Organic Chemistry Reaction Analysis: Claisen-Schmidt Condensation and Epoxidation - Prof. , Lab Reports of Organic Chemistry

Download Organic Chemistry Reaction Analysis: Claisen-Schmidt Condensation and Epoxidation - Prof. and more Lab Reports Organic Chemistry in PDF only on Docsity! 1. a. The first reaction is a Claisen-Schmidt condensation between an aldehyde and a ketone, leading to an -unsaturated ketone (C), which subsequently undergoes epoxidation to form an epoxide (D). The treatment with acid leads to a rearrangement product, a 1,2-diketone (P). The stereochemistry on the alkene is trans, while the stereochemistry on the epoxide is difficult to predict. tBu O O H + KOH R,R-JC/NaOCl Reaction 1 Reaction 2 Reaction 3 [H+] tBu O tBu O O tBu O O (A) (B) (C) (D) (P) b. The first step here is to identify the limiting reagent. Compound (A): nA = (2.00 mL*1.045 g/mL)/106.12 g/mol = 19.7 mmol Compound (B): nB = 2.0 g/100.16 g/mol = 20.0 mmol Compound (A) is the limiting reagent here, which means that 19.7 mmol of compound (C) should be formed in the reaction if the reaction would run in 100% yield. Since the yield is only 90%, the yield is 17.7 mmol or 3.34 g. c. The solvent has to be fairly polar in order to ensure that the potassium hydroxide dissolves. In addition, it should not interfere with the reaction. Like in the lab a solvent like absolute ethanol or methanol would work well for reaction 1. d. An excess of compound (B) has to be avoided in order to minimize the self condensation reaction after the aldehyde is consumed. tBu O KOH tBu tBu O This reaction is usually less favorable because the reaction of ketones is less exothermic and therefore only occurs at a reasonable rates if there is no other electrophile present in the reaction. e. The major product of reaction 1 is a trans alkene, which are generally poor substrates for epoxidation reactions using Jacobsen’s catalyst. In addition, the C=C double bond is connected to an electron-withdrawing group (COCH3), which makes it an electron-poor alkene, and therefore a weaker nucleophile. Both of these features make the alkene a poor substrate for the epoxidation (reaction 2). f. Since the alkene and the catalyst are relatively non-polar, and sodium hypochlorite (=bleach) is ionic, a two-phase system should be used here. Like in the lab, dichloromethane can be used as organic layer and a buffered bleach solution as aqueous layer. g. The easiest way to carry out the reaction is to pour the crude epoxide onto a silica or acidic alumina column without pre-treating it with a base like triethylamine. This would cause the epoxide to rearrange to form the 1,2-diketone, which would be eluted in the end. h. Extra Credit: The large peak at m/z=57 is due to a fragment [C4H9] .+ , also called tert-butyl, which is a fairly stable cation as well. The large shift of the phenolic hydrogen atoms is indirectly an indication for a very strong interaction. e. In order to determine the proper concentrations Beer’s Law is used. Assuming a 1 cm cuvette and a maximum absorbance of A=1, one obtains the maximum concentration to be c = A258/( *l) = 1/(32000*1) = 3.125*10 -5 M The second peak exhibits an absorption of A328 = 3.125*10 -5 M * 7200 * 1 = 0.225 which is within the desirable range of 0.1 < A < 1.0. f. The peak at =1631 cm -1 is due to the stretching of the C=N bond of the imine function, while the peak at =2962 cm -1 is due to the C-H(sp 3 ) of the tert.-butyl groups. g. As discussed above already, the ligand exhibits a very strong intramolecular hydrogen bond. This shows in various spectra and also in its physical properties i.e. solubility. Since the hydroxyl groups are not available for interaction with the solvent and the rest of the molecule is fairly non- polar as well, the solubility in a polar solvent like ethanol is very limited. On the other side, the ligand dissolves fairly well in medium and non-polar solvents like dichloromethane and hexane. Considering the intramolecular hydrogen bond, the “like-dissolves-like” rule still holds. 4. a. The addition of water to the reaction mixture causes often time the organic compounds that have a low polarity to precipitate as a solid or oil. This way, the polar solvents like alcohols and dimethyl sulfoxide are removed in addition to ionic compounds like sodium salts or acids. The organic compounds are subsequently extracted into an organic solvent like dichloromethane, diethyl ether or hexane. b. Most reagents used in the lab are either sensitive towards oxygen or water. For example, a drying agent like magnesium sulfate loses its activity quickly because it absorbs moisture. Many anhydrous solvents like diethyl ether or ethanol also absorb moisture and therefore are not anhydrous anymore, which causes serious problems in many reactions i.e. Grignard reaction. c. The procedure has to be downscaled by a factor of fifty. As a result, the procedure should read something like this: “…5 g of compound G and 2.4 g of compound H are dissolved in 20 mL of methanol and then refluxed for two hours. The mixture is then cooled in an ice-bath and the precipitate isolated by vacuum filtration. The filter cake is washed twice with 2 mL of ice-cold methanol. The final yield is expected to be approximately 5 g…..” The information in bold is the important information that has to be adjusted properly. Note that the reaction time remains the same. d. Even though the wetting of the column is usually recommended prior to applying the sample, it should be done with a non- or weakly polar solvent if a polar stationary phase is used for chromatography. This holds especially true if relatively weakly polar compounds are to be separated. In this case, 10% ethyl acetate in petroleum ether is enough to cause the mono- acylation product to move. Applying pure ethyl acetate to the column prior to applying the crude would block out the majority of the surface side for the acylation products, which means that they would basically just flushed through the column without being separated.  e. Saturated sodium chloride solution is used to remove very polar impurities and most of all water from the organic layer. Often times, solvents like alcohols, etc are used that increase the solubility of water in the organic layer. In essence, the extraction with saturated sodium chloride solution is a pre-drying step, which reduces the need for other drying agents like sodium sulfate, etc. f. The use of CH3OH would be a poor choice because the NMR spectrometer would have a very difficult time to lock the magnetic field. As a result the obtained data would probably be rather meaningless. In addition, the student would see a very large signal for the methanol itself in the 1 H-NMR spectrum, which would make it very difficult to detect the actual compound signals as well. g. The ground glass joints do not fit perfectly together even though they possess a standard size. In order to improve the seal and also to ensure that they can be separated later on, high-vacuum grease is used. However, the grease should be used lightly on the top third of the joint only in order to avoid the contamination of the product and grease to be all over the glassware on the outside. h. Extra credit: In the beginning, the -scale was frequently used for NMR spectroscopy. The -scale starts at =10 ppm and increases towards the right side in the spectrum (thus upfield!). As a result, =7.5 ppm correlates to =2.5 ppm. 7. a. Compound (M) has a larger dipole moment than compound (N). The intermolecular forces are stronger in the case of compound (M) resulting in an increased melting point. The mass difference between these two compounds cannot really account for the higher melting point of compound (M). b. Compound (O) is much more reactive towards water since it is an acyl chloride because the chloride ion is a better leaving group than the methoxide. As a result, the hydrolysis of the acyl chloride occurs relatively rapidly, while the hydrolysis of the ester is fairly slow (at room temperature). In both reactions the compound would be converted into the corresponding carboxylic acid. c. Compound (O) would exhibit a lower Rf-value because it is much more polar due to the presence of the nitro group. The nitro group exhibits a dipolar character (positive charge on the nitrogen atom and negative charge on the oxygen atoms), which allows for a strong interaction with polar stationary phases. Aside of that depending how much water is in the system, the acyl chloride would probably hydrolyze anyway (see c.). d. Hexane would be the best solvent for recrystallization for Compound (N) since it is the least polar. Compound (O) and compound (M) is are too polar due to the nitro groups, which means that they would exhibit a very limited solubility even at higher temperature. 8. a. Cyclopentadiene tends to dimerize relatively quickly at room temperature. As a result, it has to be prepared fresh from dicyclopentadiene by a cracking process. H H ~180 oC H H H H + This process can be understood as a Retro-Diels-Alder reaction. The obtained monomer should be kept at low temperatures (T~-78 o C) until needed in order to avoid redimerization. b. The main difference between cyclopentene and cyclopentadiene is that the resulting anions are very different in stability. While the anion formed from cyclopentene is fairly unstable, the anion formed from cyclopentadiene is very stable since it is aromatic (6 -electron). As a result a relative weak base like potassium hydroxide is sufficient to deprotonate the diene. H H + KOH K + H2O c. The use of the hexamine complex provides an additional driving force for the reaction. [Co(NH3)6]Cl2 + 2 NaCp CoCp2 + 2 NaCl + 6 NH3 THF Aside of the desired cobaltocene, ammonia is formed that leaves the system as a gas, which results in a large entropy increase. Using THF as solvent here causes the sodium chloride to precipitate in addition, which also shifts the equilibrium. d. The difference between these two compounds is that ferrocene has a full valence shell (18 VE), while cobaltocene has 19 VE, which makes it very susceptible towards oxidation. Cobaltocenium (CoCp2 + ) on the other hand are very stable and commonly used as counter ions for large anions. 9. a. The degree of unsaturation is D.B.E.= (2*11 + 2 -13 + 1)/2 = 6 b. The most important peaks in the infrared spectrum are (in cm -1 ): 3294 ( (NH)), 3004-3099 ( (CH, sp 2 )), 2873-2959 ( (CH, sp 3 )), 1714 (C=O, ketone), 1668 (C=O, amide), 1514, 1607 (C=C, aromatic), 818 (oop, para-subst.). c. The proton spectrum shows six signals as shown in the table below. (ppm) Multiplet Integration 9.10 s, br 1 7.41 d 2 7.11 d 2 3.54 s 2 2.30 s 3 2.29 s 3 The signal at =9.1 ppm is due to an amide proton, while the two doublets at =7.11 and 7.41 ppm are indicative of a para-substituted benzene ring. The singlet at =3.54 ppm is due to a methylene group next two weakly deshielding groups. Finally, the two singlets around =2.3 ppm are due to methyl groups on a benzene ring or a carbonyl function. d. The 13 C-NMR shows nine signals for eleven carbons total, which means that there is some kind of symmetry in the molecule. The signals at =204 and 164 ppm are due to a ketone and an amide, respectively (both are quaternary carbons). The signals at =134 and 135 ppm are quaternary carbon atoms and the signals at =120 and 129 ppm indicate a disubstitution on the ring. The signal at =51 ppm is a result of a methylene function, while the remaining two signals are due to methyl groups. e. Based on the discussion above, the compound Z is N-(p-tolyl)acetoacetamide. N H O O