EAS mechanism

This is a good synthesis to know, because vinyl benzene is a good starting point for many synthesis problems you will encounter down the road.

An ethyl group can be added to benzene via Friedel-Crafts alkylation. 

To add that alkene, we you will have to do some sort of elimination reaction. The easiest way to do this is via bromination. Free radical bromination (with NBS or Br₂/hν) will add a bromine to the position of the most stable radical, which is the benzylic position (due to resonance).

Then adding a strong base like potassium tert-butoxide will do an E2 reaction to form the alkene.

Note that some textbooks don't draw EAS electrophiles with formal positive charges.

Some textbooks instead draw electrophiles with partial positive charges and a leaving group. It's just a convention, and doesn't affect the product of an EAS reaction. See the problem 595 for an explanation of the difference.

The electrophile we need for this EAS reaction is HO⁺. We can generate it by protonating one of the oxygens in peroxide, which causes it to act as a leaving group.

On the left side of this image I make a note about the structure of the electrophile. Some textbooks use electrophiles with formal positive charges, such as HO⁺. Other textbooks don't use formal positive charges, and instead use partial positive charges with a leaving group, such as HO-OH₂⁺. Using HO⁺ as the electrophile is analogous to an S_N1 reaction, while using HO-OH₂⁺. is analogous to an S_N2 reaction. When you draw mechanisms, you should use whichever convention your textbook uses. Both lead to the same product.

Friedel-Crafts alkylation is prone to carbocation rearrangement. In this case, alkylation produced a 1º carbocation which rearranged to a 3º carbocation, leading to compound B.

We can avoid this by instead doing Friedel-Crafts acylation. The intermediate in acylation is the acylium ion, which is stabilized by resonance and so won't rearrange.

But after the acylation reaction we have to get rid of the carbonyl (C=O) group, so we do a Wolff-Kishner reduction (N₂H₄/NaOH, heat).

a) The species on the left are much more stable than the species on the right because they have complete octets.

Nitrogen is less electronegative than oxygen, so a positive nitrogen (with a full octect) is more stable than a positive oxygen; oxygen is electronegative and so doesn't "like" being positive (have low electron density). 

The carbocation and the "nitrocation" are both very unstable because they only have 6 valence electrons. Because nitrogen is more electronegative than carbon, the nitrocation is even less stable than the carbocation, because its worse at handling a positive charge.

b) The cation intermediates that result from substitution at C-3 only contain carbocations, while the intermdiates from C-2 or C-4 substitution contain a "nitrocation," which is less stable than a regular carbocation, and so are not favored. So pyridine undergoes EAS reactions at C-3.

One more note: For substitution at C-2 and C-4, it's tempting to draw a resonance form where the positive charge is on the nitrogen, which is flanked by two double bonds. But this cyclic compound is too strained to exist, and so doesn't contribute to this analysis.

Substitution at C-1 and C-2 both lead to many resonance forms. But the resonance forms that arrise from substitution at C-2 are not aromatic, while the resonance forms from substitution at C-1 are aromatic (the benzene ring is maintained).

So the C-1 resonance forms are more stable than the C-2 resonance forms, so substitution occurs at C-1 instead of C-2.

Electrophilic substitution at C-2 leads to a carbocation intermediate with three resonance forms, while substitution at C-3 leads to a carbocation intermediate with only two resonance forms.

The C-2 intermediate has more resonance forms than the C-3 intermediate, and so is more stable. Therefore, EAS occurs at C-2.

Short answer: -NO₂ is an meta director because the cyclohexadienyl intermediates that result from meta addtition are more stable than those that result from ortho or para addition.

Long answer: Addition at both the ortho or para positions lead to a cyclohexadienyl cation that contains two adjacent positive charges. This is very unstable, and so addition at the ortho or para positions is not favored.

The cyclohexadienyl cations that result from meta addition don't have this problem.

This will be the base for any substituent that has a positive charge (or partial positive charge). So -NO₂, -C(O)R, -CF₃, -NH₃⁺, -CN, and -SO₃R are all meta directors.

Short answer: -OR is an ortho/para director because the cyclohexadienyl intermediates that result from ortho and para addtitions are more stable than those that result from meta addition.

Long answer: The cyclohexadienyl intermediates from ortho and para addition include a resonance form where the oxygen is adjacent to the carbocation, while the cyclohexadienyl intermediates from a meta addition do not.

When an oxygen (or anything with a lone pair) is adjacent to a carbocation, it can share its lone pair and stabilize the positive charge. This is only possible when the electrophile adds to the ortho or para positions, so those positions are favored with -OR as a substituent. Hence, -OR is an ortho/para director.

Note that this logic holds for any substituent with a lone pair, so -OH, -OR, -NH₂, -NR₂, -F, -Cl, -Br, and -I are all ortho/para directors.