Corrected exercises: aromatic compounds (202-SNU-RE)

The compound must be CYCLIC; it must be PLANAR, otherwise the $p$ orbitals cannot overlap side by side in a continuous way; it must be fully CONJUGATED, that is, every atom of the ring must carry an available $p$ orbital; and it must have $4n+2$ delocalized $\pi$ electrons, with $n$ a nonnegative integer.

Benzene is cyclic and planar, and each of its six carbons is $sp^{2}$, hence carries a pure $p$ orbital perpendicular to the plane: conjugation is complete around the whole ring. It has 6 $\pi$ electrons, coming from the three double bonds. Solving $4n+2=6$ gives $n=1$: the condition is satisfied. Benzene is AROMATIC.

Cyclobutadiene has 4 $\pi$ electrons. Now $4n+2=4$ would give $n=0.5$, which is not an integer: the rule is not satisfied. Since $4=4\times 1$, it has $4n$ electrons and so falls in the ANTIAROMATIC case: it is less stable than the corresponding open butadiene, and in fact it exists only at very low temperature. Cyclooctatetraene has 8 $\pi$ electrons, that is $4n$ with $n=2$: the same reasoning would make it antiaromatic if it were planar.

Yes. The five-carbon ring carries four $sp^{2}$ carbons coming from the two double bonds, and the fifth, after loss of a proton, carries a lone pair in a $p$ orbital that joins the conjugation. That makes $4+2=6$ $\pi$ electrons, so $n=1$: the anion is AROMATIC. The effect on acidity is spectacular: cyclopentadiene has a $pK_{a}$ close to 16, comparable to that of water, whereas an ordinary alkane exceeds 50. Its conjugate base is stabilized by aromaticity, which makes removing the proton much easier. This is the clearest illustration of what aromaticity is worth in energy.

It changes the category, not the lesson. By folding into a tub shape, cyclooctatetraene BREAKS planarity: its $p$ orbitals no longer overlap continuously, cyclic conjugation is interrupted, and it is therefore neither aromatic NOR antiaromatic, but simply NON-aromatic. It then behaves like an ordinary polyene, with clearly alternating single and double bonds, and undergoes the additions that benzene refuses. It is a fine example of a molecule that distorts itself to ESCAPE antiaromaticity, which is energetically unfavourable.

If the three double bonds were independent, each would release the same energy as an isolated double bond, that of cyclohexene: $3\times\left(-120\right)=-360$ kJ/mol.

The measured value is only $-208$ kJ/mol. Benzene therefore releases $360-208=152$ kJ/mol LESS than predicted. This difference is the RESONANCE ENERGY: it is the amount by which real benzene already lies below the hypothetical localized-bond model. In other words, it started lower, so there is less left to release.

For cyclohexa-1,3-diene, two independent double bonds would give $2\times\left(-120\right)=-240$ kJ/mol, whereas the measurement gives $-232$: the difference is only 8 kJ/mol. This small stabilization is that of simple CONJUGATION between two neighbouring double bonds, which exists in every conjugated diene. The comparison is revealing: 8 kJ/mol for an open conjugation against 152 for benzene. Aromaticity is therefore not a slightly better conjugation, it is a phenomenon of a completely different magnitude, about twenty times more stabilizing.

Because the heat of hydrogenation measures the energy DROP between the starting compound and the product. All these compounds give the SAME product, cyclohexane, hence the same end point. If the drop is shorter, the starting point must have been LOWER, hence more stable. That is the whole trick of the method: by fixing the final state, a comparison of heats of reaction becomes a comparison of the stabilities of the reactants.

NO. Cyclohexene decolourizes bromine water immediately by addition across its double bond; benzene does not react under these conditions. The reason is energetic: adding $Br_{2}$ to benzene would break the cyclic conjugated system and so destroy the 152 kJ/mol of resonance energy. The gain from forming two $C-Br$ bonds does not make up for this loss, and the reaction is not favourable. This is precisely why benzene undergoes SUBSTITUTIONS, which restore aromaticity at the end, rather than additions, which would destroy it permanently. This bromine water test is, incidentally, a simple way to tell an aromatic compound from an alkene in the laboratory.

First step: the electrophile $E^{+}$ is attacked by the $\pi$ cloud of the ring, which forms a cyclic CARBOCATION called the arenium ion, or sigma complex. This cation is no longer aromatic, since the attacked carbon has become $sp^{3}$ and breaks the conjugation, but it is stabilized by resonance over three positions. Second step: a base, often the $HSO_{4}^{-}$ ion, removes the proton carried by this $sp^{3}$ carbon, which becomes $sp^{2}$ again and RESTORES aromaticity.

Because adding a nucleophile to the carbocation would give an ADDITION product, in which two carbons would be $sp^{3}$ and aromaticity permanently lost. Losing a proton, on the other hand, costs almost nothing and recovers the 152 kJ/mol of resonance energy. The system therefore takes this route every time. This is the whole explanation of the behaviour of aromatic compounds, and it comes down to this energy comparison.

Sulfuric acid, the stronger acid, protonates nitric acid, which then loses a water molecule to produce the NITRONIUM ion $NO_{2}^{+}$, the true electrophile of the reaction: $HNO_{3}+H_{2}SO_{4}\rightarrow NO_{2}^{+}+HSO_{4}^{-}+H_{2}O$. Without it, nitric acid alone is too weak an electrophile to attack so unreactive a ring.

HALOGENATION, by $Cl_{2}$ or $Br_{2}$ with a Lewis acid such as $FeCl_{3}$ or $FeBr_{3}$, which polarizes the halogen molecule. Friedel-Crafts ALKYLATION, by an alkyl halide with $AlCl_{3}$, which generates a carbocation. Friedel-Crafts ACYLATION, by an acyl chloride with $AlCl_{3}$, which generates an acylium ion. In all three cases, the catalyst serves to create or strengthen an electrophile that is strong enough.

The FIRST step is rate-determining. It is the one that destroys aromaticity to form the carbocation, which requires crossing a high barrier; its activation energy is by far the larger of the two. The second, the loss of the proton, is on the contrary very fast and strongly exothermic since it restores aromaticity. This asymmetry has a major practical consequence: the rate of the reaction, like the orientation of the product, is entirely governed by the stability of the intermediate carbocation. That is why drawing this carbocation is enough to predict the directing effects, the subject of exercise 4.

ACTIVATING: $-CH_{3}$, weakly, by an electron-donating inductive effect; $-OH$ and $-NH_{2}$, strongly, by the electron-donating resonance effect of the lone pair. DEACTIVATING: $-NO_{2}$ and $-COOH$, strongly, by an electron-withdrawing effect; $-Cl$, moderately.

ORTHO-PARA directors: $-CH_{3}$, $-OH$, $-NH_{2}$ and $-Cl$. META directors: $-NO_{2}$ and $-COOH$. The general rule pairs activating with ortho-para on the one hand, and deactivating with meta on the other; the halogens are the only family that escapes it.

A donor group enriches the ring in electrons, which stabilizes the positively charged intermediate carbocation. But this stabilization is possible only if the positive charge can end up ON the carbon that bears the donor group. When we draw the resonance forms of the arenium ion, the charge is spread over the ortho and para positions relative to the point of attack. Attacking ortho or para to the donor therefore places the charge, in one of the resonance forms, precisely on the substituted carbon, where the donor can neutralize it: this extra form lowers the energy considerably. Meta attack does not allow this. The ortho-para path is therefore faster.

Because chlorine exerts TWO effects in opposite directions. By its INDUCTIVE effect, it is very electronegative and draws electrons out of the ring, which depletes it and SLOWS every substitution: hence its deactivating character. By its RESONANCE effect, it has lone pairs that it can donate to the $\pi$ system, which stabilizes the carbocation when the charge sits on the carbon bearing it, that is, for ortho and para attack. The inductive effect wins out on the overall RATE; the resonance effect decides the ORIENTATION. Since the two questions are independent, there is no contradiction: the first concerns the height of the barrier, the second which of the three barriers is the lowest.

For 1,3-DINITROBENZENE, the two groups are meta to each other. Nitrate first, then nitrate again: the first $-NO_{2}$, a meta director, sends the second to position 3. The order is imposed and there is no alternative. For 4-NITROTOLUENE, the methyl and nitro groups are para to each other. You must ALKYLATE FIRST, to install the methyl, then nitrate: the methyl, an ortho-para director, sends the nitro group to position 4. The reverse order would fail twice over, because the nitro group would send the methyl meta, and above all because Friedel-Crafts alkylation does not work on a strongly deactivated ring. The general lesson is that the order of the steps is never a matter of indifference in aromatic synthesis.

With the acyl chloride: $CH_{3}CH_{2}COCl+AlCl_{3}\rightarrow CH_{3}CH_{2}CO^{+}+AlCl_{4}^{-}$. The electrophile is the acylium ion, here the propanoylium ion. With the alkyl halide: $CH_{3}CH_{2}CH_{2}Cl+AlCl_{3}\rightarrow CH_{3}CH_{2}CH_{2}^{+}+AlCl_{4}^{-}$. The electrophile is a carbocation, here the primary propyl cation. In both cases, aluminum chloride merely captures the chloride ion: it is regenerated at the end and does act as a catalyst, even though it is used in large quantity.

The propyl cation formed in a) is PRIMARY, hence very unstable. A hydrogen on the neighbouring carbon migrates with its electron pair, in what is called a hydride shift, and turns the primary cation into the isopropyl cation, which is secondary and far more stable. It is this rearranged cation that attacks the ring, hence the isopropylbenzene. The rearrangement is practically complete here because it is very favourable and much faster than attack on benzene, a mediocre nucleophile. In acylation, nothing of the kind can happen: the acylium ion is stabilized by resonance, the positive charge being shared between carbon and oxygen, which gives it some carbon-oxygen triple-bond character. It therefore has no reason to rearrange, and the skeleton introduced is exactly that of the reagent.

An alkyl group is a DONOR by inductive effect and by hyperconjugation: it activates the ring. The monoalkylated product is therefore more reactive than the starting benzene, and it gets attacked a second and then a third time before all the benzene has reacted. The result is a mixture, and a large excess of benzene is needed to limit the damage. An acyl group, on the contrary, is strongly WITHDRAWING: the ketone formed is markedly less reactive than the starting benzene, so it protects itself and the reaction stops by itself at the monosubstituted product. This is a rare and valuable case where the deactivation of the product is an advantage.

The rearrangement is bypassed by going through acylation, which is faithful to the skeleton, and then reducing the ketone group. First step: benzene plus propanoyl chloride in the presence of aluminum chloride, giving 1-phenylpropan-1-one cleanly and without polysubstitution. Second step: Clemmensen reduction, with zinc amalgam in hydrochloric acid, or Wolff-Kishner reduction under basic conditions, which convert the carbonyl group into a $CH_{2}$ group without touching the ring. The result is pure propylbenzene. This acylation-then-reduction sequence is the standard method for installing a straight alkyl chain on an aromatic ring. As for nitrobenzene, it carries a very strongly deactivating group: the ring is too electron-poor to attack an electrophile as weak as a carbocation, and in addition the nitrogen and oxygen of the nitro group complex aluminum chloride and neutralize it. No Friedel-Crafts reaction works on a ring bearing a strong deactivator, which fixes the order of the steps in any synthesis: acylate or alkylate BEFORE nitrating, never after.

The mechanism has a first step in which the ring attacks the electrophile, giving the arenium ion, also called the sigma complex, then a second step in which the proton carried by the attacked carbon is lost, restoring aromaticity. The $C-H$ bond is broken only in the SECOND step. A ratio $k_{H}/k_{D}$ equal to 1.0 means that replacing the hydrogens by deuteriums changes the rate not at all: breaking this bond therefore does not occur in the slow step. It is thus the FIRST step, the formation of the arenium ion, that is rate-determining. The experiment settles the question unambiguously, and it would have been just as conclusive the other way: a ratio close to 7 would have forced the opposite conclusion.

Yes, the energy profile made it predictable. The first step destroys aromaticity: the ring loses its resonance energy, about 150 kJ/mol for benzene, which makes it strongly endothermic and gives it a high barrier. The second step, on the contrary, RESTORES aromaticity: it recovers that same resonance energy, so it is very exothermic and its barrier is low. On the diagram, the first peak is clearly higher than the second, and it is the one that controls the overall rate. All of the reasoning on directing effects follows from this: we discuss the stability of the arenium ion because it is the intermediate whose formation is the slow step, and the Hammond postulate allows us to carry its stability over to that of the transition state leading to it.

Sulfonation is the only one of the usual electrophilic substitutions that is reversible under accessible conditions, because the $SO_{3}H$ group is a good leaving group in the form of sulfur trioxide, at once hydrated to sulfuric acid by the water of the medium, and because the energy balance of the reaction is close to zero. The others, nitration, halogenation and alkylation, form much stronger bonds and are practically irreversible. The equilibrium is shifted by Le Chatelier's principle, by adjusting the water: in concentrated or fuming sulfuric acid, water is absent and the equilibrium is pushed toward the sulfonated arene; in dilute acid and hot, the excess water pulls the equilibrium the other way and the sulfonic acid group leaves. The same reagent, at two different dilutions, thus performs the reaction and its opposite.

This reversibility makes the sulfonic acid group a temporary BLOCKING GROUP: you install it where you do not want substitution, carry out the real reaction elsewhere, then remove it. First step: sulfonation of toluene with concentrated sulfuric acid. The methyl group directs ortho and para, and the very bulky sulfonic acid group goes almost exclusively para. The position that bromination would have attacked is now occupied. Second step: bromination with dibromine in the presence of iron(III) bromide. With the para position blocked, the electrophile has only the positions ortho to the methyl group left, and 3-bromo-4-methylbenzenesulfonic acid is obtained. Third step: heating in dilute sulfuric acid, which hydrolyzes the carbon-sulfur bond and releases the sulfonic acid group. What remains is pure 2-bromotoluene. The general approach, protect, transform, deprotect, is one of the most fruitful in all of organic synthesis, and sulfonation is its most economical example, since the protecting group costs only sulfuric acid.

In order of increasing ease of removal, that is of decreasing bond energy: primary 423, secondary 412, tertiary 400, benzylic 375, allylic 372. The allylic and benzylic hydrogens are by far the easiest to remove, and the difference with an ordinary primary hydrogen is $423-375=48$ kJ/mol for the benzylic one. The reason is the same in both cases: the radical formed is not localized on a single carbon. The allylic radical is delocalized over three carbons by its two resonance forms, the benzylic radical over the benzylic carbon and the ortho and para positions of the ring, four resonance forms in all. A delocalized unpaired electron is far more stable than a confined one, and the bond that released it is weakened accordingly.

Light or a peroxide initiates a FREE-RADICAL reaction, not an ionic one. The bromine radical removes the weakest hydrogen available, that is, a benzylic hydrogen, and never a ring hydrogen, whose aromatic $C-H$ bond is very strong. The benzylic radical formed then reacts with dibromine and gives benzyl bromide while regenerating a bromine radical: it is a chain reaction. Bromination of the RING requires, on the contrary, ionic conditions, with a Lewis acid such as iron(III) bromide to polarize the dibromine molecule and generate an electrophile, and darkness. The same toluene and the same dibromine thus give two entirely different products depending on whether you shine light or add a catalyst: light attacks the chain, the Lewis acid attacks the ring. This is one of the cases where the experimental conditions are more informative than the reagents themselves.

Hot permanganate oxidation cuts the side chain right at the ring and converts the benzylic carbon into a carboxyl group, whatever the length of the chain: everything beyond it is oxidized to carbon dioxide. Toluene and propylbenzene therefore both give benzoic acid, which incidentally means that this reaction erases the information about the original chain. The indispensable condition is at least one hydrogen on the benzylic carbon, because oxidation begins by removing it. Now tert-butylbenzene has a benzylic carbon fully substituted by three methyl groups, without a single hydrogen: the reaction cannot start and the compound remains intact. This difference in behaviour is, moreover, used as a test.

para-Xylene $C_{8}H_{10}$ has two methyl groups, both oxidized: $C_{8}H_{10}+3\,O_{2}\rightarrow C_{8}H_{6}O_{4}+2\,H_{2}O$, terephthalic acid being $HOOC-C_{6}H_{4}-COOH$, of formula $C_{8}H_{6}O_{4}$. The carbon count is conserved, 8 on each side, which is normal since no carbon is lost: the methyl groups carry no excess carbon to eliminate. For benzyl alcohol, we exploit the weakness of the benzylic $C-H$ bond established in a). First step: free-radical bromination of toluene with N-bromosuccinimide in the presence of a peroxide, which gives benzyl bromide $C_{6}H_{5}CH_{2}Br$ without touching the ring. Second step: substitution of the bromine by hydroxide ion in aqueous solution; the benzylic carbon is primary and little hindered, and its transition state is stabilized by the ring, so the substitution is fast and gives benzyl alcohol. Direct oxidation with permanganate never stops at the alcohol: as in c), it carries the benzylic carbon through to the carboxyl group. The route therefore goes through the alkyl halide, which installs the functional group at the desired oxidation state.

First step: nitration of benzene with the mixture of nitric and sulfuric acids, giving nitrobenzene. Second step: reduction of the nitro group to an amino group, with tin in hydrochloric acid, with iron and hydrochloric acid, or by catalytic hydrogenation, giving aniline. Third step: diazotization, with sodium nitrite in hydrochloric acid, giving benzenediazonium chloride. The low temperature is essential because the diazonium salt is thermally unstable: above about ten degrees, it spontaneously loses its dinitrogen, which produces an extremely reactive phenyl cation and destroys the reagent before it can be used. These salts are therefore prepared in an ice bath and used immediately, without ever being isolated.

With $CuCl$ we get chlorobenzene, with $CuBr$ bromobenzene: these are the Sandmeyer reactions. With $CuCN$, benzonitrile. With $KI$, iodobenzene, with no copper catalyst. With hot water, phenol. With hypophosphorous acid, benzene itself, the group simply being replaced by a hydrogen: this is a deamination. Of these six products, chlorobenzene and bromobenzene were obtainable by direct halogenation, but the other four were not. There is no electrophilic substitution that installs an iodine, a hydroxyl group or a nitrile group on a ring, and deamination obviously has no equivalent. This is what gives the diazonium relay its value: it gives access to substituents that ring chemistry refuses.

The lone pair on the nitrogen of the amino group is conjugated with the ring: it delocalizes into it and massively enriches the ortho and para positions in electrons. The ring becomes so nucleophilic that it no longer needs a powerful electrophile: an unpolarized dibromine molecule is enough, whereas bare benzene requires a Lewis acid to create a much more reactive electrophile. The activation is even excessive from the chemist's point of view: the reaction does not stop at the monosubstituted product, because each bromination leaves a ring that is still strongly activated, and all three ortho and para positions are occupied at once. 2,4,6-Tribromoaniline is therefore obtained in a single operation, cold, with immediate precipitation. Controlling this reaction is, incidentally, a classic problem, solved by first acetylating the amine to temper its donating power.

Direct bromination of benzene is hopeless for this target. Bromine is deactivating, certainly, but it is an ORTHO-PARA director: the first bromine installed directs the second ortho or para on its own, never meta. You would get 1,2- and 1,4-dibromobenzene, then 1,2,4-trisubstituted derivatives, but never the 1,3,5 arrangement, which requires each bromine to be meta to the other two. No amount of repetition can fix this, because the orientation is fixed by the nature of the substituent. The solution is to obtain the desired arrangement with a group that imposes it, then erase that group. First step: nitration of benzene, giving nitrobenzene. Second step: reduction to aniline. Third step: bromination with bromine water, cold, giving 2,4,6-tribromoaniline, exactly the 1,3,5 arrangement sought, the amino group occupying the sixth position. Fourth step: cold diazotization with sodium nitrite in hydrochloric acid. Fifth step: treatment with hypophosphorous acid, which replaces the diazonium group by a hydrogen and leaves 1,3,5-tribromobenzene. The amino group does not appear in the final product: it served only to impose the orientation, then it was removed. It is a temporary directing group, a cousin of the blocking group seen earlier with sulfonation.

The $-OH$ group is strongly ACTIVATING and an ORTHO-PARA director. The major products are therefore 2-nitrophenol and 4-nitrophenol, the latter generally predominating because it is less sterically hindered. The reaction is MUCH faster than with benzene, so much so that it proceeds with dilute nitric acid and no sulfuric acid, whereas benzene requires a concentrated, heated nitrating mixture.

The $-NO_{2}$ group is strongly DEACTIVATING and a META director. The major product is 1,3-dinitrobenzene. The reaction is MUCH slower than with benzene and requires harsher conditions, fuming acid and prolonged heating. This is what makes the mononitration of benzene easy to control: as soon as a first nitro group is installed, it protects the ring against a second attack.

Target: a $-COOH$ and a $-NO_{2}$ meta to each other. The group present at the time of nitration must therefore be a META director. First OXIDIZE the methyl group to a carboxylic acid, for example with hot potassium permanganate, then NITRATE. The $-COOH$, a meta director, then sends the nitro group to position 3, and 3-nitrobenzoic acid is obtained.

Target: the two groups para to each other. This time the group present at the time of nitration must be an ORTHO-PARA director. NITRATE toluene FIRST: the methyl group, an ortho-para director, leads mainly to 4-nitrotoluene, which is separated from its ortho isomer. Then OXIDIZE the methyl group to a carboxylic acid, the oxidation not changing the positions of the substituents. 4-Nitrobenzoic acid is obtained.

The two syntheses start from the SAME reagent and use the SAME two reactions, in the reverse order, to give two different isomers. The general rule is that, in the aromatic series, the group ALREADY PRESENT dictates the position of the next one: the order of the steps is chosen according to the orientation wanted, not for convenience. In practice, you reason BACKWARD from the target, asking, for each substituent, which one must have been installed last and which director had to be present at that moment.

For naphthalene, $4n+2=10$ gives $n=2$: the condition is satisfied. For anthracene, $4n+2=14$ gives $n=3$: it is satisfied too. Both compounds are moreover cyclic, planar and fully conjugated, all their carbons being $sp^{2}$. They are therefore AROMATIC.

The degree of unsaturation is $\dfrac{2\times 10+2-8}{2}=\dfrac{22-8}{2}=7$. This number breaks down as two rings plus five double bonds, which agrees with the structural formula of naphthalene, where two fused hexagons are drawn with five double bonds in total and not six, the two rings sharing one bond.

Twice the resonance energy of benzene would be $2\times 152=304$ kJ/mol, whereas naphthalene reaches only 255. The stabilization per ring is therefore LOWER: about 128 kJ/mol per ring instead of 152. Interpretation: the two carbons at the ring fusion are shared, so delocalization is not as efficient as in two independent benzenes. Practical consequence: naphthalene is more REACTIVE than benzene. It can be hydrogenated or substituted more easily, because the sacrifice of aromaticity is less costly there.

These molecules are PLANAR, rigid and hydrophobic, with a $\pi$ cloud extending over both faces. These characteristics allow them to INTERCALATE between two successive base pairs of the DNA double helix, which are themselves flat aromatic rings stacked 340 pm apart. Intercalation locally deforms the helix and disrupts replication. For the larger ones, such as benzo[a]pyrene, there is also metabolic activation by the liver into highly electrophilic epoxides that bind covalently to the bases, causing mutations. Planarity, a property that comes directly from aromaticity, is thus the structural cause of the toxicity.

Because they form by PYROLYSIS: when organic matter burns incompletely, at high temperature and with a shortage of oxygen, carbon fragments recombine into aromatic rings, the most stable form of carbon under these conditions. This is the case for fat that drips onto the coals and rises with the smoke, for burning tobacco, and for any charred food. The simple action: avoid direct contact of the food with the flame and with blackened parts, by keeping the grill away from the coals, limiting fat dripping, and removing charred areas before eating. Cooking at a lower temperature, or precooking, greatly reduces the formation of these compounds.

The five carbons and the nitrogen are $sp^{2}$, each contributing one electron in a $p$ orbital: 6 $\pi$ electrons, $4n+2$ with $n=1$, so pyridine is aromatic. The nitrogen's lone pair occupies an $sp^{2}$ orbital IN the plane of the ring, perpendicular to the $\pi$ system: it takes no part in it and remains available.

The four carbons of the two double bonds contribute 4 electrons; the nitrogen, bonded to three atoms, places its lone pair in a conjugated $p$ orbital: $4+2=6$ $\pi$ electrons. Pyrrole is aromatic, and the nitrogen's lone pair is essential to this aromaticity.

$10^{5.2-(-3.8)}=10^{9}$: one billion. Protonating pyrrole forces its lone pair into an $N-H$ bond: it leaves the $\pi$ system, which falls to 4 electrons, and aromaticity is lost. Protonating pyridine uses a lone pair that did not take part in the ring: aromaticity is intact.

Just one. Oxygen, like the nitrogen of pyrrole, places one lone pair in a conjugated $p$ orbital, which gives $4+2=6$ $\pi$ electrons; its second lone pair stays in the plane of the ring, outside the $\pi$ system. Furan is aromatic, but less so than benzene because oxygen holds its electrons tightly.

The lone pair of piperidine is in an $sp^{3}$ orbital, 25% $s$ character; that of pyridine is in an $sp^{2}$ orbital, 33%, closer to the nucleus and less available: hence the factor $10^{5.9}$. In imidazole, the pyrrole-type nitrogen, $NH$, cannot be protonated without destroying aromaticity; it is the pyridine-type nitrogen, with its lone pair in the plane, that is protonated. Its $pK_{a}$ of 7.0 makes imidazole, present in histidine, an ideal buffer at the pH of living systems.

Both groups are ortho-para directors, and their para positions are occupied by each other. The methoxy group would direct toward its ortho positions, the methyl toward its own. The methoxy group, a resonance donor, is far more activating than the methyl: it decides. The nitro group goes ortho to the methoxy, and the major product is 1-methoxy-4-methyl-2-nitrobenzene.

By symmetry, positions 4 and 6 are equivalent; what remain are position 2, between the methyl groups, and position 5, meta to both: 3 different positions. Position 4 is ortho to one methyl and para to the other, and the two effects add up there; position 5 is not activated by either. Major product: 1-bromo-2,4-dimethylbenzene.

The methyl group directs toward its ortho positions, 2 and 6; the nitro group, a meta director, directs toward the positions meta to it, which are exactly the same. The effects reinforce each other: a single product, 2-bromo-1-methyl-4-nitrobenzene.

The methoxy group, at position 4, directs toward its ortho positions, 3 and 5; the carboxyl group, at position 1, directs toward its meta positions, which are also 3 and 5. The two effects reinforce each other: the nitro group goes to position 3, and 4-methoxy-3-nitrobenzoic acid is obtained.

Position 2 is flanked by the two methyl groups: the electrophile, and then the group installed, would be squeezed between two neighbours there. Electronically favourable but sterically penalized, it loses out to position 4, which is also activated by both methyl groups but open on one side.

Step 1: $NH_{2}^{-}$ removes a hydrogen ortho to the chlorine, forming an aryl carbanion. Step 2: the chloride leaves, and the two neighbouring carbons form a strained triple bond: benzyne. This is an ELIMINATION. Step 3: $NH_{2}^{-}$ adds to one of the two carbons of the triple bond, and the carbanion formed picks up a proton from ammonia: this is an ADDITION. The mechanism is an elimination-addition.

In benzyne, the two carbons of the triple bond are the labelled carbon and its neighbour, equivalent by symmetry. The amide attacks one or the other with equal probability: 50% of the aniline carries the amino group on the labelled carbon, 50% on the neighbouring carbon. This fifty-fifty split, observed by Roberts in 1953, is the proof of the symmetrical intermediate.

The chlorine is at position 4; its two neighbours, carbons 3 and 5, are equivalent. Benzyne connects carbons 3 and 4, and the amide adds to one or the other: 2 products, 4-methylaniline and 3-methylaniline.

The two carbons next to the bromine each carry a methyl group and no hydrogen. The amide finds no ortho hydrogen to remove: benzyne cannot form, and the reaction does not take place. This is a direct test of the mechanism.

Here the nucleophile adds to the carbon bearing the chlorine itself, forming an anionic intermediate stabilized by the two nitro groups, then the chloride is eliminated: an addition-elimination. The nucleophile always arrives on the carbon that carried the leaving group. The benzyne mechanism, by contrast, can move it to the neighbouring carbon, and it requires no electron-withdrawing group, but a very strong base.

$DI=\frac{2(8)+2-10}{2}=4$: exactly one benzene ring, the rest is saturated. Two carbons outside the ring: either one ethyl group, ethylbenzene, or two methyl groups, ortho, meta or para: the o-, m- and p-xylenes. Four isomers.

p-Xylene: the four ring hydrogens are equivalent, 1 derivative. o-Xylene: two pairs of equivalent hydrogens, positions 3 and 6 on one hand, 4 and 5 on the other, 2 derivatives. m-Xylene: positions 4 and 6 are equivalent, while positions 2 and 5 are each unique, 3 derivatives. Ethylbenzene: ortho, meta and para, 3 derivatives.

A single derivative means maximum symmetry: X is p-xylene. Two derivatives: W is o-xylene.

Permanganate converts every chain bearing a benzylic hydrogen into a $COOH$ group. A single carboxyl group, benzoic acid, means a single chain: Y is ethylbenzene. A diacid means two chains: Z is m-xylene, oxidized to isophthalic acid.

$M(C_{8}H_{10})=106.16$ g/mol, $n=\frac{2.12}{106.16}=0.0200$ mol. $M(C_{8}H_{6}O_{4})=166.13$ g/mol, theoretical mass $0.0200\times 166.13=3.32$ g. Yield: $\frac{2.99}{3.32}=90.1$ %.

The isobutyl group is an alkyl group, activating and an ortho-para director. The ortho positions are hindered by this bulky group, and the acylium ion is itself bulky: the para product is by far the major one. In addition, acylation, unlike alkylation, stops at monosubstitution.

The primary isobutyl cation would immediately rearrange, by a hydride shift, to the tert-butyl cation, which is tertiary: tert-butylbenzene would be obtained and not isobutylbenzene. This is why the route goes through an acylation followed by a reduction, which keep the skeleton.

$M=13\times 12.01+18\times 1.008+2\times 16.00=206.27$ g/mol. The carbon that bears the methyl group, the hydrogen, the ring and the carboxyl group is asymmetric: 1 asymmetric carbon, 2 stereoisomers, (R) and (S).

Reactants: $134.22+102.09+2.016+28.01=266.34$ g/mol. Atom economy: $\frac{206.27}{266.34}=77.4$ %. The only mass lost is acetic acid, $266.34-206.27=60.07$ g/mol, which is recovered and sold: the old six-step process peaked around 40%.

$n=\frac{0.200}{206.27}=9.70\times 10^{-4}$ mol. Racemic: half of each enantiomer, so 100 mg of (S). The body converts part of the (R) into (S), which explains why the racemic mixture is still sold.