3. Describe how the 'H NMR of the product from Part A would be different from...

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3. Describe how the H NMR of the product from Part A would be different from Lidocaine (see following pages). Show how the p

23 Synthesis of Lidocaine Reaction Part A CH3 na la ACOH CH3 H NH2 i N a 2) sodium acetate, H,O CH3 CH3 a-chloro-2,6-N-(dimet

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Answer #1

Lidocaine contains -N(CH2CH3)2 group compared to -Cl group in product obtained in part "A" This structural difference brings two additional signals in its (Lidocaine) both 1HNMR (triplet for 6H's and a quartet for 4H's) and 13CNMR.

Amide group in lidocaine is a good candidate for hydrogen bonding interactions. The >N-H----O=C< type intermolecular H-bonding can be shown as

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