The reaction of hydroxylamine hydrochloride with sodium acetate leads to the formation of hydroxylamine, which subsequently reacts with the ketone to form a mixture of ketoximes. The anti-form is usually the major product here.
b. The other product is the syn-form, where the OH group and the larger R-group are on the same side of the C=N bond. In this case, this is the minor product due to the steric hindrance of the methyl group in ortho-position.
c. Even though the unreacted ketone can probably be separated from oximes by recrystallization, the better way would be column chromatography, which would also allow for the two oximes to be separated from each other. Typically, alumina is used as stationary phase here, and a medium polar solvent like ethylacetate. Under these conditions, the unreacted ketone would elute first, and the two ketoximes later.
d. The oximes are more polar than the ketone due the presence of the OH group, thus travel less up the plate than the ketone. Usually, they cannot be separated on silica.
e. The carbonyl peak should disappear (~1680 cm-1) and three peaks appear at ~3200 cm-1 (OH), ~1620 cm-1 (C=N) and ~920 cm-1 (N-O).
f. Oximes are used as starting materials in the Beckman rearrangement to form amides.
Note that the R-group trans to the OH group migrates to the nitrogen atom.
2. a. The (S)-a-Methylbenzylamine clearly the best choice here, since only one set of diastereomeric ammonium salts would be formed here. The 1,2-diaminocyclohexane would have to be resolved first itself, and even then a more complex mixture of different diastereomeric salts would be formed.
b. [a] = a/(c*l) = -2.0o/(0.055 *0.5) = -72.7o
optical purity = (-72.7o)/(-75.0o)* 100% = 97%
c. Yes. The concentration is not so much different (3% vs. 5.5%) and the polarity of the solvent is similar as well (CH2Cl2 vs. CHCl3).
d. Theoretical Yield: 0.35 * 12.0 g = 4.2 g
Yield recovery: 3.6 g/4.2 g *100 % = 85.7 % = ~86%
3. a. The use of sodium metal is not a good idea since alkali metals react violently with halogenated solvents. Alkali metals initiate radical reactions that lead to explosions. The argument that the compounds have a similar boiling point is completely irrelevant in this context.
b. This is not a good idea either since this process is reversible, which means that a certain amount of water remains in the solution (see reader).
c. The student probably overheated the sample which lead to oxidation of the alcohol, the elimination of water from the product, and the formation of polymers due to the condensation of the unreacted formaldehyde with the elimination product.
d. The solvent should dissolve the sample easily, but not really interact with it, which would cause shifts in the peak maxima. The solvent should be transparent in the range to be measured. And finally, it should not react with the cuvette material.
e.
e310= A/(c*l) = 0.9/ (5*10-3 M * 1 cm) = 180 /M*cm
e495= A/(c*l) = 0.15/ (5*10-3 M * 1 cm) = 30 /M*cm
f. Sodium bicarbonate is a weaher base and also a weaker nucleophile than sodium hydroxide. Thus, it only reacts with the acids and does not destroy the ester.
HCO3- + H+ ---- > H2O + CO2
g. The higher reactivity is a result of the better leaving group. However, it also reacts with water, which means that the quality deteriorates quickly if not properly handled.
RCOCl + H2O ---- > RCOOH + HCl
4. a. In order to control the temperature, the ester solution was cooled in an ice bath. In addition, the nitration mixture was added slowly to allow the reaction mixture to dissipate the heat.
b.
c. Sulfuric acid is used as solvent for the ester (protonation of the carbonyl function) and as reagent to generate the electrophile (see b.)
d. The meta isomer is obtained in this reaction because the ester group is meta directing group, because it is an electron-withdrawing when bonded to the benzene ring via the carbonyl carbon atom.
e. The direct nitration is not possible, because first an acid-base reaction would occur yielding a NH3+ function, which would strongly deactivate the ring. In addition, this group would be strongly meta directing. In order to prevent this reaction from occurring, the amino group has to be protected using a acetyl function. The nitration can then be performed and the protection group be removed afterwards.
5.
The salicylic aldehyde is the least polar compound in the mixture, thus travels the farthest up the TLC plate. The spot in the middle is probably due to the unreacted phenol from the synthesis of the benzylic alcohol. The benzylic alcohol has the highest polarity and therefore travels the least up the plate.
b. The reaction is completed since the benzylic alcohol disappeared from the sample lane. Only the phenol and the salicylic aldehyde are present in the sample.
c. The polarity of the solvent goes down, thus all spots would move less, particularly the polar compounds.
6.
a. Cis-alkenes generally react faster than trans-alkenes due to the lower degree of steric hindrance. In addition, an electron-withdrawing group like –COOEt makes the alkene a poorer nucleophile, which slows down the reaction as well.
b. The ligand can be used to fine-tune the redox properties and also stabilize relatively unstable oxidation states. In addition, it can also be used to introduce stereochemistry into the reaction i.e. Jacobsen’s catalyst. These things are not or only in very limited way possible for metal ions or oxometallates.
c. A different oxidant has to be used i.e. mCPBA or PhIO because bleach would not be a liquid at –78oC. In addition, the reaction should be conducted under inert gas to avoid the condensation of water.
d. The selectivity would increase because the reaction is kinetically controlled. The lower reactivity means that the system is more discriminative about different pathways. This also means that the reaction would be much slower, which requires longer reaction times.
7. a. Since the molecular mass is odd, the compound contains an odd number of nitrogen atoms. In addition, the compound contains 14 carbon atoms (=15.5%:1.1%).
b. The cluster observed (ratio: 1:2:1) is due to the presents of two bromine atoms.
c. The main factor governing the intensity of a peak is the stability of the radical cation. Tropylium cations (C7H7, m/z=91) and acetyl (C2H3O, m/z=43) are very stable due resonance stabilization. As a result, they are usually relatively intense in the spectrum.
d. Extra Credit:
EI stands for “Electron Impact”, which is a hard form of ionization using higher energy electrons. CI stands for “Chemical Ionization”, which is a much softer technique that uses ionized gases like ammonia or methane.
8. a. The C-F bond is much stronger than a C-Br or C-I bond, and therefore very difficult to break. In addition, fluorinated compounds are relatively expensive as well.
b. The water reacts with the Grignard reagent to form ether-insoluble Mg(OH)Br. The compound precipitates out and covers the Mg-surface. This prevents the formation of any new Grignard reagent, because this insulating layer blocks the electron transfer.
PhMgBr + H2O ---- > Ph-H + Mg(OH)Br
c. The chipped ice is added to control the temperature during the acid-base reaction, which is highly exothermic. The sulfuric acid is added to convert the magnesium salt into the benzoic acid, which is soluble in ether.
PhCOO-MgBr+ + H2SO4 ---- > PhCOOH + Mg(HSO4)Br
d. The next step is the esterification, and water is one of the products, which means that the presence of water would decrease the yield of the ester.
9.
a. The two main parameter that determine the location of a peak in the IR spectrum are the bond strength and the mass of the atoms involved (n~(F/m)1/2). If the bond strength increases, the peak shifts to the left (higher wavenumbers). If the masses increase, the peaks shifts to the right (lower wavenumbers).
C=C > C=C > C-C
2200 > 1650 > 1100
C-H > C-D
~3000 > ~2200
b. Potassium bromide is usually transparent in the range that it measured in the lab (400-4000 cm-1). Dichloromethane is an organic compound that possesses covalent bonds which results in strong absorbance peaks in the IR spectrum as well. It is very difficult to subtract the solvent peak entirely out in the spectrum.
c. The peaks at ~2350 cm-1 originate from carbon dioxide in air. Usually, the background is run after the instrument was sitting there for a while. A drying cartridge in the instrument absorbs the water and the carbon dioxide.
d. NaCl plates: cheap, but brittle and sensitive towards many polar solvents e.g. water
AgCl plates: more expensive, light sensitive, but less sensitive towards polar solvents
10. a. D.U.=(2*7+2-8)/2 = 4
b. Important peaks in the IR are found at ~3120 cm-1 (CH(sp2)), 2950 cm-1 (CH(sp3)), 1720 cm-1 (C=O), 1580 cm-1 (C=C, alkene), 1460, 1385 (d(CH2)), 1190, 1110 cm-1 (C-O)
c. The triplet at d=1.35 ppm (“t”, 3H) is due to a methyl group next to a methylene group. The signal at d=4.35 ppm (q, 2H) is a result of a methylene group next to a methyl group and an oxygen atom. Finally, the signals at d=6.5 ppm (dd, 1H), 7.2 ppm (d, 1H) and 7.6 ppm (d, 1H) indicate some kind of alkene system, either a monosubstituted alkene or a C3H3-unit.
d. The signals at 159 ppm is a quartenary carbon, most likely a the carbonyl function that has already been identified in the IR spectrum. The range from 110-145 ppm contains three CH-functions and one quartenary carbon (smaller one). The signal at 60 ppm is due to a CH2 function connected to an oxygen atom, while the last signal at 14 ppm is due to a methyl group.
e. Based on the discussion above the molecule is Ethyl 2-furoate
