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Schaum's Outline of Organic Chemistry

George Hademenos, George Hademenos

Chapter 13

ALCOHOLS AND THIOLS - all with Video Answers

Educators

Chapter Questions

Problem 1

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Problem 1

Write equations to show why alcohols cannot be used as solvents with Grignard reagents or with $\mathrm{LiAlH}_4$.
Strongly basic $\mathrm{R} \bar{\ddagger}$ and $\mathrm{H} \bar{\ddagger}$ react with weakly acidic alcohols.
$$
\begin{aligned}
\mathrm{CH}_3 \mathrm{OH}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{MgCl}^{+} \mathrm{Cl} & \longrightarrow \mathrm{CH}_3 \mathrm{CH}_3+\left(\mathrm{CH}_3 \mathrm{O}\right)^{-}\left(\mathrm{MgCl}^{+}\right. \\
4 \mathrm{CH}_3 \mathrm{OH}+\mathrm{LiAlH}_4 & \longrightarrow 4 \mathrm{H}_2+\mathrm{LiAl}\left(\mathrm{OCH}_3\right)_4
\end{aligned}
$$

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03:00

Problem 1

(a) What is the expected product from catalytic hydrogenation of acetophenone $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$ ? (b) One of the products of the reaction in (a) is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3$. Explain its formation.
(a) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHOHCH}_3$. (b) The initial product, typical of benzylic alcohols,
<smiles>[R]C(O)c1ccccc1</smiles>
can be further reduced with $\mathrm{H}_2$. This reaction is a hydrogenolysis (bond-breaking by $\mathrm{H}_2$ ).
$$
\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHOHCH}_3+\mathrm{H}_2 \stackrel{\mathrm{Pd}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{H}_2 \mathrm{O}
$$

Niamat Khuda
Niamat Khuda
Numerade Educator
01:47

Problem 2

Name the following alcohols by the IUPAC method.
(a)
<smiles>CCCC(O)(CC)CC</smiles>
(c)
<smiles>OCC(Cl)Cl</smiles>
(e)
(b)
<smiles>C=CC(C)O</smiles>
(d)
<smiles>CCCC(O)(CC)c1ccccc1</smiles>
<smiles>O[C@H]1CCCC[C@H]1Br</smiles>
(a) 3-Ethyl-3-hexanol
(c) 2,2-Dichloroethanol
(e) cis-2-Bromocyclohexanol
(b) 3-Buten-2-ol
(d) 3-Phenyl-3-hexanol

Note that in IUPAC the $\mathrm{OH}$ is given a lower number than $\mathrm{C}=\mathrm{C}$ or $\mathrm{Cl}$.

Sima Sarker
Sima Sarker
Numerade Educator
01:10

Problem 3

Explain why (a) propanol boils at a higher temperature than the corresponding hydrocarbon; $(b)$ propanol, unlike propane or butane, is soluble in $\mathrm{H}_2 \mathrm{O}$; (c) n-hexanol is not soluble in $\mathrm{H}_2 \mathrm{O}$; (d) dimethyl ether $\left(\mathrm{CH}_3 \mathrm{OCH}_3\right)$ and ethyl alcohol $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\right)$ have the same molecular weight, yet dimethyl ether has a lower boiling point $\left(-24^{\circ} \mathrm{C}\right)$ than ethyl alcohol $\left(78^{\circ} \mathrm{C}\right)$.
(a) Propanol can H-bond intermolecularly. interaction.
<smiles>CCCCOCOCCCC</smiles>
There is also a less important dipole-dipole
(b) Propanol can H-bond with $\mathrm{H}_2 \mathrm{O}$ :
<smiles>CCCOCO</smiles>
(c) As the $\mathrm{R}$ group becomes larger, $\mathrm{ROH}$ resembles the hydrocarbon more closely. There is little $\mathrm{H}$-bonding between $\mathrm{H}_2 \mathrm{O}$ and $n$-hexanol. When the ratio of $\mathrm{C}$ to $\mathrm{OH}$ is more than 4 , alcohols have little solubility in water.
(d) The ether $\mathrm{CH}_3 \mathrm{OCH}_3$ has no $\mathrm{H}$ on $\mathrm{O}$ and cannot $\mathrm{H}$-bond; only the weaker dipole-dipole interaction exists.

Amy Jiang
Amy Jiang
Numerade Educator
06:39

Problem 4

Problem 13.4 The ir spectra of trans- and cis-1,2-cyclopentanediol show a broad band in the region $3450-3570 \mathrm{~cm}^{-1}$. On dilution with $\mathrm{CCl}_4$, this band of the cis isomer remains unchanged, but the band of the trans isomer shifts to a higher frequency and becomes sharper. Account for this difference in behavior.

The $\mathrm{OH}$ 's of the cis isomer participate in intramolecular H-bonding, Fig. 13-1(a), which is not affected by dilution. In the trans isomer, the H-bonding is intermolecular, Fig. 13-I(b), and dilution breaks these bonds, causing disappearance of the broad band and its replacement by a sharp $\mathrm{OH}$ band at higher frequency.

Zubair Abdulla
Zubair Abdulla
Numerade Educator
01:47

Problem 5

Give structures and IUPAC names of the alcohols formed from $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}=\mathrm{CH}_2$ by reaction with (a) dilute $\mathrm{H}_2 \mathrm{SO}_4 ;($ b $) \mathrm{B}_2 \mathrm{H}_6$, then $\mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}^{-}$; (c) $\mathrm{Hg}\left(\mathrm{OCOCH}_3\right)_2, \mathrm{H}_2 \mathrm{O}$, then $\mathrm{NaBH}_4$.
(a) The expected product is 3-methyl-2-butanol, $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCHOHCH}_3$, from a Markovnikov addition of $\mathrm{H}_2 \mathrm{O}$. However, the major product is likely to be 2-methyl-2-butanol, $\left(\mathrm{CH}_3\right)_2 \mathrm{COHCH}_2 \mathrm{CH}_3$, formed by rearrangement of the intermediate $\mathrm{R}^{+}$.
$$
\begin{aligned}
& \left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}=\mathrm{CH}_2 \stackrel{+\mathrm{H}^{+}}{\left(\mathrm{H}_2 \mathrm{O}\right)}\left(\mathrm{CH}_3\right)_2 \mathrm{CHC}_{\mathrm{C}} \mathrm{HCH}_3 \stackrel{\sim \mathrm{H}:}{\longrightarrow}\left(\mathrm{CH}_3\right)_2 \stackrel{+}{\mathrm{C}} \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{+\mathrm{H}_2 \mathrm{O}}{-\mathrm{H}^{+}}\left(\mathrm{CH}_3\right)_2 \mathrm{COHCH}_2 \mathrm{CH}_3 \\
& \left(2^{\circ}\right) \\
&
\end{aligned}
$$
(b) Anti-Markovnikov $\mathrm{HOH}$ addition forms $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{CH}_2 \mathrm{OH}$, 3-methyl-1-butanol.
(c) Markovnikov $\mathrm{HOH}$ addition with no rearrangment gives $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCHOHCH}_3$, 3-methyl-2-butanol.

Sima Sarker
Sima Sarker
Numerade Educator
02:06

Problem 6

Give the structure and IUPAC name of the product formed on hydroboration-oxidation of 1methylcyclohexene.
$\mathrm{H}$ and $\mathrm{OH}$ add cis and therefore $\mathrm{CH}_3$ and $\mathrm{OH}$ are trans.

Vasu Makani
Vasu Makani
Numerade Educator
01:10

Problem 7

Give 3 combinations of $\mathrm{RMgX}$ and a carbonyl compound that could be used to prepare.
<smiles>CCC(C)(O)Cc1ccccc1</smiles>
This $3^{\circ}$ alcohol is made from $\mathrm{RMgX}$ and a ketone, $\mathrm{R}^{\prime} \mathrm{COR}^{\prime \prime}$. The possibilities are:
(1)
<smiles>CCC(C)(O)c1ccccc1</smiles>
from $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{MgCl}$ and
<smiles>CCC(C)=O</smiles>
(2)
<smiles>CCC(C)(O)Cc1ccccc1</smiles>
from
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{MgBr}$ and
<smiles>CC(=O)Cc1ccccc1</smiles>
(3)
<smiles>CCC(=O)Cc1ccccc1</smiles>

Narayan Hari
Narayan Hari
Numerade Educator
01:01

Problem 8

Give four limitations of the Grignard reaction.
(1) The halide cannot possess a functional group with an acidic $\mathrm{H}$, such as $\mathrm{OH}, \mathrm{COOH}, \mathrm{NH}, \mathrm{SH}$, or $\mathrm{C} \equiv \mathrm{C}-\mathrm{H}$, because then the carbanion of the Grignard group would remove the acidic $\mathrm{H}$ and be reduced. For example:
$$
\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{Mg} \longrightarrow \underset{\text { unstable }}{[}\left[\mathrm{HOCH}_2 \mathrm{CH}_2 \mathrm{MgBr}\right] \longrightarrow(\mathrm{BrMg})^{+-} \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{H}
$$
unstable
(2) If the halide also has a $\mathrm{C}=\mathrm{O}$ (or $\mathrm{C}=\mathrm{N}-, \mathrm{C}=\mathrm{N}, \mathrm{N}=\mathrm{O}, \mathrm{S}=\mathrm{O}, \mathrm{C}=\mathrm{CH}$ ) group, it reacts inter- or intramolecularly with itself.
(3) The reactant cannot be a vic-dihalide, because it would undergo dehalogenation:
$$
\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{Br}+\mathrm{Mg} \longrightarrow \mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{MgBr}_2
$$
(4) A ketone with two bulky $\mathrm{R}$ groups, e.g., $-\mathrm{C}\left(\mathrm{CH}_3\right)_3$, would be too sterically hindered to react with an organometallic compound with a bulky $\mathrm{R}^{\prime}$ group.

Narayan Hari
Narayan Hari
Numerade Educator
04:17

Problem 9

Prepare 1-butanol from $(a)$ an alkene, (b) 1-chlorobutane, (c) 1-chloropropane and (d) ethyl bromide.
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2 \frac{1 .\left(\mathrm{BH}_3\right)_2}{2 \cdot \mathrm{H}_2 \mathrm{O}_2, \mathrm{OH}}-\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$
(b) $\mathrm{HO}^{-}+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl} \stackrel{\mathrm{H}_2 \mathrm{O}}{-} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}+\mathrm{Cl}^{-}\left(\mathrm{S}_{\mathrm{N}} 2\right)$
(c) 1-Chloropropane has one less $\mathrm{C}$ than the needed $1^{\circ}$ alcohol. The Grignard reaction is used to lengthen the chain by adding $\mathrm{H}_2 \mathrm{C}=\mathrm{O}$ (formaldehyde).
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl} \underset{\text { ether }}{\stackrel{\mathrm{Mg}}{\longrightarrow}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{MgCl} \underset{\text { I. } \mathrm{HCH}=\mathrm{O}}{2 . \mathrm{H}_3 \mathrm{O}^{+}}-\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}
$$
(d) 1-Butanol is a $1^{\circ}$ alcohol with two C's more than $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$. Reaction of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{MgBr}$ with ethylene oxide followed by hydrolysis gives 1-butanol.

Ronald Prasad
Ronald Prasad
Numerade Educator
01:02

Problem 10

For the following pairs of halides and carbonyl compounds, give the structure of each alcohol formed by the Grignard reaction. (a) Bromobenzene and acetone. (b) $p$-Chlorophenol and formaldehyde. (c) Isopropyl chloride and benzaldehyde. $(d)$ Chlorocyclohexane and methyl phenyl ketone.
(a) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Br} \underset{\text { ether }}{\stackrel{\mathrm{Mg}}{\longrightarrow}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{MgBr}$
<smiles></smiles>
(b) The weakly acidic $\mathrm{OH}$ in p-chlorophenol prevents formation of the Grignard reagent.
(c)
<smiles>CC(C)C(O)c1ccccc1</smiles>
(d)
<smiles>ClC1CCCCC1</smiles>
<smiles>CC(C)(O)[C+]1CCCC(C(C)(O)c2ccccc2)C1</smiles>
Methylcyclohexylphenylcarbinol

Narayan Hari
Narayan Hari
Numerade Educator
02:12

Problem 11

How do the alcohols from $\mathrm{LiAlH}_4$ or catalytic reduction of ketones differ from those derived from aldehydes?
Ketones yield $2^{\circ}$ alcohols while aldehydes give $1^{\circ}$ alcohols,

Vasu Makani
Vasu Makani
Numerade Educator
01:41

Problem 13

Reduction of $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCHO}$ with $\mathrm{NaBH}_4$ gives a product different from that of catalytic hydrogenation $\left(\mathrm{H}_2 / \mathrm{Ni}\right)$. What are the products?

Sima Sarker
Sima Sarker
Numerade Educator
05:22

Problem 14

Supply equations for the formation from phosphoric acid of $(a)$ alkyl phosphate esters by reactions with an alcohol, in which each acidic $\mathrm{H}$ is replaced and $\mathrm{H}_2 \mathrm{O}$ is eliminated; (b) phosphoric anhydrides on heating to eliminate $\mathrm{H}_2 \mathrm{O}$.

Arjun Tibrewal
Arjun Tibrewal
Numerade Educator
01:14

Problem 15

Alkyl esters of di- and triphosphoric acid are important in biochemistry because they are stable in the aqueous medium of living cells and are hydrolyzed by enzymes to supply the energy needed for muscle contraction and other processes. (a) Give structural formulas for these esters. (b) Write equations for the hydrolysis reactions that are energy-liberating.

Narayan Hari
Narayan Hari
Numerade Educator
02:52

Problem 16

How does sulfonate ester formation from sulfonyl chloride resemble nucleophilic displacements of alkyl halides?
The alcohol acts as a nucleophile and halide ion is displaced.

Sulfonyl chlorides are prepared from the sulfonic acid or salt with $\mathrm{PCl}_5$.
$$
\left.\begin{array}{l}
\mathrm{RSO}_2 \mathrm{OH} \\
\mathrm{RSO}_2 \mathrm{ONa}
\end{array}\right\}+\mathrm{PCl}_5 \longrightarrow \mathrm{RSO}_2 \mathrm{Cl}+\mathrm{POCl}_3+\left\{\begin{array}{l}
\mathrm{HCl} \\
\mathrm{NaCl}
\end{array} .\right.
$$

Aromatic sulfonyl chlorides are also formed by ring chlorosulfonation with chlorosulfonic acid, $\mathrm{HOSO}_2 \mathrm{Cl}$.
$$
\mathrm{ArH}+\mathrm{HOSO}_2 \mathrm{Cl} \longrightarrow \mathrm{ArSO}_2 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O}
$$
a sulfonyl chloride

Travis Maslanik
Travis Maslanik
Numerade Educator
01:33

Problem 18

Give the main products of reaction of 1-propanol with (a) alkaline aq. $\mathrm{KMnO}_4$ solution during distillation; (b) hot $\mathrm{Cu}$ shavings; (c) $\mathrm{CH}_3 \mathrm{COOH}, \mathrm{H}^{+}$.
(a) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$. Since aldehydes are oxidized further under these conditions, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$ is also obtained. Most of the aldehyde is removed before it can be oxidized.
(b) $\mathrm{Ch}_3 \mathrm{CH}_2 \mathrm{CHO}$. The aldehyde can't be oxidized further.
(c)
<smiles>[C-]#[13C]CCCOC(C)=O</smiles>

Raghvendra Singh
Raghvendra Singh
Numerade Educator
06:44

Problem 19

Explain the relative acidity of liquid $1^{\circ}, 2^{\circ}$, and $3^{\circ}$ alcohols.
The order of decreasing acidity of alcohols, $\mathrm{CH}_3 \mathrm{OH}>1^{\circ}>2^{\prime \prime}>3^{\circ}$, is attributed to electron-releasing R's. These intensify the charge on the conjugate base, $\mathrm{RO}^{-}$, and destabilize this ion, making the acid weaker.

Tom Rutherford
Tom Rutherford
Numerade Educator
03:05

Problem 20

Give simple chemical tests to distinguish (a) I-pentanol and $n$-hexane; $(b) n$-butanol and $t$-butanol; (c) 1-butanol and 2-buten-1-ol; (d) 1-hexanol and 1-bromohexane.
(a) Alcohols such as 1-pentanol dissolve in cold $\mathrm{H}_2 \mathrm{SO}_4$. Alkanes such as $n$-hexane are insoluble. (b) Unlike $t$ butanol (a $3^{\circ}$ alcohol), $n$-butanol (a $1^{\circ}$ alcohol) can be oxidized under mild conditions. The analytical reagent is chromic anhydride in $\mathrm{H}_2 \mathrm{SO}_4$. A positive test is signaled when this orange-red solution turns a deep green because of the presence of $\mathrm{Cr}^{3+}$. (c) 2-Buten-1-ol decolorizes $\mathrm{Br}_2$ in $\mathrm{CCl}_4$ solution; 1-butanol does not. (d) 1-Hexanol reduces orange-red $\mathrm{CrO}_3$ to green $\mathrm{Cr}^{3+}$; alkyl halides such as I-bromohexane do not. The halde on warming with $\mathrm{AgNO}_3(\mathrm{EtOH})$ gives $\mathrm{AgBr}$.

Aadit Sharma
Aadit Sharma
Numerade Educator
02:33

Problem 21

Draw the structure of $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$ if the compound: (1) reacts with $\mathrm{Na}$ but fails to react with a strong oxidizing agent such as $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$; (2) gives a negative iodoform test; and (3) gives a positive Lucas test in 4 minutes.

(1) Because the compound reacts with $\mathrm{Na}$, it must be an alcohol. Furthermore, because the compound does not react with a strong oxidizing agent, it must be a tertiary $\left(3^{\circ}\right)$ alcohol. Therefore, the structure of $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$ is tert-butyl alcohol:
<smiles>CC(C)(C)O</smiles>
(2) A negative iodoform test would occur for the primary four-carbon alcohol, $n$-butyl alcohol:
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}
$$
(3) In the Lucas test, the Lucas reagent reacts with $1^{\circ}, 2^{\circ}$, and $3^{\circ}$ alcohols. The alcohols are distinguished by their reactivity with the Lucas reagent: $3^{\circ}$ alcohols react immediately; $2^{\circ}$ alcohols react within 5 minutes; and $1^{\circ}$ alcohols react poorly at room temperature. Because the compound reacts with the Lucas reagent in 4 minutes, then the structure of $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$, a $2^{\circ}$ alcohol, is sec-butyl alcohol:
<smiles>CCC(C)O</smiles>

Shahina -
Shahina -
Numerade Educator
07:18

Problem 22

Write balanced ionic equations for the following redox reaction:
$$
\mathrm{CH}_3 \mathrm{CHOHCH}_3+\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4 \stackrel{\text { heat }}{\longrightarrow} \mathrm{CH}_3-\mathrm{CO}-\mathrm{CH}_3+\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3+\mathrm{H}_2 \mathrm{O}+\mathrm{K}_2 \mathrm{SO}_4
$$

Write partial equations for the oxidation and the reduction. Then: (1) Balance charges by adding $\mathrm{H}^{+}$in acid solutions or $\mathrm{OH}^{-}$in basic solutions. (2) Balance the number of $\mathrm{O}$ 's by adding $\mathrm{H}_2 \mathrm{O}$ 's to one side. (3) Balance the number of H's by adding H's to one side. The number added is the number of equivalents of oxidant or reductant.
(a)
$$
\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH} \longrightarrow\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}
$$

REDUCTION
$$
\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}
$$
(1) In acid balance charges with $\mathrm{H}^{+}$:
(no change)
$$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+8 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}
$$
(2) Balance $\mathrm{O}$ with $\mathrm{H}_2 \mathrm{O}$ :
(no change)
$$
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+8 \mathrm{H}^{++} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$$
(3) Balance $\mathrm{H}$ :
$$
\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH} \longrightarrow\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}+2 \mathrm{H} \quad \mathrm{Cr}_2 \mathrm{O}_7^{2-}+8 \mathrm{H}^{+}+6 \mathrm{H} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}
$$
(4) Balance equivalents:
$$
3\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH} \longrightarrow 3\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}+6 \mathrm{H}
$$
(5) Add:
$$
\frac{\mathrm{Cr}_2 \mathrm{O}_7^{2-}+8 \mathrm{H}^{+}+6 \mathrm{H} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}}{3\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+8 \mathrm{H}^{+} \longrightarrow 3\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}}
$$

Shazia Naz
Shazia Naz
Numerade Educator
01:59

Problem 23

How can the difference in reactivity of $1^3, 2^{\circ}$, and $3^{\circ}$ alcohols with $\mathrm{HCl}$ be used to distinguish among these kinds of alcohols, assuming the alcohols have six or less $\mathrm{C}$ 's?
The Lucas test uses conc. $\mathrm{HCl}$ and $\mathrm{ZnCl}_2$ (to increase the acidity of the acid).

The above reaction is immediate; a $2^{\circ} \mathrm{ROH}$ reacts within $5 \mathrm{~min}$; a $1^{\circ} \mathrm{ROH}$ does not react at all at room temperature.

Nicholas Sacco
Nicholas Sacco
Numerade Educator
05:32

Problem 24

$\ln \mathrm{CCl}_4$ as solvent, the nmr spectrum of $\mathrm{CH}_3 \mathrm{OH}$ shows two singlets. In $\left(\mathrm{CH}_3\right)_2 \mathrm{SO}$, there is a doublet and a quartet. Explain in terms of the "slowness" of nmr detection.

In $\mathrm{CCl}_4, \mathrm{CH}_3 \mathrm{OH} \mathrm{H}$-bonds intermolecularly, leading to a rapid interchange of the $\mathrm{H}$ of $\mathrm{O}-\mathrm{H}$. The instrument senses an average situation and therefore there is no coupling between $\mathrm{CH}_3$ and $\mathrm{OH}$ protons. In $\left(\mathrm{CH}_3\right)_2 \mathrm{SO}, \mathrm{H}$-bonding is with solvent, and the $\mathrm{H}$ stays on the $\mathrm{O}$ of $\mathrm{OH}$. Now coupling occurs. This technique can be used to distinguish among $\mathrm{RCH}_2 \mathrm{OH}, \mathrm{R}_2 \mathrm{CHOH}$ and $\mathrm{R}_3 \mathrm{COH}$, whose signals for $\mathrm{H}$ of $\mathrm{OH}$ are a triplet, a doublet and a singlet, respectively.

Zubair Abdulla
Zubair Abdulla
Numerade Educator
03:21

Problem 25

How are thiols prepared in good yield?
The preparation of thiols by $\mathrm{S}_{\mathrm{N}} 2$ attack of nucleophilic $\mathrm{HS}^{-}$on an alkyl halide gives poor yields because the mercaptan loses a proton to form an anion, $\mathrm{RS}^{-}$, which reacts with a second molecule of alkyl halide to form a thioether.
thioether

Dialkylation is minimized by using an excess of : $\mathrm{SH}^{-}$, and avoided by using thiourea to form an alkylisothiourea salt that is then hydrolyzed.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
03:16

Problem 26

Show steps in the synthesis of ethyl ethanesulfonate, $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{SO}_2 \mathrm{OCH}_2 \mathrm{CH}_3$, from $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$ and any inorganic reagents.
The alkyl sulfonic acid is make by oxidizing the thiol, which in turn comes from the halide.

Nima Gharibi
Nima Gharibi
Numerade Educator
01:31

Problem 27

Why are mercaptans $(a)$ more acidic $\left(K_a \approx 10^{-11}\right)$ than alcohols $\left(K_a \approx 10^{-17}\right)$ and $(b)$ more nucleophilic than alcohols?
(a) There are more and stronger $\mathrm{H}$-bonds in alcohols, thus producing an acid-weakening effect. Also, in the conjugate bases $\mathrm{RS}^{-}$and $\mathrm{RO}^{-}$the charge is more dispersed over the larger $\mathrm{S}$, thereby making $\mathrm{RS}^{-}$the weaker base and RSH the stronger acid (Section 3.11). (b) The larger $\mathrm{S}$ is more easily polarized than the smaller $\mathrm{O}$ and therefore is more nucleophilic. For example, RSH participates more rapidly in $\mathrm{S}_{\mathrm{N}} 2$ reactions than $\mathrm{ROH}$. Recall that among the halide anions, nucleophilicity also increases as size increases: $\mathrm{F}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}^{-}$.

Ly Tran
Ly Tran
Numerade Educator
03:41

Problem 28

Offer mechanisms for
$$
\begin{aligned}
\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\mathrm{H}_2 \mathrm{~S} . \mathrm{H}^{+}}{\stackrel{\mathrm{H}_3 \mathrm{~S}}{\mathrm{hV}}}-\mathrm{CH}_3 \mathrm{CH}(\mathrm{SH}) \mathrm{CH}_3 \\
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{SH}
\end{aligned}
$$

Acid-catalyzed addition has an ionic mechanism (Markovnikov):

The peroxide- or light-catalyzed reaction has a free-radical mechanism (anti-Markovnikov):
$$
\begin{gathered}
\mathrm{H}-\mathrm{S}-\mathrm{H} \stackrel{h v}{\longrightarrow} \mathrm{HS} \cdot+\cdot \mathrm{H} \\
\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2+\cdot \mathrm{SH} \longrightarrow \mathrm{CH}_3 \mathrm{CHCH}_2 \mathrm{SH} \\
\mathrm{CH}_3 \mathrm{CHCH}_2 \mathrm{SH} \stackrel{\mathrm{H}_2 \mathrm{~S}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{SH}+\cdot \mathrm{SH}
\end{gathered}
$$

Ian Kaigh
Ian Kaigh
Numerade Educator
01:47

Problem 29

Give the IUPAC names for each of the following alcohols. Which are $1^{\circ}, 2^{\circ}$, and $3^{\circ}$ ?
(a) 2-Methyl-1-butanol, $1^{\circ}$. (b) 2-Methyl-3-phenyl-1-propanol, $1^{\circ}$. (c) 1-Methyl-1-cyclopentanol, $3^{\circ}$. (d) 3Methyl-3-pentanol, 3 . (e) 5-Chloro-6-methyl-6-(3-chlorophenyl)-2-hepten-1-ol. (The longest chain with $\mathrm{OH}$ has seven C's and the prefix is hept-. Numbering begins at the end of the chain with $\mathrm{OH}$; therefore, -1-ol. The aromatic ring substituent has $\mathrm{Cl}$ at the 3 position counting from the point of attachment, and is put in parentheses to show that the entire ring is attached to the chain at $\mathrm{C}^6 . \mathrm{Cl}$ on the chain is at $\mathrm{C}^5$.) $1^2$.

Sima Sarker
Sima Sarker
Numerade Educator
03:59

Problem 30

Write condensed structural formulas and give IUPAC names for $(a)$ vinylcarbinol, (b) diphenylcarbinol, (c) dimethylethylcarbinol, (d) benzylcarbinol.
(a) $\mathrm{H}_2 \mathrm{C}=\mathrm{CHCH}_2 \mathrm{OH}$ 2-Propen-1-ol (Allyl alcohol)
(b) $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{CH}-\mathrm{OH}$ Diphenylmethanol (Benzhydrol)
(c)
<smiles>CCC(C)(C)O</smiles>
2-Methyl-2-butanol
(d) $\mathrm{C}_6 \mathrm{H}_5-\stackrel{\beta}{\mathrm{C}} \mathrm{H}_2 \stackrel{\alpha}{\mathrm{C}} \mathrm{H}_2 \mathrm{OH}$ 2-Phenylethanol ( $\beta$-Phenylethanol)

Benjamin Angeles
Benjamin Angeles
Numerade Educator
11:40

Problem 31

The four isomeric $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}$ alcohols are
(i) $\left(\mathrm{CH}_3\right)_3 \mathrm{COH}$
(ii) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$
(iii) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{OH}$
(iv) $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CH}_3$

Synthesize each, using a different reaction from among a reduction, an $\mathrm{S}_{\mathrm{N}} 2$ displacement, a hydration, and a Grignard reaction.

Synthesis of the $3^{\circ}$ isomer, (i), has restrictions: the $\mathrm{S}_{\mathrm{N}} 2$ displacement of a $3^{\circ}$ halide cannot be used because elimination would occur; nor is there any starting material that can be reduced to a $3^{\circ}$ alcohol. Either of the two remaining methods can be used; arbitrarily the Grignard is chosen.
$$
\mathrm{CH}_3 \mathrm{MgI}+\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O} \underset{\text { hydrolysis }}{\stackrel{\text { after }}{\longrightarrow}}\left(\mathrm{CH}_3\right)_3 \mathrm{COH}
$$

The $\mathrm{S}_{\mathrm{N}} 2$ displacement on the corresponding $\mathrm{RX}$ is best for $1^{\circ}$ alcohols such as (ii) and (iii); let us choose (ii) for this synthesis.
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}+\mathrm{OH}^{-} \stackrel{-\mathrm{Ct}^{-}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}
$$

The $1^{\circ}$ alcohol, (iii), and the $2^{\circ}$ alcohol, (iv) can be made by either of the two remaining syntheses. However, the one-step hydration with $\mathrm{H}_3 \mathrm{O}^{+}$to give (iv) is shorter than the two-step hydroboration-oxidation to give (iii).
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}^{+} \stackrel{-\mathrm{H}^{+}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CH}_3
$$

Finally, (iii) is made by reducing the corresponding $\mathrm{RCH}=\mathrm{O}$ or $\mathrm{RCOOH}$.
$$
\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{COOH} \frac{1 . \mathrm{LiAlH}_4}{2, \mathrm{H}^{+}}-\left(\mathrm{CH}_3\right)_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}
$$

Nicholas Sacco
Nicholas Sacco
Numerade Educator
09:21

Problem 32

Prepare ethyl $p$-chlorophenylcarbinol by a Grignard reaction.
Prepare this $2^{\circ}$ alcohol, $p-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CHOHCH}_2 \mathrm{CH}_3$, from $\mathrm{RCHO}$ and $\mathrm{R}^{\prime} \mathrm{MgX}$. Since the groups on the carbinol $\mathrm{C}$ are different, there are two combinations possible:
(1)
$p-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CHOHCH}_2 \mathrm{CH}_3$
(ring $\mathrm{Cl}$ has not interfered)
(2) $p-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{MgBr}$
$p-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{CHOHCH}_2 \mathrm{CH}_3$
$\mathrm{Br}$ is more reactive than $\mathrm{Cl}$ when making a Grignard of $p-\mathrm{ClC}_6 \mathrm{H}_4 \mathrm{Br}$.

Katie Miller
Katie Miller
Numerade Educator
03:21

Problem 33

Give the hydroboration-oxidation product from $(a)$ cyclohexene, $(b)$ cis-2-phenyl-2-butene, (c) trans-2-phenyl-2-butene.

Addition of $\mathrm{H}_2 \mathrm{O}$ is cis anti-Markovnikov. See Fig. 13-2. In Fig. 13-2(c) the second pair of conformations show the eclipsing of the H's with each other and of the Me's with each other. The more stable staggered conformations are not shown.

Shazia Naz
Shazia Naz
Numerade Educator
06:09

Problem 34

The following Grignard reagents and aldehydes or ketones are reacted and the products hydrolyzed. What alcohol is produced in each case? (a) Benzaldehyde $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}=\mathrm{O}\right)$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{MgBr}$. (b) Acetaldehyde and phenyl magnesium bromide. (c) Acetone and benzyl magnesium bromide. (d) Formaldehyde and cyclohexyl magnesium bromide. $(e)$ Acetophenone $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CCH}_3\right)$ and ethyl magnesium bromide.

Zubair Abdulla
Zubair Abdulla
Numerade Educator

Problem 35

Give the mechanism in each case:
(a)
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH} \stackrel{\mathrm{HCl}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}
$$
(b)
<smiles>CC(C)C(C)O</smiles>
<smiles>CCC(C)(C)Cl</smiles>
(c)
<smiles>CCC(C)(C)O</smiles>
<smiles>CCC(C)(C)Cl</smiles>
Why did rearrangement occur only in $(b)$ ?
(a) The mechanism is $\mathrm{S}_{\mathrm{N}} 2$ since we are substituting $\mathrm{Cl}$ for $\mathrm{H}_2 \mathrm{O}$ from $1^{\circ} \mathrm{ROH}_2^{+}$.
(b) The mechanism is $\mathrm{S}_{\mathrm{N}} 1$.
<smiles>CC(C)C(C)O</smiles>
<smiles>CCC1CCC1(C)C[18OH]</smiles>
<smiles>CCC(C)(Cl)C[C-]CCC(C)(C)C</smiles>
$$
2^{\circ} \mathrm{R}^{+} \text {(less stable) } \quad 3^{\circ} \mathrm{R}^{+} \text {(more stable) }
$$
(c) $\mathrm{S}_{\mathrm{N}} \mathrm{l}$ mechanism. The stable $3^{\circ}\left(\mathrm{CH}_3\right)_2 \stackrel{+}{\mathrm{C}} \mathrm{CH}_2 \mathrm{CH}_3$ reacts with $\mathrm{Cl}^{-}$with no rearrangement,

Check back soon!
02:08

Problem 36

Why does dehydration of 1-phenyl-2-propanol in acid form 1-phenyl-1-propene rather than 1phenyl-2-propene?
1-Phenyl-1-propene, $\mathrm{PhCH}=\mathrm{CHCH}_3$, is a more highly substituted alkene and therefore more stable than 1phenyl-2-propene, $\mathrm{PhCH}_2 \mathrm{CH}=\mathrm{CH}_2$. Even more important, it is more stable because the double bond is conjugated with the ring.

Susan Hallstrom
Susan Hallstrom
Numerade Educator
03:46

Problem 37

Write the structural formulas for the alcohols formed by oxymercuration-demercuration from (a) 1-heptene, (b) 1-methylcyclohexene, (c) 3,3-dimethyl-1-butene.
The net addition of $\mathrm{H}_2 \mathrm{O}$ is Markovnikov.
(a) $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4 \mathrm{CHOHCH}_3 2$-Heptanol
(b)
<smiles>CC1(O)CCCCC1</smiles>
1-Methylcyclohexanol
(c) $\left(\mathrm{CH}_3\right)_3 \mathrm{CCHOHCH}_3$ 3,3-Dimethyl-2-butanol (no rearrangement occurs)

Kendrick Buford
Kendrick Buford
Numerade Educator
02:45

Problem 38

List the alcohols and acids that compose the inorganic esters $(a)\left(\mathrm{CH}_3\right)_3 \mathrm{COCl}$ and $(b)$ $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{ONO}_2$. Name the esters.

Conceptually hydrolyze the $\mathrm{O}$ to the heteroatom bond while adding an $\mathbf{H}$ to the $\mathrm{O}$ and an $\mathbf{O H}$ to the heteroatom. (a) $\left(\mathrm{CH}_3\right)_3 \mathrm{COH}$ and $\mathrm{HOCl}$, -butyl hypochlorite. (b) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ and $\mathbf{H O N O} \mathrm{N}_2$, ethyl nitrate. Tert-butyl hypochlorite is used to chlorinate hydrocarbons by free-radical chain mechanisms.

Mystique Till
Mystique Till
Numerade Educator
05:17

Problem 39

Starting with isopropyl alcohol as the only available organic compound, prepare 2,3-dimethyl-2butanol.
This $3^{\circ}$ alcohol, $\left(\mathrm{CH}_3\right)_2 \mathrm{COHCH}\left(\mathrm{CH}_3\right)_2$, is prepared from a Grignard reagent and a ketone.

Ian Kaigh
Ian Kaigh
Numerade Educator
01:26

Problem 40

Alcohols such as $\mathrm{Ph}_2 \mathrm{CHCH}_2 \mathrm{OH}$ rearrange on treatment with acid; they can be dehydrated by heating their methyl xanthates (Tschugaev reaction). The pyrolysis proceeds by a cyclic transition state. Outline the steps using $\mathrm{Ph}_2 \mathrm{CHCH}_2 \mathrm{OH}$.

Nicole Krahulik
Nicole Krahulik
Numerade Educator
02:39

Problem 41

Methyl ketones give the haloform test:
$\mathrm{I}_2$ can also oxidize $1^{\circ}$ and $2^{\circ}$ alcohols to carbonyl compounds. Which butyl alcohols give a positive haloform test? Alcohols with the
<smiles>[R]C1(O)C2CC3CC1C3C2</smiles>
groups are oxidized to
<smiles>[R]C(C)=O</smiles>
and give a positive test. The only butyl alcohol giving a positive test is
<smiles>CCC(C)O</smiles>

Nicholas Sacco
Nicholas Sacco
Numerade Educator
01:18

Problem 42

A compound, $\mathrm{C}_9 \mathrm{H}_{12} \mathrm{O}$, is oxidized under vigorous conditions to benzoic acid. It reacts with $\mathrm{CrO}_3$ and gives a positive iodoform test (Problem 13.41). Is this compound chiral?

Since benzoic acid is the product of oxidation, the compound is a motrosubstituted benzene, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{G}$. Subtracting $\mathrm{C}_6 \mathrm{H}_5$ from $\mathrm{C}_9 \mathrm{H}_{12} \mathrm{O}$ gives $\mathrm{C}_3 \mathrm{H}_7 \mathrm{O}$ as the formula for a saturated side chain. A positive $\mathrm{CrO}_3$ test means a $1^{\circ}$ or $2^{\circ} \mathrm{OH}$. Possible structures are:
<smiles>OC(c1ccccc1)c1ccccc1</smiles>
1
<smiles>OC(c1ccccc1)c1ccccc1</smiles>
II
<smiles>OCC=Cc1ccccc1</smiles>
III
<smiles>CC(CO)c1ccccc1</smiles>
IV
Only 11 has the $-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3$ needed for a positive iodoform test. Il is chiral.

Nikhil Choudhary
Nikhil Choudhary
Numerade Educator

Problem 43

Suggest a possible industrial preparation for (a) $t$-butyl alcohol, $(b)$ allyl alcohol, (c) glycerol ( $\mathrm{HOCH}_2 \mathrm{CHOHCH}_2 \mathrm{OH}$ ).
(a)
<smiles>C=C(C)C[18N]CCC</smiles>
<smiles>CC(C)(C)O</smiles>
Isobutane
Isobutylene
(b)
<smiles>CC=CC</smiles>
$\mathrm{ClCH}_2-\mathrm{CH}=\mathrm{CH}_2$
Propene
Allyl chloride
Allyl alcohol

Check back soon!

Problem 44

Assign numbers, from 1 for the lowest to 5 for the highest, to indicate relative reactivity with $\mathrm{HBr}$ in forming benzyl bromides from the following benzyl alcohols: (a) $p-\mathrm{Cl}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{CH}_2 \mathrm{OH},(b)\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{CHOH},(c)$ $p-\mathrm{O}_2 \mathrm{~N}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{CH}_2 \mathrm{OH},(d)\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{COH},(e) \mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$.

Differences in reaction rates depend on the relative abilities of the protonated alcohols to lose $\mathrm{H}_2 \mathrm{O}$ to form $\mathrm{R}^{+}$. The stability of $\mathrm{R}^{+}$affects the $\Delta H^{\ddagger}$ for forming the incipient $\mathrm{R}^{+}$in the transition state and determines the overall rate.

Electron-attracting groups such as $\mathrm{NO}_2$ and $\mathrm{Cl}$ in the para position destablize $\mathrm{R}^{+}$by intensifying the positive charge. $\mathrm{NO}_2$ of $(c)$ is more effective since it destablizes by both resonance and induction, while $\mathrm{Cl}$ of $(a)$ destabilizes only by induction. The more $\mathrm{C}_6 \mathrm{H}_5$ 's on the benzyl $\mathrm{C}$, the more stable is $\mathrm{R}^{+}$.

Check back soon!
03:48

Problem 45

Supply structural formulas and stereochemical designations for the organic compounds (A) through $(\mathrm{H})$.
(a) (A) + conc. $\mathrm{H}_2 \mathrm{SO}_4 \stackrel{\text { cold }}{\longrightarrow}$
<smiles>CCC(C)OS(=O)(=O)O</smiles>
$\stackrel{\mathrm{H}_2 \mathrm{O}}{\mathrm{H}^{+}}$ (B)
(b) (R) $-\mathrm{CH}_3 \mathrm{CHOHCH}_2 \mathrm{CH}=\mathrm{CH}_2 \frac{\mathrm{H}_2 \mathrm{O}}{\mathrm{H}_2 \mathrm{SO}_4}-(\mathrm{C})+$ (D)
(c)
(R)-
<smiles>[CH2+]=CC(O)CC</smiles>
$(\mathrm{E})+(\mathrm{F})+(\mathrm{G})$

Vasu Makani
Vasu Makani
Numerade Educator
04:17

Problem 46

How does the Lewis theory of acids and bases explain the functions of $(a) \mathrm{ZnCl}_2$ in the Lucas reagent? $(b)$ ether as a solvent in the Grignard reagent?
(a)
(b) $\mathrm{R}^{\prime} \mathrm{MgX}$ acts as a Lewis acid because $\mathrm{Mg}$ can coordinate with one unshared electron pair of each $\mathrm{O}$ of two ether molecules to form an addition compound,

David Collins
David Collins
Numerade Educator
01:09

Problem 47

Draw a Newman projections of the conformers of the following substituted ethanols and predict their relative populations: (a) $\mathrm{FCH}_2 \mathrm{CH}_2 \mathrm{OH}$, (b) $\mathrm{H}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{OH}$, (c) $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{OH}$.
If the substituents $\mathrm{F}, \mathrm{H}_2 \mathrm{~N}$, and $\mathrm{Br}$ are designated by $\mathrm{Z}$, the conformers may be generalized as anti or gauche.
<smiles>[Z]C1CC2CC1CC2O</smiles>
$\mathrm{OH}$
anti
$\mathrm{H}^{\prime}$
<smiles>[Y]C1CCCCC1O</smiles>
gauche

For $(a)$ and $(b)$ the gauche is the more stable conformer and has a greater population because of H-bonding with $\mathrm{F}$ and $\mathrm{N}$. The anti conformer is more stable in $(c)$ because there is no $\mathrm{H}$-bonding with $\mathrm{Br}$ and dipole-dipole repulsion causes $\mathrm{OH}$ and $\mathrm{Br}$ to lie as far from each other as possible.

Raghvendra Singh
Raghvendra Singh
Numerade Educator
10:17

Problem 48

Deduce the structure of a compound, $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$, which gives the following nmr data: $\delta=0.8$ (doublet, six H's), $\delta=1.7$ (complex multiplet, one H), $\delta=3.2$ (doublet, two H`s) and $\delta=4.2$ (singlet, one H; disappears after shaking sample with $\mathrm{D}_2 \mathrm{O}$ ).

The singlet at $\delta=4.2$ which disappears after shaking with $\mathrm{D}_2 \mathrm{O}$ is for $\mathrm{OH}$ (Problem 13.28). The compound must be one of the four butyl alcohols. Only isobutyl alcohol, $\left(\mathrm{CH}_3\right)_2 \mathrm{CHCH}_2 \mathrm{OH}$, has six cquivalent $\mathrm{H}^{\prime}$ 's $\left(\mathrm{two} \mathrm{CH}_3\right.$ 's), accounting for the six- $\mathrm{H}$ doublet at $\delta=0.8$, the one- $\mathrm{H}$ multiplet at $\delta=1.7$, and the two- $\mathrm{H}$ doublet which is further downfield at $\delta=3.2$ because of the electron-attracting $\mathrm{O}$.

Ian Kaigh
Ian Kaigh
Numerade Educator
23:33

Problem 49

The attempt to remove water from ethanol by fractional distillation gives $95 \%$ ethanol, an azeotrope that boils at a constant temperature of $78.15^{\circ} \mathrm{C}$. It has a lower boiling point than either water $\left(100^{\circ} \mathrm{C}\right)$ or ethanol $\left(78.3^{\circ} \mathrm{C}\right)$. A liquid mixture is an azeotrope if it gives a vapor of the same composition. How does boiling $95 \%$ ethanol with $\mathrm{Mg}$ remove the remaining $\mathrm{H}_2 \mathrm{O}$ ?
$$
\mathrm{Mg}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_2+\underset{\text { insoluble }}{\mathrm{Mg}(\mathrm{OH})_2}
$$

The dry ethanol, called absolute, is now distilled from the insoluble $\mathrm{Mg}(\mathrm{OH})_2$.

Jennifer Stoner
Jennifer Stoner
Numerade Educator
02:50

Problem 50

Explain why the most prominent (base) peak of 1-propanol is at $m / e=31$, while that of allyl alcohol is at $m / e=57$.

$\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH}^{+}$cleaves mainly into
$$
\mathrm{CH}_3 \mathrm{CH}_2 \cdot+\stackrel{+}{\mathrm{C}} \mathrm{H}_2 \stackrel{\mathrm{O}}{\mathrm{H}} \mathrm{H} \longrightarrow \mathrm{H}_2 \mathrm{C}=\stackrel{+}{\mathrm{O}} \mathrm{H}
$$
$(m / e=31)$ rather than $\mathrm{CH}_3 \mathrm{CH}_2 \stackrel{+}{\mathrm{C}} \mathrm{HOH}+\mathrm{H} \cdot$, because $\mathrm{C}-\mathrm{C}$ is weaker than $\mathrm{C}-\mathrm{H}$. In allyl alcohol,
<smiles>C=CCO</smiles>
the $\mathrm{C}-\mathrm{H}$ bond cleaves to give
<smiles>C=C[C]O</smiles>
$(m / e=57)$. This cation is stabilized by both the $\mathrm{CH}_2=\mathrm{CH}$ and the $\mathrm{O}$.

Chloe Schroeder
Chloe Schroeder
Numerade Educator
01:48

Problem 51

Inorganic acids such as $\mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_3 \mathrm{PO}_4$, and $\mathrm{HOCl}$ (hypochlorous acid) from esters. Write structural formulas for $(a)$ dimethyl sulfate, $(b)$ tribenzyl phosphate, (c) diphenyl hydrogen phosphate, $(d) t$-butyl nitrite, (e) lauryl hydrogen sulfate (lauryl alcohol is $\left.n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{CH}_2 \mathrm{OH}\right),(f)$ sodium lauryl sulfate.
Replacing the $\mathrm{H}$ of the $\mathrm{OH}$ of an acid gives an ester.

James Irizarry
James Irizarry
Numerade Educator
04:23

Problem 52

Explain why trialkyl phosphates are readily hydrolyzed with $\mathrm{OH}^{-}$to dialkyl phosphate salts, whereas dialkyl hydrogen phosphates and alkyl dihydrogen phosphates resist alkaline hydrolysis.
The hydrogen phosphates are moderately strong acids and react with bases to form anions (conjugate bases). Repulsion between negatively charged species prevents further reaction between these anions and $\mathrm{OH}^{-}$.

Temi Ajayi
Temi Ajayi
Numerade Educator
08:19

Problem 53

The ir spectrum of RSH shows a weak $-\mathrm{S}-\mathrm{H}$ stretching band at about $2600 \mathrm{~cm}^{-1}$ that does not shift significantly with concentration or nature, of solvent. Explain the difference in behavior of $S-\mathrm{H}$ and $\mathrm{O}-\mathrm{H}$ bonds.

The $\mathrm{S}-\mathrm{H}$ bond is weaker than the $\mathrm{O}-\mathrm{H}$ bond and therefore absorbs at a lower frequency. There is little or no $\mathrm{H}$-bonding in $\mathrm{S}-\mathrm{H}$ and, unlike the $\mathrm{O}-\mathrm{H}$ bond, there is little shifting of absorption frequency on dilution.

Tianyu Li
Tianyu Li
Numerade Educator