Chapter 12, Problem 12.38P

Draw the organic products formed when cyclopentene is treated with each reagent. With some

reagents, no reaction occurs.

a. ${\text{H}}_{\text{2}}+\text{Pd-C}$

b. ${\text{H}}_{\text{2}}+$ Lindlar catalyst.

c. $\text{Na},{\text{NH}}_{\text{3}}$.

d. ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$

e. $\left[1\right]$ ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$; $\left[2\right]$ ${\text{H}}_{\text{2}}\text{O},{\text{HO}}^{-}$

f. $\left[1\right]$ ${\text{OsO}}_{\text{4}}+\text{NMO}$ $\left[2\right]$ ${\text{NaHSO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$

g. ${\text{KMnO}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$

h. $\left[1\right]{\text{LiAlH}}_{\text{4}}\text{;}\left[2\right]{\text{H}}_{\text{2}}\text{O}$.

i. $\left[1\right]{\text{O}}_3\text{;}\left[2\right]{\text{CH}}_{\text{3}}{\text{SCH}}_{\text{3}}$

j. ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$, $(-)\text{-DET}$

k. $\text{mCPBA}$

l. Product in (k); $\left[1\right]$ ${\text{LiAlH}}_{\text{4}}$; $\left[2\right]$ ${\text{H}}_{\text{2}}\text{O}$

Interpretation Introduction

(a)

Interpretation: The organic product formed when cyclopentene is treated with ${\text{H}}_{\text{2}}+\text{Pd-C}$ is to be predicted.

Concept introduction: The addition of ${\text{H}}_{\text{2}}$ in presence of catalyst is known as catalytic hydrogenation reaction. This is a syn addition. Alkyne is reduced to alkane. In this reaction two weak pi bond of alkyne and sigma bond between ${\text{H}}_{\text{2}}$ is broken and four $\text{C}-\text{H}$ bonds are formed.

Answer to Problem 12.38P

The organic product formed when cyclopentene ${\text{H}}_{\text{2}}+\text{Pd-C}$ is shown in Figure 1.

Explanation of Solution

In the given reaction, when cyclopentene is treated with ${\text{H}}_{\text{2}}+\text{Pd-C}$, the $\text{C}-\text{C}$ double bond is reduced to $\text{C}-\text{C}$ single bond. The chemical reaction is as follows,

Figure 1

Conclusion

The organic product formed when cyclopentene ${\text{H}}_{\text{2}}+\text{Pd-C}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The organic product formed when cyclopentene is treated with Lindlar catalyst is to be predicted.

Concept introduction: The addition of ${\text{H}}_{\text{2}}$ in presence of a catalyst is known as catalytic hydrogenation reaction. Lindlar catalyst consists of palladium deposited on calcium carbonate which is poisoned or deactivated by lead or sulfur. In presence of Lindlar catalyst, alkyne is reduced to $cis\text{-alkene}$.

Answer to Problem 12.38P

Cyclopentene do not react with Lindlar catalyst.

Explanation of Solution

Alkenes do not react with Lindlar catalyst. Lindlar catalyst is used to reduce alkynes to $cis\text{-alkene}$. Hence, cyclopentene do not react with Lindlar catalyst.

Conclusion

Cyclopentene do not react with Lindlar catalyst.

Interpretation Introduction

(c)

Interpretation: The product formed when the cyclopentene is treated with $\text{Na},{\text{NH}}_{\text{3}}$ is to be predicted.

Concept introduction: In presence of sodium metal in ammonia, the is alkyne is reduced to $trans\text{-alkene}$. The addition of ${\text{H}}_{\text{2}}$ takes place in anti fashion to the triple bond. The mechanism involves the addition of electron followed by protonation.

Answer to Problem 12.38P

Cyclopentene do not react with $\text{Na},{\text{NH}}_{\text{3}}$.

Explanation of Solution

Alkenes do not react with $\text{Na},{\text{NH}}_{\text{3}}$. Sodium metal in ammonia is used to reduce alkynes to $trans\text{-alkene}$. Hence, cyclopentene do not react with $\text{Na},{\text{NH}}_{\text{3}}$.

Conclusion

Cyclopentene do not react with $\text{Na},{\text{NH}}_{\text{3}}$.

Interpretation Introduction

(d)

Interpretation: The product formed when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ is to be predicted.

Concept introduction: In presence of peroxide, alkene is oxidized to epoxide. This is known as epoxidation. This is a syn addition. The weak pi bond of alkene and weak $\text{O}-\text{O}$ bond of peroxide are broken and two new $\text{C}-\text{O}$ bonds are formed. Breaking of bonds and formation of bonds takes place in single step. In the opening of epoxide ring, the nucleophile attacks from the backside of $\text{C}-\text{O}$ bond.

Answer to Problem 12.38P

The product formed when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ is shown in Figure 2.

Explanation of Solution

In the given reaction, when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$, it is oxidized to epoxide. The reaction is as follows,

Figure 2

Conclusion

The product formed when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ is shown in Figure 2.

Interpretation Introduction

(e)

Interpretation: The product formed when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ followed by ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$ is to be predicted.

Concept introduction: Anti-dihroxylation takes place in two steps epoxidation followed by hydrolysis. In presence of peroxide, alkene is oxidized to epoxide. This is known as epoxidation. This is a syn addition. The weak pi bond of alkene and weak $\text{O}-\text{O}$ bond of peroxide are broken and two new $\text{C}-\text{O}$ bonds are formed. Breaking of bonds and formation of bonds takes place in single step. In the opening of epoxide ring, the nucleophile attacks from the backside of $\text{C}-\text{O}$ bond.

Answer to Problem 12.38P

The product formed when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ followed by ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$ is shown in Figure 3.

Explanation of Solution

In the given reaction, when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$, it is oxidized to epoxide. Epoxidation is followed by ring opening with ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$. The epoxide is formed above the plane. So, the attack of nucleophile occurs from below the plane. The reaction is as follows,

Figure 3

Conclusion

The product formed when cyclopentene is treated with ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{3}}\text{H}$ followed by ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$ is shown in Figure 3.

Interpretation Introduction

(f)

Interpretation: The product formed when cyclopentene is treated with ${\text{OsO}}_{\text{4}}+\text{NMO}$ followed by ${\text{NaHSO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Addition of two hydroxyl groups on double bond to form $1,2\text{-diol}$ is known as dihydroxylation. In presence of ${\text{OsO}}_{\text{4}}$, syn dihydroxylation takes place. In this reaction, two hydroxyl groups are added on the same side of the double bond. ${\text{OsO}}_{\text{4}}$ is a toxic and expensive reagent. So, to overcome its limitation it can be used in catalytic amount along with $\text{N-methylmorpholine}N\text{-oxide}\text{(NMO)}$.

Answer to Problem 12.38P

The product formed when cyclopentene is treated with ${\text{OsO}}_{\text{4}}+\text{NMO}$ followed by ${\text{NaHSO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 4.

Explanation of Solution

In the given reaction, when cyclopentene is treated with ${\text{OsO}}_{\text{4}}+\text{NMO}$ followed by hydrolysis with ${\text{NaHSO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$, the product formed is $1,2\text{-diol}$. The reaction is as follows,

Figure 4

Conclusion

The product formed when cyclopentene is treated with ${\text{OsO}}_{\text{4}}+\text{NMO}$ followed by ${\text{NaHSO}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 4.

Interpretation Introduction

(g)

Interpretation: The product formed when cyclopentene is treated with ${\text{KMnO}}_{\text{4}}$ followed by ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$ is to be predicted.

Concept introduction: Addition of two hydroxyl groups on double bond to form $1,2\text{-diol}$ is known as dihydroxylation. In presence of ${\text{KMnO}}_{\text{4}}$, syn dihydroxylation takes place. In this reaction two hydroxyl groups are added on the same side of the double bond.

Answer to Problem 12.38P

The product formed when cyclopentene is treated with ${\text{KMnO}}_{\text{4}}$ followed by ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$ is shown in Figure 5.

Explanation of Solution

In the given reaction, when cyclopentene is treated with ${\text{KMnO}}_{\text{4}}$ followed by hydrolysis with ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$, the product formed is $1,2\text{-diol}$. The reaction is as follows,

Figure 5

Conclusion

The product formed when cyclopentene is treated with ${\text{KMnO}}_{\text{4}}$ followed by ${\text{H}}_{\text{2}}\text{O},{\text{OH}}^{-}$ is shown in Figure 5.

Interpretation Introduction

(h)

Interpretation: The product formed when cyclopentene is treated with ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: The metal hydride reagents are good reducing agents such as ${\text{LiAlH}}_{\text{4}}$. They can be used to reduce carbonyl compounds, conjugated double or triple bonds and $\text{C}-\text{X}$ bond in alkyl halides and epoxides.

Answer to Problem 12.38P

Cyclopentene gives no reaction with ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The ${\text{LiAlH}}_{\text{4}}$ does not reduce alkenes. When cyclopentene is treated with ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$, no reaction takes place.

Conclusion

Cyclopentene gives no reaction with ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(i)

Interpretation: The product formed when cyclopentene is treated with ${\text{O}}_{\text{3}}$ followed by ${\text{CH}}_{\text{3}}{\text{SCH}}_{\text{3}}$ is to be predicted.

Concept introduction: In presence of ozone and ${\text{CH}}_{\text{3}}{\text{SCH}}_{\text{3}}$, oxidative cleavage occurs. In oxidative cleavage both the sigma and pi bond of the alkene is broken and two carbonyl compounds are formed. Depending upon alkyl substituents aldehyde or ketone is formed. When ozone is added to the alkene, unstable intermediate molozonide is formed which rearranges to ozonide. This ozonide is reduced to form two carbonyl compounds in presence of ${\text{CH}}_{\text{3}}{\text{SCH}}_{\text{3}}$.

Answer to Problem 12.38P

The product formed when cyclopentene is treated with ${\text{O}}_{\text{3}}$ followed by ${\text{CH}}_{\text{3}}{\text{SCH}}_{\text{3}}$ is shown in Figure 6.

Explanation of Solution

In the given reaction, when cyclopentene is treated with ${\text{O}}_{\text{3}}$ followed by hydrolysis with ${\text{CH}}_{\text{3}}{\text{SCH}}_{\text{3}}$, the product formed is glutaraldehyde or pentanedial. The reaction is as follows,

Figure 6

Conclusion

The product formed when cyclopentene is treated with ${\text{O}}_{\text{3}}$ followed by ${\text{CH}}_{\text{3}}{\text{SCH}}_{\text{3}}$ is shown in Figure 6.

Interpretation Introduction

(j)

Interpretation: The product formed when cyclopentene is treated with ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$, $(-)\text{-DET}$ is to be predicted.

Concept introduction: Sharpless epoxidation involves the oxidation of double bond between carbon atoms to epoxide. This oxidation occurs only in allylic alcohol. This is an enantioselective oxidation, which means predominantly one enantiomer is formed. Sharpless reagents are $tert\text{-butyl}\text{hydroperoxide,}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$; titanium $\left(\text{IV}\right)$ isopropoxide, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$; and diethyl tartrate.

Answer to Problem 12.38P

The product formed when cyclopenteneis treated with ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$, $(-)\text{-DET}$ is shown in Figure 7.

Explanation of Solution

There are two different chiral diethyl tartrate isomers, $\left(\text{+}\right)\text{-DET}$ and $(-)\text{-DET}$ to indicate the direction in which they rotate the plane polarized light. The identity of isomer determines which enantiomer is the major product.

When epoxidation is done using $(-)\text{-DET}$, epoxide is formed from above the plane as shown below.

Figure 7

Conclusion

The product formed when cyclopenteneis treated with ${\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}\text{COOH}$, $\text{Ti}{\left[\text{OCH}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\right]}_{\text{4}}$, $(-)\text{-DET}$ is shown in Figure 7.

Interpretation Introduction

(k)

Interpretation: The product formed when cyclopentene is treated with $\text{mCPBA}$ is to be predicted.

Concept introduction: In presence of peroxide alkene is oxidized to epoxide this is known as epoxidation. This is a syn addition. The weak pi bond of alkene and weak $\text{O}-\text{O}$ bond of peroxide are broken and two new $\text{C}-\text{O}$ bonds are formed. Breaking of bonds and forming of bonds takes place in single step. $\text{mCPBA}$ is $\text{meta-chloroperoxybenzoic}\text{acid}$.

Answer to Problem 12.38P

The product formed when cyclopentene is treated with $\text{mCPBA}$ is shown in Figure 8.

Explanation of Solution

In the given reaction, when cyclopentene is treated with $\text{mCPBA}$ ($\text{meta-chloroperoxybenzoic}\text{acid}$), alkene is oxidized to epoxide. The reaction is as follows,

Figure 8

Conclusion

The product formed when cyclopentene is treated with $\text{mCPBA}$ is shown in Figure 8.

Interpretation Introduction

(l)

Interpretation: The product formed when cyclopentene is treated with $\text{mCPBA}$ followed by ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: In presence of peroxide, alkene is oxidized to epoxide this is known as epoxidation. This is a syn addition. The weak pi bond of alkene and weak $\text{O}-\text{O}$ bond of peroxide are broken and two new $\text{C}-\text{O}$ bonds are formed. Breaking of bonds and forming of bonds takes place in single step. $\text{mCPBA}$ is $\text{meta-chloroperoxybenzoic}\text{acid}$. The $\text{C}-\text{X}$ bond in epoxides can be reduced using metal hydride reagents. The epoxide rings open up to form alcohol.

Answer to Problem 12.38P

The product formed when cyclopentene is treated with $\text{mCPBA}$ followed by ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 9.

Explanation of Solution

In the given reaction, when cyclopentene is treated with $\text{mCPBA}$ ($\text{meta-chloroperoxybenzoic}\text{acid}$), alkene is oxidized to epoxide. The epoxide on further reaction with ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ opens up to form an alcohol. The reaction is as follows,

Figure 9

Conclusion

The product formed when cyclopentene is treated with $\text{mCPBA}$ followed by ${\text{LiAlH}}_{\text{4}},{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 9.