In our organic chemistry 2 lab i synthesized Lidocaine and also Lidocaine Bisulf

Question

In our organic chemistry 2 lab i synthesized Lidocaine and also Lidocaine Bisulfate.

I have not been able to find the solubility of lidocaine bisulfate in acetone and water. I need to find out what the sensibility of the bisulfate is to these two solvents since i was unable to get the Lidocaine Bisulfate

And finally, why did we synthesize Lidocaine Bisulfate in the first place?

Thank you

Edit: my apologies, the full information is this: After synthesizing lidocaine, it was dissolved in diethyl ether and ethanolic sulfuric acid, then crystals formed and i vacuumed them with cold acetone. Those crystals were, then, "purified" by adding "minimal ammount of water and 20x that of acetone" according to the manual. I, however, was unable to get crystals from this last step where i added water and acetone to the previous crystals. That is why i dont understand if it has anything to do with the solubility in water and acetone. I also dont know why it was necessary to synthesize Lidocaine Bisulfate since i already had synthesized Lidocaine itself.

Explanation / Answer

Lidocaine is isolated in the lab as its bisulfate salt form. The salt thus formed would have very good solubility in water and very low solubility in acetone. Thus a combination of the two solvents gives us crystals of lidocaine bisulfate product. If you do not get crystals after the bisulfate is initially formed by the reaction of H2SO4, prior to recrystallization, scratch the sides of the vessel to induce crystallization.

lidoaine as prepared may have impurities of the starting material or reagents used. In order to get a pure form of lidocaine, we synthesize the salt of it as bisulfate salt of lidocaine which is much easier to purify.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.