Organic Chemistry

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.

The allylic carbocation has two resonance forms, and so two positions where a nucleophile can attack, yielding two different products (1,2 and 1,4).

The benzylic carbocation has four resonance forms, and so in theory, four positions where a nucleophile can attack. But all but one of these resonance forms yield a product that is not aromatic, and so do not form. 

Aromatic compounds are very stable; aromatic starting materials tend to only form products that are also aromatic.

(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.)

Since chair forms can be cumbersome to deal with, always convert them to plain "overhead view" structures before doing the problem.

When in doubt, use numbering.

Chair forms can be difficult to draw, so it's a good diea to draw the "overhead view" first.

When approaching Diels-Alder problems, it's always good to user numbering: the product is always a cyclohexene compound.

In b), the trans dienophile leads to a trans product.

In c), the dienophile is an alkyne, and so becomes an alkene in the product.

This is an S_N1 reaction, but the carbocation is special because it's allylic.

If the Br^- nucleophile attacks the 3º resonance form, the 1,2 product is formed. (This is the product that would have formed if resonance was not involved.)

If the Br^- nucleophile attacks the 2º resonance form, the 1,4 product is formed.

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.