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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.

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Presentation on theme: "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."— Presentation transcript:

1 Lecture 8a Synthesis of Lidocaine (Step 3)

2 Theory I The third step of the reaction sequence is a S N 2 reaction on the CH 2 Cl function Diethylamine is moderately strong nucleophile because it is neutral The chloride ion is a moderately good leaving group in S N 2-reactions – More rigorous conditions are needed i.e., higher temperature and longer reaction time to observe a good conversion rate – 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 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 – 1 st extraction: the water removes the ammonium salt and the excess of the diethylamine – 2 nd 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: pK a =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 2 nd 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? The mixture has to boil Because of its high boiling point (111 o C) NO 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? The bottom layer pH>10 2*10 mL

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(N3-N4)=269.5 pm The short contact is due to an intramolecular H-bond – <(N3-C23-C24-N4)=13 o The amide function is almost planar – d(C23-O2)=122.8 pm The C=O bond is slightly longer than in acetone (121.3 pm) – d(C23-N3)=134.0 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 leading to 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) – (NH)=3214 cm -1 – (C=O)=1648 cm -1 –  (NH, amide II)=1537 cm -1 – Oop (1,2,3)=762 cm -1 Infrared spectrum (lidocaine) – (NH)=3260 cm -1 – (C=O)=1654 cm -1 –  (NH, amide II)=1500 cm -1 – Oop (1,2,3)=766 cm -1 – Note that the (CH, sp 3 ) peak grew compared to the NH peak! (NH) (C=O) Anilide (NH) (C=O) Lidocaine  (NH) oop  (NH) oop

11 Characterization III 1 H-NMR Spectrum (CDCl 3 ) –  (NH)=8.92 ppm –  (COCH 2 )=3.22 ppm –  (CH 2 CH 3 )=2.69 ppm –  (CH 2 CH 3 )=1.13 ppm The main difference is the presence of the two signals due to the diethylamine group Submit NMR sample (50 mg/mL CDCl 3 ) Question: Why is the NH peak so far downfield here? NH CH 2 CH 2 CH 3

12 Characterization III C=O CH 2 CO CH 2 CH 3

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 November 5, 2014 at 5 pm. Any samples that are submitted later will not receive any credit.


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