NO 2 NH 3 + Cl - (A) Reaction 1 (B) Reaction 2 (C) Reaction 3 (D)

Transcription

1 1. a. The first reaction is a Friedel-Crafts acylation. Note that the incoming acetyl group will preferentially be found in the para-position because of the steric hindrance imposed by the iso-propyl group. The second reaction is a nitration reaction. The incoming nitro group is ortho to the iso-propyl group because the para-position is blocked by the acetyl group already. The last reaction is a reduction with zinc in very acidic conditions which leads to the reduction of both, the nitro group and the carbonyl group at the same time. (CH 3 C) 2 Compound X N 2 BF 4 Solvent 2 N 2 NH Zn/conc. H (A) Reaction 1 (B) Reaction 2 (C) Reaction 3 (D) b. Compound X should be an acid like phosphoric acid. The acid acts as a catalyst to obtain the acylium ion from acetic acid anhydride (see acylation reaction in the lab). + H + H + H 3 C C c. Reaction 2 should be carried out in solvent like dichloromethane, which is able to dissolve the nitronium salts reasonable well (4 mol%) but does not react with it. The strongly ionic character of the nitronium salt makes is poorly soluble in most organic solvents. d. The electrophile in reaction 1, the acylium ion, is a weak electrophile. Thus, it only reacts with activated benzene rings in the first place. In order to have a decent rate of reaction, the reaction mixture is often times heated. n the other side, the nitronium ion is a very powerful electrophile. The reaction mixture has to be cooled to prevent polynitration and other undesirable side reactions. e. Compound D would clearly have the lowest R f -value because it is ionic and would not dissolve well in the eluent and would form very strong bonds to the polar stationary phase. Compound A would have the highest R f -value because it is pretty non-polar, which means that it does not significantly interact with the stationary phase here. f. If the ph-value would be 4-5 instead of strongly acidic, the reaction would lead to different products. The nitro group would not be reduced all the way to the amino function, but the reduction would lead to a variety of other compounds like a hydroxylamine (Ph NHH). In addition, the carbonyl group would not be reduced under these conditions either, which means that the two groups can actually react with each other as well to form an oxime.

2 g. First the limiting reagent has to be found. n A = 3.0 g/ g/mol = mol n AcAc = 5.0 g/ g/mol = mol Thus, compound A is the limiting reagent here (1:1 reaction) and mol of compound B should be formed. n B = 3.2 g/ g/mol = mol %yield= n A /n B *100 %= mol/0.025 mol * 100% = ~79% h. Extra Credit: The main issue is that the Friedel-Crafts reaction is reversible. In many cases a scrambling of the substituents is observed if strong Lewis acids like Al 3 or BF 3 are present. In this particular the reaction would also lead acetophenone and 1-(2,4-diisopropylphenyl)ethanone. acetophenone 1-(2,4-diisopropylphenyl)ethanone 2. The esterification reaction underlies an equilibrium which has a fairly low equilibrium constant (K eq =1-10). Water is one of the products in the Fischer esterification that utilizes a carboxylic acid and an alcohol. PhCH + CH 3 H PhCCH 3 + H 2 The presence of water in the reactants (the acid or the alcohol) would result in a lower yield of the ester. b. The sulfuric acid acts as a catalyst in the Fischer esterification. The strong mineral acid protonates the carbonyl group of the carboxylic acid making it a better electrophile, which is then more susceptible to the attack of the alcohol. H H

3 c. Methanol is the solvent for the reaction and as reactant. The excess of the reactant is beneficial here because it takes advantage of the Le Châtelier Principle. d. Despite the addition of the catalyst, the reaction is still fairly slow. Heating the reaction mixture to reflux increases the rate of the reaction by a factor ~20 ( ) because the temperature is raised by ~45 o C. e. The combined organic layer are extracted with to neutralize acids in the organic layer according to - HC 3 + RCH RC - + H 2 + C HC 3 + H 2 S 4 HS 4 + H 2 + C 2 The extraction is completed when 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). The experimenter has to vent the extraction vessel more frequently to release the pressure build-up caused by the carbon dioxide formation. 3. a. The reaction shown is an aldol condensation (or more accurately aisen-schmidt reaction). These reactions are often carried out using basic catalyst like potassium hydroxide and polar solvents like ethanol. Chloroform would not be a good solvent because the hydroxide ion is going to deprotonate the chloroform as well, ultimately leading to a dichlorocarbene, which can react with the product and the ketone as well. HC 3 + H - - C 3 + H 2 - C 3 :C b. The product shows characteristic peaks at = 1662 cm -1 for the highly conjugated carbonyl group (alkene function and benzene ring). In addition, the compound exhibits a fairly strong peak at = 1604 cm -1 due the newly formed C-C double bond, which is polarized by the carbonyl group. Finally, the stereochemistry of the alkene group can be established by the presence of a peak at = 985 cm -1, which is the characteristic out-ofplane bending mode for trans alkenes. c. Three of the five doublets can be attributed to the arene protons. Two of the signals belong to the alkene protons and can be identified using coupling constants. While arene protons usually exhibit coupling constants of J 3 = ~8 Hz, protons on a trans alkene show a significantly larger coupling constant J 3 = ~15-16 Hz. The -proton is more shifted due to the electrophilic nature of the -carbon atom which can be shown by a second resonance structure. - CH 3 CH 3

4 d. The lack of signals in the 13 C-NMR spectrum is a result of low concentration. The peaks that are most likely visible in the spectrum are the ortho and meta carbon atoms of both arene rings marked with the arrows below. The rationale is that they are more abundant than the other carbons due to symmetry. CH 3 e. The differences in the location for the peak maxima are a direct result of the different solvents used in the measurements because the interaction of the solvent (here: MeH) with the carbonyl group changes the orbital energies slightly. In this case, the *- transition will experience a hypsochromic shift while the n- *-transition will show a bathochromic shift. f. The product of the reaction has a large dipole moment than the two reactants. Due to the conjugation, the positive and the negative charge are more separated in terms of distance. A Spartan calculation (AM1) reveals that acetophenone has a dipole moment of =2.93 D, p-methoxybenzaldehyde =2.78 D and the product = 3.95 D, which is consistent with this picture. 4. a. In the second step of the lidocaine synthesis, 2,6-xylidine is reacted with -chloroacetyl chloride and sodium acetate to yield the chloroanilide. NH 2 HN + + NaCH 3C + Na + CH 3 CH b. The solvent used in this reaction is glacial acetic acid which is 100% acetic acid. This is necessary because -chloroacetyl chloride is very sensitive towards water (hydrolysis). The solution is usually more or less purplish in the beginning and changes to a light purple or reddish throughout the reaction. c. The addition of sodium acetate solution to the reaction mixture causes the chloroanilide to precipitate as a white solid. In acidic solution, the NH function of the amide is protonated. The acetate removes this proton. The neutral chloroanilide is not very soluble in polar solvents anymore and therefore precipitates out.

5 + H 2 N HN + CH 3 C - + CH 3 CH soluble in glacial acetic acid insoluble in glacial acetic acid Sodium hydroxide cannot be used here because it would preferentially react with the acetic acid and also deprotonate the unreacted 2,6-xylidinium salt. d. The reagent bottles should remain closed to minimize the uptake of water from the air. Glacial acetic acid is hygroscopic. In addition, -chloroacetyl chloride reacts with the water to form the corresponding carboxylic acid, which will not yield the amide under these conditions. H + H 2 + H The moisture sensitivity can be seen immediately when the bottle is opened because white fumes are visible (=H gas). 5. a. First, the student probably observed the melting point of the benzylic alcohol and not the boiling point. Secondly, the student should report a melting point range and not just one number, which does not tell the reader which value (upper limit, lower limit or average) he refers to. Due to the lack of this information it is pretty much impossible to make any definitive statements about the purity of the compound. b. The oxidant in the oxidation of the benzylic alcohol was dimethylsulfoxide. Hydrobromic acid was a catalyst and not an oxidant. The temperature cannot be above 100 o C if a water bath is used as a heat source. c. The optical purity is given in percentage and not in degree. The student stated the corrected optical rotation of his ligand in the text with the wrong sign! Taking all these things into account, the optical purity would be 95%, which is lower than it should be according to the literature (>98%). d. The alkenes used in the lab are either mono- or 1,1-disubstituted. As a result, only one stereocenter is created in the reaction (R or S). The statement implies the presence of two stereocenters, (R,R) or (S,S), which is incorrect. If two stereocenters are created, there would be four chiral compounds in the mixture, which should show up at four different retention times in the GC graph.

6 e. The potassium carbonate deprotonates the ammonium function and does not protonate the amine function in this step. Two equivalents of ligand are added in the reaction and not equimolar amounts to ensure the double displacement to occur. 6. a. The extraction with saturated sodium chloride solution is meant to remove the majority of the water in the organic layer. Thus, the amount of sodium sulfate can be kept to a minimum here. Larger amounts of sodium sulfate would be a waste and also diminish the yield because the polar alcohol absorbs on the drying agent as well. b. Water is a big problem in the Grignard experiment because it destroys the Grignard reagent and also stops the formation of new reagent due the formation of MgBr(H) that coats the Mg-surface. PhMgBr + H 2 Ph-H + MgBr(H) Even though the glassware might look dry, the polar silica surface absorbs water which can interfere with the formation of the Grignard reagent. This surface bonded water is removed with a heat gun, an industrial strength hair dryer. c. The extractions are carried out at room temperature in Chem 30CL. Thus, the compound should exhibit a high solubility in the solvent at room temperature in order to be a good solvent for extraction. However, for recrystallization the solubility at room temperature should be fairly low in order to recover a significant amount of the product in this procedure. Since a solvent cannot fulfill both requirements, it can only be used in either technique. d. The solution superheated which ultimately lead to bumping or heavy foaming. The solvent got in contact with the hot hotplate which caused it to ignite. The student should have added a boiling stick or spin bar to agitate the solution which allows for a smoother boiling. In addition, the hot plate setting has to be lower as well because ethyl acetate and petroleum ether have relatively low boiling points and are both very flammable. e. The dielectric constant of a solvent is a bulk quantity and describes how the compound behaves in an electric field, which provides some idea about the polarity of the solvent. However, the eluotropic strength depends highly on the exact interaction of the solvent with the stationary phase. Generally, solvents that have oxygen or nitrogen atoms are stronger in terms of their elution power because of their ability to form strong hydrogen bonds with the stationary phase. Thus, diethyl ether has a higher eluting power than chloroform despite its lower dielectric constant. f. Ethers are more or less hygroscopic and often contain peroxides as well. Regular drying agents like anhydrous sodium or magnesium sulfate are not good enough to dry ethers to this high degree and they do not remove peroxides either. The best way to get really clean ethers is usually to use sodium metal and benzophenone. The mixture turns dark blue when the impurities are removed from the ether. The water is destroyed chemically here (2 Na + 2 H 2 2 NaH + H 2 ) and the benzophenone forms a ketyl radical that is dark blue in color.

7 g. Extra Credit: C-200 is used as catalyst for the formation of isoxazolines in this course. It is a partially dehydrated from of barium hydroxide, Ba(H) 2 *2 H a. Simple metal carbonyl i.e. Ni(C) 4 or Cr(C) 6 are non-polar, mononuclear species. Nickel tetracarbonyl forms a tetrahedral structure while chromium hexacarbonyl exhibits an octahedral structure. Due to the low polarity the molecules experience only weak intermolecular forces, which means that they are either liquids or easy to sublime solids. b. The two bond modes are shown below. Note that the C ligand binds via the carbon atom and not via the oxygen atom. The -bond is formed by the interation of the lone pair of the carbon with an empty orbital of the metal. The -bond is usually a back-bond of filled d-orbitals of the metal into the anti-bonding orbital of the carbon-oxygen -bond. c. The starting material here is the Re 2 7 or KRe 4, which is reacted with carbon monoxide. The C acts as ligand and as reducing agent here. Re C Re 2 (C) C 2 Re 2 (C) 10 forms a double-octahedron where each Re-atom has five carbonyl ligands attached. Since this would only lead to 17 VE for each Re-atom, an additional Re-Re bond is formed to reach the magic number of 18 VE. C C C C Re C C C Re C C C d. The reactants in this process which was developed by Monsanto are methanol and carbon monoxide which are converted into acetic acid using cis-[(c) 2 RhI 2 ] - as catalyst. I C Rh I C CH 3 H + C CH 3 CH The reaction occurs via an oxidative addition of CH 3 I, the C insertion into the Rh-CH 3 bond, the addition of C and the elimination of CH 3 CI, which is subsequently hydrolyzed to form acetic acid.

8 8. a. The buffer used in this reaction is mainly Na 2 HP 4, and not NaH 2 P 4 or Na 3 P 4! The buffer is necessary to reduce the decomposition of the sensitive epoxide. It also reduces the chlorination of the alkene. Finally, the catalyst does not decompose as fast either under the given ph-conditions (ph= ~11). b. The epoxidation will not work very well in hexane. The catalyst and the active oxidant are too polar to dissolve in a non-polar solvent like hexane. The reaction would be very slow or not take place at all in the desired fashion. c. The minimum amount of 100 mg is needed because the TN of the catalyst is not very high. Some of the catalyst decomposes during the reaction, which slows down the reaction ultimately. The reaction should be done in about two hours in order to be able to work it up as well before leaving. It is not advisable to downscale the amount of alkene because the yields are generally not very high. d. Running the reaction over night is not a good idea. nce the formation of the epoxide is completed, the decomposition of the epoxide is the predominant reaction. As a result, the epoxide hydrolyzes and forms a diol, which dissolves much better in the aqueous layer. The final yield for the reaction would be very low. e. The chosen mobile phase (dichloromethane) is too polar for the epoxides generated in this course. The high R f -value is a very good indication for that. The epoxide and the alkene separate very poorly under those conditions. The observed spot is probably due to the alkene and not due to the epoxide since most of them are UV inactive. As a result, he cannot make any conclusions about the presence of the epoxide. f. The sweet smell of the crude might be an indication for the formation of the epoxide. However, wetting the column with dichloromethane is a bad move since the epoxide is less polar than the mobile phase. Since he did not pretreat the column with triethylamine solution, he also has an increased amount of rearrangement product in the final product. Using the dichloromethane as eluent is also a poor choice because there will be very little separation between the epoxide, the alkene and the aldehyde. g. As expected, the epoxide rearranged during the flash chromatography step. This is why the sweet smell disappeared. The sample isolated is probably the aldehyde (or ketone), which would only show one signal in the GC spectrum since both the aldehyde and the ketone are achiral. The signal appears a little later than the epoxide as well because the aldehyde (or ketone) has a slightly higher boiling point than the epoxides. h. Extra credit: The peak at m/z=105 is due to a benzoyl cation (Ph-C + ). Thus, the compound that he isolated in the end is probably acetophenone.

9 9. a. The degree of unsaturation is 8 (=(2* )/2). b. The most important peaks in the IR spectrum are: (CH, sp 2 ), (CH, sp 3 ), 1664 (C=, conj. ketone), 1376, 1456 ( (CH 2, CH 3 )), 482, 509 (Cp-Fe-Cp stretch) c. There are only five signals for a total of 22 hydrogen atoms in the 1 H-NMR spectrum which is indicative of a high degree of symmetry in the molecule. The triplet at =1.01 ppm (3H) is due to a methyl group next to a methylene group. The multiplet at =1.42 ppm (2 H) is due a methylene group next to a several CH x -functions. The triplet at =2.64 ppm (2H) is a methylene close by a weakly electron-withdrawing group i.e. carbonyl function. The signals at =4.48 ppm (2H) and =4.77 ppm (2H) are due to protons on a cyclopentadiene ring (see ferrocene). d. There are seven signals for a total of eighteen carbon atoms. Thus, the molecule is very symmetric. The carbon spectrum exhibits two CH 2 groups (18, 42 ppm), two CH groups (71 and 73 ppm), one CH 3 groups (14 ppm) and two quaternary carbon atoms (80, 204 ppm). e. The compound is 1,1'-dibutyrylferrocene. Fe