During synthesis of Lidocaine in step 1. The options are:
1) It reacts with the primary amine group of dimethylaniline.
2)The product will react with ethanol.
3) It reacts with alkyl chloride group of α-chloroacetyl chloride.
4) It reacts with the acyl chloride group of α-chloroacetyl chloride.
I'm confused with what this reaction does at the start of the process.
You have to use an inert solvent like Dichloromethane and a base to catch released HCl. EtOH is not a good solvent because it also will react with chloroacetyl chloride to form ethyl 2-chloroacetate (attack on alkyl chloride).
-> 3) It reacts with alkyl chloride group of α-chloroacetyl chloride.
Jürgen Wintner Thank you, that makes sense!
Q. During this same step, why does α-chloroacetyl chloride not attack α-chloro-2,6-dimethylacetanilide? What mechanism does it follow?
Q. Also during the second step, why does HCL protonate the nitrogen atom of the diethylamine group of lidocaine preferentially to that of the amido group?
a) The amide nitrogen is protected by the two methyl groups of the aromatic ring. b) The diethylamine nitrogen has a lower pKa than the amide nitrogen.
c) The two ethyl groups are electron withdrawing.
d) The amide nitrogen has a lower pKa than the diethylamine nitrogen.
Q. During step 2, what is the chemical name of the by-product that is washed away at the sink ?
Step 2: Place 4.8 g of N-(2,6-dimethylphenyl)chloroacetamide (provided dry) in a 100 mL one neck, round-bottomed flask with a small white stirrer bar, and add 25 mL of dry xylene, followed by 9 mL diethylamine. Equip the flask with a reflux condenser and bring the stirred reaction mixture to a vigorous reflux on a rotamantle. (Solids should be in solution after a few minutes at reflux.) After 20 minutes, combine approx. equal volumes of salt and ice in an ice bath, and rapidly cool the flask to about room temperature. Filter off the crystalline solid that forms by vacuum filtration, and transfer the liquid filtrate to a 100 mL separating flask. (The solid is water soluble and can be washed down the sink with plenty of water.)
Thank you for your help! I'm in the process of trying different solvents for the experiment and creating an alternative investigation by trying to justify my changes/decisions.
https://tigerweb.towson.edu/jdiscord/www/332_lab_info/summer2016chem332labs/webexpt_lidocaine_b.pdf
Q. During this same step, why does α-chloroacetyl chloride not attack α-chloro-2,6-dimethylacetanilide? What mechanism does it follow?
Amid proton must be deprotonated for reaction which leads to resonance stabilisation!
Q. Also during the second step, why does HCL protonate the nitrogen atom of the diethylamine group of lidocaine preferentially to that of the amido group?
d) The amide nitrogen has a lower pKa than the diethylamine nitrogen.
Q. During step 2, what is the chemical name of the by-product that is washed away at the sink ?
Step 2: Place 4.8 g of N-(2,6-dimethylphenyl)chloroacetamide (provided dry) in a 100 mL one neck, round-bottomed flask with a small white stirrer bar, and add 25 mL of dry xylene, followed by 9 mL diethylamine. Equip the flask with a reflux condenser and bring the stirred reaction mixture to a vigorous reflux on a rotamantle. (Solids should be in solution after a few minutes at reflux.) After 20 minutes, combine approx. equal volumes of salt and ice in an ice bath, and rapidly cool the flask to about room temperature. Filter off the crystalline solid that forms by vacuum filtration, and transfer the liquid filtrate to a 100 mL separating flask. (The solid is water soluble and can be washed down the sink with plenty of water.)
-> Diethylammonium chloride
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