lecture 8a synthesis of lidocaine (step 3). theory i the third step of the reaction sequence is a s...
TRANSCRIPT
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.