Basicity, for our purposes, involves the reaction of a lone pair with H (proton).

Nucleophilicity, for our purposes, involves the reaction of a lone pair with *any atom other than H*, by which we usually mean *carbon*.

The biggest difference is that when a lone pair is attacking any atom other than H, it generally has to be to an empty orbital, which is most often an antibonding orbital (e.g. sigma star) like it is with the SN2 reaction.

And that means approaching the backside of the C-X bond (where X is a good leaving group) which means that it can be *much* more subject to steric interactions (i.e. bulkiness).

Anything which affects steric bulk around the base is going to affect its nucleophilicity. Solvent is a perfect example – you’ve probably learned the difference between polar protic and polar aprotic solvents. The more hydrogen bonding there is around a base, the less nucleophilic it will be.

The pKa of t-BuOH is about 18 versus 16 for methanol, so for a first-order analysis, no. However, as always, “it depends”. There are certain reactions (e.g. elimination) requiring base that can undergo different pathways with bulky bases than with non sterically hindered bases. See: https://www.masterorganicchemistry.com/2012/10/24/bulky-bases-in-elimination-reactions/

True enough!

I assume you mean that you have a tertiary alkyl halide, in the presence of H2O. The first step will be loss of halide ion to give a carbocation. What happens next could either be attack of the carbocation by water (SN1) or deprotonation adjacent to the carbocation to give an alkene (elimination – E1). Both pathways will occur. However in my experience the elimination pathway is generally only going to be the major product if the reaction also mentions “Heat”.

If we are comparing neutral OH with neutral COOH, with equal steric hindrance, then the stronger nucleophile would be the group that is the most basic.

It is easier to protonate R-OH to R-OH2 (+) than it is to protonate R-CO2H to R-CO2H2 (+). We know this because the pKa of protonated R-OH is about -2.2 (CH3-OH2+) and the pKa of protonated R-CO2H is about -7.7 (Ph-CO2H2 + ) Source: http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf

If there is significant difference in steric hindrance then this would affect the relative nucleophilicity to some extent. You are correct that protic solvent would retard the nucleophilicity of both these species, and polar aprotic solvent would augment it.

What are the key nucleophilicity trends as one descends the periodic table?

Flipping??

There’s a lot of factors that go into the nucleophilicity of a species, not just electronegativity.