2. H+/H 2 O Ph. (A) Reaction 1 (B) Reaction 2 (C) Reaction 3 (P)

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1 1.a. Reaction 1 is an epoxidation that leads two different epoxides in equal quantity. The second reaction is a Lewis acid catalyzed rearrangement of the epoxide to the aldehyde. The last reaction is a Grignard reaction that produces a secondary alcohol. t BuH BF 3 /Et 2 1. MgBr/Et 2 CH 2 Cl 2 2. H+/H 2 H (A) Reaction 1 (B) Reaction 2 (C) Reaction 3 (P) H b. In order to determine the yield, the molar quantities of compound (A) and t BuH have to be determined. n A = 3.25 ml *0.909 g/ml/ g/mol = 25.0 mmol n BuH = L* 5 mol/l = 30 mmol Thus, the alkene is the limiting reagent and one expects to isolate 25 mmol of the epoxide mixture (B). n epoxide = 2.52 g / g/mol = 18.8 mmol The final yield is Yield = 18.0 mmol/25.0 mmol * 100% = 75.2% c. The driving force for reaction 2 is the release of ring strain from the epoxide and the formation of the relatively strong C= bond in the aldehyde. BF 3 BF 3 BF 3 H H H d. If water was present in reaction 2, BF 3 will react with it, which would destroy its catalytical activity. In addition, a diol would be formed as well since water attacks the cation formed as an intermediate. BF H 2 3 BF 3 H H H - H + H H

2 e. The aldehyde (C) possesses a stereocenter already and a new stereocenter is formed in reaction 3. As a result, four diastereomers are possible for compound (P). f. If the rearrangement reaction was skipped, the hydroxyl group would end up on the other carbon atom since the Grignard reagent would attack the less substituted carbon atom in the epoxide. As a result, only of stereocenter is formed, which means that only two enantiomers are formed in the end. MgBr H g. The rearrangement can be accomplished by passing the epoxide through an acidic column i.e. silica or acidic alumina. Recall, that the student tried to prevent this reaction from happening in the lab by neutralizing the Si 2 prior to the application of the crude epoxide. h. Extra Credit: The base peak is the largest peak in the spectrum which usually represents the most stable fragment. A fragment of m/z=105 is due to a C 8 H 9 fragment, which is formed from the aldehyde by -cleavage of the CH function. The fragment below is very stable because it is a secondary cation that is benzylic in nature.

3 2. a. First off, instructors appreciate students that attend to lecture and do pay attention. Water poses a major problem in the reaction and therefore has to be excluded from the reaction as much as possible. Water reacts with the Grignard reagent to form benzene and MgBr(H), which is insoluble in diethyl ether and covers the Mg-surface and therefore prevents new Grignard reagent to form. Even glassware that looks dry has water on the surface since glass is polar in nature. This water is removed by heating the glassware with the heat gun. In addition, it could not hurt to remove the white precipitate (Mg(H) 2 ) on the inside of the flask either since it holds on to water as well. MgBr + H 2 H + MgBr(H) b. The failure of the reaction to initiate can be caused by various factors. The first cause of action would be to heat the reaction mixture with the heat gun in order to get the reaction started. If this does not cause the reaction to initiate, the student should add a few iodine crystals which etch the Mg-surface and make the magnesium metal more reactive. c. The rapid addition of the bromobenzene solution causes the mixture to boil more rapidly, and be darker in the end. In addition, the student would observe a lower yield for benzoic acid because some of the Grignard reagent is consumed by the reaction with the excess bromobenzene in the mixture. MgBr + Br - + MgBr 2 This reaction is favored at higher temperatures. d. The electrophile in this reaction is carbon dioxide. It is introduced in form of dry ice.

4 3. a. Mn(Ac) 2 is significantly more stable, cheaper and easier to handle than Mn(Ac) 3. Those are the main reasons why it is used as Mn-source in the formation of the catalyst. b. The initial reflux period is meant to form the Mn(II)-salen complex, which is then oxidized to the Mn(III)-salen complex. If this step is skipped, a lot of the Mn(Ac) 2 is going to be oxidized to Mn 2 (the dark brown precipitate that does not dissolve in dichloromethane!) which is inactive when it comes to forming the catalyst. c. The addition of heptane to the solution of the catalyst in dichloromethane lowers the solvent polarity and aids the precipitation of the catalyst. The solvent polarity is further reduced by removal of dichloromethane. Both measures cause the catalyst to precipitate from solution slowly while small amount of ligand and aldehyde remain in solution. d. The lower value for (C=N) can be rationalized by the coordination of the lone pair of the nitrogen to the Mn(III) ion in the catalyst. This causes a weakening of the C=N bond, which results in a shift to lower wavenumbers for the corresponding stretching frequency. e. In order to determine the proper concentration here, the student needs to find the peak with the highest -value, which is the one at =325 nm. Thus, the maximum concentration should be c max = A/( *l) = 1.0/(17200 * 1) = 5.8 * 10-5 M (l= 1 cm, A=1.0, =17200) With this concentration, the other peaks would exhibit absorbances of A=0.52 (363 nm), 0.31 (436 nm) and 0.13 (496 nm), which means that all values are in the acceptable range of A= f. The tert.-butyl groups are meant to block the access from the front and the sides of the [Mn=] species. This forces the alkene to approach from the backside via the asymmetric cyclohexane bridge, which ultimately causes the stereoselective character of the reaction. g. The catalyst is paramagnetic, which makes it in most cases very difficult to acquire a NMR spectrum. The signals are usually very broad and also in fairly strange ranges compared to regular 1 H-NMR spectra. The optical rotation measurement is difficult because the catalyst is very dark in color. A 1% solution of the catalyst would not be transparent anymore. The detector of the polarimeter would not get enough light to be able to rotate the plane back to its original orientation.

5 4. a. The peaks at =3500 cm -1 and the =1650 cm -1 are due to moisture in the air (and not water in the KBr which would should as broad peak!). The peak at =2350 cm -1 is due to the asymmetric stretching mode of carbon dioxide. b. Generally, there is an equilibrium between a liquid and its vapor. If the vapor is removed from this equilibrium by an air stream, more of the liquid has to evaporate in order to reestablish this equilibrium. By using an air stream, the solvent can be removed faster and under milder conditions in many cases. c. Since the student only needs about one fifth of the literature procedure, the amount of material he starts with is correct. However, he cannot downscale the reaction time, because this is a questions of the kinetics of the reaction and not the scale the reaction is carried out on. By running the reaction only for 15 minutes, he terminated the reaction too early, which explains the poor yield (~25%). d. This statement contains two fundamental mistakes. First, when reporting a melting point, the melting point should be reported as range, because this also allows conclusions about the purity of the sample. Providing only one number indicates a very sharp melting point, which is inconsistent with the fact that the melting point is about 4 o C below the literature value. Aside of that a temperature difference of 4 o C does not indicate a high degree of purity. Secondly, providing a percentage error is also inappropriate since this error is highly dependent on which temperature scale is used. e. The extraction with sodium bicarbonate serves the purpose of neutralizing acids in the organic layer according to - HC 3 + RCH ---- > RC - + H 2 + C 2 - HC 3 + HX ---- > X - + H 2 + C 2 The extraction is completed once the carbon dioxide formation ceased, which means that no more gas bubbles are observed in the extraction (and/or no more pressure is build up during the shaking of the centrifuge tube). f. In order to assess the optical purity of the sample, first the specific optical rotation of the sample has to be determined. [ ] = /(c*l) = o /((0.2/5)*1)= o Then, this value is compared with the literature value ptical purity = o /12.5 o * 100% = 94%

6 The purity is not as high as desired (>99%) for enantioselective synthesis. However, the lower purity can have various causes i.e. poor resolution, wet sample (tartrate salt) or improper preparation of the solution that was measured. g. The drying process utilizes the fact that water is usually the most polar compound in the solution. However, other polar compounds i.e. alcohols absorb as well, which poses a problem if too much drying agent is used. As a result, the drying agent should be used as sparingly as possible in order to avoid loss of product. h. Extra Credit: Coupling constants are obtained by averaging out the distance of neighboring lines within a multiplet ( in ppm). This number is then multiplied with the sweep frequency of the NMR spectrometer (in MHz). The proper unit for coupling constants is Hz. Note that coupling constants are independent from the instrument used for acquisition.

7 5. a. The electrophile in the FCA reaction carried out in the lab is the acylium ion that is obtained from the reaction of acetic acid anhydride with conc. phosphoric acid. + H 3 P 4 + H 3 C C H 3 C C + + H 2 P CH 3 CH This method has the advantage that the acylium ion concentration is fairly low throughout the reaction, which makes the reaction more selective in terms of monoacylation. If CH 3 CCl/AlCl 3 is used instead, more diacylation product is observed and an anhydrous solvent (CH 2 Cl 2 ) has to be used to maintain the activity of the AlCl 3 as a catalyst. b. The diacylation is observed because the second cyclopentadiene ring is still fairly electron rich. As a result, the second acetyl group is attached to the other Cp-ring and not at the same Cp-ring because this one is deactivated already. Fe c. Note that the amine group is usually an activating group, but it most FCA reactions, it gets tied up with the catalyst and becomes converted to a NH 2 R + -group, which makes it strongly deactivating. The methoxy group is a stronger electron-donating group than the methyl group. The aldehyde function is weakly deactivating what means that this system does not undergo FCA under normal conditions. Me NH 2 CH d. Nitrobenzene can be used as solvent because the nitro group is a strongly deactivating group. The FCA occurs only under very extreme conditions or very, very slowly.

8 6. a. The salicylic aldehyde moves the furthest up the TLC plate because the aldehyde function and the hydroxyl group form an intramolecular hydrogen bond. This makes the hydroxyl group less accessible to bind to the polar stationary phase. When comparing the R f -value of the p-nitro- and p-hydroxy benzaldehyde, two factors determine the degree of movement. First, the nitro compound has a significantly lower dipole moment because both groups are electron-withdrawing. Secondly, the hydroxy compound can bind via hydrogen bonding to the stationary phase, which makes a fairly strong interaction. Thus, the p-hydroxy compound exhibits the lowest R f -value. CH CH CH H N 2 H b. The stationary phase on the TLC plate i.e. alumina or silica is usually not chiral. This means that both enantiomers interact in the same with it, resulting in identical R f -values. c. The fact that the two compounds exhibit identical mass spectra indicates that they are two enantiomers, which only show up at different retention times because the stationary phase used in the GC portion ( -cyclodextrin) is chiral. In order to obtain the optical purity, the following formula can be used e. e. Area of Area of enantiomer 1 Area of enantiomer 1 Area of enantiomer 2 *100% enantiomer e. e. *100% 52.4% The value of e.e.=52.4% is in the normal range for these type of reaction using the Jacobsen catalyst and a styrene derivative as reactant. d. The choice of solvent is based on the polarity of the stationary phase and type of compounds to be separated. If a polar stationary phase was used and compounds of medium polarity were to be separated, neither a very polar solvent like methanol nor a non-polar solvent like hexane would provide good results. The elution power of methanol is too high, which causes all compounds to move too far up the TLC plate. n the other side, the elution power of hexane is too low, which means that the compounds will not move much at all. In both cases, the separation on the TLC will not be great, since the compounds group together either on the bottom or the top of the TLC plate.

9 e. Epoxides are very sensitive towards acids and bases. Most silica is slightly acidic in nature which means that the exposure of the epoxide to the silica would cause a rearrangement (see question 1.g.). The pre-treatment with 1% NEt 3 solution reduces the acidic residues on the silica and therefore the chance for rearrangement.

10 7. a. Dichloromethane is a good solvent for extraction because it has a medium polarity that allows for many compounds to dissolve in it. Secondly, its solubility in aqueous solutions is very limited, which means it forms a second layer easily. This layer is the bottom layer often times, which requires less transfers during the extractions. n top of that, it has a low boiling point, which makes it fairly easy to remove under mild conditions, and it is non-flammable. b. The low boiling point of dichloromethane (40 o C) makes it a poor solvent for recrystallization. A temperature change from 40 o C to 0 o C (ice bath) does not cause a significant change in solubility. c. The only way to increase the rate of the reaction is by increasing the temperature in the reaction mixture. Again, the low boiling point of dichloromethane is the problem here. A 20 o C increase in temperature will only cause an increase in rate by about four, which is not a lot. d. Halogenated solvents are often contaminated by alkenes. These compounds can be removed by extraction with concentrated sulfuric acid. Water can be removed by refluxing the pre-treated dichloromethane over calcium hydride. CaH H 2 Ca(H) H 2 Drying agents like sodium metal or sodium hydride are not suitable here since they will lead to violent explosions! e. Dichloromethane is used as solvent to dissolve the samples because it has a low polarity and a low boiling point. In addition, it is used as eluent or part of the eluent since it introduces a certain degree of polarity.

11 8. a. Potassium cyclopentadienide is ionic in nature, which implies that it dissolves well in polar solvents like THF and liquid ammonia, and poorly in non-polar solvents like hexane. n the other side, cobaltocene is covalent in nature and possesses sandwich type structure like ferrocene, which means that it is non-polar and therefore dissolves well in non-polar solvents like hexane ( like-dissolves-like -rule). b. The problem here is the TiCl 4, which is a very strong Lewis acid. The reaction of TiCl 4 with THF is fairly exothermic and produces a Lewis acid-lewis base adduct, TiCl 4 (THF) 2, which is not very soluble in THF. The use of toluene as solvent prevents this reaction to occur. However, the reaction is going to be a little bit slower than normal because KCp possesses a limited solubility in toluene as well. THF is not a problem in the reaction of NiCl 2 or CoCl 2, because these compounds are much weaker Lewis acids. Side of that, most of the time the hexammine complexes are used anyway. c. Ferrocene exhibits only one singlet in the 1 H-NMR spectrum at =4.15 ppm since all protons are equivalent. The main reason why the signal shows up much lower than the aromatic protons is that ferrocene has a much higher -electron density than benzene which results in a much higher shielding. d. The three peaks are due to the asymmetric Cp-Fe-Cp (478 cm -1 ), an out-of-plane bending mode of the cyclopentadienide ring (814 cm -1 ), and the sp 2 (C-H) stretching of the cyclopentadienide ring (3077 cm -1 ).

12 9. a. The degree of unsaturation is D.B.E.= (2* )/2 = 5 b. The most important peaks in the infrared spectrum are (in cm -1 ): ( (CH, sp 2 )), ( (CH, sp 3 )), 2732, 2824 (CH), 1700 (C=, conj. aldehyde), 1606 (C=C, aromatic), 1385, 1455 ( (CH 2, CH 3 ), bend), 846 (oop, para-subst.). c. The proton spectrum shows six signals as shown in the table below. (ppm) Multiplet Integration 9.96 s d d d m d 6 The signal at =9.1 ppm is due to an aldehyde proton, while the two doublets at = 7.29 and 7.79 ppm are indicative of a para-substituted benzene ring. The doublet at = 2.54 ppm is due to a methylene group next to a CH-function and a weakly deshielding group. The multiplet at = 1.90 ppm is the center of an isobutyl group. Finally, the doublet at = 0.90 ppm is due to the two methyl groups of the isobutyl group. d. The 13 C-NMR shows eight signals for eleven carbons total, which means that there is some kind of symmetry in the molecule. The signals at = 192 ppm is due to a conjugated aldehyde). The signals at =134 and 149 ppm are quaternary carbon atoms and the tall signal at = 130 ppm is due to two CH functions in the benzene ring, indicating a disubstitution on the ring. The signal at = 51 ppm is a result of a methylene function, while the signal at = 30 ppm is another CH-function and the signal at = 22 ppm is due to a methyl group. e. Based on the discussion above, the compound Z is p-(isobutyl)benzaldehyde. CH