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Lecture 8a

Lecture 8a. Synthesis of Lidocaine (Step 3). Theory I. The third step of the reaction sequence is a S N 2 reaction on the CH 2 Cl function Diethylamine is neutral and therefore a moderately strong nucleophile The chloride ion is a moderately good leaving group in S N 2-reactions

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Lecture 8a

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  1. Lecture 8a Synthesis of Lidocaine (Step 3)

  2. Theory I • The third step of the reaction sequence is a SN2 reaction on the CH2Cl function • Diethylamine is neutral and therefore a moderately strong nucleophile • The chloride ion is a moderately good leaving group in SN2-reactions • More rigorous conditions are needed i.e., higher temperature and longer reaction time to observe a good conversion rate • The reaction usually does not go to completion in the allotted time. However, the reflux is very important to produce any reasonable amounts of lidocaine

  3. Theory II • The presence of water in the reaction mixture poses a significant problem because it reacts with the amine the amide has to be very dry! • In theory, the reaction requires two equivalents of the amine, but for practical purposes three equivalents of the amine are used • The reaction leads to the formation of diethylammonium chloride, which precipitates from the non-polar solution as a white solid

  4. Theory III • Upon completion, the reaction mixture contains lidocaine, some unreacted anilide, the excess amine and the ammonium salt • The separation of these compounds is based on the different solubilities in water and hydrochloric acid • 1st extraction: the water removes the ammonium salt and the excess of the diethylamine • 2nd extraction: the hydrochloric acid moves the lidocaine into the aqueous layer due the protonation of the diethylamine function • The unreacted anilide remains in the organic layer because it is significantly less basic than the amine (lidocaine: pKa=7.9) • The sequence of extractions is very important here! • The lidocaine is recovered by addition of a strong base (KOH) to the aqueous extract from the combined extracts from the 2nd extraction step

  5. Experiment I • Dissolve the dry anilide in toluene • Add three equivalents of the amine • Reflux the mixture vigorously for about 90 min • After cooling the mixture, extract the mixture with water • Why is toluene used here? • Can the student use more? • What does this imply? • Which observations should the student make? • How much water is used? Because of its high boiling point (111oC) NO Boiling! Usually a white precipitate forms 3*10 mL

  6. Experiment II • Extract the organic layer with 3 M hydrochloric acid • Combine the two aqueous extracts and place the solution in an ice-bath • Add 8 M KOH until the solution is strongly basic • Why is this step performed? • How much is used here? • Which layer has to be kept here? • Which pH-value is the student looking for? • How is the pH-value measured? 2*10 mL The bottom layer pH>10

  7. Experiment III • Place the mixture in an ice-bath • Isolate the white solid by vacuum filtration using a fritted funnel or fritted crucible • Wash the solid with water • Press the solid in the funnel • Why is this step performed? • What should be done if the compound does not solidify? • What is a fritted funnel? • Why is it used here? • Why is the solid pressed? Scratch the inside of the beaker The strongly basic solution disintegrates the filter paper To remove the bulk of the water

  8. Experiment IV • Dissolve the crude product in about 10 mL of hexane • Add anhydrous sodium sulfate • Reduce the volume of the solution to 3-4 mL using an air stream • Allow to the compound to crystallize • Isolate the solid by vacuum filtration • Why is the solid dissolved again? • What is the exact procedure here? • Why is the volume reduced? • Which equipment should be used here? The aqueous and organic solution have to be cleanly separated Hirsch funnel

  9. Characterization I • X-Ray Structure • Trans amide configuration (like in the chloroanilide) • The NH and CO functions are opposite of each other • d(N1-N2)=267.8 pm • The short contact is due to an intramolecular H-bond • <(N1-C9-C10-N2)=3.1o • The amide function is almost planar • d(C9-O1)=123.1 pm • The C=O bond is slightly longer than in acetone (121.3 pm) • d(C9-N1)=133.8 pm • Very short C-N bond indicative of a partial C-N double bond • d(O1-H)=214 pm • Intermolecular hydrogen bonding observed the solid leadingto the formation of a chain with alternating orientation of the aromatic ring • The distance is about 10 pm longer than in the chloroanilide resulting in a lower melting point for lidocaine compared to the chloroanilide

  10. Characterization II • Melting point • Infrared spectrum (anilide) • n(NH)=3214 cm-1 • n(C=O)=1648 cm-1 • d(NH, amide II)=1537 cm-1 • Oop (1,2,3)=762 cm-1 • Infrared spectrum (lidocaine) • n(NH)=3260 cm-1 • n(C=O)=1654 cm-1 • d(NH, amide II)=1500 cm-1 • Oop (1,2,3)=766 cm-1 • Note that the n(CH, sp3) peak grew compared to the NH peak! n(NH) d(NH) oop n(C=O) Anilide n(NH) n(C=O) oop d(NH) Lidocaine

  11. Characterization III • 1H-NMR Spectrum (CDCl3) • d(NH)=8.92 ppm • d(COCH2)=3.22 ppm • d(CH2CH3)=2.69 ppm • d(CH2CH3)=1.13 ppm • The main difference is the presence of the two signals due to the diethylamine group • Question: Why is the NH peak so far downfield here? CH2 NH CH2CH3

  12. Characterization III • 13C-NMR spectrum • d(C=O)=170 ppm • d(CH2CO)=58 ppm • d(CH2CH3)=49 ppm • The 1H-NMR and the 13C-NMR spectrum of the lidocaine for the post-lab can be found at www.chem.ucla.edu/~bacher/General/30CL/spectra/lidoH.html CH2CH3 CH2CO C=O

  13. Characterization IV • Mass spectrum (sample has to be submitted for analysis) • Question: The mass spectrum of lidocaine is dominated by a peak at m/z=86. Which fragment can this be attributed to? • The final product has to be submitted to the TA by February 7, 2014 at 5 pm. Any samples that are submitted later will not receive any credit.

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