lidocaine part2

lidocaine part2 - Experiment 4B Preparation of Lidocaine...

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1 Experiment 4B: Preparation of Lidocaine from α -Chloro-2,6-dimethylacetanilide and Diethylamine INTRODUCTION This step of the synthesis involves the reaction of α -chloro-2, 6- dimethylacetanilide, prepared in the previous part (Experiment 4A) and used without further purification, with excess diethylamine, (CH 3 CH 2 ) 2 NH. N H CH 3 CH 3 O Cl N H CH 3 CH 3 O N α -chloro-2,6-dimethylacetanilide diethylamine Lidocaine (4) + (CH 3 CH 2 ) 2 NH 2 Cl diethylammonium chloride 2 (CH 3 CH 2 ) 2 NH Diethyl amine serves three roles in this reaction: (a) as a nucleophile to displace chloride anion from –CH 2 Cl in the α -chloro-2, 6-dimethylacetanilide by an S N 2 type reaction; (b) as a base to absorb the HCl that is formed in the nucleophilic substitution reaction leading to lidocaine; and (c) as a solvent for the reaction. The reaction of α - chloro-2, 6-dimethylacetanilide with diethylamine represents another example of selection reaction because nucleophilic substitution occurs exclusively at the carbon alpha to the carbonyl. Nucleophilic attack by the amine on the carbonyl carbon of the amide group is not favored relative to reaction at the carbon alpha to the carbonyl, which bears the chlorine atom. If substitution were to occur at the carbonyl carbon, it would disrupt the resonance associated with the amide linkage. Examination of equation (4) shows that a minimum of two moles of diethylamine is required for each mole of α -chloro-2, 6-dimethylacetanilide used. One mole is needed for the S N 2 reaction and a second mole absorbs the HCl that is liberated in the substitution reaction; diethylammonium chloride, (CH 3 CH 2 ) 2 NH 2 Cl , is formed along with lidocaine, and no HCl is ever formed directly in the reaction. The mechanism for the substitution reaction between an alkyl halide and an amine may be found in lecture textbooks and/or discussed by your instructor. The reaction mixture is heated at reflux to a temperature of 60-65°C, which is above the boiling point of diethylamine (bp 55°C). As the reaction proceeds, the
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2 formation of a salt, diethylammonium chloride, is observed. After the reaction has refluxed for about 60 minutes, excess diethylamine is removed by evaporation. The compounds remaining in the conical vial are: (a) traces of diethylamine, (b) unreacted α - chloro-2, 6-dimethylacetanilide, (c) diethylammonium chloride, and (d) lidocaine, the desired product. To isolate lidocaine, the reaction mixture is extracted with several portions of aqueous HCl solution, which converts lidocaine into its water-soluble hydrochloride salt. Diethylammonium chloride, present from the initial reaction and also formed from the reaction between aqueous HCl and traces of unreacted diethylamine, is soluble in aqueous HCl solution. On the other hand, unreacted α -chloro-2, 6- dimethylacetanilide is insoluble in aqueous HCl and remains a solid that is removed from the acidic aqueous solution by vacuum filtration. N
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This note was uploaded on 12/13/2010 for the course MATH 251 taught by Professor Unknown during the Spring '08 term at Simon Fraser.

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lidocaine part2 - Experiment 4B Preparation of Lidocaine...

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