OH [H + ] KMnO 4 DME OH

Transcription

1 1.a. Reaction 1 is an elimination reaction which includes a rearrangement via a hydride shift. Reaction 2 is an oxidation reaction that leads to a cis-diol. This reaction was used to qualitatively confirm the presence of the alkene function in the lab (a color change from purple to colorless and brown precipitate are observed) OH [H + ] KMnO 4 DME OH OH (A) Reaction 1 (B) Reaction 2 (C) b. Concentrated sulfuric acid acts as a catalyst in the reaction and is necessary because the alcohol used in this reaction is a primary alcohol. In these cases fairly drastic conditions (conc. acid and high temperatures) are necessary to cause elimination of water here. OH 2 c. The first reason is that the alcohol is primary. The activation energy for the elimination of water is fairly high. Secondly, the reaction is fairly slow at room temperature and higher temperature increases the rate of the reaction. Thirdly, the reaction is entropy driven (alcohol alkene + water). A higher temperature increases G as well. Lastly, the reaction is an equilibrium reaction which means that the fact that the products (water and alkene) are distilled out at this temperature, shifts the equilibrium to the right. d. In order to determine the yield, only the molar quantities of the alcohol and the alkene are of interest. As discussed in b., sulfuric acid acts as catalyst here. n A = 4.95 ml * g/ml/ g/mol = 40.3 mmol n B = 3.27 g/ g/mol = 34.0 mmol Therefore the yield is Yield = 34.0 mmol/40.3 mmol *100 % = 84.4 %

2 e. The boiling point for compound (C) is T b =175±5 o C based on the graph below. f. Compound (A) is an alcohol that exhibits strong intermolecular hydrogen bonds in the liquid phase, while compound (B) is an alkene, where the London dispersion forces are dominating force. Since LD forces are significantly weaker than hydrogen bonding, compound (A) has a much higher boiling point than compound (B). Note that the slightly higher molecular weight of compound (A) is not really the determining factor here. The student can take advantage of this boiling point difference by removing the alkene from the equilibrium by distillation, which increases the yield. g. In order to assess the purity of the compound, the observed refractive index has to be corrected towards the literature conditions (temperature here), not vice versa. 20 n D = 22 n D + (22-20)* = * = This refractive index is more than n= ±0.001 away from the literature value of n =1.4503, which indicates a moderate degree of purity. 20 D h. Extra credit: There are two products observed in the reaction, which are enantiomers of each other. Since there is no factor that favors either product, the mixture is expected to be racemic, which means that it is not optically active.

3 2. a. The Fischer esterification is an equilibrium reaction with a relatively low K eq constant. This means that the reaction would not afford a high yield if the reactants would be used in a 1:1 ratio. The five-fold excess of alcohol serves two purposes. First, the excess of methanol pushes the reaction to the right side, producing more ester. The theoretical yield is about 95% for a five-fold excess and K eq =4, but does not significantly increase if the excess of alcohol is increased (see reader). Secondly, since benzoic acid is a solid, methanol also serves as a solvent in the reaction. b. Concentrated sulfuric acid again serves as a catalyst. It protonates the carbonyl group of the acid and makes the carbonyl carbon a better electrophile, which is now able to react with the weakly nucleophilic alcohol. OH OH c. Reflux means that the reaction mixture is boiled and the vapors formed (in this case the methanol) is condensed in an air condenser or water jacketed condenser. The reaction is carried out at a higher temperature, which increases the rate of the reaction significantly according to the Arrhenius equation (about a factor 20 here!). Even then, the reaction requires reaction times of 1-2 hours to provide a reasonable amount of product. d. Once the reaction mixture is cooled, the reaction basically stops. The addition of water causes a phase separation, because the excess methanol and the sulfuric acid dissolve in the water, which removes them from the organic layer, while the ester and the unreacted benzoic acid are in the organic layer. Note that the reverse reaction is fairly slow at room temperature even with the presence of water. e. The extraction with sodium bicarbonate serves the purpose of neutralizing the acids in the organic layer according to - HCO 3 - HCO 3 + PhCOOH ---- > PhCOO - + H 2 O + CO H 2 SO > HSO 4 + H 2 O + CO 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. Water is one of the products in the reaction, which means that its presence lowers the yield of the ester. In order to minimize water in the system, the student should make sure that his glassware is dry, that the alcohol and the acid are dry (and remain dry by closing the supply bottles) and by attaching a drying tube with CaCl 2 on the top of the air condenser to keep the water out during the reaction.

4 3. a. In order to scale the reaction correctly, all quantities have to be divided by ten. This means that 2.1 g of benzil and 2.1 g of dibenzyl ketone are dissolved in 15 ml of ethanol. A solution of 0.3 g of KOH in 1.5 ml of ethanol is added. The reaction time remains 15 minutes since this is a question of the kinetic and not the quantities. The crude product is filtered and washed with three 1 ml portions of 95% ethanol. b. As in the lab, neither solvent alone is suitable for recrystallization. Benzene dissolves the compound too well over the entire temperature range, while ethanol is a very poor solvent due to the low solubility of the compound even at high temperature. A mixture of ethanol and benzene provides a steep enough solubility curve to recover sufficient product from the recrystallization. In the lab, a mixture of toluene and 95% ethanol was used. c. First off, he probably isolated the correct compound since the melting is in the proper range given in the procedure. However, the broad melting point range indicates the presence of impurities, which makes it necessary to recrystallize the crude product. Which type of impurities is present cannot be stated for sure, but most it is most likely unreacted starting material or an intermediate.

5 4.a. The retention time for compound A is t=4.8±0.05 min and for compound C is t=5.7±0.05 min. b. First off, the spectrum exhibits two peaks that are due to the products formed in the reaction (C and D). Peak B is due to unreacted benzoin and peak A is due to an elimination product. The conclusions that the instructor can draw from this are that the student did not exclude water well enough from the reaction (presence of benzoin!) and that the student did not properly perform the extraction with sodium bicarbonate to remove the acid from the organic layer prior to removing the solvent (presence of elimination product!). c. In order to determine the ratio of the two diastereomers, the areas for peak C and peak D have to be determined: A(C) = 3 mm * 34 mm/2 = 51 mm 2 A(D) = 1.5 mm * 2.5 mm/2 = mm 2 A(total) = mm 2 The percentage of C and D are then given by P(C) = 51 mm 2 / mm 2 * 100% = 96.5% P(D) = 100% % = 3.5% d. The increase in temperature during the run leads to a faster elution of the compounds from the column. This means that the peaks are closer together and peaks C and D are going to overlap since they are fairly close together already. e. The large dilution is necessary because the GC column is a capillary column that has a fairly low capacity. A larger amount of sample would lead to tailing and very broad peaks that are not resolved in the spectrum. Thus, neither qualitative nor quantitative data can be obtained.

6 5.a The stationary phase (SiO 2 ) is polar, while the solvent (dichloromethane:petroleum ether=1:2) is very weakly polar. The first lane shows compound A only, which exhibits a fairly low R f - value due to its high polarity. Compound B, which is medium polar, has a higher R f -value than compound A. The third lane shows four spots. Since the reaction is incomplete, spots for compound A and compound B appear in the appropriate heights. In addition, spots for compound C and compound D appear further up the plate. The spot the furthest up the TLC plate is due to compound D since this compound is non-polar and has very little chance to interact with the stationary phase. b. Since the reaction has a fairly low equilibrium constant (K eq =5), the limiting reagent does not disappear if the reaction is completed. This means that there will be spots for the reactants A and B in the third lane even though the reaction is completed. In addition, all spots exhibit higher R f -values because the polarity and therefore the eluting power of the solvent increased due to the addition of dichloromethane (dichloromethane:petroleum ether=2:1) c. The student usually uses UV-light to visualize the spots. The silica has a fluorescence indicator mixed in that glows light green, and if a compound is UV active, it appears as a pink spot on the plate (long wavelength). For short wavelength, the plate appears dark and the spot glows.

7 6. a. 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. b. The drying process itself is slightly exothermic ( H<0), which means that heat is produced on the surface of the drying agent. If a low boiling solvent like diethyl ether is used, this heat can cause the solvent to boil, which looks like bubbles moving to the top. c. The ranking below is based on the fact that esters generally have higher carbonyl stretching frequencies than ketone. In addition, in camphor the C=O function is part of a smaller ring (five-membered ring) what usually results to a significantly increased stretching frequency. camphor cyclohexanone ethyl acetate (1746 cm -1 ) (1716 cm -1 ) (1743 cm -1 ) d. Since the student only measures the spectrum above =300 nm, only the peaks at =330 nm (5000) and = 480 nm (2000) are of interest. Using Beer s Law, one can find the maximum concentration to be c max = A max / *l = 1/5000*1 = 2*10-4 M where the maximum absorbance is A=1 and the cell length is l=1 cm. The peak with the larger -value is used for calibration (here at =330 nm). A 480 = 2*10-4 M * 2000 = 0.4 With this concentration, the second peak is also well within the desired range of A= e. 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. Giving only one number indicates a very sharp melting point, which is inconsistent with the fact that the melting point is about 2 o C below the literature value. Secondly, providing a percentage error is also inappropriate since this error is highly dependent on which temperature scale is used. f. The signal is due to deuterochloroform (CDCl 3 ). The signal appears as a triplet due to the fact that the carbon atom couples with the deuterium atom which has a spin of I=1. This means that it can assume three spin states I= -1, 0, 1 leading to a triplet because there is only on deuterium bonded to the carbon atom (2nI+1=number of lines).

8 g. Extra Credit: The fact that there are peaks at m/z=250, 252, 254 and 256 indicates the present of three halogens. The relative intensity of these peaks is 1:3:3:1, which means that these three halogen atoms are all bromine atoms. Taking this into account, one can deduce that the the molecule in question is most likely tribromomethane (bromoform). 7. a. Compound V dissolves fairly well in heptanes over the entire temperature range, but very poorly in water. This permits the conclusion that the compound itself is fairly nonpolar (1-1.5 on polarity scale). b. Both water and dioxane exhibit fairly steep solubility curves, which means both might potentially useful. The nature of the impurity (weakly polar) makes dioxane the better choice since the impurity would dissolve better there. On top of that, the compound also exhibits a steeper solubility curve in this solvent, which means that the recovery yield is higher as well. c. In order to dissolve the sample at 100 o C, the student needs 600 ml (=6 g/(1 g/100 ml) of solvent (pretty big waste ). Upon cooling to 0 o C, 0.6 g (=600 ml*(0.1 g/100 ml)) of the compound will remain in solution and 5.4 g (= 6.0 g 0.6 g) of the will precipitate out if the procedure is performed correctly. d. A slow crystal growth leads to better sized and shaped crystals, and most of all, a solid with a higher degree of purity, which is ultimately the goal for this purification procedure.

9 8. Spectrum 1: Compound K Characteristic peaks in cm -1 : (CH, sp 3 ), 1726 (C=O, ester), 1655 (C=C, alkene), 1377, 1437 (CH 3, bend), 1156, 1064 (COC, ester) Spectrum 2: Compound G Characteristic peaks in cm -1 : 3020 (CH, sp 2 ), (CH, sp 3 ), 1783, 1864 (C=O, anhydride), 1379, 1451 (CH 2, CH 3, bend), 1051, 1232 (COC, anhydride) Spectrum 3: Compound I Characteristic peaks in cm -1 : (OH, alcohol), 3086 (CH, sp 2 ), (CH, sp 3 ), 1515, 1613 (C=C, aromatic), 1457 (CH 3, bend), 1034, 1249 (COC, ether) Spectrum 4: Compound B Characteristic peaks in cm -1 : (CH, sp 3 ), 1699 (C=O, ketone), 1448 (CH 2, bend) Spectrum 5: Compound F Characteristic peaks in cm -1 : (OH, acid), 1669 (C=O, conj. acid), 1618 (C=C, alkene), 1366, 1449 (CH 3, bend), 1272 (C-OH, acid), 690, 760 (oop, mono-subst. arene) Spectrum 6: Compound D Characteristic peaks in cm -1 : (CH, sp 3 ), 2698, 2764 (CHO), 1728 (C=O, aldehyde), 1376, 1455 (CH 3, bend) Spectrum 7: Compound J Characteristic peaks in cm -1 : 3043 (CH, sp 2 ), (CH, sp 3 ), 1608, 1517 (C=C, aromatic), 1446, 1378 (CH 3, bend), 710, 760 (oop, mono-subst. arene) Spectrum 8: Compound C Characteristic peaks in cm -1 : (NH 2 ), (CH, sp 3 ), 1606 ( (NH 2 ), scissoring), 1362, 1451 (CH 2, bend), 841 (NH 2, bend)

10 9. a. The compound has five degrees of unsaturation (=(10* )/2=5). b. Pertinent peaks are at 3434, 3350 (NH 2 ), 3020 cm -1 (CH(sp 2 )), cm -1 (CH(sp 3 )), 1682 cm -1 (C=O, conj. ester), 1515, 1599 cm -1 (C=C, aromatic), 1378, 1455 cm -1 (CH 2, CH 3, bend), 1114, 1278 cm -1 (C-O), 742 cm -1 (oop, para-subst.) c. There are eight peaks for eleven carbons total, which indicates some symmetry, most likely in the benzene ring and somewhere in the side chain as well. d. The 1 H-NMR shows two doublets at =6.63 (d, 2H) and =7.86 (d, 2H) ppm, which are indicative of a para-disubstituted benzene ring. The triplet at =4.22 ppm (t, 2H) is a methylene group next to a methylene group and an oxygen atom. The singlet at =4.12 ppm (s, 2H) is due to the primary amine function. The multiplet at 1.76 ppm (m, 2H) is due to a methylene function between a CH 2 and a CH 3 -function. The triplet at =1.01 ppm (t, 3H) is a result of a methyl group attached to a methylene function. e. The 13 C-NMR spectrum shows three quaternary carbon atoms (120, 151 and 167 ppm (C=O)), two CH functions (132 and 114 ppm), two CH 2 functions (22 and 66 ppm) and one CH 3 -functions (10 ppm). f. The compound is propyl p-aminobenzoate. O O NH 2