+ 4 R''OH R'' + 4 H 2 R'' R'' R'' O O. Solvent 1 Solvent 2. Reaction 1 Reaction 2 Reaction NaBH4. NaCN. (COCl) 2 (A) (B) (C) (D) O O

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1 1.a. The first reaction is a benzoin condensation that is catalyzed by sodium cyanide. The resulting benzoin derivative is then reduced using sodium borohydride leading to a hydrobenzoin derivative. Reaction 3 is a Schotten-Baumann esterification using oxalyl chloride to lead to a cyclic ester. NaCN NaB4 Solvent 1 Solvent 2 (CCl) 2 Y X Reaction 1 Reaction 2 Reaction 3 (A) (B) (C) (D) b. Sodium cyanide is used instead of thiamine as a catalyst in this reaction. The cyanide ion attacks the carbonyl group of the aldehyde in a nucleophilic addition and leads to the intermediate below. NC c. Reaction 2 requires a fairly polar solvent because of the ionic nature of the sodium borohydride. Like in the lab, methanol would probably be a good choice here because it dissolved the benzoin fairly well as well. d. ne mole of sodium borohydride can reduce four moles of a ketone leading to a tetraalkylborate. Therefore 0.01 mol of NaB 4 are required to reduce 0.04 mol of compound B in theory. R R' B + 4 R R' R R' B R R' R R' owever, compound B also contains a hydroxyl group that will react with sodium borohydride as well doubling the amount of NaB 4 required. In addition, the reaction is carried out in methanol, which implies that the of solvent also reacts with the NaB 4 resulting in a need of a larger amount than 0.02 mol. B + 4 R'' R'' B R'' R'' R'' + 4 2

2 e. In order to determine the yield, first the function of the different compounds has to be assessed. The p-tolualdehyde is the only reactant in this reaction because sodium cyanide is the catalyst. n A = 2.83 ml * (1.019 g/ml)/ g/mol = mol n B = 2.30 g/ g/mol = mol Considering the fact that two moles of aldehyde are needed to form one mole of the benzoin, the theoretical yield is mol of compound B. In reality, the yield is 79.8 % (= mol /0.012 mol * 100%). f. Compound Y in reaction 3 has to be a tertiary base like triethylamine or pyridine in order to neutralize the Cl form in this reaction which would cause the reaction to reverse. In addition, the resulting salt usually precipitates from solution and therefore helps to push the reaction. + Cl Cl - N N g. The methoxy group (X=C 3 ) is a donor group which means that the electrophilic character of the aldehyde group would be decreased. The reaction with the cyanide would be slower and the overall reaction as well. The yield would probably decrease as well since the oxidation will be more of a problem now. h. Extra credit: The use of concentrated sulfuric acid would pose a problem because benzylic alcohols tend to eliminate easier than secondary alcohol because the resulting cation is resonance stabilized.

3 2. a. Fischer esterifications are equilibrium reactions with relatively low equilibrium constants. In order to improve the yield for these reactions, the Le Châtelier principle is applied in many of these reactions. In this case, an excess of alcohol is used to push the reaction towards the right side because the alcohol can serve as reactant and as solvent for the solid (m-toluic acid). b. The sulfuric acid is used as catalyst in the reaction. The carbonyl group of the acid is a relatively weak electrophile which means that the weakly nucleophilic alcohol is not strong enough to attack. The sulfuric acid protonates the carbonyl oxygen (and not the group!), which makes the carbonyl group more electrophilic. c. Despite the addition of the catalyst, the reaction is still relatively slow at room temperature. By refluxing the reaction mixture, the rate of the reaction can be increased significantly (~60 times because the temperature in the system is ~60 o C higher) in this case. Without the reflux, the reaction would proceed very slowly which many student experienced during their esterification in the lab leading to a precipitate upon cooling of the reaction mixture. d. The addition of water after the reaction is completed helps to accomplish a phase separation. The amount of water that has to be added depends often times on the amount of the alcohol left after the reaction. In the lab the addition of 4-8 ml usually was sufficient. owever some students required more because they used more alcohol to start with or their reaction did not proceed as far. The organic layer (usually bottom layer due to the higher density of the esters) contains mainly the ester, some alcohol, some unreacted carboxylic acid and some sulfuric acid, which the aqueous layer contains the bulk of the alcohol and the sulfuric acid. e. The extraction of the organic layer with sodium bicarbonate solutions serves the purpose of removing the unreacted carboxylic acid and the sulfuric acid from the organic layer. The resulting salts are more soluble in water than it the organic layer. RC + C 3 - RC C 2 2 S 4 + C 3 - S C 2 The formation of carbon dioxide is also observed during the extraction in form of bubbles and overpressure in the centrifuge tube or the separatory funnel. The student is done when the formation of the gas ceased. f. The boiling point of the ethyl ester at p=10 mm g is T b =105±5 o C based on the extrapolation (Note: The reader states a boiling point of T b =104 o C at p=10 mmg).

4 g. In order to evaluate the purity of the sample, the observe refractive index first has to be corrected for the temperature difference. n 25 D = n 22 D - (25-22)* = * = (corr.) When compared with the literature value of n 25 D = , one can make the conclusion that the compound is fairly pure since the deviation is within the range of n= ±0.001.

5 3.a. The boiling points of these compounds are: (N) 285 o C, () 220 o C, (P) 279 o C and (Q) 202 o C. Compound Q has the lowest boiling point, which is the result of the lowest dipole moment and lowest molecular weight of the compound. b. The melting points of these compounds are: (N) 172 o C, () 44 o C, (P) 115 o C and (Q) 35 o C. Compound N has the highest melting point. This can be rationalized by very strong intermolecular hydrogen bonds of both hydroxy functions and the high symmtry of the molecule (Carnelley s Rule). c. The pk a -values of these compounds are: (N) 10.9, 11.4, () 9.42, (P) 7.15, (Q) Compound P is the most acidic compound because the nitro goup on the ring helps to stabilize the negative charge of the phenolate ion. d. The solubilities in water are: (N) 80 g/l at 20 o C, () 25.7 g/l at 25 o C, (P) 15.6 g at 20 o C and (Q) 23 g at 40 o C. Compound N is most soluble in water at room temperature, which can be understood in terms of the formation of hydrogen bonds via both hydroxy functions.

6 4.a. First step would be to identify the peaks. The first peak is the solvent, the second one is due to camphor, the third one is isoborneol and the last one is due to borneol. Thus, the ratio of isoborneol to borneol is A isoborneol = 200 units A borneol = 40 units A total = 240 units Percentage(isoborneol)= 200/240 * 100% = 83.3% Percentage(borneol)= 40/240 * 100% = 16.7 % b. The conversion rate is determined by the ratio of isoborneol+borneol and the total area (isoborneol+borneol+camphor). A camphor = 20 units A total = 260 units Conversion rate= (200+40)/260 * 100% = 92.3% Camphor remaining= 20/260 * 100% = 7.7 % c. If the temperature was increased throughout the entire run, all compounds would elute earlier off the column. The problem is that the peaks are already fairly close together under the initial conditions, which means that they are not going to separate under the new conditions anymore even though they should get a little sharper. d. Under suitable conditions, the spectrum should exhibit a total of six peaks for the camphor, isoborneol and borneol since the column is chiral and can distinguish between different enantiomers as well. Recall that racemic camphor is used as reactant in the lab. e. Contrary to common belief, most stationary phases used in gas chromatography are actually high-boiling liquids and not solids. Thus, the term GLC (=gas-liquid chromatography) is more accurate in these cases.

7 5.a. Diethyl ether has a relatively low polarity and many of the compounds synthesized in this lab course are reasonable soluble in it. In addition, it forms a second layer with aqueous solutions relatively easily and is usually the top layer which minimizes the transfers during the extraction. Finally, its low boiling point also makes it fairly easy to remove later on. b. The extraction with saturated sodium chloride solution serves the purpose of removing water from the organic layer. This step is often times performed as a pre-drying step which results in a use of less solid drying agent (i.e. Na 2 S 4 ). This step is particularly helpful if alcohols or other very polar solvents were used in the reaction because they tend to increase the solubility of water in the organic layer. c. There are several reasons why the drying agent disappeared. First, he did not isolate the correct layer during the extraction, which means that he dissolved the MgS 4 in water. Secondly, even if he isolated the organic layer, the MgS 4 would dissolve in the presence of a lot of water. A very small second layer would be observed on the bottom of the container. d. Generally, recrystallization is the method of choice because it is easier to upscale and also much cheaper. owever, it does not lead to good results if the compounds are too similar in polarity. Another reason not to use recrystallization is that the compounds react in an undesirable fashion at elevated temperatures. In both cases, column chromatography is a better tool here.

8 6.a. The letter stands for stretching modes. It is usually accompanied by a subscript s or as to indicate the type of stretching mode, symmetric or asymmetric. The reason why (N) modes of amides are more intense and broader is a result of the amide resonance that increases the N- bond polarity. b. Since the student measures the range from = nm, only the peaks at = 300(15000) and = 460(300) are important. Using the larger of the peaks and the Beer-Lambert Law, one can calculate the maximum concentration to be c max = A max /( *l) = 1/(15000)*1 = 6.67*10-5 M Under these conditions, the second peak would have an absorbance of A=0.02 (=6.67*10-5 *300) which is well below the threshold of A=0.1. This means that the student has to prepare a second solution using the other molar extinction coefficient. C 460 = 1/(300)*1 = 3.33*10-3 M c. The energy gap of between non-bonding and *-orbitals is smaller than the energy gap between and *-orbitals. The wavelength is inversely proportional to the energy gap ( =h*c/( E)). The *-transition is allowed based on spectroscopic rules, while the n * transition is forbidden resulting in a low intensity peak. d. The boiling point of benzil is 347 o C, which is significantly higher than the boiling point of the solvent used in the reaction (ethyl acetate, b.p.= 78 o C). The vapor pressure of benzil at the boiling point of ethyl acetate is significantly less than 1 mm g, which makes it very unlikely that the benzil evaporated during the solvent removal. Thus, this statement is clearly wrong. e. Tetraphenylnaphthalene (TPN) exhibits a double-melting point probably due to a change in packing after the heating. The X-ray structure of the compound crystallized from isopropanol exhibits two slightly different molecules in the structure (ratio~ 2:1). It is very likely that the system equilibrates better if the compound is in the liquid form and the packing becomes a little tighter, which results in a higher melting point after the first melting process. f. The term refers to a technique that is frequently used for very reactive species that are either too difficult to isolate or unstable. This approach was used in almost every reaction i.e. the formation of the enolate in the Aldol condensation, the formation of benzyne in the Diels-Alder reaction, the generation of thiamine in Benzoin condensation, etc. g. A homogeneous solution is easier to control in general. A progress of a heterogeneous reaction depends on many more parameters i.e. size of particles, mixing speed, cooling, etc. A homogenous reaction is also easier to monitor using TLC because all reactants and products are in solution. h. Extra Credit: The ATR correction is necessary to correct for the differences of penetration depth of different wavelengths. Generally, the peaks on the left side of the spectrum will be smaller because the penetration depth is less there. Thus, the () peak and the (C, sp 3 ) peak will be faily small in size.

9 7. a. Compound V exhibits the highest absolute solubility in hexanes and a fairly decent increase in solubility in ethyl acetate, the compound is fairly non-polar (1-1.5 on polarity scale). b. The best solvent for recrystallization of compound V is ethyl acetate. The solubility curve for compound V is the steepest (increase by a factor 12) and the medium polarity impurity should dissolve in EtAc well. In hexanes and water, the compound exhibits a fairly flat solubility curve, which makes these solvent not suitable for recrystallization. c. The student will need 60 ml (=6 g/(10 g/100 ml)) of hexanes to dissolve the samples at ~80 o C. At 0 o C, 3.6 g (=60 ml * (6 g/100 ml)) will remain in solution and 2.4 g of compound V will precipitate (= 6.0 g 3.6 g). d. Boiling stones often times degrade during extended reflux, which means that some of their pieces will be part of the final product unless a filtration of the hot solution is performed. This is usually not the problem with boiling sticks, which often times suck up some of the solution. They are also easier to remove.

10 8. Spectrum 1: Compound K Characteristic peaks in cm -1 : (, broad, carboxylic acid), 1683 (C=, conj. acid), 1607 (C=C, aromatic), 1367, 1532 (N 2 ), 1295 (C-, acid), 731 (oop, ortho-subst. arene) Spectrum 2: Compound A Characteristic peaks in cm -1 : (C, sp 2 ), (C, sp 3 ), 2216 (C N, nitrile) 1475, 1609 (C=C, aromatic), 1382, 1439 (C 3, bend), 862 (oop, 1, 2, 3, 5-tetrasubst. arene) Spectrum 3: Compound C Characteristic peaks in cm -1 : (C, sp 2 ), (C, sp 3 ), 1676 (C=, conj. ketone), 1504, ~1600 (C=C, aromatic), 1382, 1427 (C 3, bend), 837 (oop, para-subst. arene) Spectrum 4: Compound J Characteristic peaks in cm -1 : 3403 (N, sec. amine), (C, sp 2 ), (C, sp 3 ), 1488, 1584 (C=C, aromatic), 1386, 1450 (C 3, bend) Spectrum 5: Compound E Characteristic peaks in cm -1 : 3064 (C, sp 2 ), (C, sp 3 ), 1673 (C=C, alkene), 1377, 1450 (C 2, C 3, bend), 877 (oop, 1,1-disubst. alkene) Spectrum 6: Compound F Characteristic peaks in cm -1 : 3028 (C, sp 2 ), (C, sp 3 ), 1717 (C=, conj. ester), 1639 (C=C, alkene), 1496, 1596 (C=C, aromatic), 1451 (C 3, bend), 1172, 1204 (CC, ester), 983 (oop, trans alkene) Spectrum 7: Compound I Characteristic peaks in cm -1 : (, alcohol), (C, sp 3 ), 1448 (C 2, bend), 1031 (C-, alcohol) Spectrum 8: Compound B Characteristic peaks in cm -1 : 3278, 3372 (N 2, prim. amine), (C, sp 2 ), (C, sp 3 ), 1603 (N2, scissoring), 1496, 1584 (C=C, aromatic), 1453 (C 2, bend), 700, 747 (oop, mono-subst. arene)

11 9. a. The degree of unsaturation is 5 (=(2* )/2). b. The most important peaks in the IR spectrum are: (, alcohol), (C, sp 2 ), (C, sp 3 ), 1736 (C=, ester), 1496, 1603 (C=C, aromatic), 1369, 1454 ( (C 2, C 3 )), 1058, 1186 (CC, ester) and 698, 733 (oop, mono-subst. arene) c. There are only five signals for a total of twelve hydrogen atoms in the 1 -NMR spectrum which is indicative of some kind of symmetry in the molecule. The triplet at =1.20 ppm (3) is due to a methyl group next to a methylene group. The broad signal at =3.62 ppm (1 ) can be attributed to the -function in the molecule. The quartet at =4.19 ppm (2 ) is due to a methylene function attached to an oxygen atom and is located next to a methyl group. The sharp singlet at =5.14 ppm (1 ) is a result of a C-function next to several electronegative groups i.e., phenyl, etc. Finally, the signal group at = ppm (5 ) is a result of a monosubstituted aromatic ring that has no strong group attached. d. There are eight signals for a total of ten carbon atoms. Thus, the molecule possesses a moderate degree of symmetry. The carbon spectrum exhibits one C 2 group (62 ppm), four C groups (73, , 128.6), one C 3 group (14) and two quaternary carbon atoms (139, 174 ppm (C=)). e. The compound is ethyl hydroxyphenylacetate.