Chapter 9, Problem 9.38P

Answer each question using the ball-and-stick model of compound A.

a. Give the IUPAC name for A, including $R,S$ designations f or stereogenic centers.

b. Classify A as a $1^{\text{ο}},2^{\text{ο}},\text{or}{3}^{\text{ο}}$ alcohol.

c. Draw a stereoisomer for A and give its IUPAC name.

d. Draw a constitutional isomer that contains an $\text{OH}$ group and give its IUPAC name.

e. Draw a constitutional isomer that contains an ether and give its IUPAC name.

f. Draw the products formed (including stereochemistry) when A is treated with each reagent: $\left[1\right]$ $\text{NaH}$; $\left[2\right]$ ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$; $\left[3\right]$ ${\text{POCl}}_{\text{3}}$, pyridine; $\left[4\right]$ $\text{HCl}$; $\left[5\right]$ ${\text{SOCl}}_{\text{2}}$, pyridine; $\left[6\right]$ $\text{TsCl}$, pyridine.

Interpretation Introduction

(a)

Interpretation: The IUPAC name for A, including $R,S$ designation for stereogenic centers is to be stated.

Concept introduction: One should follow the given steps to give the IUPAC name of cyclic alcohol. The first step is naming of ring that contains the carbon bonded to the hydroxyl group. While naming, the -e ending of parent cycloalkane must changed to the suffix -ol. The second step is numbering of carbon chain by providing lowest number to the $-\text{OH}$ group, and applying all other rules of nomenclature.

Answer to Problem 9.38P

The IUPAC name for A (including $R,S$ designation for stereogenic centers) is $\left(1R,2R\right)\text{-2-isobutylcyclopentanol}$.

Explanation of Solution

The given structure of alcohol is in the form of ball-and-stick model. It is converted into skeletal structure by replacing black ball with $\text{C}$, red ball with $\text{O}$, and gray ball with $\text{H}$. The skeletal structure of A is,

Figure 1

One should follow the given steps to give the IUPAC name of cyclic alcohol. The first step is naming of ring that contains the carbon bonded to the hydroxyl group. While naming, the -e ending of parent cycloalkane must changed to the suffix -ol. The second step is numbering of carbon chain by providing lowest number to the $-\text{OH}$ group, and applying all other rules of nomenclature.

The above structure of cyclic alcohol shows that the cyclic ring consists of $\text{5}\text{C's}$ on which the hydroxyl group is bonded to $\text{C1}$, and the isobutyl group is bonded to $\text{C2}$.Therefore, the IUPAC name for A depicted in the ball-and-stick model is $\text{2-isobutylcyclopentanol}$.

There are two stereogenic centers in the given structure (shown by $*$). Both $\text{C1}$ and $\text{C2}$ possess $\left(R\right)$ configuration as shown below.

Figure 2

Thus, the IUPAC name for A (including $R,S$ designation for stereogenic centers) is $\left(1R,2R\right)\text{-2-isobutylcyclopentanol}$.

Conclusion

The IUPAC name for A (including $R,S$ designation for stereogenic centers) is $\left(1R,2R\right)\text{-2-isobutylcyclopentanol}$.

Interpretation Introduction

(b)

Interpretation: The given alcohol is to be classified as a ${\text{1}}^{\text{ο}}$, ${\text{2}}^{\text{ο}}$, or ${\text{3}}^{\text{ο}}$.

Concept introduction: Alcohols involve a hydroxyl group $\left(-\text{OH}\right)$ bonded to an $s{p}^3$ hybridized $\text{C}$ atom. They are classified as $1^{\text{ο}},2^{\text{ο}},\text{or}{3}^{\text{ο}}$ on the basis of number of $\text{C}$ atoms bonded to the carbon with the hydroxyl group.

Answer to Problem 9.38P

The given alcohol is classified as a ${\text{2}}^{\text{ο}}$.

Explanation of Solution

Alcohols involve a hydroxyl group $\left(\text{OH}\right)$ bonded to an $s{p}^3$ hybridized $\text{C}$ atom. They are classified as $1^{\text{ο}},2^{\text{ο}},\text{or}{3}^{\text{ο}}$ on the basis of number of $\text{C}$ atoms bonded to the carbon with the hydroxyl group.

In the given structure the carbon atom of alcohol (A) is bonded to two $\text{C}$ atoms. Therefore, the given alcohol is classified as a ${\text{2}}^{\text{ο}}$.

Conclusion

The given alcohol is classified as a ${\text{2}}^{\text{ο}}$.

Interpretation Introduction

(c)

Interpretation: A stereoisomer for ‘A’ is to be drawn, and its IUPAC name is to be stated.

Concept introduction: Isomers that differ to each other only in the way of orientation of atoms in space are called stereoisomers. They possess same functional group and same IUPAC name except for the prefixes (like cis, trans, $R$ or $S$).

Answer to Problem 9.38P

A stereoisomer for ‘A’ (including its IUPAC name) is drawn in Figure 3.

Explanation of Solution

Isomers that differ to each other only in the way of orientation of atoms in space are called stereoisomers. They possess same functional group and same IUPAC name except for the prefixes (like cis, trans, $R$ or $S$).

A stereoisomer for ‘A’ (including its IUPAC name) is drawn in Figure 3.

Figure 3

Conclusion

A stereoisomer for ‘A’ (including its IUPAC name) is drawn in Figure 3.

Interpretation Introduction

(d)

Interpretation: A constitutional isomer for A that contains an $\text{OH}$ group is to be drawn, and its IUPAC name is to be stated.

Concept introduction: Isomers that differ in the way of connectivity of atoms to each other are called constitutional isomers. They may possess same or different functional groups. However, their IUPAC names are entirely different.

Answer to Problem 9.38P

A constitutional isomer for ‘A’ that contains an $\text{OH}$ group is drawn in Figure 4. Its IUPAC name is $\text{3-isobutylcyclopentanol}$.

Explanation of Solution

Isomers that differ in the way of connectivity of atoms to each other are called constitutional isomers. They may possess same or different functional groups. However, their IUPAC names are entirely different.

A constitutional isomer for ‘A’ that contains an $\text{OH}$ group is drawn below.

Figure 4

The above structure of cyclic alcohol shows that the cyclic ring consists of $\text{5}\text{C's}$ on which the hydroxyl group is bonded to $\text{C1}$, and the isobutyl group is bonded to $\text{C3}$. Therefore, its IUPAC name is $\text{3-isobutylcyclopentanol}$.

Conclusion

A constitutional isomer for ‘A’ that contains an $\text{OH}$ group is drawn in Figure 4. Its IUPAC name is $\text{3-isobutylcyclopentanol}$.

Interpretation Introduction

(e)

Interpretation: A constitutional isomer for A that contains an ether is to be drawn, and its IUPAC name is to be stated.

Concept introduction: Isomers that differ in the way of connectivity of atoms to each other are called constitutional isomers. They may possess same or different functional groups. However, their IUPAC names are entirely different.

Answer to Problem 9.38P

A constitutional isomer for ‘A’ that contains an ether is drawn in Figure 5. Its IUPAC name is $\text{isobutoxycyclopentane}$.'

Explanation of Solution

Isomers that differ in the way of connectivity of atoms to each other are called constitutional isomers. They may possess same or different functional groups. However, their IUPAC names are entirely different.

A constitutional isomer for ‘A’ that contains ether is drawn below.

Figure 5

The above structure of cyclic alcohol shows that the cyclic ring consists of $\text{5}\text{C's}$ on which the isobutoxy group is bonded to $\text{C1}$. Therefore, its IUPAC name is $\text{isobutoxycyclopentane}$.

Conclusion

A constitutional isomer for ‘A’ that contains an ether is drawn in Figure 5. Its IUPAC name is $\text{isobutoxycyclopentane}$..

Interpretation Introduction

(f)

Interpretation: The products (including stereochemistry) formed from the treatment of A with the each reagent are to be stated.

Concept introduction: In proton transfer reactions, the stereochemistry of the product is retained. The $\text{E1}$ reaction involves carbocation rearrangement, thus mixture of products are obtained, whereas the $\text{E2}$ reaction does not involve carbocation rearrangement, thus only one product is obtained.

Answer to Problem 9.38P

The products formed from the treatment of A with the reagent $\left[1\right]$ are shown in Figure 6. The products formed from the treatment of A with the reagent $\left[2\right]$ are shown in Figure 7. The products formed from the treatment of A with the reagent $\left[3\right]$ are shown in Figure 8. The products formed from the treatment of A with the reagent $\left[4\right]$ are shown in Figure 9. The products formed from the treatment of A with the reagent $\left[5\right]$ are shown in Figure 10. The products formed from the treatment of A with the reagent $\left[6\right]$ are shown in Figure 11.

Explanation of Solution

Products formed from reagent $\left[1\right]$:

The given reagent is $\text{NaH}$.

An alkoxide salt is required to prepare ether. The alkoxide salts are prepared from alcohols through the Bronsted-Lowry acid-base reaction. In this reaction, $\text{NaH}$ or ${\text{NaNH}}_{\text{2}}$ are especially utilized, due to the by-product of the reaction $\left({\text{H}}_{\text{2}}\text{or}{\text{NH}}_{\text{3}}\right)$. Both ${\text{H}}_{\text{2}}$ and ${\text{NH}}_{\text{3}}$ are gases that bubbles out rapidly from the reaction mixture.

The reaction involves only transfer of proton. Since the reaction does not involve cleavage of the $\text{C}-\text{O}$ bond, the stereochemistry of the product is retained. Thus, the products formed from the treatment of A with the given reagent are,

Figure 6

Products formed from reagent $\left[2\right]$:

The given reagent is ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Alcohols undergo dehydration reaction in the presence of strong acids like ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ or $\text{p-toluenesulfonic}\text{acid}\left(\text{TsOH}\right)$ and yield alkenes. The dehydration of secondary and tertiary alcohols occurs in three steps and it is $\text{E1}$ reaction. The first step is protonation of $\text{OH}$ to form good leaving group. The second step is heterolytic cleavage of $\text{C}-\text{O}$ bond to form a carbocation (carbocation rearrangement is possible). The third step is abstraction of proton by a base $\left({\text{HSO}}_{\text{4}}{}^{-}\text{or}{\text{H}}_{\text{2}}\text{O}\right)$ from a carbon adjacent to the carbocation to result the new $\text{π}$ bond.

Thus, the products formed from the treatment of A with the given reagent are,

Figure 7

Mixture of products is obtained due to carbocation rearrangement, and having two $\text{β}$ carbon atoms.

Products formed from reagent $\left[3\right]$:

The given reagent is ${\text{POCl}}_{\text{3}}$, pyridine.

Alcohols undergo dehydration reaction in the presence of ${\text{POCl}}_{\text{3}}\text{,}\text{pyridine}$, and yield alkenes. The mechanism of the reaction is $\text{E2}$ and it does not involve carbocation rearrangement. This dehydration reaction involves three steps. The first step is protonation of $\text{OH}$ to form good leaving group. The second step is abstraction of proton by base, pyridine. The third step is removal of the $\text{β}$ proton and thereby loss of the leaving group.

Thus, the products formed from the treatment of A with the given reagent are,

Figure 8

Products formed from reagent $\left[4\right]$:

The given reagent is $\text{HCl}$, a halogen acid.

The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain ${\text{1}}^{\text{ο}}\text{,}{\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The more substituted alcohols generally react rapidly with halogen acids. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{1}$ for ${\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alcohols and ${\text{S}}_{\text{N}}\text{1}$ for ${\text{1}}^{\text{ο}}$ alcohols. The ${\text{S}}_{\text{N}}\text{1}$ reaction yields racemic mixture products, whereas the ${\text{S}}_{\text{N}}2$ reaction yields product with inversion of configuration at stereogenic center.

Since the given alcohol, A is ${\text{2}}^{\text{ο}}$; the reaction yields racemic mixture of products.

Thus, the products formed from the treatment of A with the given reagent are,

Figure 9

Products formed from reagent $\left[5\right]$:

The given reagent is ${\text{SOCl}}_{\text{2}}$, pyridine.

Alkyl chlorides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ alcohols with thionyl chloride $\left({\text{SOCl}}_2\right)$, and pyridine. The reaction yields ${\text{SO}}_{\text{2}}$, and $\text{HCl}$ as by-products. The mechanism of the reaction involves two steps. The first step is conversion of the $\text{OH}$ group into a good leaving group. The second step is nucleophilic attack by ${\text{Cl}}^{-}$ through an ${\text{S}}_{\text{N}}2$ reaction.

Since the given alcohol, A is ${\text{2}}^{\text{ο}}$, the reaction follows ${\text{S}}_{\text{N}}2$ pathway and yields product with inversion of configuration at stereogenic center.

Thus, the products formed from the treatment of A with the given reagent are,

Figure 10

Products formed from reagent $\left[6\right]$:

The given reagent is $\text{TsCl}$, pyridine.

Alcohols are converted into alkyl tosylates by treatment with $p\text{-toluenesulfonyl}\text{chloride}\left(\text{TsCl}\right)$ in the presence of pyridine. The overall reaction converts ${}^{-}\text{OH}$ (a poor leaving group) into ${}^{-}\text{OTs}$ (a good leaving group). Since the reaction does not involve cleavage of the $\text{C}-\text{O}$ bond, the stereochemistry of the product is retained.

Thus, the products formed from the treatment of A with the given reagent are,

Figure 11

Conclusion

The products formed from the treatment of A with the reagent $\left[1\right]$ are shown in Figure 6. The products formed from the treatment of A with the reagent $\left[2\right]$ are shown in Figure 7. The products formed from the treatment of A with the reagent $\left[3\right]$ are shown in Figure 8. The products formed from the treatment of A with the reagent $\left[4\right]$ are shown in Figure 9. The products formed from the treatment of A with the reagent $\left[5\right]$ are shown in Figure 10. The products formed from the treatment of A with the reagent $\left[6\right]$ are shown in Figure 11.