ppm (t, 2H), 4.44 ppm (t, 2H), 7.42 ppm (t, 2H), 7.53 ppm (t, 1H) and 8.06 ppm (d, 2H). The last three signals allow the conclusion that there is a mono-substituted ring in the molecule. The two triplets represent two methylene groups that are connected to each other and are also connected to an oxygen atom. Finally, the large doublet and the multiplet are due to an iso-propyl group.

d. The ¹³C-NMR spectrum exhibits signals at d=22.1 ppm (CH₃), 64.6 and 66.1 ppm (CH₂ next to oxygen atom), 72.1 ppm (CH next to oxygen), 128.4, 129.7 and 132.9 ppm (CH, aromatic) and 130.5 and 166.6 ppm (quaternary carbons).

e. Based on the discussion above, the compound is 2-isopropxyethyl benzoate.