Final Exam Key Spring 2006

The hydrated form of aluminum trichloride (AlCl₃*6 H₂O) cannot be used here because it is not a Lewis acid because the six water molecules are coordinated to the aluminum, which means that it is coordinatively saturated.

d. A nitro group in compound (A) means that the first reaction will not take place. Nitro groups are strongly deactivating and a weak electrophile like the acylium ion would not react with such an electron-poor arene anymore. Consequently, the second reaction would not take place either since there is no ketone function in the intermediate anymore.

The stationary phase is polar and the mobile phase relatively non-polar. As a result, only compound A, a hydrocarbon, moves far up the TLC plate. The intermediate B is a ketone, which means that it is the most polar of these three compounds. Thus, it moves the least up the TLC plate. The final product P is less polar than the intermediate B because there is very little chance for this compound to enolize.

f. In order to determine the overall yield, the two individual steps have to be evaluated.

Total yield = yield(R1) * yield(R2) = 0.70 * 0.55 = 0.385

M.W. (A) = 106.08 g/mol

M.W. (P) = 236.12 g/mol

Moles of A: 1.75 g/106.08 g/mol = 16.5 mmol

Moles of P: 16.5 mmol * 0.385 = 6.35 mmol

Mass of P: m = 6.35 mmol * 236.12 g/mol = 1.50 g

g. Extra Credit:

Many aromatic aldehydes are contaminated with the corresponding carboxylic acid due to their oxidation in air. This is a reason why fresh aldehydes should be used for reactions where basic catalysts are used like this reaction or the benzoin condensation carried out in Chem 30BL.

2. a. In order to determine the optical purity, the following calculations have to be done.

- Concentration of sample: c=0.84 g/7 mL= 0.12 g/mL = 12 %

- Specific optical rotation of sample: [

- Comparison with literature: optical purity = 25.8o/27.5o * 100% = 93.8%

b. Student O would look at a different form of the amino acid (anionic) compared to student N (cationic) and therefore observe entirely different values. The reference value in a. would be absolutely useless for him.

c. The student could convert the amino acid into diastereomeric ester or amides, and then measure the ¹H-NMR spectrum. If done properly, the NMR spectra for the two diastereomers are different. This can be used for to analyze the composition by integration of suitable signals. A compound used for this purpose is Mosher’s Acid.

d. Since he isolated the amino acid from a natural source, the L-form was obtained like for most amino acids.

3. a.

The difference in melting point is a result of the different intermolecular forces in the two isomers. The syn isomer forms stronger intramolecular hydrogen bonds between the hydroxyl function and the hydrogen atoms on the ring, thus forms less intermolecular hydrogen bonds. As a result, the syn isomer has a lower melting point than the anti-isomer.

b. The reaction of phosphorous pentoxide with water affords phosphoric acid.

The acid then catalyzes the condensation of acetone (aldol condensation). As a result, a lot of the acetone would be destroyed this way.

c. Since the melting point is higher than the reported for the target compound, the student did not isolate the desired compound. Based on the data given, it is more likely that he isolated the starting material, which is contaminated by the target compound and possibly the byproduct. This can be justified by the fact that the melting point for the starting material is depressed compared to the literature melting point and the fact that the melting range is fairly broad.

d.. The ratio of M⁺:[M+2]⁺:[M+4]⁺ = 1:2:1 allows the conclusion that the compound contains two bromine atoms. Bromine has two isotopes (M=79 and 81) that are almost equal in abundance.

e. The best choice as a solvent here would be compound D. It has a reasonably high boiling point, which allows for an increase of the rate of the reaction by running the reaction under reflux condition. It also would be favoring the products because the compound would act as reactant and as a solvent. In addition, compound C would dissolve while compound F would precipitate.

f. The term mesh indicates the particle size for the silica particles. For flash chromatography usually the 60 mesh version is used since it has a larger surface area, which allows for better separation. However, since the packing is very dense, a pressure gradient has to be applied in order to achieve decent flow rates.

g. Sodium bicarbonate is used to neutralize acids (like sulfuric acid used as catalyst in the esterification or acid byproduct like in the Friedel Crafts acylation). The reaction leads to the formation of carbon dioxide according to

HCO₃^- + H⁺ ----- > “H₂CO₃” ----- > CO₂ + H₂O

Since a gas is formed in the reaction, the separatory funnel or extraction container has to be vented frequently to avoid the build-up of pressure.

h. The boiling point is relatively high and closer to the boiling points of the epoxides, which makes it more difficult to remove it without loosing significant amounts of the epoxide due to overheating or evaporation.

4. a. DMSO has two functions in this reaction. It is the oxidant and the solvent.

PhCH₂OH + O=SMe₂ ---- > PhCHO + H₂O + SMe₂

b. The addition of water causes the organic compounds to precipitate from DMSO which is difficult to remove by evaporation (b.p.=~190 ^oC). It also removes the majority of the hydrobromic acid, which dissolves in water.

c. The major impurity after the reaction is probably the unreacted alcohol, which is more polar than the aldehyde. Thus, a polar solvent is used that allows for the impurities to remain dissolved, while the less polar aldehyde precipitates.

d. The carbonyl stretching frequency is relatively low due to three reasons. First, the carbonyl group is connected to benzene ring which lowers the frequency already from 1715 cm^-1 (acetone) to 1685 cm^-1 (acetophenone). In addition, there is a strong resonance effect from the OH function in this position. And finally, the phenolic hydrogen forms a relatively strong intramolecular hydrogen bond to the carbonyl of the aldehyde.

e. The phenol is the least reactive while the methoxy substituted benzylic alcohol is the most reactive compound among the three compounds given. The oxidation of phenols is much more difficult to accomplish because they are not alcohols. The methoxy group stabilizes the benzylic cation better than the nitro group, which lowers the activation energy for the process significantly.

5. a. Ethanol used as solvent in this reaction because it dissolves the ligand and manganese(II) acetate reasonably well, which allows the reaction to proceed in a reasonable rate. In addition, the boiling point is about 80 ^oC, which allows the experimenter to raise the temperature in the system by about 60 ^oC (reflux). In principle, methanol would work as well, but the solubility of the ligand would be even lower and the temperature can only be raised to 64 ^oC, which means that the reaction would be slower overall.

b. Air is used as a mild oxidant in this reaction. After the initial reaction with the ligand, the Mn(II) is oxidizes to Mn(III), which can be followed by the color change. Bleach would not be a good oxidant for this reaction because it would over-oxidize the Mn(II). The resulting compounds are less stable than the Mn(III) complex.

c. After the reaction, the reaction mixture contains small amounts of ligand and the catalyst. If the solvent would just be boiled off, the ligand would remain in the final product. The addition of heptane to the dichloromethane solution only precipitates the catalyst, but not the ligand.

d. Generally, the maximum concentration is given by the largest molar extinction coefficient, which is at =288 nm. Using Beer’s Law (A=lc) and the fact that usually a 1 cm cuvette is used,

C_max(1)= 1/(*l) = 1/27600 = 3.6 * 10^-5 M

Using this concentration, the peaks at =325 nm (A=0.623) and =436 nm (A=0.196) are well within the range. However, the peak at =496 nm (A=0.080) would be out of the desired range of A=0.1 – 1.0. This means that the experiment has to prepare a second solution with a higher concentration.

C_max(2) = 1/(*l) = 1/2200 = 4.5 * 10^-4 M

6. a. The second step was the formation of the anilide from 2,6-xylidine and -chloroacetyl chloride.

b. The acid chloride is very moisture sensitive. Glacial acetic acid is used as solvent because it is anhydrous and allows for the unreacted 2,6-xylidine to remain in solution while the neutral amide precipitates.

c. Sodium acetate is used as base in this reaction. It is strong enough to deprotonate the protonated amide but not the xylidium salt. Thus, the amide because neutral and precipitates from the polar solvent.

d. 2,6-Xylidine contains a primary amine function with exhibits two peaks in the 3300-3500 cm^-1 range. The acid chloride exhibits a very carbonyl stretching frequency (1811 cm^-1). The anilide on the other side only shows one NH peak in the range above, while the carbonyl peak is shifted down to about 1660 cm^-1.

7. a. The ortho isomer has a lower boiling point (ortho: 215 ^oC vs. para: 279 ^oC) because this isomer exhibits strong intramolecular hydrogen bonding between the nitro group and the phenolic hydrogen. Thus, less intermolecular hydrogen bonding is observed, which lowers the melting and boiling point for the ortho compound.

b. The ortho isomer possesses the higher Rf-value because this isomer interacts less with the polar stationary phase due to strong intramolecular hydrogen bonding between the nitro group and the phenolic hydrogen, which makes it less accessible for the interaction with the stationary phase.

c. The para isomer is more soluble in ethanol (ortho: 24.5 g/100mL at 15 ^oC, para: 116 g/100 mL at 0 ^oC), because it is more polar. The intramolecular hydrogen bond makes the ortho compound less polar, which also means less soluble in polar solvent (“like-dissolves-like” rule).

d. The para isomer has a higher OH-stretching frequency (para: 3331 cm^-1, ortho: 3249 cm^-1), because the intramolecular hydrogen bond lower the OH-stretching frequency by about 100 cm^-1.

e. Due to the intramolecular hydrogen bond to the nitro group the hydrogen is much more deshielded in the ortho isomer than in the para isomer. Consequently, the chemical shift is higher for the ortho isomer (d=~10.5 ppm in CDCl₃) than for the para isomer (d=~6.3 ppm in CDCl₃).

8. a. The electrophile in the nitration reaction is the nitronium ion, which is obtained by the reaction of conc. nitric acid and conc. sulfuric acid. Nitric acid acts as the base here.

HNO₃ + H₂SO₄ ---- > NO₂⁺ + H₂O + HSO₄^-

b. The sulfuric acid is the acid in the mixture and also serves as solvent to protonate the carbonyl oxygen. The reaction mixture becomes homogeneous this way because the nitration mixture above is miscible with the solution of the ester in sulfuric acid.

c. The nitronium ion is a very strong electrophile, which means it also reacts with electron-poor aromatic systems. However, the activation energy for the second step is slightly higher than the activation energy for the first nitration. By controlling the temperature in the reaction mixture properly the dinitration can be minimized. Since the acylium ion is a weak electrophile, a second acylation does not occur.

d. The solvent mixture is used because the target compound dissolves well in the mixture at high temperature but not at low temperature. In addition, the unreacted ester and the dinitration product remain in solution. Hexane would be a bad choice because the dinitration product would precipitate before the mono-nitration product. Thus, no purification would be accomplished.

9. a. Degree of unsaturation= (2*C + 2 – H)/2 = 5 (phenyl ring + ?)

b. The most important IR peaks (cm-1): 3297 (OH, stretch), 1633 (C=O, conj. ketone), 1516, 1587 (C=C, aromatic), 1370, 1453 (CH3, bend), 1040 (C-O, phenol)

c. H-NMR spectrum:

Shift	Assignment
11.8, s, 2H	Two phenolic OH
7.26, t, 1H	Aromatic H, two neighbors
6.39, d, 2 H	Aromatic H, one neighbor
2.66, s, 3H	Methyl group on ring or CO function

Shift	Assignment
205.47	C=O
161.38	C (OH)
136.18	CH
110.49	C
108.47	CH
33.29	CH3

e. 2,6-Dihydroxyacetophenone

f. Extra Credit: If the spectrometer frequency is lowered the multiplets (doublet and triplet) would show a larger spacing since 1 ppm equals less Hz.