last updated Friday, October 23, 2015

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. Toluene is used as the solvent in the reaction because it displays a higher boiling point than most of the other solvents that are usually used in the lab. Its low polarity causes the ionic byproduct to precipitate.

c. If 1.20 g of the dry anilide (=6.09 mmol=1.2g/197.66 g/mol) were used, 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).

d. The solution has to boil and a reflux ring has to be observed in the lower third of the reflux condenser. In addition, the formation of a precipiate is usually observed after some time as well.

e. The extraction with water aims to remove the ammonium salt and the unreacted diethylamine, which are both water-soluble. The extraction with 3 M hydrochloric acid transfers the lidocaine into the aqueous layer (=bottom 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.

Since lidocaine displays a pK_a-value of pK_a=7.9, the pH-value of the mixture has to be above pH=10, which has to be determined using pH-paper. Often times, the lidocaine initially forms a yellow oil on top of the solution, which solidifies upon standing in the ice-bath.

g. The NMR sample is submitted in CDCl₃. The desirable concentration is 50 mg/mL. Note that the solution level should be ~5 cm in the NMR tube.