When attacking hydrogen the nucleophile acts as a base. When attacking carbon it acts as the nucleophile

When the nucleophile attacks a hydrogen the result is elimination (having an R and an X each on their own carbon will result in R-double-bonded-C-chain + HNu + X^-). When it attacks carbon the result is substitution (with the same initial substance you'll get a trans-R-Nu alkane chain + X^-)

Substitution is simpler because only 1 bond is broken and only 1 new bond is made whereas in elimation 2 bonds are broken and then made

Substitution is entropically neutral and depends on the relative strength of the C-X and C-Nu bonds as well as the stability of Nu^- vs X^-

The associative mechanism means that in order to substitute Nu^- for X^- a bond with Nu^- is made before the bond with X^- is broken. Carbon can't use this mechanism as it can't expand its outer electron shell. Heavier elements like Si can use this but not carbon

In order to substitute Nu^- for X^- the bond between X^- and the terminal carbon is broken before the bond between Nu^- and the terminal carbon is formed. Carbon can use this mechanism as a positive carbon is better than a negatively charged one and there's no need to expand its valence shell

When trying to replace an X group on a terminal carbon with Nu^- the bonds are broken and formed simultaneously. Carbon can use this mechanism as this doesn't require expansion of its valence shell and this results in no charge in the carbon

It is a one-step mechanism with one transition state, the rate law is k[alkyl halide][nucleophile] (which is second order) and therefore is a substitution, nucleophilic, 2nd order reaction (SN2 or S_N2)

The nature of the leaving group, the structure of the substrate, the nature of the nucleophile and the solvent

Calculations indicate that there is no change in charge on C at the terminal state as compared to the starting state

Steric hindrance means that atoms can't be brought too close together, otherwise the contained energy or reactivity might be affected. For the SN2 reaction this has to do with the umbrella inversion so that the previously established groups don't interfere with the new nucleophile

The nucleophile will be exposed to the in-phase HOMO in comparison to its own MOs. In the frontside attack the HOMO of the nucleophile will be exposed to an in-phase MO of the halide but will be exposed to a repulsing (antibonding) MO on the carbon

The weakest bases make the best leaving groups. The comparison goes as follows: I^-, Br^-, Cl^-, F^-

The role of the solvent is the solvate the nucleophile. Solvation of the nucleophile makes it less nucleophilic. In the absense of solvation, the strongest base is the best nucleophile

The best nucleophiles are soft (polarizable) and/or small (linear). OH^- is decent but azide is one of the best

Dipolar aprotic solvents have no protons on polar atoms so therefore they can't solvate anions. This means that the nucleophiles are unsolvated (naked) and are ready to react without desolvation

Given a good nucleophile and a good electrophile (with a good leaving group) a variety of functional groups can be prepared

It's a 2 step mechanism with 2 transition states, the first step is rate determining. The rate law = k[alkyl halide] and therefore it is a substitution, nucleophilic, 1st order reaction (SN1).

The more complex the alkyl halide the more rapid the SN1 reaction

(R)-2-butanol will result from a top attack and (S)-2-butanol will result from a bottom attack. When these occur in the same ratio (when retention of the original configuration occurs along with the inverted configuration) it's called racemization

The more polar solvent will stabilize the more positive ionization transition state for the SN1 reaction of neutral reactants. This will affect the relative rates; the amount of free energy will be decreased with the more polar solvent

They're the same as CA rearrangements in electrophilic additions to alkenes

A nucleophile will mostly attack an sp² carbon. Since a squared carbon is surrounded by less electrons it's easier for the typically electron-rich nucleophile to attack the squared carbon

No known nucleophile can displace a halide ion. Also, pi electrons may interfere

It has very low stability in SN1 reactions and none of these ion types are able to form

Only a front-side displacement can occur and this is too high in energy; the resulting products would actually have so much energy that they would likely fall back to the initial version very quickly

You would need an N₂⁺ leaving group which is one of the very best

For SN1 reactions tertiary reactions have the least change in energy and methyl groups have the most energy. In SN2 reactions this order is reversed and both peaks for the secondary alkyl halide are about the same as each other.

They can only undergo SN2 reactions

They can undergo neither SN1 nor SN2 reactions

They can undergo either SN1 or SN2 reactions

They can undergo both SN1 and SN2 reactions

They can undergo SN1 reactions only

They can only undergo SN1 reactions

SN1 reactions are stepwise mechanisms with a carbocation intermediate whereas SN2 reactons are one-step mechanisms

SN1 reactions have a unimolecular RD step whereas SN2 reactions have a bimolecular RD step

There is in SN1 reactions but not in SN2

The strength of the nucleophile doesn't affect SN1 reactions but it does affect SN2 reactions (remember the rate laws: k[alkyl halide] for SN1, k[nucleophile][alkyl halide] for SN2). In SN2 reactions the more nucleophilic the faster the reaction will proceed

In SN1 reactions tertiary then secondary alkyl halides react (but not primary or methyl groups). In SN2 reactions methyl the primary then secondary alkyl halides react

SN1 products will have both the retained and the inverted configurations compared to the reactant whereas SN2 products will always have the inverted configuration compared to the reactant

The leaving groups are exactly the same. The order goes like this: I^->Br^->Cl^->F^-

3 groups goes the fastest and a 5 membered ring is made. After that are groups amounts 4, 1, 5 and 2 (1 has the smallest entropy but it's sterically strained)

SAM is a biological methylating agent. When a methyl group is taken off of the sulfur in SAM and added to norepinephrine the resulting products are a sulfide leaving group and epinephrine