Condensed Key

1.

a. You will have to choose the pale green powder, because this is the anhydrous form. The white crystals are AlCl3*6H2O, which will be useless for the FC-acylation. Only the anydrous form is a Lewis acid, which works as a catalyst. This form smokes in air, according to

AlCl3 + 3 H2O -----> Al(OH)3 + 3 HCl

b. The acylium ion is a weak electrophil, the nitronium ion is a strong electrophil. After the addition of the first electrophil, the resulting arene system is a poorer nucleophil, which is not strong enough anymore to react with a weak electrophil like the resonance stabilized acylium ion.

c. After hydrolysis of the PhCOCH3*AlCl3 adduct, the mixture contains water insoluble basic Al-salts. These salts are dissolved by conc. HCl according to

Al(OH)3 + 3 HCl -----> [Al(H2O)6}3+ + 3 H2O + 3 Cl-

d. A polar solvent is required to dissolve all starting materials. Dichloromethane is a medium polarity solvent that does not coordinate to the AlCl3. Ether has about the same polarity, but is a much stronger Lewis base and forms coordination compounds with AlCl3. Those compounds can not act as Lewis acid anymore.

e. Amines are bases. They form an Lewis Acid-base-adduct with AlCl3, which ties up the lone pair on the nitrogen atom. The nitrogen has now a formal positive charge and acts as a deactivating group now that directs the incoming electrophil into meta position. The reaction is slower than the FC-acylation of benzene.

2.

a.

1. Water reacts with the Grignard reagent to form benzene and an ether insoluble Mg-salt that coats the Mg surface. The electron-transfer is blocked now and teh reaction can not proceed.

PhMgBr + H2O -----> PhH + MgBr(OH)

2. The esterification is an equilibrium reaction, where water is one of the products. The presence of water (usually in the PhCOOH due to insufficient drying) causes a lower yield of the ester. Show equation.

3. The presence of water lowers the concentration of the nitronium ion. The reaction slows down. Show equation.

b.

1. To convert PhCOOMgBr (after carbonation) to PhCOOH. Also to dissolve unreacted Mg-turnings.

2. It serves as a catalyst to protonate the carbonyl oxygen on the acid and make it a better electrophil for the weak nucleophil methanol.

3. As solvent for the ester (PhC(OH)OMe+) and to generate the nitronium ion.

3.

a. The NMR spectrometer needs a D-atom as a reference for the lock.

b. It is due to CDCl3. (for D: I=1, 2xI+1=3)

c. The spectrum is proton decoupled which increases the sensitivity and simplifies the spectra. The coupling information can be obtained from the DEPT spectrum.

4.

a. TLC is a micro technique. Small spots with small quantities are required in order to get good separation and avoid tailing.

b. The polarity of the compound in regard to the stationary phase is the important criterium here. Therefore, the naphthalene has the highest Rf value (non-polar), then the amide, the alcohol and finally the acid (most polar) with the lowest Rf value.

c. A UV lamp was used (fluoresence indicator on TLC plate)

5.

6.

Pathway 1 is the preferred way. The reaction of the acid with thionyl chloride leads to the acid chloride which is more reaction that the acid itself. Chloride is a much better leaving group than the hydroxy group. The reaction actually proceeds at room temperature within a reasonable amount of time (Schotten Baumann Esterification).

The problem with pathway 2 is that the tertiary alcohol will eliminate water very easily to afford butene, and not (or very little ester). Remember that the elimation for tert. alcohols can be done under much milder conditions than for prim. or sec. alcohols.

7.

a. The compound contains one Cl-atom (see M/M+2 ratio ~ 3:1). One of the main fragments that is cleaved off is m/z=43, which indicates the presence of an acetyl group. Knowing this and the molecular weight, the formula is C8H7OCl (p-chloroacetophenone).

b. m/z=135 is due to the acylium ion (ClC6C4CO+) and m/z=111 is ClC6H4+

8. IR spectrum

wavenumber assignment
2800-2900 sp3 CH
1700 C=O
1600, 1500 C=C (aromatic)
1250, 1040 C-O-C (ester)

Proton NMR

Shift Assignment
6.5-7.0 (2d, 4H) para subst. ring
3.6 (s, 3H) OCH3
2.6 ('t', 4H) X-CH2-CH2-Y
2.0 (s, 3H) X-CH3

DEPT spectrum

Shift Assignment
208 C=O (quartenary)
158, 132 aromatic C (quartenary)
130, 114 aromatic CH
55 OCH3
45 CH2
30 CH3
29 CH2

Answer:

9. There are several acceptable ways and reagents to obtain this compound. Here is one example.