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Answer :
Final answer:
The hydrogens of the CC-8 methyl group in the FAD moiety are more acidic than those of the CC-7 methyl group due to delocalization of electrons.
Explanation:
The hydrogens of the CC-8 methyl group in the FAD moiety are more acidic than those of the CC-7 methyl group due to the delocalization of electrons. When a proton is removed from the methyl group at CC-8, the electrons that are left behind can be delocalized onto the nitrogen at the 5-position or onto the nitrogen at the 10-position. This resonance stabilization of the negative charge on the carbon atom makes the hydrogens more acidic. On the other hand, when a proton is removed from the methyl group at CC-7, the electrons that are left behind can only be delocalized onto carbons, which does not provide as much stabilization.
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Final answer:
The hydrogens at the CC-8 position in FAD are more acidic due to the electron delocalization onto nearby nitrogens when a proton is removed, this leads to increased stability of the conjugate base. In contrast, the hydrogens at the CC-7 position lack this delocalization, making them less acidic.
Explanation:
The hydrogens of the CC-8 methyl group in FAD are more acidic than those of the CC-7 methyl group because of the potential for electron delocalization upon deprotonation. Specifically, statement a) is correct: When a proton is removed from the methyl group at CC-8, the electrons that are left behind can be delocalized onto the nitrogen at the 5-position or onto the nitrogen at the 10-position. This delocalization stabilizes the conjugate base, making the removal of a proton more favorable, and thus the hydrogen more acidic. On the other hand, statement e) accurately describes the methyl group at CC-7: When a proton is removed from this methyl group, the electrons that are left behind cannot be delocalized, which results in less stabilization of the conjugate base and therefore a lower acidity compared to the protons at the CC-8 position.