1. a.

Spot 1: R_f=0.1/3.8=0.03

Spot 2: R_f=1.0/3.8=0.26

Spot 3: R_f=1.8/3.8=0.47

Spot 4: R_f=3.0/3.8=0.79

b. The reaction is not completed since there is still alkene present in the product lane (=spot 4).

c. Spot 1: catalyst (very polar)

Spot 2: epoxide

Spot 3: ketone byproduct

Spot 4: alkene (non-polar)

d. The ceric staining method is used for polar compounds that do not absorb light in the near UV range. The staining solution contains molybdate which is pale yellow and Ceric sulfate (yellow). It interacts with polar compounds and turns dark blue upon heating due to the reduction to Mo(VI) to Mo(V).

e. 4-Methyl styrene: C₉H₁₀ (m.w.=118 g/mol), 4-Methyl styrene oxide: C₉H₁₀O (m.w.134 g/mol)

Moles of 4-Methyl styrene: a=0.5 g/(118 g/mol)=4.2*10^-3 mol

Moles of 4-Methyl styrene oxide: b= 0.8 * 4.2*10^-3 mol = 3.4*10^-3 mol

Moles of catalyst: c=0.15 g/(635 g/mol)=2.36*10^-4 mol

Moles of bleach: d=0.07 mol/L * 0.01 L = 7*10^-3 mol

Product isolated = 3.4*10^-3 mol * 134 g/mol = 0.45 g

Catalyst loading = 2.36*10^-4 mol / 4.2*10^-3 mol = 5.6 %

Oxidant to alkene = 7*10^-3 mol/4.2*10^-3 mol = 1.65

2. a. Ferrocene has 18 valence electrons, which is the 'magic number' for transition metal compounds. Nickelocene and cobaltocene have 20 and 19 VE, which makes them much more susceptible towards oxidation.

b. In order to prove which structure is correct, either electrophilic aromatic substitution can be performed e.g. Friedel Crafts-Acylation to form acetyl ferrocene to identify structure A, or a Diels-Alder-reaction using acetylene dicarboxylic ester (MeOOC-C=C-COOMe) as dienophile to form a DA-adduct, which proves that structure B is the right structure. Based on your lab work, you know that structure A is correct.

c. The orange color originates from d-d and LMCT-transitions in ferrocene.

d. The n(Fe-C)-peaks are located around 480 and 500 cm^-1. They are much lower than peaks in most organic compounds because the Fe-atom is much heavier.

e. See reader for equations

f. The higher p-electron density in ferrocene (1.2 vs. 1 in benzene) causes the shift to lower ppm in both spectra. The higher p-electrondensity causes better shielding of the nucleus.

3. Moles of compound = 0.01 g/(120 g/mol) = 8.33 * 10^-5 mol

Concentration of stock solution: c₁=8.33 * 10^-5 mol/0.05L = 1.66 * 10^-3 mol/L

Concentration of diluted solution: c₂=1.66 * 10^-3 mol/L/10=1.66 * 10^-4 mol/L

Absorption expected: A_calc=1.66 * 10^-4 mol/L * 2000 L/mol*cm * 1 cm = 0.333

Absorption observed: A_obs = 0.325

Purity = A_obs/A_calc=0.325/0.333 = 0.975 ~97.5%

4. a. The peak at 1630 and 1609 cm^-1 is in both cases due to the C=N function (imine) in the molecule. The shift to lower wavenumbers is due to the coordination of the nitrogen to the Mn(III), which pulls electron-density out of the C=N-bond as well.

b. The peaks are due to the n(Mn-N) and n(Mn-O) bond in the ligand. The peaks are lower to the involvement of the heavier Mn-atom into the vibrational mode.

c. There are various effects to be considered to explain the observed trends.

- The hydroxy group is a strong donor group and weakens the carbonyl function via resonance.

- The hydroxy function also can form hydrogen bondings, which weakens the the carbonyl function as well.

The ortho substituted compound has the lowest C=O stretching frequency because it experiences both effects, the resonance and the hydrogen bonding. The difference in the two para-subtituted compounds is due to the different strength of donor ability of the OH group and the methyl group.

5.[a] = 3.6^o/(0.025 g/mL *2 dm ) = 72^o

Optical purity = 72^o/73^o = 98.6%

The compound (Mosher's Acid) is used as chiral resolution agent for enantiomers since it has a chiral center.

6. a. The reagent is generated in situ from conc. nitric acid and conc. sulfuric acid. For equation see reader.

b. The activation energy of the mono-nitration and the di-nitration are not much apart from each other. Since the reaction is very exothermic, the reaction temperature has to be carefully controlled by placing the mixture in an ice-bath. Adding the nitration mixture slowly in addition will minimize the formation of the di-nitro compound as well.

c. Sulfuric acid is used as solvent for the ester (protonation of the carbonyl function) and as reagent to generate the electrophile (see a.)

d. The meta isomer is obtained in this reaction because the ester group is meta directing as an electron-withdrawing group.

7. a. For equation see reader.

b. The potassium hydroxide is used to release the free amine from the ammonium salt formed in a. (For equation see reader).

c. Acid chloride are more reactive than other carbonyl compounds. Hence the reaction can be carried out under relatively mild conditions as compared to using the acid. The reactivity also causes problems since the compound hydrolysizes very easily. The presence of water should be minimized by using dry solvents and glassware.

d. Penicillins, peptid bonds, DEET (OFF)

8. a. To neutralize any acidic residue on the silica column. These residues cause a rearrangement of the epoxide to yield ketones or aldehydes.

b. The reaction of hydrogen peroxide and Ce(OH)₃ is exothermic. In addition, H₂O₂ is not very stable in the presence of transition metals and in basic solution. H₂O₂ is dissolved in water.

c. Anhydrous K₂CO₃ is used in the first step in order to dry the basic solution. MgSO₄ or Na₂SO₄ would absorb significant amounts of the target compound.

d. Common impurities in diethyl ether are water and peroxide. They can be removed by treating the solvent with a mixture of sodium and benzophenone.

2 Na + 2 H₂O --> H₂ + 2 NaOH

The mixture is dark blue if the impurties are removed.

9. a. For catalyst structure see reader. We isolated the (R,R)-form using L-(+)-tartaric acid.

b. Area A = 1 am *2.2 cm /2 = 1.1 cm²

Area B = 0.5 cm * 1 cm/2 = 0.25 cm²

Total area = 1.35 cm²

Ratio: A = 1.1/1.35 * 100 = 81.5 %, B = 0.25/1.35 * 100 = 18.5 %

e.e.=81.5 %-18.5 %=63 %

c. For mechanism see reader. Bottom line is that there are competing mechanisms that determine the e.e.-value for a given substrate. If the reaction would only follow the concerted mechanism, only the cis-epoxide should be observed. If a conjugated system is used, the amount of the trans-epoxide increases since the radical intermediate is more stable. As a result the e.e. value decreases for these type of substrates.

10. a. The fact that the molecular mass is odd, the presence of nitrogen is very likely. After some conclusion that can be made from the other spectra, a formula of C₁₀H₁₃NO can be deducted.The degree of unsaturation is 5.

b. Peaks if significance are at 3000-3100 (CH, sp²), 2850-3000 (CH, sp³), 1600 and 1500 (C=C, aromatic), 1450 and 1370 (CH₂, CH₃, bend), 1240 and 1120 (C-O, stretch), 700 and 760 cm^-1 (oop bending, mono).

c. The H-NMR shows two triplets and one doublet in the aromatic range with five protons. This indicates a mono-substituted ring. The two triplets between 3-4 ppm are due to two X-CH₂-CH₂-Y functions in the molecule.

d.

e.

Structure