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Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 14

ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS - all with Video Answers

Educators

Chapter Questions

Problem 1

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01:22

Problem 1

Does peroxide formation occur more rapidly with $\left(\mathrm{RCH}_2\right)_2 \mathrm{O}$ or $\left(\mathrm{R}_2 \mathrm{CH}\right)_2 \mathrm{O}$ ? With $\left(\mathrm{R}_2 \mathrm{CH}\right)_2 \mathrm{O}$ because the $2^{\circ}$ radical is more stable and forms faster.

Catherine Lemar
Catherine Lemar
Numerade Educator
01:01

Problem 1

Use any needed starting material to synthesize the following ethers, selecting from among intermolecular dehydration, Williamson synthesis, and alkoxymercuration-demercuration. Justify your choice of method.
(a) $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{OCH}_2 \mathrm{CH}_3$
(b)
<smiles>CCCOC(C)CC</smiles>
(c) dicyclohexyl ether
(a) Use Williamson synthesis; $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{Cl}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} \mathrm{Na}^{+}$. Since alkoxymercuration is a Markovnikov addition, it cannot be used to prepare an ether in which both $\mathrm{R}^{\prime}$ s are $1^{\circ}$. Unless one of the R's can form a stable $\mathrm{R}^{+}$, intermolecular dehydration cannot be used to synthesize a mixed ether.
(b)
This is better than Williamson synthesis because there is no competing elimination reaction.
(c) Dehydration; Cyclohexyl- $\mathrm{OH} \stackrel{\mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}$ (Cyclohexyl $)_2 \mathrm{O}$. This is a simple ether.

Narayan Hari
Narayan Hari
Numerade Educator

Problem 2

CAN'T COPY

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09:24

Problem 2

Account for the product from the following reaction of the ${ }^{14} \mathrm{C}$-labeled chloroepoxide:
$\mathrm{S}_{\mathrm{N}} 2$ attack by $\mathrm{CH}_3 \mathrm{O}^{-}$on the (less substituted) ${ }^{14} \mathrm{C}$ gives an intermediate alkoxide, then displaces $\mathrm{Cl}^{-}$by another $\mathrm{S}_{\mathrm{N}} 2$ reaction, forming the new epoxide.
<smiles>COCC([O-])CCl</smiles>
2. $\mathrm{S}_{\mathrm{N}} 1$ Ring-Opening

In acid the protonated epoxide may undergo ring-opening to give an intermediate carbocation.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
03:21

Problem 2

(a) Give the product of the $\mathrm{S}_{\mathrm{N}}$ 2-type addition of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{MgBr}$ to ethylene oxide. (b) What is the synthetic utility of the reaction of Grignard reagents and ethylene oxide?
(a)
<smiles>Br[Mg]1(Br)CCCC1</smiles>
$\left(\mathrm{S}_{\mathrm{N}} 2\right)$
1-butanol
(b) It is a good method for extending the $\mathrm{R}$ group of the Grignard by $-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$ in one step.

Shazia Naz
Shazia Naz
Numerade Educator

Problem 2

Outline the $\mathrm{S}_{\mathrm{N}} 2$ mechanism for acid- and base-catalyzed addition to ethylene oxide and give the structural formulas of the products of addition of the following: (a) $\mathrm{H}_2 \mathrm{O}$, (b) $\mathrm{CH}_3 \mathrm{OH}$, (c) $\mathrm{CH}_3 \mathrm{NH}_2,($ d $) \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{SH}_{\text {. }}$.
In acid, $O$ is first protonated.
The protonated epoxide can also react with nucleophilic solvents such as $\mathrm{CH}_3 \mathrm{OH}$.
In base, the ring is cleaved by attack of the nucleophile on the less substituted $\mathrm{C}$ to form an alkoxide anion, which is then protonated. Reactivity is attributed to the highly strained three-membered ring, which is readily cleaved.
(a) $\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{OH}$
(b) $\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{OH}$
(c) $\mathrm{CH}_3 \mathrm{NHCH}_2 \mathrm{CH}_2 \mathrm{OH}$
(d) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{SCH}_2 \mathrm{CH}_2 \mathrm{OH}$
Base-induced ring-openings require a strong base because the strongly basic $\mathrm{O}^{-}$is displaced as part of the alkoxide. Acid-induced ring-openings are achieved with weak bases, such as nucleophilic solvents, because now the very weakly basic $\mathrm{OH}$, formed by protonation of the $\mathrm{O}$ atom, is displaced as part of the alcohol portion of the product.

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01:16

Problem 2

(a) Give the structural formula of the epoxide formed when $m$-chloroperoxybenzoic acid reacts with (i) cis-2-butene and (ii) trans-2-butene. (The epoxides are stereoisomers.) (b) What can you say about the stereochemistry of the epoxidation? $(c)$ Why is a carbocation not an intermediate?
(b) The stereochemistry of the alkene is retained in the epoxide. The reaction is a stereospecific cis addition.
(c) The cis-and trans-alkenes would give the same carbocation, which would go on to give the same product(s). The mechanism probably involves a one-step transfer of the $\mathrm{O}$ to the double bond, without intermediates.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
02:32

Problem 3

Since $\left(\mathrm{CH}_3\right)_2 \ddot{\mathrm{S}}=\ddot{\mathrm{O}}$ : is an ambident nucleophile, reaction with $\mathrm{CH}_3 \mathrm{I}$ could give $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{~S}^{+}=\mathrm{O}_{\mathrm{I}} \mathrm{I}^{-}\right.$ or $\left[\left(\mathrm{CH}_3\right)_2 \mathrm{~S}=\mathrm{O}-\mathrm{CH}_3\right] \mathrm{I}^{-}$. (a) What type of spectroscopy can be used to distinguish between the two products?
(b) Predict the major product.
(a) Use nmr spectroscopy. In $\left[\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{S}}=\mathrm{O}^{-}\right.$all H's are equivalent and a single peak is observed. Note that ${ }_{16}^{32} \mathrm{~S}$ has even numbers of protons and neutrons and so shows no nuclear spin absorption. $\left[\left(\mathrm{CH}_3^a\right)_2 \mathrm{~S}=\mathrm{O}-\mathrm{CH}_3\right] \mathrm{I}^{-}$has two different kinds of H's and therefore two peaks would be observed. (b) Since S is a far better nucleophile than $\mathrm{O}$, $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{~S}_{+}^{\mathrm{S}}=\mathrm{O}^{-} \mathrm{I}^{-}\right.$is the almost exclusive product.

Catherine Lemar
Catherine Lemar
Numerade Educator
01:02

Problem 3

What compounds would you use to prepare 2,3-diphenyl-2,3-butanediol,
<smiles>CC(C)(C(=O)c1ccccc1)C(C)(c1ccccc1)c1ccccc1</smiles>
by (a) halide hydrolysis and $(b)$ reductive dimerization of a carbonyl compound?
(a) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{ClC}\left(\mathrm{CH}_3\right) \mathrm{ClC}_6 \mathrm{H}_5$ or $\mathrm{C}_6 \mathrm{H}_5 \mathrm{C}\left(\mathrm{CH}_3\right) \mathrm{ClC}\left(\mathrm{CH}_3\right) \mathrm{OHC}_6 \mathrm{H}_5$,
(b) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
05:13

Problem 3

Specify and account for your choice of an alkoxide and an alkyl halide to prepare the following ethers by the Williamson reaction: $($ a $) \mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}\left(\mathrm{CH}_3\right)_3,($ b $)\left(\mathrm{CH}_3\right)_2 \mathrm{CHOCH}_2 \mathrm{CH}=\mathrm{CH}_2$.
(a) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{X}+\mathrm{Na}^{+-} \mathrm{OC}\left(\mathrm{CH}_3\right)_3$
(b) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHO}^{-} \mathrm{Na}^{+}+\mathrm{XCH}_2 \mathrm{CH}=\mathrm{CH}_2$

The $2^{\circ}$ and $3^{\circ}$ alkyl halides readily undergo E2 eliminations with strongly basic alkoxides to form alkenes. Hence, to prepare mixed ethers such as $(a)$ and $(b)$ the $1^{\circ}$ alkyl groups should come from $\mathrm{RX}$ and the $2^{\circ}$ and $3^{\circ}$ alkyl groups should come from the alkoxide.

Anupa Sharad Medhekar
Anupa Sharad Medhekar
Numerade Educator
04:18

Problem 4

Outline the mechanism for acid- and base-catalyzed additions to ethylene oxide and give the structural formulas of the products of addition of the following: (a) $\mathrm{H}_2 \mathrm{O}$, (b) $\mathrm{CH}_3 \mathrm{OH}$, (c) $\mathrm{CH}_3 \mathrm{NH}_2$, (d) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{SH}_{\text {. }}$

In acid, $\mathrm{O}$ is first protonated.

The protonated epoxide can also react with nucleophilic solvents such as $\mathrm{CH}_3 \mathrm{OH}$.

In base, the ring is cleaved by attack of the nucleophile on the less substituted $\mathrm{C}$ to form an alkoxide anion, which is then protonated. Reactivity is attributed to the highly strained three-membered ring, which is readily cleaved.

Anish Wadhwa
Anish Wadhwa
Numerade Educator

Problem 4

Show how dimethyl sulfate, $\mathrm{MeOSO}_2 \mathrm{OMe}$, is used in place of alkyl halides in the Williamson syntheses of methyl ethers.

Alkyl sulfates are conjugate bases of the very strongly acidic alkyl sulfuric acids and are very good leaving groups. Dimethyl sulfate is less expensive than $\mathrm{CH}_3 \mathrm{I}$, the only liquid methyl halide at room temperature. Liquids are easier to use than gases in laboratory syntheses.

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02:09

Problem 5

Suggest a mechanism consistent with the following observations for the solvomercuration of $\mathrm{RCH}=\mathrm{CH}_2$ with $\mathrm{R}^{\prime} \mathrm{OH}$ in the presence of $\mathrm{Hg}(\mathrm{OAc})_2$, leading to the formation of $\mathrm{RCH}\left(\mathrm{OR}^{\prime}\right) \mathrm{CH}_2 \mathrm{Hg}(\mathrm{OAc})$ : (i) no rearrangement, (ii) Markovnikov addition, (iii) anti addition, and (iv) reaction with nucleophilic solvents.

Absence of rearrangement excludes a carbocation intermediate. Stereoselective anti addition is reminiscent of a bromonium-ion-type intermediate, in this case a three-membered ring mercurinium ion. The anti regioselective Markovnikov addition requires an $\mathrm{S}_{\mathrm{N}} 1$-type backside attack by the nucleophilic solvent on the more substituted $\mathrm{C}$ of the ring-the $\mathrm{C}$ bearing more of the partial + charge $\left(\delta^{+}\right)$. Frontside attack is blocked by the large $\mathrm{HgOAc}$ group.
Although the mercuration step is stereospecific, the reductive demercuration step is not, and therefore neither is the overall reaction.

Alkendra Singh
Alkendra Singh
Numerade Educator
11:54

Problem 6

Give the alkene and alcohol needed to prepare the following ethers by alkoxymercurationdemercuration: $(a)$ diisopropyl ether, (b) 1-methyl-1-methoxycyclopentane, (c) 1-phenyl-1-ethoxypropane, $(d)$ di- $t$ butyl ether.
(a) $\mathrm{CH}_2=\mathrm{CHCH}_3$ and $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3$ give $\left(\mathrm{CH}_3\right)_2 \mathrm{CHOCH}\left(\mathrm{CH}_3\right)_2$.
(b)
<smiles>COC1(C)CCCC1</smiles>
(c) $\mathrm{PhCH}=\mathrm{CHCH}_3$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ give $\mathrm{PhCH}\left(\mathrm{OCH}_2 \mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3$. Although each double-bonded $\mathrm{C}$ is $2^{\circ}, \mathrm{ROH}$ preferentially bonds with the more positively charged benzylic $\mathrm{C}$.
(d) Ethers with two $3^{\circ}$ alkyl groups cannot be synthesized in decent yields because of severe steric hindrance.

Ian Kaigh
Ian Kaigh
Numerade Educator
11:54

Problem 6

Give the alkene and alcohol needed to prepare the following ethers by alkoxymercuration demercuration: $(a)$ diisopropyl ether, (b) 1-methyl-1-methyl cyclopentane, (c) 1-phenyl-1-ethoxypropane, $(d)$ di- $t$ butyl ether.
(a) $\mathrm{CH}_2=\mathrm{CHCH}_3$ and $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3$ give $\left(\mathrm{CH}_3\right)_2 \mathrm{CHOCH}\left(\mathrm{CH}_3\right)_2$.
(b)
<smiles>COC1(C)CCCC1</smiles>
(c) $\mathrm{PhCH}=\mathrm{CHCH}_3$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ give $\mathrm{PhCH}\left(\mathrm{OCH}_2 \mathrm{CH}_3\right) \mathrm{CH}_2 \mathrm{CH}_3$. Although each double-bonded $\mathrm{C}$ is $2^{\circ}, \mathrm{ROH}$ preferentially bonds with the more positively charged benzylic $\mathrm{C}$.
(d) Ethers with two $3^{\circ}$ alkyl groups cannot be synthesized in decent yields because of severe steric hindrance.

Ian Kaigh
Ian Kaigh
Numerade Educator

Problem 7

Give (a) $\mathrm{S}_{\mathrm{N}} 2$ and $(b) \mathrm{S}_{\mathrm{N}} 1$ mechanisms for formation of ROR from $\mathrm{ROH}$ in conc. $\mathrm{H}_2 \mathrm{SO}_4$.
(a)
(1)
(2)
<smiles>[R]C[R]1=CC[C@@H](C)C[C@H]1C</smiles>
<smiles>[R20][H]</smiles>
conjugate acid of an ether
(3)
(b) (1) Same as (1) of (a)
(2) $\mathrm{R}: \mathrm{OH}_2^{+} \longrightarrow \mathrm{R}^{+}+\mathrm{H}_2 \mathrm{O}$
(3)
<smiles>[R]O[R7]#[R][H]</smiles>
(4) Same as (3) of (a)

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00:40

Problem 8

Compare the mechanisms for the formation of an ether by intermolecular dehydration of $(a) 1^{\circ},(b)$ $3^{\circ}$, and (c) $2^{\circ}$ alcohols.
(a) For $1^{\circ}$ alcohols the mechanism is $\mathrm{S}_{\mathrm{N}}$, with alcohol as the attacking nucleophile and water as the leaving group. There would be no rearrangements.

(b) The mechanism for $3^{\circ}$ alcohols is $\mathrm{S}_{\mathrm{N}} 1$. However, a $3^{\circ}$ carbocation such as $\mathrm{Me}_3 \mathrm{C}^{+}$cannot react with $\mathrm{Me}_3 \mathrm{COH}$, the parent $3^{\circ}$ alcohol, or any other $3^{\circ} \mathrm{ROH}$ because of severe steric hindrance. It can react with a $1^{\circ} \mathrm{RCH}_2 \mathrm{OH}$, if such an alcohol is present.
$$
\mathrm{Me}_3 \mathrm{C}^{+}+\mathrm{RCH}_2 \mathrm{OH} \stackrel{-\mathrm{H}^*}{\longrightarrow} \mathrm{Me}_3 \mathrm{COCH}_2 \mathrm{R} \quad \text { (a mixed ether) }
$$

The $3^{\circ}$ carbocation can also readily eliminate $\mathrm{H}^{+}$to give an alkene, $\mathrm{Me}_2 \mathrm{C}=\mathrm{CH}_2$.
(c) $2^{\circ}$ alcohols react either way. Rearrangements may occur when they react by the $\mathrm{S}_{\mathrm{N}} \mathrm{l}$ mechanism, because the intermediate is a carbocation.

Nikhil Choudhary
Nikhil Choudhary
Numerade Educator
09:09

Problem 9

List the ethers formed in the reaction between concentrated $\mathrm{H}_2 \mathrm{SO}_4$, and equimolar quantities of ethanol and $(a)$ methanol, $(b)$ tert-butanol.
(a) These $1^{\circ}$ alcohols react by $\mathrm{S}_{\mathrm{N}} 2$ mechanisms to give a mixture of three ethers: $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5$ from $2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$, $\mathrm{CH}_3 \mathrm{OCH}_3$ from $2 \mathrm{CH}_3 \mathrm{OH}$, and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OCH}_3$ from $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{OH}$.
(b) This is an $\mathrm{S}_{\mathrm{N}} 1$ reaction.
$$
\begin{array}{r}
\left(\mathrm{CH}_3\right)_3 \mathrm{COH} \stackrel{\mathrm{H}^{+}}{\longrightarrow}\left(\mathrm{CH}_3\right)_3 \mathrm{COH}_2 \stackrel{+\mathrm{H}_2 \mathrm{O}}{\longrightarrow}\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{C}} \stackrel{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}}{-\mathrm{H}^{+}}\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3 \\
\text { Ethyl tert-butyl ether }
\end{array}
$$

Reaction between $\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}$and $\left(\mathrm{CH}_3\right)_3 \mathrm{COH}$ is sterically hindered and occurs much less readily.

Ronald Prasad
Ronald Prasad
Numerade Educator
02:48

Problem 11

(R)-2-Octanol and its ethyl ether are levorotatory. Predict the configuration and sign of rotation of the ethyl ether prepared from this alcohol by: $(a)$ reacting with $\mathrm{Na}$ and then $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br} ;(b)$ reacting in a solvent of low dielectric constant with concentrated $\mathrm{HBr}$ and then with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} \mathrm{Na}^{+}$.
(a) No bond to the chiral $\mathrm{C}$ of the alcohol is broken in this reaction; hence the $R$ configuration is unchanged but rotation is indeterminant. (b) These conditions for the reaction of the alcohol with $\mathrm{HBr}$ favor an $\mathrm{S}_{\mathrm{N}} 2$ mechanismof the chiral $\mathrm{C}$ is inverted. Attack by $\mathrm{RO}^{-}$is also $\mathrm{S}_{\mathrm{N}} 2$ and the net result of two inversions is retention of configuration. Again rotation is unpredictable.

Lottie Adams
Lottie Adams
Numerade Educator
04:52

Problem 12

(a) Why do ethers dissolve in cold concentrated $\mathrm{H}_2 \mathrm{SO}_4$ and separate out when water is added to the solution? $(b)$ Why are ethers used as solvents for $\mathrm{BF}_3$ and the Grignard reagent?
(a) Water is a stronger base than ether and removes the proton from the protonated ether.
(b) $\mathrm{BF}_3$ and $\mathrm{RMgX}$ are Lewis acids which share a pair of electrons on the $-\dddot{\mathrm{O}}-$ of ethers.
Notice that two ether molecules coordinate with one $\mathrm{Mg}$ atom.

Tom Rutherford
Tom Rutherford
Numerade Educator
01:07

Problem 13

Identify the ethers that are cleaved with excess $\mathrm{HI}$ to yield (a) $\left(\mathrm{CH}_3\right)_3 \mathrm{CI}$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{I}$, (b) cyclohexyl and methyl iodides, (c) $\mathrm{I}\left(\mathrm{CH}_2\right)_5 \mathrm{I}$.
(a) $\left(\mathrm{CH}_3\right)_3 \mathrm{COCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$

Narayan Hari
Narayan Hari
Numerade Educator
02:02

Problem 14

(a) Show how the cleavage of ethers with $\mathrm{HI}$ can proceed by an $\mathrm{S}_{\mathrm{N}} 2$ or an $\mathrm{S}_{\mathrm{N}} 1$ mechanism. (b) Why is $\mathrm{Hl}$ a better reagent than $\mathrm{HBr}$ for this type of reaction? (c) Why do reactions with excess $\mathrm{HI}$ afford two moles of Rl?
(a)
(b) $\mathrm{HI}$ is a stronger acid than $\mathrm{HBr}$ and gives a greater concentration of the oxonium ion
<smiles></smiles>
$\mathrm{I}^{-}$is also a better nucleophile in the $\mathrm{S}_{\mathrm{N}} 2$ reactions than is $\mathrm{Br}^{-}$.
(c) The first-formed $\mathrm{ROH}$ reacts in typical fashion with $\mathrm{HI}$ to give $\mathrm{RI}$.

Tom Rutherford
Tom Rutherford
Numerade Educator
00:57

Problem 15

Account for the following observations:
The high polarity of the solvent $\left(\mathrm{H}_2 \mathrm{O}\right)$ in reaction (2) favors an $\mathrm{S}_{\mathrm{N}} 1$ mechanism giving the $3^{\circ} \mathrm{R}^{+}$.

The low polarity of solvent (ether) in reaction (1) favors the $\mathrm{S}_{\mathrm{N}} 2$ mechanism and the nucleophile, $\mathrm{I}^{-}$, attacks the $1^{\circ} \mathrm{C}$ of $\mathrm{CH}_3$.

Alma Victoriano
Alma Victoriano
North Carolina State University
01:37

Problem 16

Can resonance account for the preferential radical substitution at the $\alpha$-C of ethers?
The intermediate $\alpha$ radical RCHOR is not stabilized by delocalization of electron density by the adjacent $O$ through extended $\pi$ bonding. One resonance structure would have 9 electrons on $O$.

Aadit Sharma
Aadit Sharma
Numerade Educator
03:34

Problem 17

(a) Give a mechanism for the formation of the explosive solid hydroperoxides, e.g.
<smiles>[R]COC([R])O</smiles>
from ethers and $\mathrm{O}_2$. (b) Why should ethers be purified before distillation?
(a)
(b) An ether may contain hydroperoxides which concentrate as the ether is distilled and which may then explode. Ethers are often purified by mixing with $\mathrm{FeSO}_4$ solution, which reduces the hydroperoxides to the nonexplosive alcohols $(\mathrm{ROOH} \rightarrow \mathrm{ROH})$.

Tom Rutherford
Tom Rutherford
Numerade Educator
09:25

Problem 19

Give the main products of $(a)$ mononitration of $p$-methylphenetole, $(b)$ monobromination of $p$ methoxyphenol.
(a) Since $\mathrm{OR}$ is a stronger activating group than $\mathrm{R}$ groups, the $\mathrm{NO}_2$ attacks ortho to $-\mathrm{OC}_2 \mathrm{H}_5$; the product is 2nitro-4-methylphenetole. (b) $\mathrm{OH}$ is a stronger activating group than $\mathrm{OR}$ groups; the product is 2-bromo-4methoxyphenol.

Anish Wadhwa
Anish Wadhwa
Numerade Educator
01:04

Problem 20

Which alcohol would undergo dehydration to give $(a)$ THP? $(b)$ 1,4-dioxane?
(a) Since the product is a cyclic ether, the starting material must be a diol with $\mathrm{OH}$ groups on the terminal C's. The diol must have five $\mathrm{CH}_2$ groups to match the number in THP. The alcohol used in $\mathrm{HO}\left(\mathrm{CH}_2\right)_5 \mathrm{OH}, 1,5-$ pentanediol. This is an intramolecular dehydration to form an ether.
(b) 1,4-Dioxane has two ether groups, requiring dehydration between two pairs of $\mathrm{OH}$ groups. Again the starting alcohol must be a diol, but now the dehydration is intermolecular. The alcohol used is ethylene glycol, $\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{OH}$.

Narayan Hari
Narayan Hari
Numerade Educator
00:58

Problem 21

(a) With the aid of the mechanism show why DHP, unlike typical alkenes, readily undergoes the following reaction:
2,3-Dihydro-4H-pyran (DHP) a tetrahydropyranyl (THP) ether
(b) Why do THP ethers, unlike ordinary ethers, cleave under mildly aqueous acidic conditions?
(a) The $\mathrm{H}^{+}$adds to $\mathrm{C}=\mathrm{C}$ to generate a carbocation with the positive charge on the $\mathrm{C}$ that is $\alpha$ to the $-\ddot{\mathrm{O}}-$ of the ring. This is a fairly stable cation because the positive charge is stabilized by delocalization of electron density from the $\mathrm{O}$ atom.
The nucleophilic site ( $-\dddot{\mathrm{O}}-$ ) of $\mathrm{ROH}$ then bonds to the $\mathrm{C}^{+}$of the carbocation, forming an onium ion of the ether, which loses a proton to the solvent alcohol $(\mathrm{ROH})$ and becomes the ether product.
(b) Aqueous acid reverses the reaction of $(a)$, reforming the same intermediate carbocation. This loses a proton to give the $\mathrm{C}=\mathrm{C}$, rather than reacting with water to give the very unstable alcohol-analog of the ether.

Nikhil Choudhary
Nikhil Choudhary
Numerade Educator
01:19

Problem 22

Since THP ethers are, like most ethers, stable in base, their formation can be used to protect the $\mathrm{OH}$ group from reacting under basic conditions. Using this fact, show how to convert $\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{Cl}^2 \mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{D}$ via the Grignard reagent.
$\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{Cl}$ cannot be converted directly to the Grignard because of the presence of the acidic $\mathrm{OH}$ group; the product obtained would be $\mathrm{HOCH}_2 \mathrm{CH}_3$. The desired reaction is achieved by protecting the $\mathrm{OH}$ group as shown schematically:
$$
\mathrm{DHP}+\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{Cl} \longrightarrow \mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{O}-\mathrm{THP} \longrightarrow \mathrm{DCH}_2 \mathrm{CH}_2 \mathrm{O}-\mathrm{THP} \longrightarrow \mathrm{DCH}_2 \mathrm{CH}_2 \mathrm{OH}+\mathrm{DHP}
$$

To generalize on Problem 14.22 a good protecting group (i) is easily attached, (ii) permits the desired chemistry to occur, and (iii) is easily removed. Other methods for protecting $\mathrm{OH}$ groups involve benzyl and silyl ethers:
$$
\begin{aligned}
& \mathrm{ROH}+\mathrm{BrCH}_2 \mathrm{Ph} \underset{-\mathrm{AgBr}^{-}}{\stackrel{\mathrm{ROCH}_2}{2} \mathrm{Ph} \stackrel{\mathrm{H}_2 / \mathrm{Pt}}{\longrightarrow}} \mathrm{ROH}+\mathrm{CH}_3 \mathrm{Ph} \\
& \underset{\text { Chlorotrimethylsilane }}{\mathrm{ROH}}+\mathrm{Cl}-\mathrm{Si}\left(\mathrm{CH}_3\right)_3 \stackrel{\text { pyridine }}{\longrightarrow} \underset{-\mathrm{HCl}}{\mathrm{ROSi}\left(\mathrm{CH}_3\right)_3} \stackrel{\mathrm{H}_3 \mathrm{O}^{+}}{\longrightarrow} \mathrm{ROH}+\mathrm{HOSi}\left(\mathrm{CH}_3\right)_3 \\
&
\end{aligned}
$$

Crown ethers are large-ring cyclic ethers with several $O$ atoms. A typical example is 18-crown-6 ether, Fig. 14-1 $(a)$. The first number in the name is the total number of atoms in the ring, the second number is the number of $\mathrm{O}$ atoms. Crown ethers are excellent solvaters of cations of salts through formation of ion-dipole bonds. 18-Crown- 6 ether strongly complexes and traps $\mathrm{K}^{+}$, [from, e.g., $\mathrm{KF}$, as shown in Fig. 14-1 $(b)$ ].

Grigoriy Sereda
Grigoriy Sereda
Numerade Educator
02:13

Problem 23

Suggest two important synthetic uses of crown ethers.
(1) They enable inorganic salts to be used in nonpolar solvents, a media with which salts are typically incompatible. (2) The cation of the salt is complexed in the center of the crown ether, leaving the anion "bare" and enhanced in reactivity. These effects of crown ethers are similar to those achieved with phase-transfer agents. The "bare" anion is also present when polar aprotic solvents are used.

Ly Tran
Ly Tran
Numerade Educator
05:42

Problem 25

Why does trans-2-chlorocyclohexanol give a very good yield of 1,2-epoxycyclohexane, but the cis isomer gives no epoxide?

The nucleophilic $\mathrm{O}^{-}$group displaces the $\mathrm{Cl}$ atom (as $\mathrm{Cl}^{-}$) by an intramolecular $\mathrm{S}_{\mathrm{N}}$ 2-type process which requires a backside attack. In the trans isomer, the $\mathrm{O}^{-}$and $\mathrm{CI}$ are properly positioned for such a displacement, and the epoxide is formed. In the cis isomer, backside attack cannot occur and the epoxide is not formed.

This role of the $\mathrm{O}^{-}$, called neighboring-group participation, always leads to an inversion of configuration if the attacked C is a chiral center (stereocenter).

Dan Ni
Dan Ni
Numerade Educator

Problem 29

Outline mechanisms to account for the different isomers formed from reaction of
<smiles>CC(C)(C)C1CO1</smiles>
with $\mathrm{CH}_3 \mathrm{OH}$ in acidic $\left(\mathrm{H}^{+}\right)$and in basic $\left(\mathrm{CH}_3 \mathrm{O}^{-}\right)$media.
$\mathrm{CH}_3 \mathrm{O}^{-}$reacts by an $\mathrm{S}_{\mathrm{N}} 2$ mechanism attacking the less substituted $\mathrm{C}$.
Isobutylene oxide
In acid the $\mathrm{S}_{\mathrm{N}} 1$ mechanism produces the more stable $3^{\circ} \mathrm{R}^{+}$, and the nucleophilic solvent forms a bond with the more substituted $\mathrm{C}$.

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Problem 30

Account for the fact that (R) $-\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}-\mathrm{CH}_2$ reacts with $\mathrm{CH}_3 \mathrm{OH}$ in acid to give the product with inversion and very little racemization.

When the protonated epoxide undergoes ring-opening, the $\mathrm{CH}_3 \mathrm{OH}$ molecule attacks from the backside of the $\mathrm{C}^{+}$. The nearby, newly formed $\mathrm{OH}$ group hasn't moved out of the way and blocks approach from the frontside. This leads to inversion at the chiral carbon. Since there was no change in group priorities, the configuration in the product is (S).

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01:09

Problem 32

Give the products and the number of moles of $\mathrm{HIO}_4$ consumed in the reaction with 2,4-dimethyl$2,3,4,5$-hexanetetrol. Indicate the adjacencies with zigzag lines.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
05:35

Problem 33

Give the structural formula for the major product from the pinacol rearrangement of 1,1,2triphenyl-1,2-propanediol. Indicate the protonated $\mathrm{OH}$ and the migrating group
<smiles>CC(O)(c1ccccc1)C(O)(c1ccccc1)c1ccccc1</smiles>
Loss of $\mathrm{OH}^a$ yields the more stable $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{C}^{+}-\mathrm{C}(\mathrm{OH})\left(\mathrm{C}_6 \mathrm{H}_5\right) \mathrm{CH}_3, \mathrm{C}_6 \mathrm{H}_5$ rather than $\mathrm{CH}_3$ migrates to form
<smiles>CC(=O)C(c1ccccc1)c1ccccc1</smiles>
the major product. Migration of $\mathrm{CH}_3$ would give
<smiles>O=C(CC(c1ccccc1)(c1ccccc1)c1ccccc1)c1ccccc1</smiles>
which also arises from the loss of $\mathrm{OH}^b$.

Anupa Sharad Medhekar
Anupa Sharad Medhekar
Numerade Educator
05:06

Problem 34

Explain why the reaction of $\mathrm{HS}^{-}$and $\mathrm{RX}$ is little used to prepare RSH.
Sulfur atoms in molecules and ions are very good nucleophilic sites. Hence, once formed in base, RSH yields $\mathrm{RS}^{-}$, which reacts with RX to give the thioether. For this reason thiourea is used with RX to give thiols (Problem 13.25).

Ian Kaigh
Ian Kaigh
Numerade Educator
02:27

Problem 35

Give the expected principal organic products from the following reactions:
(a) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{SH}+\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \stackrel{\mathrm{OH}}{\longrightarrow}$
(b) $\mathrm{ICH}_2 \mathrm{CH}_2 \mathrm{I}+\mathrm{HSCH}_2 \mathrm{CH}_2 \mathrm{SH} \stackrel{\mathrm{OH}^{-}}{\longrightarrow}$
(c) $\mathrm{Na}_2 \mathrm{~S}(1 \mathrm{~mol})+\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \longrightarrow$

Catherine Lemar
Catherine Lemar
Numerade Educator
02:48

Problem 36

What structural features must be present in order for an individual thioether to give a single alkane on hydrogenolysis?
The thioether must have the same two $\mathrm{R}$ groups or it must be cyclic.

Shahina -
Shahina -
Numerade Educator
02:46

Problem 37

Explain why sulfonium salts and sulfoxides having different $\mathrm{R}$ or Ar groups are resolvable into enantiomers.

The $\mathrm{S}$ atom in each of these molecules has three $\sigma$ bonds and an unshared pair of $e^{-}$'s. According to the HON rule (Section 2.3), these $\mathrm{S}$ atoms use $s p^3$ HO's. If all the attached groups are different, the $\mathrm{S}$ is a chiral center. Notice in the figure below that chirality prevails even though one of the $s p^3$ HO's houses an unshared pair of $e^{-} \mathrm{s}$. The fact that these species are resolvable indicates that their molecules do not undergo inversion of configuration, notwithstanding the fact that a lone pair of $e^{-}$'s is present. Such rigidity of configuration is characteristic of third-period elements (S, P), but not of second-period elements (C, N).

Dr. Satish Ingale
Dr. Satish Ingale
Numerade Educator
01:45

Problem 39

Give the structural formula and IUPAC name for $(a) n$-propyl propenyl ether, $(b)$ isobutyl tertbutyl ether, (c) 12-crown-4 ether.

(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}=\mathrm{CHCH}_3$ 1-(n-Propoxy)-1-propene
(b)
<smiles>CC(C)COC(C)(C)C</smiles>
2-Isobutoxy-2-methylpropane

Sima Sarker
Sima Sarker
Numerade Educator
06:11

Problem 40

Give the structural formulas for (a) ethylene glycol, (b) propylene glycol, and (c) trimethylene glycol.
(a) $\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{OH}$
(b) $\mathrm{CH}_3 \mathrm{CHOHCH}_2 \mathrm{OH}$
(c) $\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$.

Nicholas Sacco
Nicholas Sacco
Numerade Educator
00:49

Problem 41

Account for the fact that the $\mathrm{C}-\mathrm{O}-\mathrm{C}$ bond angle in dimethyl ether is greater than the $\mathrm{H}-\mathrm{O}-\mathrm{H}$ bond angle in water $\left[112^{\circ}\right.$ versus $\left.105^{\circ}\right]$ ?

The repulsive van der Waals forces between the two $\mathrm{CH}_3$ groups in dimethyl ether are greater than those between the two H's in water because the methyl groups are larger than the H's and have more electrons.

Lottie Adams
Lottie Adams
Numerade Educator
02:30

Problem 42

Distinguish between an ether and an alcohol by $(a)$ chemical tests, $(b)$ spectral methods.
(a) $1^{\circ}$ and $2^{\circ}$ alcohols are oxidizable and give positive tests with $\mathrm{CrO}_3$ in acid (orange color turns green). All alcohols of moderate molecular weights evolve $\mathrm{H}_2$ on addition of $\mathrm{Na}$. Dry ethers are negative to both tests.
(b) The ir spectra of alcohols, but not ethers, show an $\mathrm{O}-\mathrm{H}$ stretching band at about $3500 \mathrm{~cm}^{-1}$. Comparing the ir spectra is the best method for distinguishing between these functional groups.

Lottie Adams
Lottie Adams
Numerade Educator
View

Problem 43

Why is di-t-butyl ether very easily cleaved by $\mathrm{HI}$ ?
On treatment with $\mathrm{HI}$, the ether is protonated. This oxonium ion cleaves readily to give $t$-butyl alcohol and the relatively stable $t$-butyl carbocation. Iodide ion adds to the carbocation, and the alcohol reacts with $\mathrm{HI}$; both give $t$ butyl iodide.

Emily Himsel
Emily Himsel
Numerade Educator
01:07

Problem 44

Give a chemical test to distinguish $\mathrm{C}_5 \mathrm{H}_{12}$ from $\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O}$.
Unlike $\mathrm{C}_5 \mathrm{H}_{12},\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O}$ is basic and dissolves in concentrated $\mathrm{H}_2 \mathrm{SO}_4$.
$$
\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{OH}^{+}+\mathrm{HSO}_4^{-}
$$

Raghvendra Singh
Raghvendra Singh
Numerade Educator
01:03

Problem 46

Supply structures for compounds (A) through (F).
$$
\begin{gathered}
\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+(\mathrm{A}) \longrightarrow \mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{OH}^{\frac{\mathrm{H}_2 \mathrm{SO}_4}{\text { heat }}} \text { (B) } \stackrel{\text { alc } \mathrm{KOH}}{\longrightarrow} \text { (C) } \\
\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}+\text { alc. } \mathrm{KOH} \longrightarrow \text { (D) } \stackrel{\text { HOC! }}{\longrightarrow} \text { (E) } \stackrel{\mathrm{NaOH}}{\longrightarrow} \text { (F) }
\end{gathered}
$$
(A) $\mathrm{HOCl}$
(B) $\mathrm{ClCH}_2 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}$
(C) $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}-\mathrm{O}-\mathrm{CH}=\mathrm{CH}_2$
(D) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2$
(E)
<smiles>CC(C)(O)CCl</smiles>
(F)
<smiles>CC1(C)CO1</smiles>
Formation of $(\mathrm{F})$, isobutylene oxide, is an internal $\mathrm{S}_{\mathrm{N}} 2$ reaction.

Narayan Hari
Narayan Hari
Numerade Educator
04:08

Problem 47

Are the $m / e$ peaks 102,87 , and 59 (base peak) consistent for $n$-butyl ether (A) or methyl $n$-pentyl ether (B)? Give the structure of the fragments which justify your answer.

The parent $\mathrm{P}^{+}$is $m / e=102$, the molecular weight of the ether. The other peaks arise as follows:
$$
102-15\left(\mathrm{CH}_3\right)=87 \quad 102-43\left(\mathrm{C}_3 \mathrm{H}_7\right)=59
$$

Fragmentations of $\mathrm{P}^{+}$ions of ethers occur mainly at the $\mathrm{C}^\alpha-\mathrm{C}^\beta$ bonds. (A) fits these data for $\mathrm{H}_2 \stackrel{+}{\mathrm{C}}-$ $\mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\left(m / e=87 ; \mathrm{C}^{x^{\prime}}-\mathrm{C}^{\beta /}\right.$ cleavage $)$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{O}-\dot{\mathrm{C}} \mathrm{H}_2\left(m / e=59 ; \mathrm{C}^x-\mathrm{C}^\beta\right.$ cleavage $)$. Cleavage of the $\mathrm{C}^\beta-\mathrm{C}^\alpha$ bonds in (B) would give a cation, $\mathrm{CH}_3 \mathrm{O}=\mathrm{CH}_2(m / e=45)$, but this peak was not observed.

Ian Kaigh
Ian Kaigh
Numerade Educator
11:40

Problem 48

Prepare the following ethers starting with benzene, toluene, phenol $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right)$, cyclohexanol, any aliphatic compound of three C's or less and any solvent or inorganic reagent: $(a)$ dibenzylether, $(b)$ di- $n$-butyl ether, $(c)$ ethyl isopropyl ether, $(d)$ cyclohexyl methyl ether, $(e) p$-nitrophenyl ethyl ether, $(f)$ divinyl ether, $(g)$ diphenyl ether.
(a)
$$
\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3 \underset{\text { light }}{\stackrel{\mathrm{Cl}}{\longrightarrow}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl} \underset{\mathrm{H}_2 \mathrm{O}}{\stackrel{\mathrm{OHH}^{-}}{\longrightarrow}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH} \underset{-\mathrm{H}_2 \mathrm{O}}{\stackrel{\mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}}\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2\right)_2 \mathrm{O}
$$
(b)
(c)
(1')
Use the $I^{\circ} \mathrm{RX}$ to minimize the competing E2 elimination reaction or use
(d) $\mathrm{C}_6 \mathrm{H}_{11} \mathrm{OH}+\mathrm{CH}_2 \mathrm{~N}_2 \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_{11} \mathrm{OCH}_3+\mathrm{N}_2$ or $\mathrm{C}_6 \mathrm{H}_{11} \mathrm{OH} \stackrel{\mathrm{N}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_{11} \mathrm{O}^{-} \mathrm{Na}^{+} \stackrel{\mathrm{CH}_3 \mathrm{I}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_{11} \mathrm{OCH}_3$
(e) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \stackrel{\mathrm{NaOH}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-} \mathrm{Na}^{+} \stackrel{\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5 \stackrel{\mathrm{HNO}_3}{\mathrm{H}_2 \mathrm{SO}_4} p-\mathrm{NO}_2 \mathrm{C}_6 \mathrm{H}_4 \mathrm{OC}_2 \mathrm{H}_5$
Williamson synthesis of an aryl alkyl ether requires the Ar to be part of the nucleophile $\mathrm{ArO}^{-}$and not the halide, since $\mathrm{ArX}$ does not readily undergo $\mathrm{S}_{\mathrm{N}} 2$ displacements. Note that since $\mathrm{ArOH}$ is much more acidic than $\mathrm{ROH}$, it is converted to $\mathrm{ArO}^{-}$by $\mathrm{OH}^{-}$instead of by $\mathrm{Na}$ as required for $\mathrm{ROH}$.
(f) See Problem 14.46, compounds (A), (B) and (C). Vinyl alcohol, $\mathrm{H}_2 \mathrm{C}=\mathrm{CHOH}$, cannot be used as a starting material because it is not stable and rearranges to $\mathrm{CH}_3 \mathrm{CHO}$. The double bond must be introduced after the ether bond is formed.
(g)
$$
\mathrm{C}_6 \mathrm{H}_6 \stackrel{\mathrm{Br}_2}{\underset{\mathrm{Fe}}{\longrightarrow}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{Br} \underset{\substack{\mathrm{Cu}\left(>200^{\circ} \mathrm{C}\right) \\ \text { no solvent }}}{\stackrel{\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-} \mathrm{Na}^{+}}{\longrightarrow}}\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{O}
$$

Phenols do not undergo intermolecular dehydration. Although aryl halides cannot be used as substrates in typical Williamson syntheses, they do undergo a modified Williamson-type synthesis at higher temperature in the presence of $\mathrm{Cu}$.

Nicholas Sacco
Nicholas Sacco
Numerade Educator
05:40

Problem 49

Prepare ethylene glycol from the following compounds: (a) ethylene, (b) ethylene oxide, (c) 1,2dichloroethane.
(a) Oxidation: $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2 \stackrel{\text { dil. aq. } \mathrm{KMnO}_4}{\longrightarrow} \mathrm{HOCH}_2-\mathrm{CH}_2 \mathrm{OH}$
(b) Acid hydrolysis
(c) Alkaline hydrolysis: $\mathrm{ClCH}_2-\mathrm{CH}_2 \mathrm{Cl} \stackrel{\mathrm{H}_2 \mathrm{O} \cdot \mathrm{OH}^{-}}{\longrightarrow} \mathrm{HOCH}_2-\mathrm{CH}_2 \mathrm{OH}$

Nicholas Sacco
Nicholas Sacco
Numerade Educator
05:29

Problem 50

Outline the steps and give the product of pinacol rearrangement of: (a) 3-phenyl-1,2-propanediol, (b) 2,3-diphenyl-2,3-butanediol.
(a)
<smiles>O=C(CCc1ccccc1)c1ccccc1</smiles>

Susan Hallstrom
Susan Hallstrom
Numerade Educator
05:26

Problem 51

Show how ethylene oxide is used to manufacture the following water soluble organic solvents:
(a) Carbitol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{OH}\right)$
(b) Diethylene glycol ( $\left.\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{OH}\right)$
(c) Diethanolamine $\left(\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{OH}\right)$
(d) 1,4-Dioxane
<smiles>C1COCCO1</smiles>

Nima Gharibi
Nima Gharibi
Numerade Educator
01:44

Problem 52

Ethers, especially thoșe with more than one ether linkage, are also named by the oxa method. The ether $\mathrm{O}$ 's are counted as $\mathrm{C}$ 's in determining the longest hydrocarbon chain. The $\mathrm{O}$ is designated by the prefix oxa-, and a number indicates its position. Use this method to name:

Lottie Adams
Lottie Adams
Numerade Educator

Problem 53

Outline mechanisms to account for the different isomers formed from reaction of
<smiles>CC(C)(C)C1CO1</smiles>
with $\mathrm{CH}_3 \mathrm{OH}$ in acidic $\left(\mathrm{H}^{+}\right)$and in basic $\left(\mathrm{CH}_3 \mathrm{O}^{-}\right)$media.
$\mathrm{CH}_3 \mathrm{O}^{-}$reacts by an $\mathrm{S}_{\mathrm{N}} 2$ mechanism attacking the less substituted $\mathrm{C}$.
<smiles>COCC(O)CCCCCCCCCCC1OC1C</smiles>
Isobutylene oxide

In acid, the $\mathrm{S}_{\mathrm{N}} 1$ mechanism produces the more stable $3^{\circ} \mathrm{R}^{+}$.

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06:11

Problem 54

Prepare mustard gas, $\left(\mathrm{ClCH}_2 \mathrm{CH}_2\right)_2 \mathrm{~S}$, from ethylene.
$$
\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{S}_2 \mathrm{Cl}_2 \longrightarrow\left(\mathrm{ClCH}_2 \mathrm{CH}_2\right) \mathrm{S}+\mathrm{S}
$$

Kevin Zaborsky
Kevin Zaborsky
Numerade Educator

Problem 55

A compound, $\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}_2$, gives a negative test with $\mathrm{HIO}_4$. List all possible structures and show how ir and nmr spectroscopy can distinguish among them. (Note that gem-diols can be disregarded since they are usually not stable.)

There are no degrees of unsaturation and hence no rings or multiple bonds. The O's must be present as $\mathrm{C}-\mathrm{O}-\mathrm{H}$ and/or $\mathrm{C}-\mathrm{O}-\mathrm{C}$. The compound can be a diol, a hydroxyether or a diether. A negative test with $\mathrm{HIO}_4$ rules out a vicdiol. Possible structures are: a diol, ${ }^l \mathrm{HOCH}_2^2 \mathrm{CH}_2^3 \mathrm{CH}_2^2 \mathrm{OH}^l$ (A); two hydroxyethers, ${ }^l \mathrm{HOCH}_2^2 \mathrm{CH}_2^3 \mathrm{OCH}_3^4(\mathrm{~B})$ and ${ }^I \mathrm{HOCH}_2^2 \mathrm{OCH}_2^3 \mathrm{CH}_3^4(\mathrm{C})$; and a diether (an acetal), $\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{OCH}_3$ (D). (D) is pinpointed by ir; it has no $\mathrm{OH}$, there is no O-H stretch and peaks are not observed at greater than $2950 \mathrm{~cm}^{-1}$. (A) can be differentiated from (B) and (C) by nmr. (A) has only three kinds of equivalent H's, as labeled, while (B) and (C) each have four. In dimethyl sulfoxide, the nmr spectrum of $(\mathrm{C})$ shows all $\mathrm{H}$ peaks to be split: $\mathrm{H}^3$, a quartet, couples $\mathrm{H}^4$, a triplet; $\mathrm{H}^2$, a doublet, couples $\mathrm{H}^I$, a triplet. The nmr spectrum of (B) in DMSO shows a sharp singlet for $\mathrm{H}^4$ integrating for three H's. Other differences may be observed but those described above are sufficient for identification. The DMSO used is deuterated, $\left(\mathrm{CD}_3\right)_2 \mathrm{SO}$, to prevent interference with the spectrum.

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