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I tried to draw the mechanism of the diazepam synthesis. Question: Do I really get the tertiary amine or does dimethyl sulfate methylate the amine twice?

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2 Answers 2

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While amines may undergo multiuple methylations to form quaternary ammonium salts, the nitrogen atom being akylated in the diazepam synthesis is an amide. Amides are different.

The lone pair on the amide nitrogen atom in diazepam is thoroughly delocalized into the carbonyl, into the aromatic ring, and all the way to the imine (see image below - some resonance contributors not shown). Due to this delocalization, that amide nitrogen atom is not nucleophilic and will not be methylated a second time. I would be much more worried about alkylating the imine nitrogen atom, which is made more nucleophilic by delocalization.

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This delocalization is the reason why you need to add a base in order to do the methylation step in the first place. The amide in N-demethyldiazepam is likewise a poor nucleophile due to delocalization.

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However, once a base deprotonates the amide, the amide becomes the better nucleophile due to the concentration of negative charge. Even though one of the lone pairs is delocalized, the other is not.

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The amine in nordiazepam is never deprotonated by the base. The conjugate base of an amine has a pka of around 35 were methoxide is only around 15. The methoxide base is to ensure that the amine never undergoes intramolecular rearrangement to afford the positively charged ammonium ion of the amine which has a pka of around 9. The methoxide in the reaction will keep that formation from being stable as it would become deprotonated by the methoxide if it tried to occur (pka of 15 for methoxide vs 9 for ammonium ion). Thus insuring that the original nordiazepam formation does not rearrage while the methylation takes place.

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