Lecture 8a

Synthesis of Lidocaine (Step 3)

Theory I

• The third step of the reaction sequence is a SN2 reaction on the CH2Cl function

• Diethylamine is moderately strong nucleophile because it is neutral

• 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 reflux is very important to produce any reasonable amounts

of lidocaine

H3C CH3

HNCl

O

H3C CH3

HNNEt2

O

+ 2 HNEt2 + H2NEt2+Cl-

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

CH3

HN

H3C

O

Cl

HNEt2

HN

+ Et2NH2+ Cl-

Lidocaine Diethylammonium chloride

HCl LidocaineKOH

Acid extration soluble in aqeous layer

H3C

O

NEt2HN

H3C

O

HNEt2+ Cl-

CH3 CH3

H2O

Et2NH2+OH-

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

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 (111oC)

NO

Usually a white precipitate forms

3*10 mL

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

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

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

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)=13o

• 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

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)

n(C=O)Anilide

n(NH)

n(C=O)Lidocaine

d(NH) oop

d(NH) oop

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

• Submit NMR sample (50 mg/mL CDCl3)

• Question: Why is the NH peak so far downfield here?

NH

CH2

CH2CH3

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

C=O

CH2CO

CH2CH3

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.