Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 10 BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS - all with Video Answers

Educators

Chapter Questions

Problem 1

Benzene is a planar molecule with bond angles of $120^{\circ}$. All six C-to-C bonds have the identical length, $0.139 \mathrm{~nm}$. Is benzene the same as $1,3,5$-cyclohexatriene?

No. The bond lengths in 1,3,5-cyclohexatriene would alternate between $0.153 \mathrm{~nm}$ for the single bond and $0.132 \mathrm{~nm}$ for the double bond. The $\mathrm{C}$-to- $\mathrm{C}$ bonds in benzene are intermediate between single and double bonds.

Adriano Chikande

Numerade Educator

01:19

Problem 2

(a) How do the following heats of hydrogenation $\left(\Delta H_h, \mathrm{~kJ} / \mathrm{mol}\right)$ show that benzene is not the ordinary triene 1,3,5-cyclohexatriene? Cyclohexene, -119.7; 1,4-cyclohexadiene, -239.3; 1,3-cyclohexadiene, -231.8 ; and benzene, -208.4 . (b) Calculate the delocalization energy of benzene. (c) How does the delocalization energy of benzene compare to that of 1,3,5-hexatriene $\left(\Delta H_h=-336.8 \mathrm{~kJ} / \mathrm{mol}\right)$ ? Draw a conclusion about the relative reactivities of the two compounds.

In computing the first column of Table 10-1, we assume that in the absence of any orbital interactions each double bond should contribute $-119.7 \mathrm{~kJ} / \mathrm{mol}$ to the total $\Delta H_h$ of the compound, since this is the $\Delta H_h$ of an isolated $\mathrm{C}=\mathrm{C}$ (in cyclohexane). Any difference between such a calculated $\Delta H_h$ value and the observed value is the delocalization energy. Since $\Delta H_h$ for 1,4-cyclohexadiene is $7.5 \mathrm{~kJ} / \mathrm{mol}$ less than that for 1,3-cyclohexadiene, conjugation stabilizes the 1,3-isomer. [Remember that the smaller (more negative) the energy, the more stable the structure.]
(a) 1,3,5-Cyclohexatriene should behave as a typical triene and have $\Delta H_h=-359.1 \mathrm{~kJ} / \mathrm{mol}$. The observed $\Delta H_h$ for benzene is $-208.4 \mathrm{~kJ} / \mathrm{mol}$. Benzene is not $1,3,5$-cyclohexatriene; in fact, the latter does not exist.
(b) See Table 10-1.
(c) The delocalization energy of benzene $(-150.7 \mathrm{~kJ} / \mathrm{mol})$ is much smaller than that of $1,3,5$-hexatriene $(-22.3 \mathrm{~kJ} / \mathrm{mol})$. Three conjugated double bonds engender a large negative delocalization energy only when they are in a ring. Since the ground-state enthalpy of benzene is much smaller in absolute value than that of the triene, the $\Delta H^{\ddagger}$ for addition of $\mathrm{H}_2$ to benzene is much greater, and benzene reacts much slower. Benzene is less reactive than open-chain trienes towards all electrophilic addition reactions.

Grigoriy Sereda

Numerade Educator

01:19

Problem 3

(a) Use the $\Delta H_h$ 's for complete hydrogenation of cyclohexene, 1,3-cyclohexadiene and benzene, as given in Problem 10.2, to calculate $\Delta H_h$ for the addition of $1 \mathrm{~mol}$ of $\mathrm{H}_2$ to (i) 1,3-cyclohexadiene, (ii) benzene. (b) What conclusion can you draw from these values about the rate of adding $\mathrm{I}$ mol of $\mathrm{H}_2$ to these three compounds? (The $\Delta H$ of a reaction step is not necessarily related to $\Delta H^t$ of the step. However, in the cases being considered in this problem, $\Delta H_{\text {reaction }}$ is directly related to $\Delta H^{\ddagger}$.) (c) Can cyclohexadiene and cyclohexene be isolated on controlled hydrogenation of benzene?

Equations are written for the reactions so that their algebraic sum gives the desired reactant, products and enthalpy.
(a) (i) Add reactions (1) and (2):

Note that reaction (1) is a dehydrogenation (reverse of hydrogenation) and that its $\Delta H$ is positive. (ii) Add the following two reactions:
(b) The reaction with the largest negative $\Delta H_h$ value is the most exothermic and, in this case, also has the fastest rate. The ease of addition of $1 \mathrm{~mol}$ of $\mathrm{H}_2$ is:
$$
\text { cyclohexene }(-119.7)>1.3 \text {-cyclohexadiene }(-112.1) \gg \text { benzene }(+23.4)
$$
(c) No. When one molecule of benzene is converted to the diene, the diene is reduced all the way to cyclohexane by two more molecules of $\mathrm{H}_2$ before more molecules of benzene react. If $1 \mathrm{~mol}$ each of benzene and $\mathrm{H}_2$ are reacted, the product is $\frac{1}{3} \mathrm{~mol}$ of cyclohexane and $\frac{2}{3} \mathrm{~mol}$ of unreacted benzene.

Grigoriy Sereda

Numerade Educator

03:10

Problem 4

The observed heat of combustion $\left(\Delta H_c\right)$ of $\mathrm{C}_6 \mathrm{H}_6$ is $-3301.6 \mathrm{~kJ} / \mathrm{mol}$.* Theoretical values are calculated for $\mathrm{C}_6 \mathrm{H}_6$ by adding the contributions from each bond obtained experimentally from other compounds; these are (in $\mathrm{kJ} / \mathrm{mol}$ ) -492.4 for $\mathrm{C}=\mathrm{C},-206.3$ for $\mathrm{C}-\mathrm{C}$ and -225.9 for $\mathrm{C}-\mathrm{H}$. Use these data to calculate the heat of combustion for $\mathrm{C}_6 \mathrm{H}_6$ and the difference between this and the experimental value. Compare the difference with that from heats of hydrogenation.
The contribution is calculated for each bond and these are totaled for the molecule.
$$
\begin{aligned}
\text { Six } \mathrm{C}-\mathrm{H} \text { bonds }=6(-225.9) & =-1355.4 \mathrm{~kJ} / \mathrm{mol} \\
\text { Three } \mathrm{C}-\mathrm{C} \text { bonds }=3(-206.3) & =-618.9 \\
\text { Three } \mathrm{C}=\mathrm{C} \text { bonds }=3(-492.4) & =-1477.2 \\
\text { TOTAL } & =-3451.5\left(\text { calculated } \Delta H_c \text { for } \mathrm{C}_6 \mathrm{H}_6\right) \\
\text { Experimental } & =-3301.6 \\
\text { DIFFERENCE } & =-149.9 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$

This difference is the delocalization energy of $\mathrm{C}_6 \mathrm{H}_6$; essentially the same value is obtained from $\Delta H_h$ (Table 10-1).

Rabia Shuaib

Numerade Educator

01:13

Problem 5

How is the structure of benzene explained by $(a)$ resonance, $(b)$ the orbital picture, $(c)$ molecular orbital theory?
(a) Benzene is a hybrid of two equal-energy (Kekulé) structures differing only in the location of the double bonds:
(b) Each C is $s p^2$ hybridized and is $\sigma$ bonded to two other C's and one H (Fig. 10-1). These $\sigma$ bonds comprise the skeleton of the molecule. Each $\mathrm{C}$ also has one electron in a $p$ orbital at right angles to the plane of the ring. These $p$ orbitals overlap equally with each of the two adjacent $p$ orbitals to form a $\pi$ system parallel to and above and below the plane of the ring (Fig. 10-2). The six $p$ electrons in the $\pi$ system are associated with all six C's. They are therefore more delocalized and this accounts for the great stability and large resonance energy of aromatic rings.
Fig. 10-1
(c) The six $p$ AO's discussed in part (b) interact to form six $\pi$ MO's. These are indicated in Fig. 10-3, which gives the signs of the upper lobes (cf. Fig. 8-3 for butadiene). Since benzene is cyclic, the stationary waves representing the electron clouds are cyclic and have nodal planes, shown as lines, instead of nodal points. See Problem 9.28 for the significance of a 0 sign. The six $p$ electrons fill the three bonding MO's, thereby accounting for the stability of $\mathrm{C}_6 \mathrm{H}_6$.

Anand Jangid

Numerade Educator

08:55

Problem 6

Account for aromaticity observed in: (a) 1,3-cyclopentadienyl anion but not 1,3-cyclopentadiene; (b) 1,3,5-cycloheptatrienyl cation but not 1,3,5-cycloheptatriene; (c) cyclopropenyl cation; (d) the heterocycles pyrrole, furan and pyridine.
(a) 1,3-Cyclopentadiene has an $s p^3$-hybridized $\mathrm{C}$, making cyclic $p$ orbital overlap impossible. Removal of $\mathrm{H}^{+}$from this $\mathrm{C}$ leaves a carbanion whose $\mathrm{C}$ is now $s p^2$-hybridized and has a $p$ orbital capable of overlapping to give a cyclic $\pi$ system. The four $\pi$ electrons from 2 double bonds plus the two unshared electrons total six $\pi$ electrons; the anion is aromatic $(n=1)$.
(b) Although the triene has six $p$ electrons in three $\mathrm{C}=\mathrm{C}$ bonds, the lone $s p^3$-hybridized $\mathrm{C}$ prevents cyclic overlap of $p$ orbitals.
Generation of a carbocation by ionization permits cyclic overlap of $p$ orbitals on each $\mathrm{C}$. With six $\pi$ electrons, the cation is aromatic $(n=1)$.
(c) Cyclopropenyl cation has two $\pi$ electrons and $n=0$.
The ions in parts $(a),(b)$, and $(c)$ are reactive but they are much more stable than the corresponding open-chain ions.
(d) Hückel's rule is extended to heterocyclic compounds as follows:
Note that dipoles are generated in pyrrole and furan because of delocalization of electrons from the heteroatoms.

James Irizarry

Numerade Educator

16:58

Problem 7

Cyclooctatetraene $\left(\mathrm{C}_8 \mathrm{H}_8\right)$, unlike benzene, is not aromatic; it decolorizes both dil. aq. $\mathrm{KMnO}_4$ and $\mathrm{Br}_2$ in $\mathrm{CCl}_4$. Its experimentally determined heat of combustion is $-4581 \mathrm{~kJ} / \mathrm{mol}$. (a) Use the Hückel rule to account for the differences in chemical properties of $\mathrm{C}_8 \mathrm{H}_8$ from those of benzene. (b) Use thermochemical data of Problem 10.4 to calculate the resonance energy. (c) Why is this compound not antiaromatic? (d) Styrene, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CH}_2$, with heat of combustion $-4393 \mathrm{~kJ} / \mathrm{mol}$, is an isomer of cyclooctatetraene. Is styrene aromatic?
(a) $\mathrm{C}_8 \mathrm{H}_8$ has eight rather than six $p$ electrons. Since it is not aromatic it undergoes addition reactions.
(b) The calculated heat of combustion is:
$$
\begin{aligned}
8 \mathrm{C}-\mathrm{H} \text { bonds }=8(-225.9) & =-1807 \mathrm{~kJ} / \mathrm{mol} \\
4 \mathrm{C}-\mathrm{C} \text { bonds }=4(-206.3) & =-825 \\
4 \mathrm{C}=\mathrm{C} \text { bonds }=4(-492.4) & =-1970 \\
\text { TOTAL } & =-4602 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$

The difference $-4602-(-4581)=-21 \mathrm{~kJ} / \mathrm{mol}$ shows small (negative) resonance energy and no aromaticity.
(c) Although the molecule has $(n=2) \pi$ electrons it is not antiaromatic, because it is not planar. It exists chiefly in a "tub" conformation (Fig. 10-4).
(d) Styrene is aromatic; its delocalization energy is $-4602-(-4393)=-209 \mathrm{~kJ} / \mathrm{mol}$. This is attributable to the presence of the benzene ring. Styrene has a more negative energy than does benzene $[-150 \mathrm{~kJ} / \mathrm{mol}]$ because its ring is conjugated to the $\mathrm{C}=\mathrm{C}$ bond, thereby extending the delocalization of the electron cloud.

Susan Hallstrom

Numerade Educator

04:11

Problem 8

Deduce the structure and account for the stability of the following substances which are insoluble in nonpolar but soluble in polar solvents. (a) A red compound formed by reaction of $2 \mathrm{~mol}_{\text {, of }} \mathrm{AgBF}_4$ with $1 \mathrm{~mol}$ of 1,2,3,4-tetraphenyl-3,4-dibromo cyclobut-1-ene. (b) A stable compound from the reaction of $2 \mathrm{~mol}$ of $\mathrm{K}$ with $1 \mathrm{~mol}$ of 1,3,5,7-cyclooctatetraene with no liberation of $\mathrm{H}_2$.

The solubility properties suggest that these compounds are salts. The stability of the organic ions formed indicates that they conform to the Hückel rule and are aromatic.
(a) Two $\mathrm{Br}^{-}$'s are abstracted by two $\mathrm{Ag}^{+}$'s to form two $\mathrm{AgBr}$ and a tetraphenyl cyclobutenyl dication.
(b) Since $\mathrm{K} \cdot$ is a strong reductant and no $\mathrm{H}_2$ is evolved, two $\mathrm{K}$ 's supply two electrons to form a cyclooctatetraenyl dianion (Fig. 10-5). This planar conjugated unsaturated monocycle has 10 electrons, conforms to the Hückel rule $(n=2)$ and is aromatic.

Marissa Turner

Numerade Educator

02:34

Problem 9

Sondheimer synthesized a series of interesting conjugated cyclic polyalkenes that he designated the $[n]$-annulenes, where $n$ is the number of C's in the ring. Account for his observation that $(a)[18]$-annulene is somewhat aromatic, [16]- and [20]-annulene are not; $(b)[18]$ annulene is more stable than [14]-annulene.
(a) The somewhat aromatic [18]-annulene has $4 n+2(n=4) \pi$ electrons; there are $4 n \pi$ electrons in the nonaromatic, nonplanar [16]- and [20]-annulenes.
(b) [14]-Annulene is somewhat strained because the H's in the center of the ring are crowded. This steric strain prevents a planar conformation, which diminishes aromaticity.

Mercedes Mazza

Numerade Educator

01:13

Problem 10

Use the Hückel rule to indicate whether the following planar species are aromatic or antiaromatic:
(a) Aromatic. There are $2 \pi$ electrons from each $\mathrm{C}=\mathrm{C}$ and 2 from an electron pair on $\mathrm{S}$ to make an aromatic sextet. (b) Antiaromatic. There are $4 n(n=2) \pi$ electrons. (c) Aromatic. There are $6 \pi$ electrons. (d) Aromatic. There are $10 \pi$ electrons and this anion conforms to the $(4 n+2)$ rule $(n=2)$. (e) Antiaromatic. The cation has $4 n(n=2) \pi$ electrons. $(f)$ and $(g)$ Antiaromatic. They have $4 n(n=1) \pi$ electrons.

Raghvendra Singh

Numerade Educator

05:01

Problem 11

The relative energies of the MO's of conjugated cyclic polyenes can be determined by the following simple polygon rule instead of using nodal planes as in Problem 10.5(c). Inscribe a regular polygon in a circle, with one vertex at the bottom of the circle and with the total number of vertices equal to the number of MO's. Then the height of a vertex is proportional to the energy of the associated MO. Vertices below the horizontal diameter are bonding $\pi$, those above are antibonding $\pi^*$, and those on the diameter are nonbonding $\pi^{\mathrm{n}}$. Apply the method to 3-, $4-, 5-, 6-, 7-$, and 8-carbon systems and indicate the character of the MO's.
See Fig. 10-6.

Vivek Singh

Numerade Educator

01:16

Problem 12

Draw a conclusion about the stability and aromaticity of naphthalene from the fact that the experimentally determined heat of combustion is $255 \mathrm{~kJ} / \mathrm{mol}$ smaller in absolute value than that calculated from the structural formula.

The difference, $-255 \mathrm{~kJ} / \mathrm{mol}$, is naphthalene's resonance energy. Naphthalene is less aromatic than benzene because a per-ring resonance energy of $\frac{1}{2}(-255)=-127.5 \mathrm{~kJ} / \mathrm{mol}$ is smaller in absolute value than that of benzene $(-150 \mathrm{~kJ} / \mathrm{mol})$.

Eileen Sullivan

Numerade Educator

05:22

Problem 13

Deduce an orbital picture (like Fig. 10-2) for naphthalene, a planar molecule with bond angles of $120^{\circ}$.

See Fig. 10-7. The C's use $s p^2$ hybrid atomic orbitals to form $\sigma$ bonds with each other and with the H's. The remaining $p$ orbitals at right angles to the plane of the C's overlap laterally to form a $\pi$ electron cloud.

Vivek Singh

Numerade Educator

16:58

Problem 14

(a) Draw three resonance structures for naphthalene. (b) Which structure makes the major contribution to the structure of the hybrid in that it has the smallest energy? (c) There are four kinds of $\mathrm{C}$-to- $\mathrm{C}$ bonds in naphthalene: $\mathrm{C}^1-\mathrm{C}^2, \mathrm{C}^2-\mathrm{C}^3, \mathrm{C}^1-\mathrm{C}^9$, and $\mathrm{C}^9-\mathrm{C}^{10}$. Select the shortest bond and account for your choice.
(b) Structure I has the smallest energy because only it has two intact benzene rings.
(c) The bond, in all of its positions, that most often has double-bond character in the three resonance structures is the shortest. This is true for $\mathrm{C}^I-\mathrm{C}^2$ (also for $\mathrm{C}^3-\mathrm{C}^4, \mathrm{C}^5-\mathrm{C}^6$, and $\mathrm{C}^7-\mathrm{C}^8$ ).
Anthracene and phenanthrene are isomers $\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)$ having three, fused benzene rings;
As the number of fused rings increases, the delocalization energy per ring continues to decrease in absolute value, and compounds become more reactive, especially toward addition. The delocalization energies per ring for anthracene and phenanthrene are -117.2 and $-126.8 \mathrm{~kJ} / \mathrm{mol}$, respectively.

Susan Hallstrom

Numerade Educator

08:48

Problem 15

Name the compounds:
(a) p-Aminobenzoic acid. (b) $m$-Nitrobenzenesulfonic acid. (c) $m$-isopropylphenol. (d) 2-Bromo-3-nitro-5 hydroxybenzoic acid. (Named as a benzoic acid rather than a phenol because $\mathrm{COOH}$ has priority over $\mathrm{OH}$.) $(e$ ) 3,4'-Dichlorobiphenyl; the notational system in biphenyl is:

Dr. Satish Ingale

Numerade Educator

01:45

Problem 16

Give the structural formulas for (a) 2,4,6-tribromoaniline, (b) $m$-toluenesulfonic acid, (c) $p$ bromobenzalbromide, $(d)$ di-o-tolylmethane, $(e)$ trityl chloride.

Mystique Till

Numerade Educator

01:09

Problem 17

Name the following compounds:
(a) 1-naphthalenesulfonic acid or $\alpha$-naphthalenesulfonic acid, (b) 1-naphthaldehyde or $\alpha$-naphthaldehyde, (c) 8 bromo-1-methoxynaphthalene.

Lijeesh Krishnan

Numerade Educator

16:58

Problem 18

(a) Write equations for the reductions of (i) naphthalene, (ii) anthracene and (iii) phenanthrene. (b) Explain why naphthalene is reduced more easily than benzene. (c) Explain why anthracene and phenanthrene react at the $\mathrm{C}^9-\mathrm{C}^{10}$ double bond and go no further.

(b) Each ring of naphthalene is less aromatic than the ring of benzene and therefore is more reactive.
(c) These reactions leave two benzene rings having a combined resonance energy of $2(-150)=-300 \mathrm{~kJ} / \mathrm{mol}$. Were attack to occur in an end ring, a naphthalene derivative having a resonance energy of $-255 \mathrm{~kJ} / \mathrm{mol}$ would remain. Two phenyls are less energetic (more stable) than one naphthyl.

Susan Hallstrom

Numerade Educator

02:32

Problem 19

In the Birch reduction benzene is reduced with an active metal ( $\mathrm{Na}$ or $\mathrm{Li})$ in alcohol and liquid $\mathrm{NH}_3\left(-33^{\circ} \mathrm{C}\right)$ to a cyclohexadiene that gives only $\mathrm{OCHCH}_2 \mathrm{CHO}$ on ozonolysis. What is the reduction product?

Since the diene gives only a single product on ozonolysis, it must be symmetrical. The reduction product is $1,4-$ cyclohexadiene.

George Bennett

Numerade Educator

02:50

Problem 20

Typical of mechanisms for reductions with active metals in protic solvents, two electrons are transferred from the metal atoms to the substrate to give the most stable dicarbanion, which then accepts two $\mathrm{H}^{+} \mathrm{s}$ from the protic solvent molecules to give the product. (a) Give the structural formula for the dicarbanion formed from $\mathrm{C}_6 \mathrm{H}_6$ and $(b)$ explain why it is preferentially formed.

Lottie Adams

Numerade Educator

01:58

Problem 21

Oxidation of 1-nitronaphthalene yields 3-nitrophthalic acid. However, if 1-nitronaphthalene is reduced to $\alpha$-naphthylamine and if this amine is oxidized, the product is phthalic acid.

How do these reactions establish the gross structure of naphthalene?
The electron-attracting $-\mathrm{NO}_2$ stabilizes ring $\mathrm{A}$ of 1-nitronaphthalene to oxidation, and ring $\mathrm{B}$ is oxidized to form 3-nitrophthalic acid. By orbital overlap, $-\mathrm{NH}_2$ releases electron density, making ring $\mathrm{A}$ more susceptible to

Zubair Abdulla

Numerade Educator

02:47

Problem 22

Like naphthalene, anthracene and phenanthrene are readily oxidized to a quinone. Suggest the products and account for your choice.
Oxidation at $\mathrm{C}^{\varphi}$ and $\mathrm{C}^{10}$ leaves two stable intact benzene rings. (See Problem $10.18(c)$.)

Nima Gharibi

Numerade Educator

01:41

Problem 23

Use the Diels-Alder reaction to synthesize benzoic acid, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$.$\mathrm{S}$ and $\mathrm{Se}$ can be used in place of $\mathrm{Pt}$, and $\mathrm{H}_2 \mathrm{~S}$ and $\mathrm{H}_2 \mathrm{Se}$ are then the respective products.

Shazia Naz

Numerade Educator

01:00

Problem 24

(a) Draw two Kekulé structures for 1,2-dimethylbenzene (o-xylene). (b) Why are these structures not isomers? What are they? (c) Give the carbonyl products formed on ozonolysis.
(b) These structures differ only in the position of the $\pi$ electrons and therefore are contributing (resonance) structures, not isomers.
(c) Although neither contributing structure exists, the isolated ozonolysis products arising from the resonance hybrid are those expected from either one:

Lottie Adams

Numerade Educator

01:13

Problem 25

What are the necessary conditions for $(a)$ aromaticity and $(b)$ antiaromaticity?
(a) (1) A planar cyclic molecule or ion. (2) Each atom in the ring must have a $p$ AO. (3) These $p$ AO's must be parallel, so that they can overlap side-by-side. (4) The overlapping $\pi$ system must have $(4 n+2) \pi$ electrons (Hückel).
(b) In (a) change $(4 n+2)$ to $4 n$.

Raghvendra Singh

Numerade Educator

02:51

Problem 26

Design a table showing the structure, number of $\pi$ electrons, energy levels of $\pi$ MO's and electron distribution, and state of aromaticity of: $(a)$ cyclopropenyl cation, $(b)$ cyclopropenyl anion, $(c)$ cyclobutadiene, $(d)$ cyclobutadienyl dication, $(e)$ cyclopentadienyl anion, $(f)$ cyclopentadienyl cation, $(g)$ benzene, $(h)$ cycloheptatrienyl anion, $(i)$ cyclooctatetraene, $(j)$ cyclooctatetraenyl dianion.
See Table 10-2. (H's are understood to be attached to each doubly bonded C.)

Kevin Chimex

Numerade Educator

01:10

Problem 27

Explain aromaticity and antiaromaticity in terms of the MO's of Problem 10.26.
Aromaticity is observed when all bonding MO's are filled and nonbonding MO's, if present, are empty or completely filled. Hückel's rule arises from this requirement. A species is antiaromatic if it has electrons in antibonding MO's or if it has half-filled bonding or nonbonding MO's, provided it is planar.

Anand Jangid

Numerade Educator

04:04

Problem 28

Name the monobromo derivatives of $(a)$ anthracene, $(b)$ phenanthrene.
(a) There are 3 isomers: 1-bromo-, 2-bromo-, and 9-bromoanthracene.
(b) There are 5 isomers: 1-bromo-, 2-bromo-, 3-bromo-, 4-bromo-, and 9-bromophenanthrene.

Lottie Adams

Numerade Educator

05:01

Problem 29

What is the Diels-Alder addition product of anthracene and ethene? Reaction occurs at the (most reactive) $\mathrm{C}^9$ and $\mathrm{C}^{10}$ positions.

Ian Kaigh

Numerade Educator