Textbook: Loudon 5th Ed. (2009)

This is an S_N1 reaction. We know this because none of the reagents have charges (H₂O is neutral; if it were HO^-, it would probably be S_N2 or E2). So a carbocation will be formed in this reaction, and the compound whose carbocation is the most stable will react the fastest. So the 3º bromide will react fastest with water.

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.

The first step in this reaciton, like many reactions, is an acid-base reaction- when you see an H⁺ (acid), the first step is usually something getting protonated. This starting material is a thiol (sulfur), but the same thing would happen with an alcohol or ether (oxygen).

When you have a protonated thiol (or alcohol, ether, etc.), you know the reaction isn't finished yet- products are rarely charged. There are two legal moves to get that positive charge off of the protonated thiol. Either the thiol deprotonates, or it takes off as a neutral leaving group. It can't deprontate because that would be going backwards, so your only "legal move" is for the HSCH₂CH₃ to act as a leaving group (See problem 518).

But then you're left with a carbocation, which is definitely not the final product. How can we get rid of it? (See problem 335 for general ways carbocations react). If the HSCH₂CH₃ attacked in an S_N1 type reaction that would be OK, but it would be going backwards! So the only way to get rid of the carbocation is to do a beta-elminiation (E1).

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.

This problem is similar to problem 536. The only difference among these compound is their leaving group, so the compound with the best leaving group will undergo substitution reactions most rapidly.

Good leaving groups are stable. Larger ions tend to be more stable than smaller atoms (due to a smaller charge:size ratio. See problem 288), so when going down the periodic table, stability increases. I^- is more stable than Br^-, which is more stable than Cl^-, etc. So I^- is the best leaving group, and 2-iodobutane will react the fastest with a nucleophile.

Each of these ions has a charge of -1, and none of them has any resonance forms. So the two things to consider are electronegativity and size. Electronegativity is relevant as we go from left to right across the periodic table. But here we are going up to down on the periodic table, so size is relavant. Larger ions are more stable than smaller ions (due to a smaller charge:size ratio), so I^- is the most stable of the group.

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.

You know this is an elimination reaction because no nucleophile is present; H₃PO₄ and H₂SO₄ are non-nucleophilic acids. IF the reagent were HCl or HBr on the other hand, this would be a substitution reaction.

Since acid is present, a carbocation will probably form, so we know that this is an E1 mechanism.

The 2º carbocation that is initially formed will undergo a 1,2-hydride shift to become a 3° carbocation, but that doesn't affect the final product of the reaction; with either the 2° or 3° carbocation, the most stable alkene product is 2-methyl-2-butene.