Chapter 5, Problem 5.28AP

Interpretation Introduction

(a)

Interpretation:

Principal organic product obtained on reaction of $\text{1-ethylcyclopentene}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ solvent is to be stated.

Concept introduction:

In an electrophilic addition reaction, there is an addition of an electrophilic reagent on a nucleophilic group. The double bond and triple bond containing compounds act as a nucleophile. The addition of electrophilic reagent on nucleophilic compound leads to the dissociation of $\text{π}$ bond and formation of two $\text{σ}$ bonds.

Answer to Problem 5.28AP

The product formed on reaction of $\text{1-ethylcyclopentene}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ solvent is shown below.

Explanation of Solution

In the reaction of $\text{1-ethylcyclopentene}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ solvent, $\text{1-ethylcyclopentene}$ acts as a nucleophile while bromine acts as an electrophile. This is an electrophilic addition reaction. In this reaction, $\text{1-ethylcyclopentene}$ donates its $\text{π}$ electrons in the antibonding orbital of bromine and this leads to the generation of brominiumcation and bromide anion. The brominiumcation attacks alkene, $\text{1-ethylcyclopentene}$ from below the plane. The bromide anion now attacks the less substituted carbon atom as it has less charge density from above the plane.There will be less steric hindrance above the plane. Therefore, bromineshows electrophilic addition reaction on the double bond of alkene. The product formed on reaction of $\text{1-ethylcyclopentene}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ solvent is shown below.

Figure 1

The product formed by reaction of $\text{1-ethylcyclopentene}$ with ${\text{Br}}_{\text{2}}$ in ${\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}$ solvent is $\text{1, 2-dibromo-1-ethylcyclopentane}$.

Conclusion

The reaction between $\text{1-ethylcyclopentene}$ and bromine gives $\text{1, 2-dibromo-1-ethylcyclopentane}$.

Interpretation Introduction

(b)

Interpretation:

Principal organic product obtained on reaction of $\text{1-ethylcyclopentene}$ with ${\text{O}}_{\text{3}}$ at $\text{-78°C}$ is to be stated.

Concept introduction:

In the reactionof alkene with ozone, the oxidation of alkene takes place which leads to formation of molozonide product. This reaction takes place through $1,3-$ cycloaddition. In this, the ozonide formed is quite unstable in nature.

Answer to Problem 5.28AP

The product formed by the reaction of $\text{1-ethylcyclopentene}$ with ${\text{O}}_{\text{3}}$ at $\text{-78}\text{°C}$ is shown below.

Explanation of Solution

In the reaction, $\text{1-ethylcyclopentene}$ reacts with ozone to form molozonide. In this $1,3-$ cycloaddition takes place.

The general view of $1,3-$ cycloaddition is given below.

Figure 2

In the same manner, $\text{1-ethylcyclopentene}$ gets oxidized with ozone to form molozonide as shown below.

Figure 3

Conclusion

The product formed by the reaction between $\text{1-ethylcyclopentene}$ and ${\text{O}}_{\text{3}}$ at $\text{-78°C}$ is shown in Figure 3.

Interpretation Introduction

(c)

Interpretation:

Principal organic product obtained on the reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{S}$ on the product obtained by the reaction of $\text{1-ethylcyclopentene}$ with ${\text{O}}_{\text{3}}$ at $\text{-78°C}$ is to be stated.

Concept introduction:

In the reactionof alkene with ozone, the oxidation of alkene takes place which leads to form molozonide product. This reaction takes place through $1,3-$ cycloaddition. In this, the ozonide formed is quite unstable in nature.

Answer to Problem 5.28AP

The product formed by the reaction of $\text{1-ethylcyclopentene}$ and ozone followed by ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{S}$ is $\text{5-oxo-heptanal}$.

Explanation of Solution

The compound $\text{1-ethylcyclopentene}$ gets oxidized by $1,3$ dipolar cycloaddition with ozone to form molozonide. The reaction is given below.

Figure 3

The molozonide formed is unstable in nature, therefore, it gets easily reduced by ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{S}$ to form carbonyl group containing compound as shown below.

Figure 4

Therefore, the product formed by the reaction between $\text{1-ethylcyclopentene}$ and ozone followed by reaction of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{S}$ is $\text{5-oxo-heptanal}$.

Conclusion

The reaction between $\text{1-ethylcyclopentene}$ and ozone followed by ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{S}$ gives $\text{5-oxo-heptanal}$.

Interpretation Introduction

(d)

Interpretation:

Principal organic product obtained on the reaction of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ on the product obtained by the reaction of $\text{1-ethylcyclopentene}$ with ${\text{O}}_{\text{3}}$ at $\text{-78}\text{°C}$ is to be stated.

Concept introduction:

In the reactionof alkene with ozone, the oxidation of alkene takes place which leads to form molozonide product. This reaction takes place through $1,3-$ cycloaddition. In this the ozonide formed is quite unstable in nature.

Answer to Problem 5.28AP

The product formed by the reaction of $\text{1-ethylcyclopentene}$ and ozone followed by ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\text{5-oxo-heptanoic}\text{acid}$.

Explanation of Solution

The compound $\text{1-ethylcyclopentene}$ gets oxidized by $\text{1, 3}$ dipolar cycloaddition with ozone to form molozonide. The reaction is given below.

Figure 3

The molozonide formed is unstable in nature, therefore, it gets easily oxidized by ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to form carbonyl group. It converts aldehyde group to carboxylic acid group.

Figure 5

Therefore, the product formed by the reaction between $\text{1-ethylcyclopentene}$ and ozone followed by ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\text{5-oxo-heptanoic}\text{acid}$.

Conclusion

The reaction between $\text{1-ethylcyclopentene}$ and ozone followed by ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ gives $\text{5-oxo-heptanoic}\text{acid}$.

Interpretation Introduction

(e)

Interpretation:

Principal organic product on reaction of $\text{1-ethylcyclopentene}$ with ${\text{O}}_2$ flame is to be stated.

Concept introduction:

The reaction between unsaturated alkene with ${\text{O}}_2$ flame is a combustion reaction. It is highly exothermic reaction which produces large amount of heat and the reaction continues till the fuel is used up completely.

Answer to Problem 5.28AP

The product formed by the reaction of $\text{1-ethylcyclopentene}$ with ${\text{O}}_2$ flame is carbon dioxide and water.

Explanation of Solution

The reaction between $\text{1-ethylcyclopentene}$ and oxygen leads to the formation of carbon dioxide and water. This reaction produces large amount of heat, therefore, it is an exothermic reaction.

Figure 6

Therefore, the reaction between $\text{1-ethylcyclopentene}$ with ${\text{O}}_2$ flame leads to formation of carbon dioxide and water.

Conclusion

The reaction between $\text{1-ethylcyclopentene}$ with ${\text{O}}_2$ flame leads to formation of carbon dioxide and water.

Interpretation Introduction

(f)

Interpretation:

Principal organic product on reaction of $\text{1-ethylcyclopentene}$ with $\text{HBr}$ is to be stated.

Concept introduction:

In markovnikov’s addition of $\text{HBr}$ on alkene, the more substituted carbon gives the most stable carbocation. This carbocation is attacked by the nucleophile ${\text{Br}}^{-}$ anion.

Answer to Problem 5.28AP

The product formed by the reaction of $\text{1-ethylcyclopentene}$ with $\text{HBr}$ is $\text{1-bromo-1-ethylcyclopentane}$

Explanation of Solution

In the reaction of $\text{1-ethylcyclopentene}$ with $\text{HBr}$, $\text{1-ethylcyclopentene}$ acts as a nucleophile while $\text{HBr}$ acts as an electrophile. This is an electrophilic addition reaction. In this reaction, $\text{1-ethylcyclopentene}$ donates its $\text{π}$ electrons in the antibonding orbital of $\text{HBr}$ and this leads to the dissociation of bond between hydrogen and bromine atoms. There is formation of planar carbocation. In this unsymmetrical alkene, the carbocation formed is more substituted. The Markovnikov’s addition of $\text{HBr}$ takes place which leads to the formation of $\text{1-bromo-1-ethylcyclopentane}$.

Figure 7

Therefore, the product formed by the reaction between $\text{1-ethylcyclopentene}$ and $\text{HBr}$ is $\text{1-bromo-1-ethylcyclopentane}$.

Conclusion

Thereaction between $\text{1-ethylcyclopentene}$ and $\text{HBr}$ gives $\text{1-bromo-1-ethylcyclopentane}$.

Interpretation Introduction

(g)

Interpretation:

Principal organic product on reaction of $\text{1-ethylcyclopentene}$ with ${\text{I}}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

.In an electrophilic addition reaction, there is an addition of an electrophilic reagent on a nucleophilic group. The double bond and triple bond containing compounds act as nucleophile. The addition of electrophilic reagent on nucleophilic compound leads to the dissociation of $\text{π}$ bond and formation of two $\text{σ}$ bonds

Answer to Problem 5.28AP

The product formed by the reaction of $\text{1-ethylcyclopentene}$ with iodine in presence of ${\text{H}}_{\text{2}}\text{O}$ is $\text{2-iodo-1-ethylcyclopentanol}$.

Explanation of Solution

In the reaction first reaction between $\text{1-ethylcyclopentene}$ and iodine takes place. In this reaction, $\text{1-ethylcyclopentene}$ acts as a nucleophile while iodine acts as an electrophile. This is an electrophilic addition reaction. In this reaction, $\text{1-ethylcyclopentene}$ donates its $\Pi$ electrons in the antibonding orbital of iodine and this leads to the dissociation of bond between two iodine atoms. There is formation of carbocation in which the more substituted carbon atom carries the positive charge. Then the attack of water takes place on the carbocation present. The product formed by the reaction of $\text{1-ethylcyclopentene}$ with iodine in presence of ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 8

Therefore, the reaction between $\text{1-ethylcyclopentene}$ and iodine in presence of ${\text{H}}_{\text{2}}\text{O}$ gives $\text{2-iodo-1-ethylcyclopentanol}$.

Conclusion

The reaction between $\text{1-ethylcyclopentene}$ and iodine in presence of ${\text{H}}_{\text{2}}\text{O}$ gives $\text{2-iodo-1-ethylcyclopentanol}$.

Interpretation Introduction

(h)

Interpretation:

Principal organic product on reaction of $\text{1-ethylcyclopentene}$ with ${\text{H}}_{\text{2}}$ in presence of $\text{Pt}/\text{C}$ is to be stated.

Concept introduction:

In acatalytic hydrogenation reaction, palladium is used as a catalyst which accelerates the rate of reaction. This method is used to convert alkene into alkanes by the addition of hydrogen atom.

Answer to Problem 5.28AP

The product formed by the reaction of $\text{1-ethylcyclopentene}$ with ${\text{H}}_{\text{2}}$ in presence of $\text{Pt}/\text{C}$ is $\text{ethylcyclopentane}$.

Explanation of Solution

In this reaction, palladium is used as a catalyst which helps in accelerating the rate of hydrogenation reaction. This reaction leads to conversion of $\text{ethylcyclopentene}$ to $\text{ethylcyclopentane}$.

The product formed by the reaction between $\text{1-ethylcyclopentene}$ with ${\text{H}}_{\text{2}}$ in presence of $\text{Pt}/\text{C}$ is shown below.

Figure 9

Conclusion

Theproduct formed by the reaction between $\text{1-ethylcyclopentene}$ with ${\text{H}}_{\text{2}}$ in presence of $\text{Pt}/\text{C}$ is $\text{ethylcyclopentane}$.

Interpretation Introduction

(i)

Interpretation:

Principal organic product on reaction of $\text{1-ethylcyclopentene}$ with $\text{HBr}$ in presence of peroxides is to be stated.

Concept introduction:

In Anti-Markovnikov’s addition of $\text{HBr}$ on alkene, the less substituted carbon is the most stable and that carbocation is formed. This formed carbocation is attacked by the nucleophile ${\text{Br}}^{-}$ anion.

Answer to Problem 5.28AP

The product formed on reaction of $\text{1-ethylcyclopentene}$ with $\text{HBr}$ in presence of peroxides is $\text{1-bromo-2-ethylcyclopentane}$.

Explanation of Solution

In thereaction, in presence of peroxides, $\text{HBr}$ shows Anti-Markovnikov’s addition on $\text{1-ethylcyclopentene}$. This leads to attack of bromine on more substituted carbon atom and attack of hydrogen on less substituted carbon atom.

The product formed by the reaction between $\text{1-ethylcyclopentene}$ with $\text{HBr}$ in presence of peroxides is shown below.

Figure 10

Conclusion

The product formed by the reaction between $\text{1-ethylcyclopentene}$ with $\text{HBr}$ in presence of peroxidesis $\text{1-bromo-2-ethylcyclopentane}$.

Interpretation Introduction

(j)

Interpretation:

Principal organic product on reaction of with ${\text{BH}}_{\text{3}}$ in tetrahydrofuran(THF) is to be stated.

Concept introduction:

In hydroboration reaction, antimarkovnikov’s addition of ${\text{BH}}_{\text{3}}$ takes place onalkene. In ${\text{BH}}_{\text{3}}$, hydrogen atom carries negative charge while ${\text{BH}}_{\text{2}}$ carries positive charge as it is Lewis acid. This isrepresented as $\stackrel{-\delta }{\text{H}}-\stackrel{+\delta }{{\text{BH}}_{\text{2}}}$. Hydrogen attacks on the more substituted carbocation and ${\text{BH}}_{\text{2}}$ attacks on less substituted carbon atom in hydroboration reaction.

Answer to Problem 5.28AP

The product formed by thereaction of $\text{1-ethylcyclopentene}$ with ${\text{BH}}_{\text{3}}$ in tetrahydrofuran(THF) is shown below.

Explanation of Solution

The reaction between $\text{1-ethylcyclopentene}$ with ${\text{BH}}_{\text{3}}$ intetrahydrofuran (THF) is an Anti-Markovnikov’s addition of ${\text{BH}}_{\text{3}}$ on $\text{1-ethylcyclopentene}$. It is a hydroboration reaction. In ${\text{BH}}_{\text{3}}$, hydrogen atom carries negative charge while ${\text{BH}}_{\text{2}}$ carries positive charge as it is lewis acid.

The product formed by thereaction between $\text{1-ethylcyclopentene}$ with ${\text{BH}}_{\text{3}}$ intetrahydrofuran(THF) is shown below.

Figure 11

Conclusion

The product formed by the reaction between $\text{1-ethylcyclopentene}$ and ${\text{BH}}_{\text{3}}$ in tetrahydrofuran (THF) is shown in Figure 1.

Interpretation Introduction

(k)

Interpretation:

Principal organic product formed by the reaction of product of $\text{1-ethylcyclopentene}$ with ${\text{BH}}_{\text{3}}$ in tetrahydrofuran(THF) with $\text{NaOH}$ in presence of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is to be stated.

Concept introduction:

In hydroboration reaction, antimarkovnikov’s addition of ${\text{BH}}_{\text{3}}$ takes place on alkene. In ${\text{BH}}_{\text{3}}$, hydrogen atom carries negative charge while ${\text{BH}}_{\text{2}}$ carries positive charge as it is Lewis acid. This is represented as $\stackrel{-\delta }{\text{H}}-\stackrel{+\delta }{{\text{BH}}_{\text{2}}}$. Hydrogen attacks on the more substituted carbocation and ${\text{BH}}_{\text{2}}$ attacks on less substituted carbon atom in hydroboration reaction.

The compound ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is responsible for oxidation reaction of boron hydride group.This complete reaction leads to antimarkovnikov’s addition of water on alkene.

Answer to Problem 5.28AP

The product formed on reaction of boron substitued $\text{1-ethylcyclopentene}$ with $\text{NaOH}$ in presence of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\text{2-ethylcyclopentanol}$.

Explanation of Solution

In the reaction,.the hydroxide group of $\text{NaOH}$ in presence of hydrogen peroxide replace the ${\text{-BH}}_{\text{2}}$ group on $\text{1-ethylcyclopentane}$ with the hydroxyl group. This is an oxidation reaction. This complete reaction leads to antimarkovnikov’s addition of water on alkene.

The product formed by the reaction between boron substitued $\text{1-ethylcyclopentene}$ with $\text{NaOH}$ in presence of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is shown below.

Figure 12

Conclusion

The product formed by thereaction between boron substitued $\text{1-ethylcyclopentene}$ with $\text{NaOH}$ in presence of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ gives $\text{2-ethylcyclopentanol}$.

Interpretation Introduction

(l)

Interpretation:

Principal organic product on reaction of $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

Inan oxymercuration reaction, $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ acts as an electrophile and accepts electron from pi-electrons of the alkene. It leads to the dissociation of one acetyl group and forms a cyclic carbocation in which water attacks on more substituted carbon atom.

Answer to Problem 5.28AP

The product formed on reaction of $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Explanation of Solution

This is an oxymercuration reaction in which $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ acts as an electrophile and accepts electron from pi- electrons of the alkene. It leads to dissociation of one acetyl group.forms a cyclic carbocation in which water attacks on more substituted carbon atom.

The product formed by the reaction between $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 13

Conclusion

The product formed by thereaction between $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ is shown in Figure 1.

Interpretation Introduction

(m)

Interpretation:

Principal organic product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ with ${\text{NaBH}}_{\text{4}}/\text{NaOH}$ is to be stated.

Concept introduction:

In an oxymercuration reaction, $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ acts as an electrophile and accepts electron from pi- electrons of the alkene. It leads to dissociation of one acetyl group. Forms a cyclic carbocation in which water attacks on more substituted carbon atom.

In demercuration reaction, ${\text{NaBH}}_{\text{4}}/\text{NaOH}$ helps in removal of mercuric acetate. It replaces mercuric acetate with hydrogen.

Answer to Problem 5.28AP

The Principal organic product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ with ${\text{NaBH}}_{\text{4}}/\text{NaOH}$ is $\text{1-ethylcyclopentanol}$.

Explanation of Solution

The product formed by the reaction between $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ is shown in Figure 13.

In demercuration reaction, ${\text{NaBH}}_{\text{4}}/\text{NaOH}$ helps in removal of mercuric acetate. It replaces mercuric acetate with hydrogen. The reaction of product formed in Figure 13 with ${\text{NaBH}}_{\text{4}}/\text{NaOH}$ is shown below.

Figure 14

The product formed on reaction is $\text{1-ethylcyclopentanol}$.

Conclusion

The product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{Hg}{\left(\text{OAc}\right)}_{\text{2}}$ in presence of ${\text{H}}_{\text{2}}\text{O}$ with ${\text{NaBH}}_{\text{4}}/\text{NaOH}$ is $\text{1-ethylcyclopentanol}$.

Interpretation Introduction

(n)

Interpretation:

Principal organic product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{HI}$ is to be stated.

Concept introduction:

In an electrophilic addition reaction, there is an addition of an electrophilic reagent on a nucleophilic group. The double bond and triple bond containing compounds act as a nucleophile. The addition of electrophilic reagent on nucleophilic compound leads to the dissociation of $\text{π}$ electrons and formation of two $\text{σ}$ bonds.

Answer to Problem 5.28AP

The product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{HI}$ is $\text{1-ethyl-1-iodocyclopentane}$.

Explanation of Solution

The reaction between $\text{1-ethylcyclopentene}$ and $\text{HI}$ is electrophilic addition reaction. This is Markovnikov’s addition reaction in which the more substituted carbocation is formed.It is then attacked by an anion. The product formed on the reaction between $\text{1-ethylcyclopentene}$ and $\text{HI}$ is shown below.

Figure 15

Conclusion

Principal organic product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{HI}$ is $\text{1-ethyl-1-iodocyclopentane}$.

Interpretation Introduction

(o)

Interpretation:

Principal organic product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{HI}$ in the presence of $\text{AIBN}$ is to be stated.

Concept introduction:

The reaction between an alkene, $\text{HI}$ in the presence of $\text{AIBN}$ catalyst proceeds through the free- radical mechanism. $\text{AIBN}$, acts as an initiator for free radical mechanism. The free radical formed is unstable in nature, it makes the reaction faster.

Answer to Problem 5.28AP

The Principal organic product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{HI}$ in presence of $\text{AIBN}$ is $\text{1-ethyl-1-iodocyclopentane}$.

Explanation of Solution

The reaction between $\text{1-ethylcyclopentene}$ and in presence of $\text{AIBN}$ is a free-radical mechanism. In which $\text{AIBN}$ acts as an initiator for the generation of free- radical. The product formed on the reaction between $\text{1-ethylcyclopentene}$ and $\text{HI}$ in presence of $\text{AIBN}$ is shown below.

Figure 15

Conclusion

The product formed by the treatment of product of reaction of $\text{1-ethylcyclopentene}$ with $\text{HI}$ in presence of $\text{AIBN}$ is $\text{1-ethyl-1-iodocyclopentane}$.