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Short link: mendelset.com/chapters/1114
Use curved arrows to draw a mechanism for the generic electrophilic aromatic substitution (EAS) reaction below.
Notice that the electrophile takes the place of the hydrogen.
Indicate the eletrophile formed by each set of reagents/conditions below.
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.
Draw the structure of the major organic product from each reaction sequence.
The purpose of this problem is to illustrate that there are often several routes to prepare a single molecule.
For the left route:
For the right route:
Note that both methods add a carbon chain to benzene, a step that is quite common in synthesis problems!
Rationalize the follwing pKa values. Explain your answer in terms of the stabilites of the conjugates bases of each acid.
Note: the lower the pKa, the stronger the acid.
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:
The cyclopentadienyl anion meets all of these criteria, and so is aromatic, and very stable.
Pyrrole is an example of a heteroaromatic compound: it contains a heteroatom (atom that is not carbon or hydrogen, such as N, O, S, etc.), and is aromatic.
Because pyrrole is aromatic, we should be able to draw many resonance forms- usually as many resonance forms as sides (in this case, five sides, so five resonane forms).
Draw all resonance forms for pyrrole. (I've started you off.)
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.
Imidazole (shown below) has two nitrogen atoms, N-1 and N-3. Which nitrogen is more basic?
To answer this problem, draw the product after each nitrogen protonates, and compare their stabilities. Explain your reasoning.
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.
Phenol can be prepared from benzene and hydrogen peroxide in the presence of a really strong acid. Propose a mechanism for this reaction.
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 SN1 reaction, while using HO-OH2+. is analogous to an SN2 reaction. When you draw mechanisms, you should use whichever convention your textbook uses. Both lead to the same product.
Show how to prepare vinyl benzene from benzene.
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.
A chemist tried to prepare compound A from benzene via Friedel-Crafts alkylation and instead produced compound B.
Why did this happen? How could the chemist prepare compound A?
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).
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