Chapter 21

Question 1

Benzene Resonance

Answer 1

•We often represent benzene as a hybrid of two equivalent Kekulé structures.
–Each makes an equal contribution to the hybrid and thus the C-C bonds are neither double nor single, but something in between.
•The concepts of hybridization of atomic orbitals and the theory of resonance, developed in the 1930s, provided the first adequate description of benzene’s structure.
–The carbon skeleton is a planar regular hexagon.
–All C-C-C and H-C-C bond angles 120°
•Resonance energy:Resonance energy:The difference in energy between a resonance hybrid and the most stable of its hypothetical contributing structures in which electrons are localized on particular atoms and in particular bonds.
–One way to estimate the resonance energy of benzene is to compare the heats of hydrogenation of benzene and cyclohexene.

Question 2

pi system of benzene

Answer 2

–(a) The carbon framework with the six 2p orbitals.
–(b) Overlap of the parallel 2p orbitals forms one torus above the plane of the ring and another below it
–this orbital represents the lowest-lying pi-bonding molecular orbital.

Question 3

Huckel’s rule

Answer 3

1. The compound must be cyclic.
2. It must have one p orbital on each atom of the ring.
3. It should be planar or nearly planar so that there is continuous or nearly continuous overlap of all p orbitals.
4. It must have a closed loop of (4n+2) p electrons in the cyclic arrangement of p orbitals.

Question 4

aromaticity chemical shift

Answer 4

between 6.5 and 9
–The magnetic field induced by circulation of  electrons in an aromatic ring deshields the hydrogens of the aromatic ring and shifts their signal to higher frequency.

Question 5

Aromatic Hydrocarbons

Answer 5

• Cyclobutadiene, Benzene and Cyclooctatetraene are the first members of a family called the annulenes.
–What is remarkable relative to [10]annulene is that if the two hydrogens facing inward toward the center of the ring are replaced by a methylene (CH2) group, the ring is able to assume a conformation close enough to planar that it becomes aromatic.

Question 6

Annulene

Answer 6

A cyclic hydrocarbon with a continuous alternation of single and double bonds.

Question 7

Antiaromatic Hydrocarbons

Answer 7

•Antiaromatic hydrocarbon:Antiaromatic hydrocarbon:A monocyclic, planar, fully conjugated hydrocarbon with 4n pi electrons (4, 8, 12, 16, 20…).
–An antiaromatic hydrocarbon is especially unstable relative to an open-chain fully conjugated hydrocarbon of the same number of carbon atoms.
•Cyclobutadiene is antiaromatic.
–In the ground-state electron configuration of this molecule, two electrons fill the 1 bonding MO.
–The remaining two electrons lie in the 2 and 3nonbonding MOs.

Question 8

Cyclobutadiene

Answer 8

–The ground state of planar cyclobutadiene has two unpaired electrons, which make it highly unstable and reactive.

Question 9

Cyclooctatetraene

Answer 9

–Cyclooctatetraene, with 8 pi electrons is not aromatic; it shows reactions typical of alkenes.
–X-ray studies show that the most stable conformation is a nonplanar “tub” conformation.
–Although overlap of 2p orbitals occurs to form pi bonds, there is only minimal overlap between sets of 2p orbitals because they are not parallel.

Question 10

frost circles

Answer 10

•Frost circle:Frost circle: A graphic method for determining the relative order of pi MOs in planar, fully conjugated monocyclic compounds.
–Inscribe in a circle a polygon of the same number of sides as the ring to be examined such that one of the vertices of the polygon is at the bottom of the circle.
–The relative energies of the MOs in the ring are given by where the vertices of the polygon touch the circle.
•Those MOs
–Below the horizontal line through the center of the ring are bonding MOs.
–on the horizontal line are nonbonding MOs.
–above the horizontal line are antibonding MOs.
–Frost circles describing the MOs for monocyclic, planar, fully conjugated four-, five-, and six-membered rings.

Question 11

Heterocyclic Aromatics

Answer 11

Heterocyclic Aromatics

Question 12

pyridine

Answer 12

–The nitrogen atom of pyridine is sp2 hybridized.
–The unshared pair of electrons lies in an sp2hybrid orbital and is not a part of the six pi electrons of the aromatic system (the aromatic sextet).
–Resonance energy of pyridine is 134 kJ (32 kcal)/mol.

Question 13

Furan and Pyrrole

Answer 13

–The oxygen atom of furan is sp2 hybridized.
–one unshared pairs of electrons on oxygen lies in an unhybridized 2p orbital and is a part of the aromatic sextet.
–The other unshared pair lies in an sp2 hybrid orbital and is not a part of the aromatic system
–The resonance energy of furan is 67 kJ (16 kcal)/mol.

Question 14

Aromatic Hydrocarbon Ions

Answer 14

•Any neutral, monocyclic, unsaturated hydrocarbon with an odd number of carbons must have at least one CH2 group and, therefore, cannot be aromatic.
–Cyclopropene, for example, has the correct number of pi electrons to be aromatic, 4(0) + 2 = 2, but does not have a closed loop of 2p orbitals.

Question 15

Cyclopropenyl Cation

Answer 15

–If, however, the CH2 group of cyclopropene is transformed into a CH+ group in which carbon is sp2 hybridized and has a vacant 2p orbital, the overlap of orbitals is continuous and the cation is aromatic.
–When 3-chlorocyclopropene is treated with SbCl5, it forms a stable salt.
–If planar cyclopentadienyl cation were to exist, it would have 4 pi electrons and be antiaromatic.
–Note that we can draw five equivalent contributing structures for the cyclopentadienyl cation. Yet this cation is not aromatic because it has only 4 pi electrons.

Question 16

Cyclopentadienyl Anion

Answer 16

•To convert cyclopentadiene to an aromatic ion, it is necessary to convert the CH2 group to a CH group in which carbon becomes sp2 hybridized and has 2 electrons in its unhybridized 2p orbital.
–As seen in the Frost circle, the six pi electrons of cyclopentadienyl anion occupy the 1, 2, and 3 molecular orbitals, all of which are bonding.
•The pKa of cyclopentadiene is 16.
–In aqueous NaOH, it is in equilibrium with its sodium salt.
–It is converted completely to its anion by very strong bases such as NaNH2 , NaH, and LDA.
•Cycloheptatriene forms an aromatic cation by conversion of its CH2 group to a CH+ group with its sp2 carbon having a vacant 2p orbital.

Question 17

nomenclature

Answer 17

•Monosubstituted alkylbenzenes are named as derivatives of benzene.
–Many common names are retained.

Question 18

Disubstituted Benzenes

Answer 18

•Locate two groups by numbers or by the locators orthoortho(1,2-), metameta (1,3-), and parapara (1,4-).–Where one group imparts a special name, name the compound as a derivative of that molecule.
–Where neither group imparts a special name, locate the groups and list them in alphabetical order.

Question 19

Polysubstituted Derivatives

Answer 19

–If one group imparts a special name, name the molecule as a derivative of that compound
–If no group imparts a special name, list them in alphabetical order, giving them the lowest set of numbers.

Question 20

phenols

Answer 20

–Phenol was first isolated from coal tar in 1834 by Friedlieb Runge (maybe most famous for his isolation of caffeine).
–It is the famous Carbolic Acid used by Sir Joseph Lister that led to aseptic surgical techniques.
–Phenol is the active ingredient in some oral .analgesics, such as Carmax and Chloraseptic spray.
–Phenol was the most popular lethal injection used by the Nazis (ca. 1 g injected i.v. causes death).
–It contributes in small quantities to the flavor of some whiskeys.
•The functional group of a phenol is an -OH group bonded to a benzene ring.
–Hexylresorcinol is a mild antiseptic and disinfectant.
–Eugenol is used as a dental antiseptic and analgesic.
–Urushiol is the main component of the oil of poison ivy.

Question 21

Acidity of Phenols

Answer 21

•Phenols are significantly more acidic than alcohols, compounds that also contain the OH group.
–Most phenols do not react with weak bases such as NaHCO3; they do not dissolve in aqueous NaHCO3.
–Carbonic acid is a stronger acid than phenol. Therefore, the position of this equilibrium lies far to the left
•Phenols are weak acids and react with strong bases to form water-soluble salts.
–Water-insoluble phenols dissolve in NaOH(aq).
•Separation of water-insoluble phenols from water-insoluble alcohols.
–The greater acidity of phenols compared with alcohols is due to the greater stability of the phenoxide ion relative to an alkoxide ion.
•Alkyl and halogen substituents effect acidities by inductive effects:
–Alkyl groups are electron-releasing.
–Halogens are electron-withdrawing.
–Nitro groups increase the acidity of phenols by both an electron-withdrawing inductive effect and a resonance effect.
–Part of the acid-strengthening effect of -NO2 is due to its electron-withdrawing inductive effect.–In addition, -NO2 substituents in the ortho and para positions help to delocalize the negative charge.

Question 22

Alkyl-Aryl Ethers

Answer 22

•Alkyl-aryl ethers can be prepared by the Williamson ether synthesis:–but only using phenoxide salts and haloalkanes.–haloarenes cannot be used because they are unreactive to SN2 reactions.

Question 23

Kolbe Carboxylation

Answer 23

•Phenoxide ions react with carbon dioxide to give a carboxylate salt.
–The mechanism begins by nucleophilic addition of the phenoxide ion to a carbonyl group of CO2.

Question 24

Quinones

Answer 24

• Because of the presence of the electron-donating -OH group, phenols are susceptible to oxidation by a variety of strong oxidizing agents.
• Perhaps the most important chemical property of quinones is that they are readily reduced to hydroquinones.

Question 25

Coenzyme Q

Answer 25

•Coenzyme Q is a carrier of electrons in the respiratory chai

Question 26

Vitamin K

Answer 26

–Both natural and synthetic vitamin K (menadione) are 1,4-naphthoquinones.

Question 27

Benzylic Oxidation

Answer 27

•Benzene is unaffected by strong oxidizing agents such as H2CrO4 and KMnO4
–Halogen and nitro substituents are also unaffected by these reagents.
–An alkyl group with at least one hydrogen on its benzylic carbon is oxidized to a carboxyl group.
–If there is more than one alkyl group on the benzene ring, each is oxidized to a -COOH group

Question 28

Benzylic Chlorination

Answer 28

•Chlorination and bromination occur by a radical chain mechanism.

Question 29

Benzylic Reactions

Answer 29

•Benzylic radicals (and cations also) are easily formed because of the resonance stabilization of these intermediates.–The benzyl radical is a hybrid of five contributing structures.

Question 30

Benzylic Halogenation

Answer 30

–Benzylic bromination is highly regioselective.

–Benzylic chlorination is less regioselective.

Question 31

Hydrogenolysis

Answer 31

•Hydrogenolysis:Hydrogenolysis:Cleavage of a single bond by H2–Among ethers, benzylic ethers are unique in that they are cleaved under conditions of catalytic hydrogenation.

Question 32

Benzyl Ethers

Answer 32

•The value of benzyl ethers is as protecting groups for the OH groups of alcohols and phenols.–To carry out hydroboration/oxidation of this alkene, the phenolic -OH must first be protected; it is acidic enough to react with BH3 and destroy the reagent.