last updated Wednesday, November 05, 2014

Meeting 9 (Condensed Key)

1. a. Water acts as an acid in the reaction mixture because it reacts with the amine, which is a base. The resulting ammonium salt does not act as a nucleophile anymore (because it does not have a lone pair on the nitrogen atom anymore) but the hydroxide does act as a completing nucleophile in the reaction. :-(

b. Three equivalents of diethylamine are used in the experiment because two of them are needed in the reaction (one acts as nucleophile and the second one as a base). The third equivalent is used to react with the water in the system and also speed up the reaction.

c. Thus, 1.2 g of the anilide is equivalent to 6.09 mmol (=1.2g/197.66 g/mol). The student will need to use 18.3 mmol of the amine, which is equivalent to 1.89 mL (=18.3 mmol*73.14 g/mol/0.707 g/mol). If the reaction proceeds according to plan, the student should see the formation of a precipitate (H₂NEt₂⁺Cl^-).

d. The extraction with water aims to remove the ammonium salt and the unreacted diethylamine, which are both water-soluble.

e. The extraction with 3 M hydrochloric acid transfers the lidocaine into the aqueous layer because the diethylamine group was protonated (R=2,6-(CH₃)₂C₆H₃NHCOCH₂-).

The unreacted chloroanilide is signifcantly less basic and remains in the organic layer at this point.

f. The addition of potassium hydroxide to the aqueous layer from e. causes the lidocaine to be released in an acid-base reaction.

The pH-value of the mixture should be above pH=10 which has to be determined using pH-paper. The lidocaine initially forms a yellow oil on top of the solution and solidifies upon standing.

g. The NMR sample is submitted in CDCl₃. The desirable concentration is 50-100 mg/mL.