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Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 11

AROMATIC SUBSTITUTION. ARENES - all with Video Answers

Educators

Chapter Questions

Problem 1

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Problem 1

Name the product and account for the orientation in the following electrophilic substitution reactions:
(a) 1-Methylnaphthalene $+\mathrm{Br}_2, \mathrm{Fe}$
(b) 2-Ethylnaphthalene $+\mathrm{Cl}_2, \mathrm{Fe}$
(c) 2-Ethylnaphthalene $+\mathrm{C}_2 \mathrm{H}_5 \mathrm{COCl}+\mathrm{AlCl}_3$
(d) 1-Methylnaphthalene $+\mathrm{CH}_3 \mathrm{COCl}, \mathrm{AlCl}_3\left(\mathrm{CS}_2\right)$
(e) 2-Methoxynaphthalene $+\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{SO}_4$
(f) 2-Nitronapththalene $+\mathrm{Br}_2, \mathrm{Fe}$
(a) 1-Methyl-4-bromonaphthalene. Br substitutes in the more reactive $\alpha$ position of the activated ring.
(b) 1-Chloro-2-ethylnaphthalene. $\mathrm{C}^{\prime}$ (ortho and $\alpha$ ) is activated by $\mathrm{C}_2 \mathrm{H}_5$ since $\mathrm{C}^4$, which is also $\alpha$, is meta to $\mathrm{C}_2 \mathrm{H}_5$.
(c) I-(2-Ethylnaphthyl) ethyl ketone. Same reason as in $(b)$. (d) 4-(1-Methylnaphthyl) methyl ketone. Same reason as in $(a)$. (e) 1-Nitro-2-methoxynaphthalene. Same reason as in $(b)$. ( $f$ ) 1-Bromo-6-nitronaphthalene and 1-bromo-7-nitronaphthalene. $\mathrm{NO}_2$ deactivates its ring and bromination occurs at the $\alpha$ positions of the other ring, more at $\mathrm{C}^5$ that bears no delocalized $\delta^{+}$.

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01:57

Problem 2

For each electrophilic aromatic substitution in Table 11-1, give equations for formation of $\mathrm{E}^{+}$and indicate what is $\mathrm{B}^{-}$or $\mathrm{B}$ (several bases may be involved). In reaction $(c)$ the electrophile is a molecule, $\mathrm{E}$.
(a) $\mathrm{X}_2+\mathrm{FeX}_3 \longrightarrow \mathrm{X}^{+}\left(\mathrm{E}^{+}\right)+\mathrm{FeX}_4^{-}\left(\mathrm{B}^{-}\right)$(forms $\left.\mathrm{HX}+\mathrm{FeX}_3\right)$
(b) $\mathrm{H}_2 \mathrm{SO}_4+\mathrm{HONO}_2 \longrightarrow \mathrm{HSO}_4^{-}\left(\mathrm{B}^{-}\right)+\mathrm{H}_2 \stackrel{+}{\mathrm{ONO}} \mathrm{N}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{NO}_2^{+}\left(\mathrm{E}^{+}\right)$
(c) $2 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{HSO}_4^{-}\left(\mathrm{B}^{-}\right)+\mathrm{SO}_3(\mathrm{E})$
(d)

Mercedes Mazza
Mercedes Mazza
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06:53

Problem 3

Write structural formulas for the principal monosubstitution products of the indicated reactions from the following monosubstituted benzenes. For each write an $\mathbf{S}$ or $\mathbf{F}$ to show whether reaction is SLOWER or FASTER than with benzene. (a) Monobromination, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CF}_3$. (b) Mononitration, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOCH}_3$. (c) Monochlorination, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OCH}_3$. (d) Monosulfonation, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{I}$. (e) Mononitration, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{C}_6 \mathrm{H}_5$. ( f ) Monochlorination, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CN}$. $(g)$ Mononitration, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NHCOCH}_3$. $(h)$ Monosulfonation, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3$.

Anish Wadhwa
Anish Wadhwa
Numerade Educator
01:04

Problem 3

How does the absence of a primary isotope effect prove experimentally that the first step in aromatic electrophilic substitution is rate-determining?
$\mathrm{A} \mathrm{C}-\mathrm{H}$ bond is broken faster than is a $\mathrm{C}-\mathrm{D}$ bond. This rate difference (isotope effect, $k_{\mathrm{H}} / k_{\mathrm{D}}$ ) is observed only if the $\mathrm{C}-\mathrm{H}$ (or $\mathrm{C}-\mathrm{D}$ ) bond is broken in the rate-determining step. If no difference is observed, as is the case for most aromatic electrophilic substitutions, $\mathrm{C}-\mathrm{H}$ bond-breaking must occur in a fast step (in this case the second step). Therefore, the first step, involving no $\mathrm{C}-\mathrm{H}$ bond-breaking, is rate-determining. This slow step requires the loss of aromaticity, the fast second step restores the aromaticity.

Grigoriy Sereda
Grigoriy Sereda
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11:00

Problem 4

How is $\mathrm{E}^{+}$generated, and what is the base, in the following reactions? $(a)$ Nitration of reactive aromatics with $\mathrm{HNO}_3$ alone. (b) Chlorination with $\mathrm{HOCl}$ using $\mathrm{HCl}$ as catalyst. (c) Nitrosation (introduction of a $\mathrm{NO}$ group) of reactive aromatics with HONO in strong acid. (d) Deuteration with DCl.
(a) $\mathrm{HNO}_3+\mathrm{H}-\mathrm{O}-\mathrm{NO}_2 \longrightarrow \mathrm{NO}_3^{-}+\left[\begin{array}{c}\mathrm{H} \\ 1 \\ \mathrm{H}-\stackrel{+}{\mathrm{O}}-\mathrm{NO}_2\end{array}\right] \rightarrow \mathrm{H}_2 \mathrm{O}$ (Base) $+\mathrm{NO}_2\left(\mathrm{E}^{+}\right)$
unstable
Nitronium ion
(b) $\mathrm{H}^{+}+\mathrm{H}-\mathrm{O}-\mathrm{Cl} \longrightarrow$
(c)
Nitrosonium ion
(b) $\mathrm{D}^{+}$, transferred by $\mathrm{DCl}$ to benzene. Base is $\mathrm{Cl}^{-}$.
$$
\begin{aligned}
& \mathrm{C}_6 \mathrm{H}_6+\mathrm{DCl} \longrightarrow\left[\mathrm{C}_6 \mathrm{H}_5^{\prime}{ }_{\mathrm{D}}^{\prime}\right]^{\mathrm{H}}+\mathrm{Cl}^{-} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{D}+\mathrm{HCl} \\
& \text { base }_1 \operatorname{acid}_2 \quad \text { acid }_1 \quad \text { base }_2 \\
&
\end{aligned}
$$

Tom Rutherford
Tom Rutherford
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Problem 4

Outline practical laboratory syntheses from benzene or toluene and any needed inorganic reagents of: (a) p-chlorobenzal chloride, (b) 2,4-dinitroaniline, (c) m-chlorobenzotrichloride, (d) 2,5-dibromonitrobenzene.

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01:01

Problem 5

Since the initial step of aromatic electrophilic substitution is identical with that of alkene addition, explain why $(a)$ aromatic substitution is slower than alkene addition; (b) catalysts are needed for aromatic substitution; (c) the intermediate carbocation eliminates a proton instead of adding a nucleophile.
(a) The intermediate benzenonium ion is less stable than benzene; hence its formation has a high $\Delta H^{\ddagger}$ and the reaction is slowed. Loss of aromaticity is energetically more unfavorable than loss of $\pi$ bond.
(b) The catalysts are acids which polarize the reagent and make it more electrophilic.
(c) The addition reaction would be endothermic and would produce the less stable cyclohexadiene. Loss of a proton, on the other hand, produces a stable aromatic ring.

Narayan Hari
Narayan Hari
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01:23

Problem 6

Sulfonation resembles nitration and halogenation in being an electrophilic substitution, but differs in being reversible and in having a moderate primary kinetic isotope effect. Illustrate with diagrams of enthalpy $(H)$ versus reaction coordinate.
In nitration (and other irreversible electrophilic substitutions) the transition state (TS) for the reaction wherein
loses $\mathrm{H}^{+}$has a considerably smaller $\Delta H^{+}$than does the TS for the reaction in which $\mathrm{NO}_2^{+}$is lost. In sulfonation the $\Delta H^{\ddagger}$ for loss of $\mathrm{SO}_3$ from
is only slightly more than that for loss of $\mathrm{H}^{+}$.
In terms of the specific rate constants
(2)
$k_2$ is about equal to $k_{-1}$. (For nitration, $k_2 \gg k_{-1}$.) Therefore, in sulfonation the intermediate can go almost equally well in either direction, and sulfonation is reversible. Furthermore, since the rate of step (2) affects the overall rate, the substitution of $\mathrm{D}$ for $\mathrm{H}$ decreases the rate because $\Delta H^{\ddagger}$ for loss of $\mathrm{D}^{+}$from
is greater than $\Delta H^{\ddagger}$ for loss of $\mathrm{H}^{+}$from the protonated intermediate. Hence, there is a modest primary isotope effect.
Groups other than $\mathrm{H}^{+}$can be displaced during electrophilic aromatic attack. The acid-catalyzed reversal of sulfonation (desulfonation) exemplifies such a reaction; here, $\mathrm{H}^{+}$displaces $\mathrm{SO}_3 \mathrm{H}$ as $\mathrm{SO}_3$ and $\mathrm{H}^{+}$:

Grigoriy Sereda
Grigoriy Sereda
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Problem 7

Use the principle of microscopic reversibility (Problem 6.21) to write a mechanism for desulfonation.

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03:01

Problem 8

(a) Give the delocalized structure (Problem 11.1) for the 3 benzenonium ions resulting from the common ground state for electrophilic substitution, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{G}+\mathrm{E}^{+}$. (b) Give resonance structures for the parabenzenonium ion when $\mathrm{G}$ is $\mathrm{OH}$. (c) Which ions have $\mathrm{G}$ attached to a positively charged $\mathrm{C}$ ? $(d)$ If the products from this reaction are usually determined by rate control (Section 8.5 ), how can the Hammond principle be used to predict the relative yields of $o p$ (i.e., the mixture of ortho and para) as against $m$ (meta) products? $(e)$ In terms of electronic effects, what kind of $\mathrm{G}$ is a (i) $o p$-director, (ii) $m$-director? $(f)$ Classify $\mathrm{G}$ in terms of its structure and its electronic effect.
(c) The ortho and para. This is why $\mathrm{G}$ is either an op-or an m-director.
(d) Because of kinetic control, the intermediate with the lowest-enthalpy transition state (TS) is formed in the greatest amount. Since this step is endothermic, the Hammond principle says that the intermediate resembles the TS. We then evaluate the relative energies of the intermediates ( $o p$ vs. $m$ ) and predict that the one with the lowest enthalpy has the lowest $\Delta H^{\ddagger}$ and is formed in the greatest yield.
(e) (i) An electron-donating $\mathrm{G}$ can better stabilize the intermediate when it is attached directly to positively charged (op) C's. Such G's are op-directing. (ii) An electron-withdrawing G destabilizes the ion to a greater extent when attached directly to positively charged $(o p) \mathrm{C}$ 's. They destabilize less when attached meta and are thus $m$ directors.
( $f$ ) Electron-donating (op-directors): (i) Those that have an unshared pair of electrons on the atom bonded to the ring, which can be delocalized to the ring by extended $\pi$ bonding.
Other examples are $-\ddot{\mathrm{O}}-,-\ddot{\mathrm{X}}$ : (halogen) and $-\ddot{\mathrm{S}}-$.
(ii) Those with an attached atom participating in an electron-rich $\pi$ bond, e.g.
(iii) Those without an unshared pair, which are electron-donating by induction or by hyperconjugation (absence of bond resonance), e.g., alkyl groups.
Electron-withdrawing ( $m$-directors): The attached atom has no unshared pair of electrons and has some positive charge, e.g.

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator
02:08

Problem 9

Explain: (a) All $m$-directors are deactivating. (b) Most $o p$-directing substituents make the ring more reactive than benzene itself--they are activating. (c) As exceptions, the halogens are $o p$-directors but are deactivating.
(a) All $m$-directors are electron-attracting and destabilize the incipient benzenium ion in the TS. They therefore diminish the rate of reaction as compared to the rate of reaction of benzene.
(b) Most $o p$-directors are, on balance, electron-donating. They stabilize the incipient benzenium ion in the TS, thereby increasing the rate of reaction as compared to the rate of reaction of benzene. For example, the ability of the - $\ddot{\mathrm{O}} \mathrm{H}$ group to donate electrons by extended $p$ orbital overlap (resonance) far outweighs the ability of the $\ddot{\mathrm{OH}}$ group to withdraw electrons by its inductive effect.
(c) In the halogens, unlike the $\mathrm{OH}$ group, the electron-withdrawing inductive effect predominates and consequently the halogens are deactivating. The $o-, p-$, and $m$-benzethonium ions each have a higher $\Delta H^t$ than does the cation from benzene itself. However, on demand, the halogens contribute electron density by extended $\pi$ bonding.
and thereby lower the $\Delta H^{\ddagger}$ of the ortho and para intermediates but not the meta cation. Hence the halogens are $o p$-directors, but deactivating.

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator

Problem 10

Compare the activating effects of the following op-directors:
(a) $-\ddot{\mathrm{O}} \mathrm{H},-\ddot{\mathrm{O}}=$ and
<smiles>CC(=O)CO</smiles>
(b) $-\mathrm{NH}_2$ and
<smiles>CNC(C)=O</smiles>
Explain your order.
(a) The order of activation is $-\mathrm{O}^{-}>-\mathrm{OH}>-\mathrm{OCOCH}_3$. The $-\mathrm{O}^{-}$, with a full negative charge, is best able to donate electrons, thereby giving the very stable uncharged intermediate
<smiles>O=C1C=CC(F)C=C1</smiles>
In $-\mathrm{OCOCH}_3$ the $\mathrm{C}$ of the $\stackrel{s+}{\mathrm{C}}=\stackrel{s n}{\mathrm{O}}$ group has + charge and makes demands on the $-\ddot{O}-$ for electron density, thereby diminishing the ability of this - - to donate electrons to the benzenonium ion.
(b) The order is $-\mathrm{NH}_2>-\mathrm{NHCOCH}_3$ for the same reason that $\mathrm{OH}$ is a better activator than $-\mathrm{OCOCH}_3$.
Table 11-2 extends the results of Problem 11.10.

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02:38

Problem 11

(a) Draw enthalpy-reaction diagrams for the first step of electrophilic attack on benzene, toluene (meta and para) and nitrobenzene (meta and para). Assume all ground states have the same energy. (b) Where would the para and meta substitution curves for $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$ lie on this diagram?
(a) Since $\mathrm{CH}_3$ is an activating group, the intermediates and TS's from $\mathrm{PhCH}_3$ have less enthalpy than those from benzene. The para intermediate has less enthalpy than the meta intermediate. The TS and intermediates for $\mathrm{PhNO}_2$ have higher enthalpies than those for $\mathrm{C}_6 \mathrm{H}_6$, with the meta at a lower enthalpy than the para. See Fig. 11-2.
(b) They would both lie between those for benzene and $p$-nitrobenzene, with the para lower than the meta.

Adriano Chikande
Adriano Chikande
Numerade Educator
05:36

Problem 12

(a) Explain in terms of the reactivity-selectivity principle (Section 4.4) the following yields of meta substitution observed with toluene: $\mathrm{Br}_2$ in $\mathrm{CH}_3 \mathrm{COOH}, 0.5 \% ; \mathrm{HNO}_3$ in $\mathrm{CH}_3 \mathrm{COOH}, 3.5 \% ; \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$ in $\mathrm{GaBr}_3$,

$21 \%$. (b) In terms of kinetic vs. thermodynamic control, explain the following effect of temperature on isomer distribution in sulfonation of toluene: at $0^{\circ} \mathrm{C}, 43 \% o-$ and $53 \% p-;$ at $100^{\circ} \mathrm{C}, 13 \% o-$ and $79 \% p-$.
(a) The most reactive electrophile is least selective and gives the most meta isomer. The order of reactivity is:
$$
\mathrm{CH}_3 \mathrm{CH}_2^{+}>\mathrm{NO}_2^{+}>\mathrm{Br}_2\left(\mathrm{Br}^{+}\right)
$$
(b) Sulfonation is one of the few reversible electrophilic substitutions and therefore kinetic and thermodynamic products can result. At $100^{\circ} \mathrm{C}$ the thermodynamic product predominates; this is the para isomer. The ortho isomer is somewhat more favored by kinetic control at $0^{\circ} \mathrm{C}$.

Sana Riaz
Sana Riaz
Numerade Educator
00:45

Problem 13

$\mathrm{PhNO}_2$, but not $\mathrm{C}_6 \mathrm{H}_6$, is used as a solvent for the Friedel-Crafts alkylation of $\mathrm{PhBr}$. Explain. $\mathrm{C}_6 \mathrm{H}_6$ is more reactive than $\mathrm{PhBr}$ and would preferentially undergo alkylation. $-\mathrm{NO}_2$ is so strongly deactivating that $\mathrm{PhNO}_2$ does not undergo Friedel-Crafts alkylation or acylations.

Alkendra Singh
Alkendra Singh
Numerade Educator
04:52

Problem 14

Account for the percentages of $m$-orientation in the following compounds: $(a) \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3(4.4 \%$ ), $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl}(15.5 \%), \mathrm{C}_6 \mathrm{H}_5 \mathrm{CHCl}_2(33.8 \%), \mathrm{C}_6 \mathrm{H}_5 \mathrm{CCl}_3(64.6 \%) ;\left(\right.$ b) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3(100 \%), \mathrm{C}_6 \mathrm{C}_5 \mathrm{CH}_2 \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3$ $(88 \%), \mathrm{C}_6 \mathrm{H}_5\left(\mathrm{CH}_2\right)_2 \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3(19 \%)$.
(a) Substitution of the $\mathrm{CH}_3 \mathrm{H}$ 's by $\mathrm{Cl}$ 's causes a change from electron-release $\left(\leftarrow \mathrm{CH}_3\right)$ to electron-attraction $\left(\rightarrow-\mathrm{CCl}_3\right)$ and $m$-orientation increases.
(b) ${ }^{+} \mathrm{NMe}_3$ has a strong electron-attracting inductive effect and is $m$-orienting. When $\mathrm{CH}_2$ groups are placed between this $\mathrm{N}^{+}$and the ring, this inductive effect falls off rapidly, as does the $m$-orientation. When two $\mathrm{CH}_2$ 's intercede, the electron-releasing effect of the $\mathrm{CH}_2$ bonded directly to the ring prevails, and chiefly op-orientation is observed.

Khoobchandra Agrawal
Khoobchandra Agrawal
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10:15

Problem 15

Predict and explain the reaction, if any, of $(a)$ phenol (PhOH), (b) $\mathrm{PhH}$ and (c) benzenesulfonic acid with $\mathrm{D}_2 \mathrm{SO}_4$ in $\mathrm{D}_2 \mathrm{O}$.
(a) $\mathrm{D}_2 \mathrm{SO}_4$ transfers $\mathrm{D}^{+}$, an electrophile, to form 2,4,6-trideuterophenol. Reaction is rapid because of the activating $-\mathrm{OH}$ group. The meta positions are deactivated. (b) PhH reacts slowly to give hexadeuterobenzene. $(c)$ The sulfonic acid does not react, because $-\mathrm{SO}_3 \mathrm{H}$ is too deactivating.

Nima Gharibi
Nima Gharibi
Numerade Educator
05:36

Problem 16

Indicate by an arrow the position(s) most likely to undergo electrophilic substitution in each of the following compounds. List the number of the above rule(s) used in making your prediction. (a) $m$-xylene, (b) $p$ nitrotoluene, (c) $m$-chloronitrobenzene, $(d) p$-methoxytoluene, $(e) p$-chlorotoluene, $(f) m$-nitrotoluene, $(g) o$ methylphenol (o-cresol).

Matthew Lueckheide
Matthew Lueckheide
Numerade Educator
02:09

Problem 17

Account for $(a)$ formation of the $x$-isomer in nitration and halogenation of naphthalene, $(b)$ formation of $\alpha$-naphthalenesulfonic acid at $80^{\circ} \mathrm{C}$ and $\beta$-naphthalenesulfonic acid at $160^{\circ} \mathrm{C}$.
(a) The mechanism of electrophilic substitution is the same as that for benzene. Attack at the $\alpha$ position has a lower $\Delta H^*$ because intermediate I, an allylic $\mathrm{R}^{+}$with an intact benzene ring, is more stable than intermediate Il from $\beta$ attack.
In II the + charge is isolated from the remaining double bond and hence there is no direct delocalization of charge to the double bond without involvement of the stable benzene ring. In both I and II the remaining aromatic ring has the same effect on stabilizing the + charge. Since I is more stable than II, $\alpha$-substitution predominates.
(b) $\quad \alpha$-Naphthalenesulfonic acid is the kinetic-controlled product [see part (a)]. However, sulfonation is a reversible reaction and at $160^{\circ} \mathrm{C}$ the thermodynamic-controlled product, $\beta$-naphthalenesulfonic acid, is formed.

Alkendra Singh
Alkendra Singh
Numerade Educator

Problem 19

From $\mathrm{C}_6 \mathrm{H}_6(\mathrm{PhH})$ or $\mathrm{PhCH}_3$ synthesize: (a) $p$ - $\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{NO}_2$, (b) $m-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{NO}_2$, (c) $p-$ $\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH},(d) m-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH}$.

In the synthesis of disubstituted benzenes, the first substituent present determines the position of the incoming second. Therefore the order of introducing substituents must be carefully planned to yield the desired isomer.
(a) Since the two substituents are para, it is necessary to introduce the op-directing $\mathrm{Cl}$ first.
$$
\mathrm{PhH} \underset{\mathrm{Cl}_2}{\stackrel{\mathrm{Fe}}{-}} \mathrm{PhCl} \underset{\mathrm{H}_2 \mathrm{SO}_4}{\stackrel{\mathrm{HNO}_3}{-}} p-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{NO}_2
$$
(b) Since the substituents are meta, the $m$-directing $\mathrm{NO}_2$ is introduced first.
$$
\mathrm{PhH} \underset{\mathrm{H}_2 \mathrm{SO}_4}{\stackrel{\mathrm{HNO}_3}{\longrightarrow}} \mathrm{PhNO}_2 \underset{\mathrm{Fe}}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}}-m-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{NO}_2
$$
(c) The $\mathrm{COOH}$ group is formed by oxidation of $\mathrm{CH}_3$. Since $p-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH}$ has two $m$-directing groups, the $\mathrm{NO}_2$ must be added while the op-directing $\mathrm{CH}_3$ is still present.
$$
\mathrm{PhCH}_3 \frac{\mathrm{HNO}_3}{\mathrm{H}_2 \mathrm{SO}_4}-p-\mathrm{CH}_3 \mathrm{C}_6 \mathrm{H}_4 \mathrm{NO}_2+o-\mathrm{CH}_3 \mathrm{C}_6 \mathrm{H}_4 \mathrm{NO}_2
$$
The para isomer is usually easily separated from the op mixture.
$$
p-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{CH}_3 \underset{\mathrm{H}^{+}}{\stackrel{\mathrm{KMnO}_4}{\longrightarrow}} p-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH}
$$
(use phase transfer catalysts, Prob.7.26.)
(d) Now the substituents are meta, and $\mathrm{NO}_2$ is introduced when the $m$-directing $\mathrm{COOH}$ is present.
$$
\mathrm{PhCH}_3 \underset{\mathrm{H}^{+}}{\stackrel{\mathrm{KMnO}_4}{-}} \mathrm{PhCOOH} \frac{\mathrm{HNO}_4}{\mathrm{H}_2 \mathrm{SO}_4}-m-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{COOH}
$$

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06:03

Problem 20

Show steps in the synthesis of $(a) o$-chlorotoluene and $(b)$ 1,3-dimethyl-2-ethylbenzene.
(a) In this synthesis $-\mathrm{SO}_3 \mathrm{H}$ is the blocking group. In the second step the op-directing $\mathrm{CH}_3$ and the $m$-directing $\mathrm{SO}_3 \mathrm{H}$ reinforce each other.
<smiles>Cc1ccccc1Cl</smiles>
(b) In this synthesis $-\mathrm{C}\left(\mathrm{CH}_3\right)_3$ is the blocking group. In the second step, although $\mathrm{C}_2 \mathrm{H}_5$ and $\mathrm{C}\left(\mathrm{CH}_3\right)_3$ are competing $o p$-directors, the bulkiness of the latter group inhibits attack ortho to itself.
In the reaction with $\mathrm{HF}$, the electrophile $\mathrm{H}^{+}$replaces $\mathrm{C}\left(\mathrm{CH}_3\right)_3^{+}$, which forms $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$.

Ian Kaigh
Ian Kaigh
Numerade Educator
03:21

Problem 21

Account for the product in the reaction.
$\mathrm{CN}^{-}$is a nucleophile. $\mathrm{NO}_2$ 's activate the ring toward nucleophilic substitution at op-positions by withdrawing the electron density and placing charge on the $\mathrm{O}$ 's of $\mathrm{NO}_2$ :

When not sterically hindered, the ortho position may be more reactive. $\mathrm{CN}^{-}$is a "thin" nucleophile and its insertion ortho to each $\mathrm{NO}_2$ is not hindered.

A good leaving group, such as halide ion $\left(\mathrm{X}^{-}\right)$, is more easily displaced than $\mathrm{H}^{-}$from a benzene ring by nucleophiles. Electron-attracting substituents, such as $\mathrm{NO}_2$ and $\mathrm{CN}$, in ortho and para positions facilitate the nucleophilic displacement of $\mathrm{X}$ of aryl halides. The greater the number of such ortho and para substituents, the more rapid the reaction and the less vigorous the conditions needed.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
00:57

Problem 22

Write resonance structures to account for activation in addition-elimination aromatic nucleophilic substitution from delocalization of the charge of the intermediate carbanion by the following para substituent groups: (a) $-\mathrm{NO}_2,(b)-\mathrm{CN},(c)-\mathrm{N}=\mathrm{O},(d) \mathrm{CH}=\mathrm{O}$.
Only the resonance structures with the negative charge on the para $\mathrm{C}$ are written to show delocalization of charge from ring $\mathrm{C}$ to the para substituent.

Lottie Adams
Lottie Adams
Numerade Educator
07:05

Problem 23

Compare addition-elimination aromatic nucleophilic and electrophilic substitution reactions with aliphatic $\mathrm{S}_{\mathrm{N}} 2$ reactions in terms of $(a)$ number of steps and transition states, $(b)$ character of intermediates.
(a) Nucleophilic and electrophilic aromatic substitutions are two-step reactions, having a first slow and rate determining step followed by a rapid second step. Aliphatic $\mathrm{S}_{\mathrm{N}} 2$ reactions have only one step. There are 2 transition states for the aromatic and 1 for the aliphatic substitution. (b) $\mathrm{S}_{\mathrm{N}} 2$ reactions have no intermediate. In electrophilic aromatic substitution the intermediate is a carbocation, while that in nucleophilic substitution is a carbanion.

Nicholas Sacco
Nicholas Sacco
Numerade Educator
02:56

Problem 24

Why do the typical $\mathrm{S}_{\mathrm{N}} 2$ and $\mathrm{S}_{\mathrm{N}} 1$ mechanisms not occur in nucleophilic aromatic substitution?

The $S_N 2$ backside attack cannot occur, because of the high electron density of the delocalized $\pi$ cloud of the benzene ring. Furthermore, inversion at the attacked $\mathrm{C}$ is sterically impossible. The $\mathrm{S}_{\mathrm{N}} 1$ mechanism does not occur, because the intermediate $\mathrm{C}_6 \mathrm{H}_5^{+}$, with a + charge on an $s p$-hybridized $\mathrm{C}$, would have a very high energy. The ring would also have a large ring strain.

Temi Ajayi
Temi Ajayi
Numerade Educator
03:21

Problem 25

How do the following observations support the benzyne mechanism? (a) Compounds lacking ortho $\mathrm{H}$ 's, such as 2,6-methylchlorobenzene, do not react. (b) 2,6-Deutero Bromobenzene reacts more slowly than bromobenzene. (c) o-Bromoanisole, $o$ - $\mathrm{CH}_3 \mathrm{OC}_6 \mathrm{H}_4 \mathrm{Br}$, reacts with $\mathrm{NaNH}_2 / \mathrm{NH}_3$ to form $m$ - $\mathrm{CH}_3 \mathrm{OC}_6 \mathrm{H}_4 \mathrm{NH}_2 .(d)$ Chlorobenzene with $\mathrm{Cl}$ bonded to ${ }^{14} \mathrm{C}$ gives almost $50 \%$ aniline having $\mathrm{NH}_2$ bonded to ${ }^{14} \mathrm{C}$ and $50 \%$ aniline with $\mathrm{NH}_2$ bonded to an ortho $\mathrm{C}$.

(a) With no $\mathrm{H}$ ortho to $\mathrm{Cl}$, vicinal elimination cannot occur.
(b) This primary isotope effect (Problem 7.28) indicates that a bond to $\mathrm{H}$ is broken in the rate-determining step, which is consistent with the first step in the benzyne mechanism being rate-determining.
(c) $\mathrm{NH}_2^{-}$need not attack the $\mathrm{C}^2$ from which the $\mathrm{Br}^{-}$left; it can add at $\mathrm{C}^3$.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
05:40

Problem 26

Account for the observation that $\mathrm{NaOH}$ reacts at $300^{\circ} \mathrm{C}$ with $p$-bromotoluene to give $m$ - and $p$ cresols, while $m$-bromotoluene yields the three isomeric cresols.

The benzyne intermediate from $p$-bromotoluene has a triple bond between $\mathrm{C}^3$ and $\mathrm{C}^4$; both $\mathrm{C}^{\prime}$ are independently attacked by $\mathrm{OH}^{-}$, giving a mixture of $m$ - and $p$-cresols $\left(\mathrm{HOC}_6 \mathrm{H}_4 \mathrm{CH}_3\right)$. Two isomeric benzynes are formed from $m$ bromotoluene, one with a $\mathrm{C}^2-$ to- $-\mathrm{C}^3$ triple bond and the other with a $\mathrm{C}^3$-to- $\mathrm{C}^4$ triple bond. Hence, this mixture of benzynes reacts with $\mathrm{OH}^{-}$at all three $\mathrm{Cs}$, giving the mixture of three isomeric cresols.

Matthew Lueckheide
Matthew Lueckheide
Numerade Educator
01:16

Problem 27

Supply systematic and, where possible, common names for:
(a) p-Isopropyltoluene (p-cymene). (b) 1,3,5-Trimethylbenzene (mesitylene). (c) p-Methylstyrene. (d) 1,4Diphenyl-2-butyne (dibenzylacetylene). (e) (Z)-1,2-Diphenylethene (cis-stilbene).

Susan Hallstrom
Susan Hallstrom
Numerade Educator
02:09

Problem 28

Arrange the isomeric tetramethylbenzenes, prehnitene $(1,2,3,4-)$ and durene $(1,2,4,5-)$ in order of decreasing melting point and verify this order from tables of melting points.

The more symmetrical the isomer, the closer the molecules are packed in the crystal and the higher is the melting point. The order of decreasing symmetry, durene $>$ prehnitene, corresponds to that of their respective melting points, $+80^{\circ} \mathrm{C}>-6.5^{\circ} \mathrm{C}$.

Freddie Montague
Freddie Montague
Numerade Educator
04:07

Problem 29

Explain the following observations about the Friedel-Crafts alkylation reaction. $(a)$ In monoalkylating $\mathrm{C}_6 \mathrm{H}_6$ with $\mathrm{RX}$ in $\mathrm{AlX}_3$ an excess of $\mathrm{C}_6 \mathrm{H}_6$ is used. (b) The alkylation of $\mathrm{PhOH}$ and $\mathrm{PhNH}_2$ gives poor yields. (c) $\mathrm{Ph}-\mathrm{Ph}$ cannot be prepared by the reaction
$$
\mathrm{PhH}+\mathrm{PhCl} \stackrel{\mathrm{AlCl}_3}{\varkappa} \mathrm{Ph}-\mathrm{Ph}+\mathrm{HCl}
$$
(d) At $0^{\circ} \mathrm{C}$
$$
\mathrm{PhH}+3 \mathrm{CH}_3 \mathrm{Cl} \stackrel{\mathrm{AlCl}_3}{\longrightarrow} \text { 1,2,4-trimethylbenzene }
$$
but at $100^{\circ} \mathrm{C}$ one gets 1,3,5-trimethylbenzene (mesitylene). (e) The reaction
$$
\mathrm{PhH}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl} \stackrel{\mathrm{AlCl}_3}{\longrightarrow} \mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{HCl}
$$
gives poor yield, whereas
$$
\mathrm{PhH}+\mathrm{CH}_3 \mathrm{CHClCH}_3 \stackrel{\mathrm{AlCl}_3}{\longrightarrow} \mathrm{PhCH}\left(\mathrm{CH}_3\right)_2+\mathrm{HCl}
$$
gives very good yield.
(a) The monoalkylated product, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{R}$, which is more reactive than $\mathrm{C}_6 \mathrm{H}_6$ itself since $\mathrm{R}$ is an activating group, will react to give $\mathrm{C}_6 \mathrm{H}_4 \mathrm{R}_2$ and some $\mathrm{C}_6 \mathrm{H}_3 \mathrm{R}_3$. To prevent polyalkylation an excess of $\mathrm{C}_6 \mathrm{H}_6$ is used to increase the chance for collision between $\mathrm{R}^{+}$and $\mathrm{C}_6 \mathrm{H}_6$ and to minimize collision between $\mathrm{R}^{+}$and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{R}$.
(b) $\mathrm{OH}$ and $\mathrm{NH}_2$ groups react with and inactivate the catalyst.
(c) $\mathrm{PhCl}+\mathrm{AlCl}_3 \rightarrow \mathrm{Ph}^{+}+\mathrm{AlCl}_4^{-}, \mathrm{Ph}^{+}$has a very high enthalpy and doesn't form.
(d) The alkylation reaction is reversible and therefore gives the kinetic-controlled product at $0^{\circ} \mathrm{C}$ and the thermodynamic-controlled product at $100^{\circ} \mathrm{C}$.
(e) The $\mathrm{R}^{+}$intermediates, especially the $1^{\circ} \mathrm{RCH}_2^{+}$can undergo rearrangements. With $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}$ we get
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2^{+} \stackrel{\sim \mathrm{H}}{\longrightarrow} \mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{HCH}_3
$$
and the major product is $\mathrm{PhCH}\left(\mathrm{CH}_3\right)_2$.

Vasu Makani
Vasu Makani
Numerade Educator
03:09

Problem 30

Prepare $\mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ from $\mathrm{PhH}$ and any open-chain compound.
$$
\mathrm{PhH}+\mathrm{ClCH}_2 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{AlCl}_3}{\longrightarrow} \mathrm{PhCH}_2 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{H}_2 / \mathrm{Pt}}{\longrightarrow} \mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
or
$$
\mathrm{PhH}+\mathrm{ClCOCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{AlCl}_3}{\longrightarrow} \mathrm{PhCOCH}_2 \mathrm{CH}_3 \frac{\mathrm{Zn} / \mathrm{Hg}, \mathrm{HCl}}{\text { (Clemmensen reduction) }} \longrightarrow \mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}$ cannot be used because the intermediate $1^{\circ} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2^{+}$rearranges to the more stable $2^{\circ}$ $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}^{+}$, to give $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{CH}_3\right)_2$ as the major product. The latter method is for synthesizing $\mathrm{PhCH}_2 \mathrm{R}$.

Benjamin Angeles
Benjamin Angeles
Numerade Educator

Problem 31

Give the structural formula and the name for the major alkylation product:
(a) $\mathrm{C}_6 \mathrm{H}_6+\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{Cl} \stackrel{\mathrm{AlCl}_7}{\longrightarrow}$
(b) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3+\left(\mathrm{CH}_3\right)_3 \mathrm{CCH}_2 \mathrm{OH} \stackrel{\mathrm{BF}_3}{\longrightarrow}$
(c) $\mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \mathrm{Cl} \underset{100{ }^{\circ} \mathrm{C}}{\stackrel{\mathrm{AlCl}_3}{-}}$
(d) $m$-xylene $+\left(\mathrm{CH}_3\right)_3 \mathrm{CCl} \frac{\mathrm{AlCl}_3}{1000^{\circ} \mathrm{C}}$
(a)
<smiles>CC(C)CC1CCCCC1(C)CC(C)C(C)C</smiles>
(b)
(c)
(d)
<smiles>Cc1cc(C)cc(C(C)(C)C)c1</smiles>
Thermodynamic product; it has less steric strain and is more stable than the kinetic-controlled isomer having a bulky $t$-butyl group ortho to a $\mathrm{CH}_3$.

Check back soon!
07:32

Problem 32

$\mathrm{PhCH}_3$ reacts with $\mathrm{Br}_2$ and $\mathrm{Fe}$ to give a mixture of three monobromo products. With $\mathrm{Br}_2$ in light, only one compound, a fourth monobromo isomer, is isolated. What are the four products? Explain the formation of the light-catalyzed product.

With $\mathrm{Fe}$, the products are $o-, p-$, and some $m-\mathrm{BrC}_6 \mathrm{H}_4 \mathrm{CH}_3$. In light the product is benzyl bromide, $\mathrm{PhCH}_2 \mathrm{Br}$. Like allylic halogenation (Section 6.5), the latter reaction is a free-radical substitution:
(1)
$$
\mathrm{Br}_2 \stackrel{\text { uv }}{\longrightarrow} 2 \mathrm{Br} \text {. }
$$
(2) $\mathrm{Br}^*+\mathrm{PhCH}_3 \longrightarrow \mathrm{PhCH}_2+\mathrm{HBr}$
(3) $\mathrm{PhCH}_2+\mathrm{Br}_2 \longrightarrow \mathrm{PhCH}_2 \mathrm{Br}+\mathrm{Br}$.

Steps (2) and (3) are the propagating steps.

Anish Wadhwa
Anish Wadhwa
Numerade Educator
03:06

Problem 33

Which is more reactive to radical halogenation, $\mathrm{PhCH}_3$ or $p$-xylene? Explain. p-Xylene reactivity depends on the rate of formation of the benzyl-type radical. Electron-releasing groups such as $\mathrm{CH}_3$ stabilize the transition state, producing the benzyl radical on the other $\mathrm{CH}_3$ and thereby lowering the $\Delta H^{ \pm}$and increasing the reaction rate.

Temi Ajayi
Temi Ajayi
Numerade Educator
11:12

Problem 34

Outline a synthesis of 2,3-dimethyl-2,3-diphenylbutane from benzene, propylene, and any needed inorganic reagents.
The symmetry of this hydrocarbon makes possible a self-coupling reaction with 2-bromo-2-phenylpropane.

Susan Hallstrom
Susan Hallstrom
Numerade Educator

Problem 35

Give all possible products of the following reactions and underline the major product.
(a) $\mathrm{PhCH}_2 \mathrm{CHOHCH}\left(\mathrm{CH}_3\right)_2 \stackrel{\mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}$
(b) $\mathrm{PhCH}_2 \mathrm{CHBrCH}\left(\mathrm{CH}_3\right)_2 \underset{\mathrm{KOH}}{\stackrel{\text { alc. }}{\mathrm{KOH}}}$
(c) $\mathrm{PhCH}=\mathrm{CHCH}_3+\mathrm{HBr} \longrightarrow$
(d) $\mathrm{PhCH}=\mathrm{CHCH}_3+\mathrm{HBr} \stackrel{\text { peroxide }}{\longrightarrow}$
(e) $\mathrm{PhCH}=\mathrm{CHCH}=\mathrm{CH}_2+\mathrm{Br}_2$ (equimolar amounts) $\longrightarrow$
(a) $\mathrm{PhCH}_2 \mathrm{CH}=\mathrm{C}\left(\mathrm{CH}_3\right)_2+\mathrm{PhCH}=\mathrm{CHCH}\left(\mathrm{CH}_3\right)_2$. The major product has the $\mathrm{C}=\mathrm{C}$ conjugated with the benzene ring and therefore, even though it is a disubstituted alkene, it is more stable than the minor product, which is a trisubstituted nonconjugated alkene.
(b) Same as part $(a)$ and for the same reason.
(c) $\mathrm{PhCH}_2 \mathrm{CHBrCH}_3+\mathrm{PhCHBrCH}_2 \mathrm{CH}_3 . \mathrm{H}^{+}$adds to $\mathrm{C}=\mathrm{C}$ to give the more stable benzyl-type $\mathrm{PhCHCH}_2 \mathrm{CH}_3$. Reaction with $\mathrm{Br}^{-}$gives the major product. The benzyl-type cation $\mathrm{PhCHR}$ (like $\mathrm{CH}_2=\mathrm{CHCH}_2^{+}$) can be stabilized by delocalizing the + to the $o p$-positions of the ring:
<smiles>CC/C=C1/[CH+][CH+][C@H]2C[C@H]1C2</smiles>
(d) $\mathrm{PhCH}_2 \mathrm{CHBrCH}_3+\mathrm{PhCHBrCH}_2 \mathrm{CH}_3 . \mathrm{Br}$ - adds to give the more stable benzyl-type $\mathrm{PhC} \dot{\mathrm{C}} \mathrm{CHBrCH}_3$ rather than $\mathrm{PhCHBr}+\mathrm{HCH}_3$. We have already discussed the stability of benzylic free radicals.
(e) $\mathrm{PhCHBrCHBrCH}=\mathrm{CH}_2+\mathrm{PhCH}=\mathrm{CHCHBrCHBr}+\mathrm{PhCHBrCH}=\mathrm{CHCH}_2 \mathrm{Br}$. The major product is the conjugated alkene, which is more stable than the other two products [see part $(a)$ ].

Check back soon!
03:01

Problem 36

Explain the following observations. (a) A yellow color is obtained with $\mathrm{Ph}_3 \mathrm{COH}$ (trityl alcohol) is reacted with concentrated $\mathrm{H}_2 \mathrm{SO}_4$, or when $\mathrm{Ph}_3 \mathrm{CCl}$ is treated with $\mathrm{AlCl}_3$. On adding $\mathrm{H}_2 \mathrm{O}$, the color disappears and a white solid is formed. (b) $\mathrm{Ph}_3 \mathrm{CCl}$ is prepared by the Friedel-Crafts reaction of benzene and $\mathrm{CCl}_4$. It does not react with more benzene to form $\mathrm{Ph}_4 \mathrm{C}$. (c) A deep-red solution appears when $\mathrm{Ph}_3 \mathrm{CH}$ is added to a solution of $\mathrm{NaNH}_2$ in liquid $\mathrm{NH}_3$. The color disappears on adding water. $(d) \mathrm{A}$ red color appears when $\mathrm{Ph}_3 \mathrm{CCl}$ reacts with $\mathrm{Zn}$ in $\mathrm{C}_6 \mathrm{H}_6 . \mathrm{O}_2$ decolorizes the solution.
(a) The yellow color is attributed to the stable $\mathrm{Ph}_3 \mathrm{C}^{+}$, whose + is delocalized to the op-positions of the 3 rings.
$$
\begin{aligned}
& \mathrm{Ph}_3 \mathrm{COH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Ph}_3 \mathrm{C}^{+}+\mathrm{H}_3 \mathrm{O}^{+}+\mathrm{HSO}_4^{-} \\
& \mathrm{Ph}_3 \mathrm{CCl}+\mathrm{AlCl}_3 \longrightarrow \mathrm{Ph}_3 \mathrm{C}^{+}+\mathrm{AlCl}_4^{-} \\
& \mathrm{Ph}_3 \mathrm{C}^{+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ph}_3 \mathrm{COH}+\mathrm{H}_3 \mathrm{O}^{+} \\
& \begin{array}{l}
\text { Lewis Lewis } \\
\text { base }
\end{array} \\
& \begin{array}{ll}
\text { Lewis Lewis } & \text { white } \\
\text { acid base } & \text { solid }
\end{array} \\
&
\end{aligned}
$$
(b) With $\mathrm{AlCl}_3, \mathrm{Ph}_3 \mathrm{CCl}$ forms a salt, $\mathrm{Ph}_3 \mathrm{C}^{+} \mathrm{AlCl}_4^{-}$, whose carbocation is too stable to react with benzene. $\mathrm{Ph}_3 \mathrm{C}^{+}$ may also be too sterically hindered to react further.
(c) The strong base $::_{\mathrm{N}} \mathrm{H}_2^{-}$removes $\mathrm{H}^{+}$from $\mathrm{Ph}_3 \mathrm{CH}$ to form the stable, deep red-purple carbanion $\mathrm{Ph}_3 \mathrm{C}^{-}$, which is then decolorized on accepting $\mathrm{H}^{+}$from the feeble acid $\mathrm{H}_2 \mathrm{O}$.
$$
\begin{aligned}
& \mathrm{Ph}_3 \mathrm{CH}+: \mathrm{NH}_2^{-} \longrightarrow \mathrm{H}: \mathrm{NH}_2+\mathrm{Ph}_3 \mathrm{C}^{-} \\
& \text {acid }_1 \text { base }_2 \quad \text { acid }_2 \text { base, (deepred) } \\
& \mathrm{Ph}_3 \mathrm{C}^{-}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ph}_3 \mathrm{CH}+\mathrm{OH}^{-} \\
& \text {base, } \text { acid }_2 \quad \text { acid }_1 \text { base }_2 \\
&
\end{aligned}
$$

The $\mathrm{Ph}_3 \mathrm{C}^{-}$is stabilized because the - can be delocalized to the op-positions of the three rings (as in the corresponding carbocation and free radicals).
(d) $\mathrm{Cl}$ - is removed from $\mathrm{Ph}_3 \mathrm{CCl}$ by $\mathrm{Zn}$ to give the colored radical $\mathrm{Ph}_3 \mathrm{C}$, which decolorizes as it forms the peroxide in the presence of $\mathrm{O}_2$.
$$
\begin{array}{r}
2 \mathrm{Ph}_3 \mathrm{CCl}+\mathrm{Zn} \\
2 \mathrm{Ph}_3 \mathrm{C}+2 \mathrm{Ph}_3 \mathrm{C}+\mathrm{ZnCl}_2 \\
\longrightarrow \mathrm{Ph}_3 \mathrm{C}: \ddot{\mathrm{O}}: \mathrm{CPh}_3
\end{array}
$$

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator
01:01

Problem 37

In the following hydrocarbons the alkyl H's are designated by the Greek letters $\alpha, \beta, \gamma$, etc. Assign each letter an Arabic number, beginning with I for LEAST, in order of increasing ease of abstraction by a Br:

Brittany Carnathan
Brittany Carnathan
Numerade Educator
01:05

Problem 38

Use + and - signs for positive and negative tests in tabulating rapid chemical reactions that can be used to distinguish among the following compounds: $(a)$ chlorobenzene, benzyl chloride and cyclohexyl chloride; $(b)$ ethylbenzene, styrene, and phenylacetylene.
See Tables 11-4(a) and $11-4(b)$.

Alkendra Singh
Alkendra Singh
Numerade Educator
03:06

Problem 40

Which xylene is most easily sulfonated?
$m$-Xylene is most reactive and sulfonates at $\mathrm{C}^4$ because its $\mathrm{CH}_3$ 's reinforce each other [Rule 1; Problem I1.16(a)].

Temi Ajayi
Temi Ajayi
Numerade Educator
07:23

Problem 41

Write structures for the principal mononitration products of $(a) o$-cresol (o-methylphenol), $(b)$ $p$ - $\mathrm{CH}_3 \mathrm{CONHC}_6 \mathrm{H}_4 \mathrm{SO}_3 \mathrm{H}$, (c) m-cyanotoluene (m-toluonitrile).

Nicholas Sacco
Nicholas Sacco
Numerade Educator
01:02

Problem 42

Assign numbers from 1 for LEAST to 5 for MOST to designate relative reactivity to ring monobromination of the following groups.
(a) (l) $\mathrm{PhNH}_2$,
(II) $\mathrm{PhNH}_3^{+} \mathrm{Cl}^{-}$,
(III) $\mathrm{PhNHCOCH}_3$,
(IV) $\mathrm{PhCl}$,
(V) $\mathrm{PhCOCH}_3$.
(b) (I) $\mathrm{PhCH}_3$,
(II) $\mathrm{PhCOOH}$,
(III) $\mathrm{PhH}$
(c) (l) p-xylene,
(II) $p-\mathrm{C}_6 \mathrm{H}_4(\mathrm{COOH})_2$,
(III) $\mathrm{PhMe}$,
(IV) $p-\mathrm{CH}_3 \mathrm{C}_6 \mathrm{H}_4 \mathrm{COOH}$,
(V) $m$-xylene.
See Table 11-5.

Nidhi Singhi
Nidhi Singhi
Numerade Educator
11:02

Problem 43

Use $\mathrm{PhH}, \mathrm{PhMe}$, and any aliphatic or inorganic reagents to prepare the following compounds in reasonable yields: (a) $m$-bromobenzenesulfonic acid, (b) 3-nitro-4-bromobenzoic acid, (c) 3,4-dibromonitrobenzene, (d) 2,6-dibromo-4-nitrotoluene.

Matthew Lueckheide
Matthew Lueckheide
Numerade Educator
02:29

Problem 44

Supply structures for organic compounds (A) through (O).
(a) $\mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{Br}_2 \stackrel{\text { uv }}{\longrightarrow}$ (A) $\frac{\text { alc. }}{\mathrm{KOH}}$
(B) $\frac{\text { cold dil. }}{\mathrm{KMnO}_4}$
(C) $\frac{\mathrm{hot}}{\mathrm{KMnO}_4}$
(b) $\mathrm{PhBr}+\mathrm{Mg} \stackrel{\mathrm{F}_2 \mathrm{O}}{\longrightarrow}$
(E) $\stackrel{\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{Br}}{\longrightarrow}$
(F)
(c) $\mathrm{Ph}-\mathrm{C}=\mathrm{CH}+\mathrm{CH}_3 \mathrm{MgX} \longrightarrow$ (I) $\stackrel{\mathrm{ArCH}_2 \mathrm{Cl}}{\longrightarrow}$ (J) $\stackrel{\mathrm{Li}, \mathrm{NH}_3}{\longrightarrow}(\mathrm{K})$
(d) $p-\mathrm{CH}_3 \mathrm{C}_6 \mathrm{H}_4 \mathrm{C}=\mathrm{CPh}+\mathrm{H}_2 / \mathrm{Pt} \longrightarrow$ (L) $\stackrel{\mathrm{HBr}}{\longrightarrow}$ (M)
(e)
<smiles>C/C=C\c1ccccc1</smiles>
$$
+\mathrm{Br}_2(\mathrm{Fe}) \longrightarrow(\mathrm{N}) \frac{\mathrm{HBr}}{\text { peroxide }}-(\mathrm{O})
$$

Raghvendra Singh
Raghvendra Singh
Numerade Educator
02:36

Problem 45

Assign numbers from 1 for LEAST to 3 for MOST to the Roman numerals for the indicated compounds to show their relative reactivities in the designated reactions.
(a) $\mathrm{HBr}$ addition to (I) $\mathrm{PhCH}=\mathrm{CH}_2$, (II) $p-\mathrm{CH}_3 \mathrm{C}_6 \mathrm{H}_4 \mathrm{CH}=\mathrm{CH}_2$, (III) $p-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{CH}=\mathrm{CH}_2$.
(b) Dehydration of (I) $p-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{CHOHCH}_3$, (II) $p-\mathrm{H}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{CHOHCH}_3$, (III) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHOHCH}_3$.
(c) Dehydration of
(I)
<smiles>CCC(C)(O)c1ccccc1</smiles>
(II)
<smiles>CC(CO)c1ccccc1</smiles>
(III) $\mathrm{PhCHOHCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ -
(d) Solvolysis of (I) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl}$,
(II) $p-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{Cl}$,
(III) $p-\mathrm{CH}_3 \mathrm{OC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{Cl}$.
See Table 11-6.

Monisha Shukla
Monisha Shukla
Numerade Educator
08:37

Problem 46

Show the syntheses of the following compounds from benzene, toluene and any inorganic reagents or aliphatic compounds having up to three $\mathrm{C}$ 's:
(a) $p-\mathrm{BrC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{Cl}$
(b) $p-\mathrm{BrC}_6 \mathrm{H}_4 \mathrm{CH}=\mathrm{CH}_2$
(c)
<smiles>CC(C)(O)c1ccccc1</smiles>
(d) $p-\mathrm{O}_2 \mathrm{NC}_6 \mathrm{H}_4 \mathrm{CH}_2 \mathrm{Ph}$
(e) $\mathrm{Ph}_2 \mathrm{CHCH}_3$

Zubair Abdulla
Zubair Abdulla
Numerade Educator
01:22

Problem 47

Deduce the structural formulas of the following arenes. (a) (i) Compound $\mathrm{A}\left(\mathrm{C}_{16} \mathrm{H}_{16}\right)$ decolorizes both $\mathrm{Br}_2$ in $\mathrm{CCl}_4$ and cold aqueous $\mathrm{KMnO}_4$. It adds an equimolar amount of $\mathrm{H}_2$. Oxidation with hot $\mathrm{KMnO}_4$ gives a dicarboxylic acid, $\mathrm{C}_6 \mathrm{H}_4(\mathrm{COOH})_2$, having only one monobromo substitution product. (ii) What structural feature is uncertain? $(b)$ Arene $\mathrm{B}\left(\mathrm{C}_{10} \mathrm{H}_{14}\right)$ has five possible monobromo derivatives $\left(\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Br}\right)$. Vigorous oxidation of $\mathrm{B}$ yields an acidic compound, $\mathrm{C}_8 \mathrm{H}_6 \mathrm{O}_4$, having only one mononitro substitution product, $\mathrm{C}_8 \mathrm{H}_5 \mathrm{O}_4 \mathrm{NO}_2$.
(a) (i) Compound $\mathrm{A}$ has one $\mathrm{C}=\mathrm{C}$ since it adds one $\mathrm{H}_2$. The other 8 degrees of unsaturation mean the presence of two benzene rings. Since oxidative cleavage gives a dicarboxylic acid, $\mathrm{C}_6 \mathrm{H}_4(\mathrm{COOH})_2$, each benzene ring must be disubstituted. Since $\mathrm{C}_6 \mathrm{H}_4(\mathrm{COOH})_2$ has only one monobromo derivative, the $\mathrm{COOH}$ 's must be para to each other.

Lottie Adams
Lottie Adams
Numerade Educator
01:34

Problem 49

Assign numbers from 1 for LEAST to 3 for MOST to show the reiative reactivities of the compounds with the indicated reagents: (a) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$ (I), $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHBrCH}_3$ (II) and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{CHBr}$ (III) with alcoholic $\mathrm{AgNO}_3 ;$ (b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$ (I), $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl}$ (II) and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$ (III) with $\mathrm{KCN}$; (c) $m$-nitrochlorobenzene (I), 2,4dinitrochiorobenzene (II) and $p$-nitrochlorobenzene (III) with sodium methoxide.
See Table 11-7.

Nadir Iqbal
Nadir Iqbal
Numerade Educator
01:04

Problem 50

Which $\mathrm{Cl}$ in 1,2,4-trichlorobenzene reacts with ${ }^{-} \mathrm{OCH}_2 \mathrm{COO}^{-}$to form the herbicide "2,4-D"? Give the structure of "2,4-D."
$\mathrm{Cl}$ 's are electron-withdrawing and activate the ring to nucleophilic attack. The $\mathrm{Cl}$ at $\mathrm{C}^l$ is displaced because it is ortho and para to the other Cl's.

Narayan Hari
Narayan Hari
Numerade Educator
04:36

Problem 51

Explain these observations: $(a) p$-Nitrobenzenesulfonic acid is formed from the reaction of $p$ nitrochlorobenzene with $\mathrm{NaHSO}_3$, but benzenesulfonic acid cannot be formed from chlorobenzene by this reaction.
(b) 2,4,6-Trinitroanisole with $\mathrm{NaOC}_2 \mathrm{H}_5$ gives the same product as 2,4,6-trinitrophenetole with $\mathrm{NaOCH}_3$.
(a) Nucleophilic aromatic substitution occurs with $p$-nitrochlorobenzene, but not with chlorobenzene, because $\mathrm{NO}_2$ stabilizes the carbanion [Problem $11.22(a)$ ].
(b) The product is a sodium salt formed by addition of alkoxide.

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator