1. a.

1.b.

mw(A) = 174.63 g/mol

mw(B) = 120.15 g/mol

mw(P) = 240.28 g/mol

Hence,

B: 2.37 mL = 0.02 mol

A: 1.75 g = 0.01 mol

Compound A is the limiting reagent in the reaction. We expect 0.085 mol of the product, which means 2.04 g of P.

1.c.

d. The color the higher degree of conjugation causes a red shift of the absorbance peaks. The compound is probably orange in color.

e. An increased conjugation causes a further shift of the absorbance to higher wavelengths. Hence the compound is orange-red in color.

f. Sodium bicarbonate is a weak base compared to potassion carbonate and potassium hydroxide. Since ammononium salts are weak acids, they will not get deprotonated, which means that the reaction will not take place in the desired fashion.

2. a.

The chosen solvent is non-polar the polar compounds in this mixture (B abd P) will not move much up the plate, which means that they will be poorly separated. The aldehyde, which is less polar, will have a much higher Rf-value. Since the reaction is incomplete, the sample lane should show three spots.

b.

Ethyl acetate is much more polar than hexane, which means that all polar compounds will move higher up the TLC plate. Since the reaction is completed, the amount of compound A in the sample should be very small (the diagram does not show a spot for A in the sample lane).

c. Since compound A and compound P are colored, these compounds can be observed directly. Compound B is a colorless liquid which can be detected using UV-light. Ceric staining is not required here.

3. a. The most important peaks to look for in the IR spectrum is the imine function (C=N) at 1630 cm^-1, an dthe presence of the two NO₂ peaks around 1350 and 1540 cm^-1.

b. The H-NMR spectrum should show a peak around 8-8.5 ppm for the imine proton, a singlet at 2.3 ppm for the methyl group and four doublets, a singlet and a triplet in the aromatic range (6.5-7.5 ppm).

c. Twelve signals should be observed for the product molecule in the C-NMR spectrum. Ten of them are in the aromatic range (110-140 ppm), one of them in the alkyl range (~25 ppm) and one around 160 ppm (imine carbon)

4.

5.

a. The most common impurity in aromatic aldehydes is the corresponding carboxylic acid due to the oxidation in air (aldehyde is a liquid, the acid is a solid) . The acid can be removed using a sodium bicarbonate solution.

b. The acid would neutralize some of the potassium hydroxide which releases the amine from the ammonium salt. The yield would decrease accordingly.

6. a.

b. In the experiment, three equivalents of the amine are used in order to form the lidocaine from the chloroanilide. The first equivalent ends up in the compound, the second one is used as a based to scavenge the byproduct (HCl) and one in order to drive the reaction towards the product.

c. Hydrochloric acid is used to protonate the lidocaine and make it water soluble. The unreacted chloroanilide remains in teh organic layer in this step.

d. The isolated product is not very pure since the melting point range is larger than 1 ^oC and the melting point lower than the literature value.

7.a. PhCO⁺ < CH₃CO⁺ < NO₂⁺

b.

c.

The reaction of ferrocene is much easier because the aromatic ring is very electron-rich (1.2 p-elctrons per carbon atom). Acetophenone does not undergo Friedel-Crafts acylations since it is considered electro-poor.

8. a. The absorption of water is reversible. At lower temperatures, the formation of the hydrate is preferred while at higher temperature the reverse reaction (release of water) is favored to the increase of entropy.

b. THF (like most ethers) also contains peroxides and dissolved gases e.g. oxygen. In order to remove these impurities, usually the solvent is refluxed over sodium metal (and benzophenone) as indicator. The solution turns dark blue when water, peroxides and oxygen are not present anymore.

c. Dry ethers are used in Grignard reactions and other organometallic reactions e.g. synthesis of metallocenes (MCp_x).

d. CaSO₄ (2) < CaCl₂ (6) < MgSO₄ (7) < Na₂SO₄ (10)

9. a. In the lab L-(+)-tartaric acid was used to isolate the (R,R)-form of the amine, which ultimately lead to the (R,R)-form of the Jaconsen ligand. In order to obtain the (S,S)-form of the ligand, the (S,S)-form of the diamine has to be first isolated. This can be accomplished by adding D-(-)-tartaric acid, which forms a stable salt with the (S,S)-form of the diamine. The separation is done based on different solubilities of the diastereomeric salts.

b. The column used in this experiment (HP-5) is not chiral, which means that the the two peaks cannot be the two different enantiomers of the epoxide. The first peak shoudl be assigned to the epoxides, and second one to the aldehyde/ketone, which is formed by the acid-catalyzed rearrangement of the epoxide (the approximately ratio here is 82%:18%).

c. TLC was used to monitor the reaction, and to optimize the solvent mixture for the purification step using flash chromatography.

d. For diagram see reader. The triethylamine solution is applied in order to remove an acidic residue from the column, which minimizes the cid-catalyzed rearrangement of the epoxide on the column.

10. a. D.U. = 5

b. The IR spectrum shows peaks at 3000-3100 cm^-1, which can be assigned to sp²(CH) functions, and also peaks in the range from 2850-2950 cm^-1, which are due to sp³(CH) groups. The peaks at 1610 and 1510 cm^-1 are due to aromatic C=C-bonds. The strong peaks at 1240 and 1025 cm^-1 are due to a C-O-C function (ether?). Finally, the peak at 820 cm^-1 originates from a para disubstituted benzene ring.

c.

d.

e. Structure