The only differences among these compounds is the substitution of the alpha carbon (methly, 1º, 2º, or 3º).

This is an S_N2 reaction. We know this because NaI is an S_N2 reagent- charges give it away; when we see charges (Na⁺ is positive and I^- is negative) it's probably an S_N2 reaction. So we want reactants that are less substituted: methyl is more reactive than 1º, which is more reactive than 2º, etc. So methyl bromide reacts the fastest with NaI.

Charged molecules are generally less table than neutral ones, and each of the molecules below has a negatively charged oxygen.

But not all negative charges are equal; some oxygens are "closer to neutral" than others. How? Because resonance stabilizes charges by sharing electron density over multiple atoms (this is called electron delocalization).

Hydroxide (HO^-) doesn't have resonance, so the oxygen has 100% of the negative charge to itself. On the other hand, the sulfonate ester (SO₃R^-) has three resonance forms, so each oxygen only has ~33% of a negative charge. So the sulfonate ester is the most stable anion.

In general, the more resonance forms a molecule has, the more stable it is.

The only differences among these three molecules is their leaving groups, so whichever has the best leaving group will react fastest with the nucleophile (azide, N₃^-). Stable molecules make good leaving groups, so the best leaving group of the three is SO₃R^-, which has has three resonance forms (see problem 535). So the third compound will react the fastest with NaN₃.

Nucleophilicity increases with electron density; negatively charged molecules are more nucleophilic than neutral molecules. So HS^- is the best nucleophile in this group.

Nucleophilicity also increases with size. This is because the larger an atom, the more polarizable it is. Size increases as we go down the periodic table, so H₂S is a better nucleophile than H₂O.

The delta (Δ) in the reaciton arrows means that heat is being added to this reaction, which tends to favor elimination over substitution. Also, the reaction is using a non-nucleophilic acid (H₂SO₄), which tends to favor elimination reactions (H₃PO₄ is another common reagent for E1 reactions, while HCl or HBr tend to go S_N1).

Because this reaction is taking place in acid, a carbocation is likely to form, so this is an E1 reaction. Since water is lost over the course of the reaction, this is a dehydration, which is a type of elimination reaction.

Predicting S_N1/S_N2/E1/E2 competition reactions tends to drive students crazy, but it's not so bad once you notice the general pattern:

• basic conditions (positive and negative charges) tend to go S_N2 or E2 (no carbocation)
• neutral or acidic conditions tend to go S_N1 or E1 (carbocation is formed).

That's how you determine a SN1/E1 reaction from an SN2/E2 reaction. But how to decide between substitution or elimination? General things to watch for are bulk, nucleophilicity, and heat:

• If you see heat (or Δ), the reaction will go elimination.
• If you see a big, bulky compound, the reaction will go elmination.
• If you see a strong base, the reaction will go elimination. Strong base is anything stronger than RO^-.

The exception: if everything is primary, it will probably go S_N2.

These rules probably seem confusing, so let's go through these eight examples and see how they apply.

a) NaCN is charged! (Na⁺ and CN^-), so it's S_N2 or E2. CN is not a strong base, so it's S_N2.

b) KOtBu (potassium tert-butoxide) is charged, so it's S_N2 or E2. ^-OtBu is a strong base, so if anything is more bulky than 1º it will go E2. ^-OtBu is 3º, so it will definitely go E2 (KOtBu is a classic E2 reagent).

c) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. ^-OMe is 1º (actually, not even 1º), but the alkyl halide is 2º, so it will go E2.

d) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. But in this case there is no bulk whatsoever- nothing is >1º! NaOMe is 1 and the alkyl halide is also 1º, so it will go S_N2.

e) Methanol (MeOH) is neutral so probably E1 or S_N1. Methanol is a weak base and there's no bulk, so S_N1. In general water and alcohol do a mixture of S_N1 and E1 with alkyl halides (mostly S_N1).

f) H₂SO₄ is acidic so probably E1 or S_N1. Can't be S_N1 though because there is no nucleophile in H₂SO₄. (HSO₄^- is a very weak nucleophile). An alcohol with H₂SO₄ or H₃PO₄ is a dehydration reaction- E1.

g) H₂SO₄ is acidic so probably E1 or S_N1. In this case we have a nucleophile- Cl^-, so it will go S_N1.

h) Amines are neutral but they don't so S_N1/E1- they tend to go S_N2/E2, because they are basic (an amine solution has a basic pH). This amine is really bulky so it will go E2.

These are substitution reactions. In each case we're replacing the -OH with either -CN or -I, so the nucleophiles will be NaCN or NaI (S_N2 conditions).

Hydroxide (HO^-) is a poor leaving group, so the first step in each of these reactions will be to convert the alcohol into a better leaving group. Two good leaving groups are Br^- and SO₃R^-, but which one to use?

Every time an S_N2 reation takes place the wedge on the alpha carbon becomes a dash (and vice-versa).

For a), the wedge remains a wedge, so we have to do two S_N2 reactions (wedge to dash to wedge again). So we use PBr₃ to turn the OH into a Br. (Bromination of an alcohol with PBr₃ is an S_N2 reaction and so inverts stereochemistry).

For b), the wedge becomes a dash, so we can only do one SN2 reaction. So instead of using PBr₃ to make the OH a better leaving group, we use RSO₂Cl, which doesn't break the carbon-oxygen bond and so doesn't invert the stereochemistry.

This is two separate problems. Many textbooks describe how to mentally convert " zig-zag" (wedge/dash) structures to Fischer projections, but I've never met a student who can do this without making mistakes. So you should convert the name ((2R,3S) 2-bromo-3-chlorobutane) to a zig-zag structure, and then convert the name to a Fischer projection. Never try to convert a zig-zag structure directly to a Fischer projection.

To draw the zig-zag structure, first draw a structure with each halogen (the highest priority substituents) as a wedge, see the R/S configuration we drew, and then adjust as necessary. Why wedges? Because that puts the hydrogen as a dash, so R/S is easy to assign. With two wedges, the structure is (2R, 3R). So the 2R position is fine, and we just switch the wedge at C-3 to a dash, and the structure is correct (2R,3S).

To draw the Fischer projection, we do something similar- arbitaritly draw a structure, check R/S, and then adjust as necessary. In the structure below, I drew (arbitarily) 2S,3S, but we need 2R,3S, so I just switch C-2 and we have the correct structure.

The methyl group is bulky and so is most stable in the equatorial position.

The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.

So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.

The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.

Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.

The starting material (2-aminobutane) is a racemic; it has equal amounts of R and S enantiomers.

Enantiomers have the same the same chemical and physical properties, so we can't separate them using common lab techniques, such as recrystallization, chromatography, etc. But diastereomers can have different properties, and we can exploit this fact to separate otherwise inseparable compounds.

Adding an optically pure acid to the amine produces a diastereomeric mixture of salts which can be separated. Adding a base (NaOH) "breaks" the salt and allows us to isolate the pure R or S amine.

We are able to separate the (R,R) and (R,S) salts because they are diastereomers and so have different chemical and physical properties.