LDA will not attach, but it may be strong enough to form an aryne. For example one could deprotonate C-H and elimination of the best leaving group (Cl) would give an aryne, which could then undergo attack by a nucleophile. When the nucleophile does attack, it will likely do so such that the negative charge is placed closer to the electron withdrawing fluoro groups.
See this artice on arynes – possibly helpful. https://www.masterorganicchemistry.com/2018/09/17/nucleophilic-aromatic-substitution-2-benzyne/

For molecules with an axis of chirality, one looks along the chiral axis and numbers the substituents on the near carbon (1) or (2) according to the CIP rules. One then looks at the substituents on the far carbon and numbers them (3) or (4) according to the CIP rules. One then notes whether 1,2,3 goes clockwise (r) or counterclockwise (s). See this article (wikipedia) for an example:
https://en.wikipedia.org/wiki/Axial_chirality

Thanks for the question! Short answer is that the nitrogen in NADH is very poorly basic here because the lone pair on nitrogen is conjugated with the two adjacent pi bonds. Protonation of these types of functional groups (look up “enamines”) requires pretty strong acid (at least below pH 4). Even if it was protonated at physiological pH, protonation is very reversible, and it would represent a dead-end that just quickly equilibrates back into NADH.

Furthermore, if protonation were to occur, it would happen primarily on carbon, rather than on nitrogen. This is because the nitrogen lone pair donates a lot of electron density into the adjacent alkenes, and these carbons become quite basic (relatively speaking). Enamines protonate primarily on carbon.

There is no partial aromaticity in NADH. The ring in NADH is not aromatic, nor is it even partially aromatic. That’s because the carbon with two H’s attached has no orbital capable of pi-bonding with the adjacent atoms. In order for aromaticity to occur, there must be a continuous ring of atoms capable of pi-bonding. When NADH donates hydride (H-) this breaks one of the C-H bonds, and results in formation of a new C-N pi bond. This allows conjugation of pi bonds all around the ring, and formation of an aromatic ring provides the driving force for this reaction.

I hope this answers your question! James

I hope this answers your question

I am wondering if the pyridine ring has any partial aromaticity?

And to what extent is that more or less important than steric and electron-withdrawing effects of the ring itself and the nearby ribose?

Fixed – thank you for spotting the typo and letting me know. Please let me know of any other ones you see! James