15 and Ultraviolet Spectroscopy

Predict whether the compound will be aromatic, antiaromatic, or non-aromatic by determining whether Huckel's rule applies to a given structure.

Show how to make a system similar to benzene and cyclobutadiene.

Predict if a structure will be aromatic.

If nitrogen's lone pairs are used in the aromatic system, you can determine whether the nitrogen atom is weakly basic or strongly basic.

The structures of aromatic compounds can be determined using IR, NMR, UV, and mass spectrums.

Predict the distinguishing features of the compound's spectrum.

The breaking strength of Zylon is 1.6 times greater than that of Kevlar.

It's highly resistant to stretching and is also heat resistant.

The structure is composed of aromatic rings with single bonds.

Michael Faraday isolated a pure compound of boiling point 80 degC from the oily mixture that came from illuminating gas.

An empirical formula of CH was found in the hydrogen-to-carbon ratio of1:1.

The same compound was created in 1834 by heating benzoic acid, isolated from gum benzoin, in the presence of lime.

The empirical formula was CH.

In the 19th century there were many compounds that were related to benzene.

These compounds had low hydrogen-to-carbon ratios and could be converted to benzene.

Friedrich Kekule proposed a structure for ben zene with three double bonds.

The structure with alternating single and double bonds was considered odd because it was only recently proposed.

Only one of the different 1,2-dichlorobenzenes is known to exist.

Kekule's hypothesis is that the resonance picture of benzene is natural.

The double bonds are longer than the single bonds.
The bonds of the benzene ring are the same length as the ring itself.
The ring and carbon nuclei are positioned at the same distances, so there is only one difference.

Benzene is similar to the two Kekule structures.

Because the pi bonds this problem and has a dream in which atoms are delocalized over the ring, we often inscribe a circle in the hexagon.

Three double bonds were twisted in a snake-like motion.

This representation helps us remember that there is no local and that one of the snakes seized a single or double bond, which prevents us from drawing a different tail.
The isomers of Kekule differ in the placement of double bonds.
The movement with three double bonds is explained by individual pairs of electrons.

A more realistic representation of ben zene can be drawn using this resonance picture.

Two carbon atoms bond to one hydrogen atom.
All the carbon-carbon bonds are the same length and angle.

We will look at how aromatic compounds differ from aliphatic compounds.

An aromatic structure confers extra stability and we can predict aromaticity in some interesting and unusual compounds.

The Kekule structure and the resonance-delocalized picture show that benzene is a long-term exposure.

The reactions of polyenes might be expected of benzene.
Cause leukemia is characterized by its reactions that are quite unusual.
An alkene reacts to form a glycol and decreases the number of red blood ganate.
There is a purple permanganate color and an increase in the number of cells.
White blood cells are malfunctioning when permanganate is added.

Red and white blood cells are made of H2O.

Most alkenes decolorize solutions of bromine in carbon tetrachloride.

As bromine adds to the double bond, the red color disappears.

There is no reaction when bromine is added to benzene.

The H 120deg bond angles are exactly the same as the 1.397 A carbon-carbon bond lengths.

The bromine color disappears slowly when ferric bromide is added to the mixture.

Benzene's reluctance to undergo typical alkene reactions suggests that it must be stable.

We can get a quantitative idea of hydrogenation's stability by comparing its molar heats.
All hydrogenate is used to form cyclohexane.

It takes 120 kJ>mol to oxidize cyclohexene.

The heat of hydrogenation of cyclohexene is twice that of exothermic hydrogenation.

The isolated double bonds have zero resonance energy.

The value for cyclohexene is twice as high as that for cyclohexa-1,3-diene.

A resonance energy of 8 kJ is typical for a diene.

Higher pressures of hydrogen and a more active catalyst are required for hydrogenation of benzene.

The value for cyclohexene is three times that of this hydrogenation.

The resonance energy of benzene can't be explained by the conjugate effects alone.

The heat of hydrogenation for benzene is not as hot as it is for cyclohexa-1,3-diene.
It is difficult to stop the reaction after 1 mole of H2 because the product hydrogenates more easily than benzene.
The benzene ring is notreactive.

For a long time, chemists assumed that ben zene's large resonance energy came from two stable resonance structures.

They thought that the stability of other hydrocarbons would be similar.

There are cyclobutadiene, cyclooctatetraene, and larger annulenes named similarly.

Two Kekule-like structures that seem to show a benzene-like reso bonds can be drawn if an annulene is assumed to be have alternating single and double planar.

The compounds were supposed to be aromatic.

The results imply that the simple resonance picture is not correct.

Cyclobutadiene has never been isolated.

It undergoes a very fast Diels-Alder dimerization.
To avoid the Diels-Alder reaction, cyclobutadiene has been prepared at low concentrations in the gas phase.

The evidence shows that cyclooctatetraene is not as stable as benzene.

Structural studies have shown that cyclooctatetraene is not a planar substance.
Poor overlap between adjacent pi bonds makes it most stable in a "tub" conformation.

The carbon and hydrogen atoms are shown in the resonance forms of benzene, cyclobutadiene, and cyclooctatetraene.

Two hybrid carbon atoms form continuous rings above and below the plane of the carbon atoms.

The stability of the aromatic ring cannot be fully explained by the idea of benzene being a resonance hybrid of two Kekule structures.

The key to understanding aromaticity and predicting which compounds will have the stability of an aromatic system is provided by molecular orbital theory.

Linear systems such as buta-1,3-diene and the allyl system are Cyclic systems.

There is a chance of two distinct MOs having the same energy in a two-dimensional system.
We can still follow the same principles in the representation of benzene.

There must be six orbitals.

The lowest-lying MO in this is benzene.

There is a possibility of nonbonding MOs in some cases.

From above, you can see the six p molecular orbitals of benzene.

At the second energy level, both p2 and p3 have one nodal plane.

For a total of two net bonding interactions, p2 has four bonding interactions and two antibonding interactions.
p3 has two bonding interactions and four nonbonding interac tions.
There are no antibonding interactions in p3 but there is no electron density on the two carbon atoms.
Although we can't use the number of bonding and antibonding interactions as a measure of an orbital's energy, it is clear that p2 and p3 are bonding more strongly than p1.

Both 4 and p5 have two nodal planes in them.

There are two antibonding interactions and four nonbonding interactions in the p*4 orbital.

p*5 has four antibonding interactions and two bonding interactions, for a total of two antibonding interactions.

p2 and p3 are bonding, but 4 and p5 are not.

The p*6 has three nodal planes.

The lowest and highest in energy are 1 and p6 respectively.

There are three pi bonds in the Kekule structure for benzene.

The unusual stability of benzene is explained by this electronic configuration.

The first MO is stable and all-bonding.

Experiments show that cyclobutadiene is unstable.

The all-bonding MO is p1.

The p2 and p3 are both symmetrically situated nodal planes.

There are two bonding interactions and two antibonding interactions.
The net bonding order is zero.
p*4 has two nodal planes and is antibonding.

The four cyclobutadiene MOs are shown in Figure 16-7.

The lowest-lying MO is strongly bonding, and the highest-lying is equally antibonding.

All antibonding cyclobutadiene has pi molecular orbitals.

The lowest-lying orbital has two electrons filling it.

There are different degenerate orbitals with unpaired spins.

Its highest-lying electrons are in nonbonding orbitals.

The dramatic stability difference between benzene and cyclobutadiene was predicted by the molecular orbital theory.

The lowest-lying MO is the only one that doesn't have a nodes, until only one highest-lying MO suggests that the true remains.

The energy diagram in benzene looks like a ring.

The pattern in cyclobutadiene looks like a diamond.

The center of the polygon is given a horizontal direction by the polygon rule.

The pi electrons are filled into the orbitals by filling the lowest-energy ones first.

Our definition of aromatic compounds includes compounds with double bonds with large resonance energies.

We can be more specific about the properties that are required for a compound to be aromatic.

The structure must have some number of pi bonds.

For effective overlap to occur, the structure must be nearly planar.

Open-chain structures are more stable than aromatic structures.

benzene is more stable than hexa-1,3,5-triene.

Buta-1,3-diene is more stable than cyclobutadiene.

It has the same electronic energy as the open-chain counterpart.

The CH3 was developed in 1931 to predict which compounds are aromatic and which are antiaromatic.

We need to be certain that the compound under consideration meets the criteria for an aromatic or antiaromatic system.

The system is aromatic.

The rule predicts benzene to be aromatic.

The rule predicts cyclobutadiene to be antiaromatic.

The classical structure has eight pi electrons and four double bonds.

If the rule was applied to cyclooctatetraene, it would show antiaromaticity.
It doesn't show the high reactivity associ Huckel's rule, which is commonly used with antiaromaticity.
The reactions are typical of alkenes.

Its double bonds are unfavorable.

It's possible to apply cyclooctatetraene instead of cyclobutadiene.
The system assumes a nonplanar "tub" that avoids most of the overlap between nonaromatic and nonplanar.

The rule does not apply.

Make a model of cyclooctatetraene.

They all react as polyenes.

The molecule can adopt the necessary planar conformation.

An excessive amount of angle strain is required in the all-cis [10]annulene.
Two hydrogen atoms interfere with each other, so the isomer with two trans double bonds cannot be a planar one.
The partially trans isomer of aromatic compounds can be removed.

Predicting planarity is difficult.

pi electrons can achieve certain structures.

The following two compounds have aromatic properties.

The following compounds are classified as aromatic, antiaromatic, or nonaromatic.

One of the compounds is more stable than the other two.

Benzene is aromatic because it has equal-energy orbitals.

The p2 and p3 are filled with electrons.
Cyclobutadiene has an open shell of electrons.
There are two half-filled orbitals that can accept and donate electrons.
Under general conditions, we must show what a filled shell of orbitals is.

There is a pattern of orbitals in the system.

The lowestlying MO is filled with two electrons.
The additional shells have two MOs and space for four electrons.
It will have a filled shell.

There is a pattern of MOs in the system.

The lowest-lying MO is always filled with two electrons.
Four electrons are required to fill a shell after two steps.

If the system has more than one ring.

There is a half-filled shell at the bottom.

The energy of one of the MOs is determined by the height of each vertex.

Draw a nonbonding line.

This is an electronic configuration.

The antibonding MOs are difficult to draw and are usually empty in stable to draw.

The system has pi electrons.

We have talked about aromaticity using the examples of the annulenes.

There are even numbers of carbon atoms with alternating single and double bonds.

The rule applies to systems with odd numbers of carbon atoms.
We now look at aromatic ion and their antiaromatic counterparts.

The system would be neutral if there were five pi electrons, but it would be radical because there are not enough pairs.

The system is predicted to be antiaromatic with four pi electrons.
The rule predicts aromaticity with six pi electrons.

The cyclopentadienyl anion is stable because it is aromatic.

It can be formed by abstracting a protons from cyclopentadiene.
It is nearly as acidic as water and more acidic than alcohol.

The non aromatic diene is converted to the aromatic cyclopentadienyl anion by the loss of a protons.

When we say that the cyclopentadienyl anion is aromatic, we don't mean that it is stable like benzene.

The cyclopentadienyl anion is a carbanion.
This ion is more stable than the open-chain ion because it is aromatic.

The cyclopentadienyl cation has four pi electrons.

The cyclopentadienyl cation is not easy to form.
Even in concentrated sulfuric acid, cyclopenta-2,4-dien-1-ol does not lose water.
The cation is too unstable.

If we use a simple resonance approach, we might not expect the cyclopentadienyl ion to be stable.

The resonance less stable more stable approach is not a good predictor of stability.

The Huckel's rule is a better predictor of stability for aromatic and antiaromatic systems.

There is a misleading suggestion of stability given by the resonance picture.

The cyclopropenyl case requires the drawing of the molecular orbitals.

Add the nonbonding line to each MO to label them as bonding, nonbonding, or antibonding.

It goes through the average of the MOs.

Adding electrons to your energy diagram will show the configuration of the cyclopropenyl cation and the cyclopropenyl anion.

To confirm the electronic configurations of the cyclopentadienyl cation and anion, fill in the electrons and draw the energy diagram.

The anion has eight pi electrons, while the cation has six.

We can draw resonance forms that show either the positive charge of the cation or the negative charge of the anion over the seven atoms of the ring.

We know that the six-electron system is aromatic and the eight-electron system is antiaromatic.

There is a misleading suggestion of stability given by the resonance picture.

The cycloheptatrienyl cation can be formed by treating the alcohol with a small amount of sulfuric acid.

This is the first example of a stable cation.

The tropylium salts can be isolated and stored for a long time.

The tropylium ion is not as stable as benzene.
It means that the ion is more stable than less stable.

The result agrees with the prediction that the cycloheptatrienyl anion is antiaromatic.

The cyclopentadienyl anion is stable because of aromatic stabilization.

Dianions of hydrocarbons are very difficult to form.
An aromatic dianion is formed when cyclooctatetraene reacts with potassium metal.

The aromatic cyclooc tatetraene dianion is easy to prepare.

Explain why each compound should be aromatic.

AgBF4 is used to treat 3-chlorocyclopropene.

The organic product can be obtained in a variety of ways.
The original 3-chlorocyclopropene is regenerated when the material is dissolved in nitromethane.
Determine the structure of the material and write equations for its formation and reaction.

The stability of cyclopropenone and cycloheptatrienone is better than anticipated.

Cyclopentadienone is unstable and undergoes a Diels-Alder dimerization very quickly.

The applications of the rule are summarized in this list.

The aromatic 2, 6, and 10 p electron systems are classified according to the number of pi electrons, while the 4 and 8 p electron systems are classified according to the number of pi electrons.

The atoms of other elements can also be aromatic.

Nitrogen, oxygen, and sulfur make up the majority of the aromatic compounds.

Pyridine is an aromatic nitrogen analogue.

The ring has six pi electrons.
H units of benzene and the nonbonding pair of electrons on nitrogen replace benzene's bond to a hydrogen atom.
The pi system does not overlap with them.

The characteristics of aromatic compounds are shown by Pyridine.

It has a reso nance energy of 113 kJ>mol and it usually undergoes substitution rather than addition.
pyridine is basic because it has a pair of nonbonding electrons.

There are six delocalized electrons in Pyridine.

They don't interact with the pi electrons of the ring.

N+ H + OH electrons can be used to abstract a particle.

The pyridinium ion is still aromatic.

Pyrrole has only four pi elec storage of oxygen.

The nitrogen atom has a single pair of electrons.

The resonance energy of Pyrrole is 92 kJ>mol.

Explain how pyrrole works.

The pyrrole structure has a charge distribution.

The structure of the pyrrole is to blame for the 2CH2COO difference.

One of the electron pairs in the aromatic sextet is needed for a N bond to be made.

The pyrrole is not aromatic.

At the 2-position, CH3 and myoglobin make pyrrole by binding it to one of the carbon atoms of the ring.

Pyrrole is a weak base.

The nitrogen atoms are 888-609- 888-609- 888-609- 888-609-

The lone pairs are not needed for the aromatic sextet.

The lone pair is basic and not involved in the aromatic system.

The aromatic sextet is part of the 2 orbital bond to hydrogen.

Nitrogen is not very basic.
The two nitrogens become equivalent once imidazole is protonsated.
Nitrogen can lose a protons and return to an imidazole molecule.

There are three basic nitrogen atoms and one pyrrole-like nitrogen in Purine.

Pyrimidine and purine derivatives are used to specify the genetic code.

Imidazole derivatives increase the activity of enzymes.
Chapters 23 and 24 will show us more about these important derivatives.

Purine and pyrimidine analogs can be used as anti-cancer drugs.

The chemical shifts shown are due to the blocking of the The proton NMR spectrum of 2-pyridone by 5-fluorouracil.

Many cancer cells as well as some healthy cells are killed by the key base in DNA.

You can use resonance forms to explain your answer.

One of the bases in DNA is thymine.

The box at the side of the page shows the structure of 5-fluorouracil.

Furan is an aromatic five-membered Heterocycle similar to pyrrole, but in furan the to acid rain, because of the release of SO2 from the coal.

Heteroatom is the difference between nitrogen and oxygen.
The oxygen atom has two lone pairs of electrons, as shown in the classical structure for furan.

67 kJ>mol cleaner-burning product has a resonance energy of Furan.

Pyrrole, furan, and thiophene are isoelectronic.

The resonance energy of thiophene is 121 kJ>mol.

Explain why each compound is aromatic.

Pyrrole and furan are examples.

B3N3H6 is a stable compound.

Explain why borazole is aromatic.

The simplest fused aromatic compound is Naphthalene.

We use one of the Kekule resonance structures or the circle notation for the aromatic rings to represent naphthalene.

There are two aromatic rings in naphthalene.

Two iso lated aromatic rings have 6 pi electrons in them.
The smaller amount of electron density gives naphthalene less resonance energy than benzene.

The compounds become more reactive as the number of rings increases.

There is a resonance energy of 347 kJ>mol or 84 kcal>mol per aromatic ring.
There is a slightly higher resonance energy of 381 kJ>mol for phenanthrene.

Each of these compounds has only 14 pi electrons in its three aromatic rings.

Anthracene and phenan threne can undergo addition reactions that are more characteristic of their polyene relatives because they are not as stable as benzene.

Anthracene undergoes 1,4-addition at the 9- and 10-positions to give a product with two isolated, fully aromatic benzene rings.
phenanthrene is added at the 9- and 10-positions to give a product with two aromatic rings.

There are two additions shown.

9-bromophenanthrene results when the product from (c) is heated.

There is a mechanism for this.

The black material in diesel exhaust has more fused rings than anthracene and phenanthrene, and they have less resonance because of small particles that are rich energy per ring.

Most of the large PAHs must be drawn with polynuclear aromatic hydrocarbons.

Three compounds are present in tobacco smoke.

These compounds are so hazardous that laboratories must install special containment facilities to work with them, yet smokers expose their lung tissues to them every time they smoke a cigarette.

Its effects appear to be caused by its exposure to arene oxides, which can be attacked by DNA.

The resulting derivatives can't be transcribed.
They cause errors that cause the genes to change.

One of the oldest forms of pure carbon is graphite.

The stability of the old and new forms of carbon is dependent on aromaticity.

We don't usually think of carbon as an organic compound.

"Amorphous carbon" means charcoal, soot, coal, and carbon black.

Most of the mate rials are microcrystalline.
Small particle sizes and large surface areas are what they are characterized by.
The small particles absorb gases and solutes from solution and form strong, stable dispersions in polymers, such as the dispersion of carbon black in tires.

The hardest naturally occurring substance is diamond.

Diamond has a three-dimensional lattice with carbon atoms linked together.
The lattice extends throughout the crystal so that a diamond is one giant molecule.
They are unable to carry a current.

Diamond is a lattice of carbon atoms.

There are layers of aromatic rings.
A single layer of Graphene is one atom thick.

The distance between layers is 3.35 A, which is twice the van der Waals radius for carbon, suggesting there is little or no bonding between layers.

The layers can slide across each other, making it a good lubricant.
It is a good electrical conductor parallel to the layers, but it resists electrical currents when they are parallel to the layers.

We think of each layer of graphite as a lattice of aromatic rings.

There are no bonds between layers and all the valences are satisfied.

The layers are held together by only van der Waals forces.
The pi electrons within a layer can conduct electrical currents parallel to the layer, but electrons cannot easily jump between layers, so they are resistive to the layers.

The conversion of diamond to graphite is very slow for those who have invested in it.

The higher density of diamond suggests that it might be con verted to diamond under high pressures.
Small industrial diamonds can be made using catalysts such as Cr and Fe, which can be made using pressures over 125,000 atm.

The University of Manchester's Andre Geim and Konstantin Novoselov won the 2010 Nobel Prize in physics for their work on Graphene, which is a single layer of graphite one atom thick.

They pulled one layer away from the surface of the piece.
Graphene is an excellent conductor.
If it can ever be mass-produced in large sheets, it holds great promise for touch-screen monitors.

A molecule of formula C60 was isolated from the soot produced by using a laser and an electric arcs.

There are only two types of bonds in C60, and it has only one type of carbon atom.

A soccer ball has the same structure as C60, with each atom representing a carbon atom.

The carbon atoms are the same.
Two six-membered rings and one five-membered ring are served by each carbon.
There are two types of bonds, the bonds that are shared by a five-membered ring and a six-membered ring, and the bonds shared between two six-membered rings.
It appears that the six-membered rings are aromatic, but the double bonds are partially hidden between the six-membered rings.
Some of the addition reactions of alkenes can be seen in the double bonds.

Half of the C60 sphere is fused to a cylinder made of six- membered rings.

The strength-to-weight ratio and the fact that they are only along the length of the tube have aroused interest.
Each year, thousands of tons of nanotubes are produced.
They are added to the cured polymers to make them stronger.
They are used to promote bone growth in tissue cultures and as tips for atomic force microscope probes.

A five-membered ring and two six-membered rings are bridgehead carbons.

A cylinder made of aromatic six-membered rings is called a nanotube.

Half of a C60 sphere is at the end of the tube.

The structure to curve at the end of the tube is caused by the five membered buckyball.

Purine is a compound with rings that share two atoms and a bond between them.

The properties of fused-ring Heterocycles are similar to those of the simple Heterocycles.

In nature, fused compounds are used to treat a wide variety of illnesses.

A member of the fluoroquinolone class is ciprofloxacin.

Benzene derivatives have been used as industrial reagents for over 100 years.

Many of their names are from the past.

Many compounds are derivatives of benzene, with their substituents being attached to an alkane.

The substitution in disubstituted benzenes can be specified with numbers.

Numbers are used to indicate the positions of three or more substituents.

If the substitution pattern is functional group that defines the base name, a structure to be C1 is assumed.

Many disubstituted benzenes have historical names.

There is no obvious connection between the structure of the molecule and some of these.

Name all the benzenes that have between one and six chlorine atoms.

Table 16-1 contains the melting points, boiling points, and densities of benzene and some derivatives.

Benzene derivatives have higher melting points than aliphatic compounds because they are more symmetrical.

Para disubstituted benzenes pack better into crystals and have higher melting points than the ortho and meta isomers.

The boiling points of many benzene derivatives are related to their dipole moments.

The lowest boiling point for Dichlorobenzene and the lowest boiling point for the Mothballs is zero.

The two compounds are boiling.

Dichlorobenzene has the lowest boiling point because it is the highest melting point of the dichlorobenzenes.

The benzenes are denser than water.

C stretch around 1600 cm-1 is a characteristic of aromatic compounds.

The aromatic bond order is only about 1 12 so this is a lower C stretching frequency.

The aromatic bond vibrates at a lower Frequency because it is less stiff than a normal double bond.

H is stretching just above 3000 cm-1.

The compounds labeled Compounds 4, 5, and 7 show aromatic rings.

The aromatic ring current protects the 1H NMR signals around d 7 to d 8.

The aromatic protons absorb around 7.2 d in benzene.

The signals may be moved further downfield bydrawing groups such as carbonyl, nitro, or cyano groups.

The aromatics that are ortho or meta split.

The spin-spin splitting constants are used for ortho protons and meta protons.

Carbon atoms absorb between 120 and 150 in the 13C spectrum.

The presence of an aromatic ring is usually confirmed by the combination of 13C NMR with 1H NMR or IR spectroscopy.

A resonance-stabilized benzylic cation can be given by the cleavage of a benzylic bond.

The aromatic tropylium ion may be given by rearranging the benzyl cation.

The ultraviolet spectrum of aromatic compounds is different from that of nonaromatic polyenes.

There are three absorptions in the ultraviolet region for benzene: an intense band at lmax of 184 nm, a moderate band at lmax of 204 nm, and a characteristic low-intensity band of multiple absorptions centered around 254 nm.

There are three major bands in the benzene spectrum.

If benzene were always an unperturbed, perfectly hexagonal structure, the weaker band at 204 nm would correspond to a "forbidden" transition.

The molar absorptivities are usually 200 to 300.

Most of the characteristics of benzene are shown in simple benzene derivatives, which are in the moderate band and benzenoid band.

The values of lmax are increased by about 5 nm by the use of alkyl and halogen substituents.

The isomer of this compound is rearranged when it is treated with a strong acid.

Suggest a structure for the isomeric product.

An organic compound is not aromatic.

Different forms of carbon are referred to as allotropic forms.

There are alternating single and double bonds.

The compound is characterized by a large resonance energy.

The benzene ring is a structural unit for aromatic hydrocarbons.

The aromatic group that remains after taking a hydrogen atom off an aromatic ring is the generic alkyl group.

The band is characterized by multiple sharp absorptions.

The arrangement of five-membered and six-membered rings is similar to a dome.

Orbitals have the same energy.

According to Marilyn Monroe, a girl's best friend.

A generic term for carbon clusters similar to C60 and compounds related to them.

One or more of the ring atoms in a compound is not carbon.

A compound that has a large resonance energy and is aromatic.

A classical formula for an aromatic compound.

A carbon tube is a cylinder of six-membered rings and half a C60 sphere.

There is a relationship on a ring.

There is a relationship on a benzene ring.

There is a relationship on a benzene ring.

The benzene ring is a substituent on another molecule.

The energy diagram of the MOs of a regular, completely conjugated system has the same shape as the compound, with one edge at the bottom.

There is a polynuclear aromatic Heterocycle.

The stabilization provided by delocalization is more than that provided by a local structure.

The extra stabilization provided by the delocalization of the electrons in the aromatic ring is called resonance energy.

Each skill is followed by problem numbers.

If you use the Huckel's rule, you can predict whether a given ion or annulene will be a problem.

If nitrogen's lone pairs are used in the aromatic system, you can determine whether the nitrogen atom is weakly basic or strongly basic.

The theory of aromatic compounds can be used to explain the properties of fused aromatic systems.

Draw their structures from the names.

Predict the properties of aromatic compounds and the effects that aromatic rings have on neighboring parts of the molecule.

To determine the structures of aromatic compounds, use IR, NMR, UV, and mass spectrums.

Problems 16-38, 44, 45, 46, 48 are given an aromatic compound.

The structure of each compound can be drawn.

Four compounds are shown.

One of the compounds reacts more quickly, or with a more favorable equilibrium constant, in each pair.
Explain the enhanced reactivity.

One of the hydrocarbons is more acidic than the others.

Explain why it is acidic.

There was no proof that benzene was a six-membered ring in Kekule's time.

The structure was determined using the known numbers of monosubstituted and disubstituted benzenes and the knowledge that benzene did not react similarly to a normal alkene.

There are six hydrogen atoms in each structure.

Draw all the possible monobrominated derivatives that could result from a random substitution of one hydrogen with a bromine.

Benzene had only one monobromo derivative.

Draw all the possible dibromo derivatives for the structures that only had one monobromo derivative.

At the time resonance theory was unknown, Benzene was known to have three dibromo derivatives.

Determine which structure was most consistent with what was known about benzene at that time: Benzene gives one monobrominated derivative and three dibrominated derivatives, and it gives negative chemical tests for alkene.

The ion and molecule are grouped by similar structures.

Give the number of pi electrons in the ring to the aromatic species.

Nitrogen atoms are included in each of the following Heterocycles.

Classify the nitrogen atom according to the availability of its lone pair of electrons.

Some of the compounds have aromatic properties, but others do not.

Explain why they are aromatic.

Predict which nitrogen atoms are more basic than water.

Simple alkyl intermediates are more stable than benzylic cations, anions, and radicals.

Use resonance forms to show the delocalization of the positive charge, unpaired electron, and negative charge of the benzyl cation, radical, and anion.

In the presence of light, toluene reacts with bromine to give benzyl bromide.

There is a mechanism for this reaction.

You can use a drawing of the transition state to explain your answer.

Adding a third group and figuring out how many isomers are formed is one of the methods used by Korner.

Two isomers are formed when xylene is nitrated.

A chemist isolated an aromatic compound from a formula.

He nitrated this compound and made three isomers.

The structure should be consistent with the spectrum and the additional information provided.

The formula shown in the analysis is C8H7OCl.

There is a moderate absorption at 1602 cm-1 and a strong absorption at 1690 cm-1 in the IR spectrum.

The Diels-Alder reaction requires a Kekule structure that shows how the reactive positions of anthracene are at the end of a diene.

A common organic lab experiment is the Diels-Alder reaction of anthracene with maleic anhydride.

Biphenyl has a structure.

dianions of hydrocarbons are very rare.

A dianion of formula [C9H9]2 is formed by the reaction of the following hydrocarbons with two equivalents of butyllithium.
Give a structure for this dianion and explain why it forms so quickly.

The structure of a ribonucleoside is shown here.

The four bases are uracil, guanine, and adenine.

Determine which bases are aromatic.

Which nitrogen atoms are basic?

A student found an old bottle in the stockroom.

She obtained the following mass after smelling a pleasant odor.
On shaking with D2O, the peak of the NMR disappears.
Show how your structure is consistent with the spectrum.
The resulting ion is stable.

It has been separated into enantiomers.

The optical rotation is enormous.

Explain why the rotation is large and speculate as to why it is so active.

To see the relative energies of all the MO's, draw the energy diagram and show which orbitals the electrons would occupy in the ground state.

Predict whether the ion is aromatic or not.

The hydrogens in pyridine are shown.

The ortho protons are deshielded to d 8.60, which is a typical aromatic chemical shift.

A family of compounds called chlorophyll is present in green plants.

The energy in the sun can be used to convert carbon dioxide and water into sugars.
The chlorin is a large-ring magnesium complex.
The large pi system makes it aromatic.

Many properties, including aromaticity, have been probed with the use of NMR.