Concept (explain why this is so..)

Imidazole is aromatic. When N-3 protonates, the product is still aromatic.

But when N-1 protonates, the product is no longer aromatic (and therefore significantly less stable).

Because of this, N-3 is much more basic than N-1. Another way of thinking of this is that the lone pair on N-1 is involved in the aromatic circuit, and so is not available to pick up a proton.

One of the rules for aromaticity is that all atoms shoudl be sp² hybridized. But the nitrogen in pyrrole is sp³ hybridized, so how is it still aromatic? Because in 4/5 of its resonance forms the nitrogen is sp² hybridized; the real picture of pyrrole looks more like the structure on the left (dashed circle) than any individual resonance form.

The benzylic proton (middle compound) is more acidic than the allylic proton (left compound) because its conjugate base is more stable. This is because it has more resonance forms.

Cyclopentadiene (right compound) is the strongest of the three because it has the most stable conjugate base. Why is it the most stable? Because it's aromatic! To be aromatic, a compound must:

1. be cyclic and planar
2. be sp² hybridized
3. Have a Huckel number of pi electrons- 2, 6, 10, 14, etc.

The cyclopentadienyl anion meets all of these criteria, and so is aromatic, and very stable.

(This is similar to problem 569, which involved the allylic position.)

Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.

(Radical resonance uses hooks instead of arrows and so is a little different.)

a) is a free-radical bromination. The radical will be formed at the most stable position, which will be the allylic position due to resonance (see problem 569). There are two allylic positions in this molecule, so the more substituted one will be the more stable; the bromine will add to the 3º allylic carbon.

b) is an E2 reaction. There are two types of beta protons and so two possible alkene products. One is an isolated diene and the other is a conjugated diene. Conjugated dienes are more stable, and so that will be the major product.

To undergo a Diels-Alder reaction, a dienophile must be in s-cis conformation. 

The bicylic compound below is locked into s-trans conformation; it can never rotate into s-cis conformation and so can't undergo a Diels-Alder reaction.

Of the other two compounds, the middle compound most easily rotates into s-cis conformation, and so will undergo a Diels-Alder reaction the fastest. The compound on the right has steric strain when in s-cis conformation, and so won't do Diels-Alder as easily.

Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.

(Radical resonance uses hooks instead of arrows and so is a little different.)

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 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.

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.