Aromatic Chemistry

Question 1

Criteria for Aromaticity

Answer 1

• Cyclic
• Planar
• Conjugated (alternating single and double bonds)
• 4n+2 Pi Electrons

Question 2

Criteria for Anti-Aromaticity

Answer 2

• Cyclic
• Planar
• Conjugated (alternating single and double bonds)
• 4n Pi Electrons

Question 3

Benzene

Answer 3

• Discovered in 1825 by Michael Faraday as a liquid out of gas lamps.
• Found by physical chemistry techniques to have the formula C6H6.
• All 6 positions were equivalent so mono substituted benzene has only one isomeric form.
• Disubstituted benzene has only three isomeric forms (ortho, meta, para)

Question 4

Stability of Benzene

Answer 4

• Thermodynamically, benzene is 150 kJ mol-1 more stable that would be expected based on it’s formula alone. This is explained by Hückel’s Rule.

Question 5

Hückel’s Rule

Answer 5

Hückel’s Rule states that a planar cyclic fully conjugated compound with 4n+2 pi electrons benefits from a special stability as a result of the electron delocalisation and is described as ‘aromatic”.

Question 6

Anti-Aromatic Compounds

Answer 6

For planar, cyclic, conjugated molecules with 4n pi electrons delocalisation of electrons would be destabilising resulting in an “anti-aromatic” structure which normally distorts to be non-planar to avoid delocalisation and to remain “non-aromatic”.

Question 7

Ionic Molecules and Aromaticity

Answer 7

Ionic molecules can also benefit from aromaticity. Some examples are the cyclopentadienide and cycleheptatrienium ions.

Question 8

NMR Effects - Aromatic Systems

Answer 8

The delocalisation of electrons in an aromatic system can be picked up in chemical shift measurements. The external magnetic field applied during NMR measurements causes the electrons in the molecule to circulate in a diamagnetic ring current and set up a secondary magnetic field which enhances the external one outside the ring and opposes it inside the ring.

Question 9

NMR Effects - Anti Aromatic Systems

Answer 9

The delocalisation of electrons in a anti-aromatic system can be picked up in chemical shift measurements. The external magnetic field applied during NMR measurements causes the electrons in the molecule to circulate to give a paramagnetic ring current which sets up a secondary magnetic field which will then enhance the external one when the electrons are inside the ring and oppose it when they are outside the ring.

Question 10

Heterocyclic Compounds - Cyclopentadienide Anion

Answer 10

By replacing the C- of the cyclopentadienide anion with a heteroatom such as O, N or S we also get a 6 pi aromatic system. Note the lone pair of the heteroatom is needed since the 4 carbons can only contribute 4 electrons.

Question 11

Heterocyclic Compounds - Benzene

Answer 11

If we replace one or more benzene ring atoms with a heteroatom, the molecule remains aromatic. The heteroatom only contributes one electron from its lone pair to the conjugated system in order to fulfil the criteria for aromaticity.

Question 12

Naphthalene

Answer 12

Naphthalene is a white solid found in coal tar with a pungent smell used for “mothballs”. It could be regarded as a 10 pi system, or, more accurately, as two 6 pi systems. It is aromatic but the stabilisation energy is slightly lower than that of twice benzenes stabilisation energy. The bonds lengths are all equal, as is consistent with the resonance forms.

Question 13

Azulene

Answer 13

• first prepared in 1936, it is an isomer of naphthalene, C10H8.
• As well as the two uncharged resonance forms which are [10]-annulenes, there is an especially favourable form which combines a delocalised cyclopentadienide anion and a cycloheptatrienium cation, two 6 pi systems. This results is azulene having a small dipole moment of 0.8 Debye.
  There is also small charge transfer band in the visible region which explains its very intense blue colour.
  It has a stabilisation energy of 180 kJ mol-1.

Question 14

Reactivity of Aromatics

Answer 14

Because of the aromatic stabilisation, it is not favourable for reactions to take place that result in the permanent loss of the 6-pi system. Benzene undergoes aromatic substitution but as the intermediate is not necessarily aromatic, these reactions can have a high activation energy and are often associated with a catalyst.

Question 15

Treatment of benzene with chlorine

Answer 15

Treatment of benzene with chlorine results in substitution to give chlorobenzene and HCl.

Question 16

Electrophilic Aromatic Substitution

Answer 16

This is important as it provides a way to functionalist aromatics. In substitution reactions, the positive charge can be delocalised over three positions (o- and p- to the electrophile).

Question 17

Nitration of Benzene

Answer 17

Reaction with nitric acid and concentrated sulphuric acid converts benzene to nitrobenzene. The linear electrophile is formed by protonation of HNO3 followed by loss of water. Sulfuric acid is acting as a catalyst.

Question 18

Sulfonation

Answer 18

Treatment of benzene with concentrated sulfuric acid under severe conditions or with a solution of SO3 in H2SO4 under milder conditions gives benzene sulfonic acid. If the product is treated with more dilute sulphuric acid the processed is reversed to regenerate the benzene.

Question 19

Halogenation

Answer 19

Elemental halogens react to give halobenzenes. However this is only common for Cl and Br. I2 is too unreactive while F2 is too dangerous and aryl fluorides can be made other better ways. A Lewis acid is required, AlCl3 for chlorination and FeBr3 for bromination.

Question 20

Friedel Crafts Alkylation

Answer 20

Reaction of benzene with an alkyl in the present of a Lewis acid such as AlCl3 leads to substitution. However, there may be complications with rearrangement of the alkyl group. Adding one alkyl group activates the system towards further reaction so there may be multiple alkylations. The reaction can work well with groups that cannot rearrange (e.g. methyl or text-butyl halides), however, in general the reaction is problematic.

Question 21

Friedel Crafts Acylation

Answer 21

Reaction of benzene with an acyl halide in the presence of a Lewis acid such as AlCl3 leads to substitution. Adding an acyl group deactivates the compound, meaning it stops at mono substitution and thus the reaction works well with no isomerism. Since there are convenient ways to get from RC=O to RCH2, this also provides a way to monoalkylate benzene- first acylate it and then reduce the ketone to get to the target.

Question 22

Inductive Effects

Answer 22

Inductive effects are electron withdrawing (-I) or electron donating (+I) effects purely due to the electronegativity of the substituting and are transmitted via the single Ar-X bond.

Question 23

Mesomeric Effects

Answer 23

Mesomeric effects are electron withdrawing (-M) or electron donating (+M) effects via resonance and involving at some stage a double bond between Ar and X.

Question 24

Substituent Effects: NH2,NHR, NR2

Answer 24

Electronic Effects: +M>-I
Directing: o-/p-
Reactivity: Activating

Question 25

Substituent Effects: OH, OR

Answer 25

Electronic Effects: +M>-I
Directing: o-/p-
Reactivity: Activating

Question 26

Substituent Effects: NHCOR, OCOR

Answer 26

Electronic Effects: +M>-I
Directing: o-/p-
Reactivity: Activating

Question 27

Substituent Effects: Ph, -CH=CH2

Answer 27

Electronic Effects: +M>-I
Directing: o-/p-
Reactivity: Activating

Question 28

Substituent Effects: R

Answer 28

Electronic Effects: +I
Directing: o-/p-
Reactivity: Activating

Question 29

Substituent Effects: H

Answer 29

Electronic Effects: -
Directing: -
Reactivity: -

Question 30

Substituent Effects: Cl, Br, I

Answer 30

Electronic Effects: -I>+M
Directing: o-/p-
Reactivity: Deactivating

Question 31

Substituent Effects: CHO, COR

Answer 31

Electronic Effects: -I>-M
Directing: m-
Reactivity: Deactivating

Question 32

Substituent Effects: CO2H, CO2R, CN

Answer 32

Electronic Effects: -I>-M
Directing: m-
Reactivity: Deactivating

Question 33

Substituent Effects: SO3H

Answer 33

Electronic Effects: -I>-M
Directing: m-
Reactivity: Deactivating

Question 34

Substituent Effects: NO2

Answer 34

Electronic Effects: -I>-M
Directing: m-
Reactivity: Deactivating

Question 35

Strongly Activating Groups

Answer 35

Both phenol and aniline contain strongly activating groups and therefore undergo direct tri-substitution with many reagents. They both reaction with bromine at RT (no catalyst) to get the trisubstituted product. To get mono substitution, the reactivity can be reduced by acetylation to give acetanilide which is then well behaved.

Question 36

Nitration Conditions

Answer 36

Nitration of highly activating compounds such as o-dimethoxybenzene can be done using dilute nitric acid alone at RT while an electron withdrawing substituent needs stronger acids and formation of polynitroarenes needs both stronger acids and severe conditions.

Question 37

Ortho vs. Para Selectivity

Answer 37

Mixtures of isomers are commonly found from o-/p- directing groups. Where the existing substituent or the electrophile are large, para substituent may be favoured due to steric hinderance, however even with large groups both isomers are still formed - the selectivity is never complete. The isomeric product often differ markedly in their properties and can usually be easily seperated by distillation or recrystallisation.

Question 38

Combination of substituent effects

Answer 38

In cases where the directing effects are conflicting the most strongly activating group dominates and determines where substitution occurs.

Question 39

Use of removable directing groups

Answer 39

Sulfonation is reversible with oleum or strong sulfuric acid causing the forward reaction, but more dilute sulfuric acid causing the reverse. This gives a way of directing substitution exclusively to the ORTHO position.

Question 40

Electrophilic Substitution on Naphthalene

Answer 40

There are two possibilities for substitution onto naphthalene both of which offer five possible resonance forms, however the ones which maintain a complete benzene ring are favourable. 1-Substitution (near join of two rings) offers two resonance forms in which a complete benzene ring is maintained, whereas 2-substitution (further from join of two rings) offers only one resonance form in which a complete benzene ring is present and for this reason 1-substitution dominates.

Question 41

Sulfonation of Naphthalene

Answer 41

As expected, naphthalene-1-sulfonic acid is formed under normal conditions, however this is the kinetic product and at higher temperatures naphthalene-2-sulfonic acid can be produced. This is more stable since it avoids the 1,8 peri interactions.

Question 42

Effect of Substituents on Naphthalene Substitution

Answer 42

Substitution follows the expected trends, however, the activating/deactivating effect will control which RING reacts. In the case of a deactivating group, substitution occurs at the other ring in the two alpha positions regardless of where on the other ring the substituent is. If an activating group is present, substitution occurs on the same ring in the 2- and 4- positions when a 1-substituent is present but almost exclusively in the 1- position in the case of a 2-substituent.

Question 43

Nucleophilic Aromatic Substitution

Answer 43

This is less common than electrophilic substitution and only occurs in a few special cases. The first is where a halo benzene has one or more powerfully electron-withdrawing (usually nitro) groups in the o-/p- positions.
The substitution occurs by an addition/elimination mechanism and the intermediate is stabilised by delocalisation of the negative charge by the nitro groups. In many cases the intermediates, “Meisenheimer Complexes” are stable enough to be isolated.
The initial nucleophilic attack is the rate determining step and this explains the trend of reactivity among the halogens (more electronegative gives faster reaction as it favours attack by the anion).

Question 44

Sangers Reagent

Answer 44

1-Fluro-2,4-dinitrobenzene (FDNB) formed the bases of an important early method for determining the N-terminal residue of peptides. Treatment of an unknown peptide with the reagent results in a nucleophilic aromatic substitution by the N-terminal NH2 group giving the N-2,4-dinitrophenyl derivative. Hydrolysis of the whole peptide then gave a mixture of amino acids with only the terminal one as the dinitrophenyl derivative. It could then be easily identified by chromatographic comparison with an authentic sample. This played a decisive role in the determination of the structure of insulin.

Question 45

Benzyne and other Arynes

Answer 45

Reaction of aryl halides with strongly basic nucleophiles such as NaNH2, sodium amide, leads to apparent nucleophilic substitution. However, this occurs in a unusual way as easily shown by having another substituent present. In an early experiment 14C labelling was used to show the new nucleophile is not necessarily on the same carbon as the leaving group had been.
The formation of isomeric products arises from the addition of the nucleophile to the symmetrical neutral intermediate “benzene” so this is an elimination-addition mechanism.
If there is an additional substituent present, the resulting unsymmetrical aryne reacts selectively to give the m-product to avoid steric hinderance and due to the stabilisation of the intermediate negative charge by the -I effect of the substituent.

Question 46

Biphenylene

Answer 46

Biphenylene is a stable solid which exists mainly as two resonance forms which have complete benzene rings separated by an elongated rectangular cyclobutadiene ring. This benefits from the benzene ring aromaticity but avoids the instability of anti-aromatic cyclobutadiene. (No hint of any contribution from the form with a cyclobutadiene but not complete benzene rings).

Question 47

Diazonium Salts

Answer 47

Readily formed from aromatic amines and allow replacement of the original amino acid group by a wide variety of nucleophiles: F, Cl, Br, I, CN, OH or H. In addition coupling to phenols or aromatic amines gives the ado dyes of great industrial importance.

Question 48

Formation of diazonium salts

Answer 48

They are formed by reaction of an aniline with sodium nitrate and an acid. This process is call “diazotisation”. They are rather unstable and are formed and used in dilute aqueous solution taking care to keep the temperature just above 0°C. In the pure state, most diazonium salts are dangerously explosive so they are never isolated but rather used in situ for the next reaction.

Question 49

Diazonium Salt - reaction with loss of N2

Answer 49

Heating the aqueous solution or suspension of a diazonium salt carefully with a suitable nucleophile leads to loss of nitrogen gas and interception of Ar+ in an overall nucleophilic substitution. In the absence of any added nucleophile water reacts to give the phenol. However, where the phenol is the desired product it is best to use sulfuric acid for the diazotisation otherwise with hydrochloric acid some chlorobenzene can be formed. The new group ends up where the NH2 started.

Question 50

Diazo Coupling - a reaction with retention of N2

Answer 50

Treatment of a diazonium salt with tin(II) chloride or sodium sulfite results in reduction to the phenyl hydrazine. However, the diazonium group can also behave as an electrophile and cause electrophilic aromatic substitution on an added aromatic compound. It is weakly electrophilic so this reaction works best with highly activating groups as in anilines and phenols. For steric reasons the p-directing effect predominates and we get mainly the azo compounds shown. 2-Naphthol is also an important diazo coupling agent and couples at the 1-position.

Question 51

Uses of Azo Compounds

Answer 51

The azo function absorbs in the visible region and azo dyes and pigments are among the most important synthetic colouring agents. Some pH indicators are azo dyes such as methyl red.

Question 52

Benzidine Rearrangement

Answer 52

Benzidine is a dangerous carcinogen formerly used in the dye industry. The reaction is intramolecular and second order in [H+]. The remarkable mechanism involves homiletic fission of the N-N bond in the deprotonated form.

Question 53

Five-membered Heterocyles

Answer 53

Replacement of a CH=CH unit of benzene by a heteroatom gives a five metered heterocycle. These are aromatic and isoelectronic with cyclopentadienide anion. The three most important ones are thiophene (S), pyrrole (NH) and furan (O).

Question 54

Properties of five-membered heterocyles

Answer 54

The properties of these compounds are dominate by the fact that the heteroatom’s lone pair is needed for the 6-pi system. This means the heteroatom has lost full control of it’s electrons and becomes relatively electron poor, while the four carbon atoms enjoy a share of 6-pi electrons in a five atom ring and therefore relatively electron rich. For this reason these systems are described as pi-excessive heterocycles. The extra electron density results in shielding in NMR giving lower than expected chemical shifts for an aromatic compound.

Question 55

Basicity/Acidity of Pyrrole

Answer 55

Because of it’s lone pair being required for the 6-pi system, it is not available to act as a base and pyrrole is not at all basic although it might seem to have a secondary amine structure. The NH is however significantly acidic and can be removedly a strong base to give an anion isoelectronic with cyclopentadiene anion which still retains it’s aromaticity.

Question 56

Five-mebered heterocycles - electrophilic aromatic substitution

Answer 56

The higher electron density favours attack of E+ and these compounds are observed to undergo substitution very readily. Rate enhancement can be spectacular and pyrrole in particular is “super reactive” with bromination even under mild conditions proceeding immediately to the tetrabromo.

Question 57

Acid Sensitivity of Pyrrole

Answer 57

Treatment of pyrrole with concentrated nitric acid results in oxidative degradation, nitration has to be carried out using milder conditions.

Question 58

Five-mebered heterocycles - substitution position

Answer 58

As not all position are equivalent we have to consider directing effects. Substitution tends to occur on the 2- and 5- positions on these compounds but on 3- and 4- positions is also possible. This can be rationalised by the greater number of resonance forms for the intermediate in the former case.

Question 59

Benzo fuesd five-mebered heterocycles

Answer 59

Indoles are particularly important as there are many biologically active indole compounds including indole alkaloids, the amino acid tryptophan and the neurotransmitter serotonin.

Question 60

Indole Properties

Answer 60

As with pyrrole, indole is acidic and deprotonation and N-alkylation are possible. Again indole is completely non-basic since the lone pair is required for aromaticity and it not available.

Question 61

Benzo fuesd five-mebered heterocycles - electrophilic aromatic substitution

Answer 61

Substitution occurs exclusively on the activated pi-excessive heterocyclic ring rather than the benzene ring. The directing effect means that substitution occurs mainly at the 3-position. We again need to consider the possible resonance forms but discount any that do not include a complete benzene ring (as these are unstable). These compounds do not undergo nucleophilic aromatic substitution.

Question 62

Six membered ring heterocycles

Answer 62

If we replace =CH- by a heteroatom we get the six membered ring heterocycle pyridine (can’t use O or S or ring would be charged). The resulting compound is aromatic but importantly the lone pair of N is not needed for the 6-pi system meaning that compared to benzene the N atom is electron rich while the C atoms are relatively electron poor, for this reason pyridine is referred to as a pi-defficient heterocycle.

Question 63

Pyridine Properties

Answer 63

As it’s lone pair is not involved in the pi-system, pyridine is basic (pka 5.2) and readily forms salts with acid or other electrophiles.

Question 64

Pyridine - electrophilic aromatic substitution

Answer 64

Because of the pi-deficient character, electrophilic aromatic substitution occurs very reluctantly or not at all, with the alternative possibility of addition to the N atom instead of attack at C. To get around this, we can use pyridine N-odixe (which uses up the lone pair on N in bonding to oxygen) which has stable resonance forms with C- on the ring meaning electrophilic substitution goes readily (exclusively on the 4-position). The compound can then be deoxygenated.

Question 65

Pyridine - nucleophilic aromatic substitution

Answer 65

Nucleophilic aromatic substitution occurs readily is there is a suitable leaving group such as a halogen and it is placed so that the negative charge in the intermediate can be stabilised by being on the N. This means the halogen must be on the 2- or 4-position.

Question 66

Rings with two nitrogens

Answer 66

Replacing two C atom by N atoms gives further aromatic compounds:
- pyridazine (N directly beside each other)
- pyrimidine (N meta to each other)
- pyrazine (N para to each other)
In al cases the N lone pairs are not required for the 6-pi system and these are pi deficient systems behaving like pyridine.

Question 67

Benzo-fused six membered heterocycles

Answer 67

We can fuse benzene onto pyridine in two ways to get to isomer compounds quinoline (N in 1-position) and isoquinoline (N in 2-position). These are both well known, biologically active alkaloids.

Question 68

Benzo-fused six membered heterocycles - electrophilic aromatic substitution

Answer 68

Substitution occurs exclusively on the benzene ring rather than the strongly deactivated pyridine ring. The two rings behave essentially independently so the substitution is positioned neared the ring junction (5- and 8-position) rather than further from the ring (6- and 7-) for the same reasons we saw in naphthalene.

Question 69

Imidazole

Answer 69

Imidazole is a benzene with one CH=CH unit and one =C atom replaced by N. This compounds is a stable solid and behaves more like pyridine than pyrrole. It undergoes electrophilic substitution quite reluctantly and is strongly basic, even more so than expected. The main reason for this is that protonation gives a symmetrical resonance stabilised cation.