Chapter 17, Problem 17.45AP

Interpretation Introduction

(a)

Interpretation:

The reaction in which $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ reacts with deuterated water is to be completed.

Concept introduction:

Grignard reagents are organometallic compounds which are prepared using alkyl halides in the presence of magnesium metal in dry ether. These reagents act as strong nucleophiles and bases.

Answer to Problem 17.45AP

The complete reaction between $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ and ${\text{D}}_{\text{2}}\text{O}$ is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 1

The complete reaction is shown below.

Figure 2

The hydrolysis of Grignard reagent is shown in Figure 2 in the presence of a solvent ${\text{D}}_{\text{2}}\text{O}$. Thus, $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ on treatment with deuterated water gives an alkene in which one hydrogen atom is replaced by deuterium.

Therefore, the product formed is shown in Figure 2.

Conclusion

The complete reaction between $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ and ${\text{D}}_{\text{2}}\text{O}$ is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The reaction between $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ and epoxide is to be completed.

Concept introduction:

Grignard reagents are organometallic compounds which are prepared using alkyl halides in the presence of magnesium metal in dry ether. These reagents act as strong nucleophiles and bases.

Answer to Problem 17.45AP

The complete reaction between $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ and epoxide is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 3

The complete reaction is shown below.

Figure 4

The Grignard reagent on treatment with epoxides in the presence of ${\text{H}}_3{\text{O}}^{+}$ gives primary alcohol as a product. The $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ on reaction with epoxide leads to the opening of epoxide ring. As a result, $\text{hepta-4, 6-dien-1-ol}$ is formed as a product.

Therefore, the product formed is shown in Figure 4.

Conclusion

The complete reaction between $\text{penta-2, 4-dienylmagnesium}\text{bromide}$ and epoxide is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The reaction between $\text{trans-1-bromo-4-methylpent-2-ene}$ and sodium ethanethiolate is to be completed.

Concept introduction:

Nucleophiles are electron-rich species. The nucleophilic substitution reactions are the reactions in which nucleophile attacks the electrophilic center and eliminates another group.

Answer to Problem 17.45AP

The complete reaction between $\text{trans-1-bromo-4-methylpent-2-ene}$ and sodium ethanethiolate is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 5

The complete reaction is shown below.

Figure 6

The above figure shows the reaction between ethyl sulfide ion and $\text{trans-1-bromo-4-methylpent-2-ene}$ gives thioether as a product. The species, ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{CH}}_2{\text{S}}^{-}$ acts as a nucleophile and attacks at the carbon adjacent to the double bond. As a result, $\text{Br}$ leaves as a leaving group. Therefore, the product is formed by the removal of $\text{NaBr}$ is shown in Figure 6.

Conclusion

The complete reaction between $\text{trans-1-bromo-4-methylpent-2-ene}$ and sodium ethanethiolate is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The reaction between $\text{trans-1-bromo-4-methylpent-2-ene}$ and ethanol is to be completed.

Concept introduction:

Nucleophiles are electron-rich species. The nucleophilic substitution reactions are the reactions in which nucleophile attacks the electrophilic center and eliminates another group.

Answer to Problem 17.45AP

The complete reaction between $\text{trans-1-bromo-4-methylpent-2-ene}$ and ethanol is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 7

The complete reaction is shown below.

Figure 8

Figure 8 shows the reaction between $\text{trans-1-bromo-4-methylpent-2-ene}$ and ethanol which gives $\text{trans-1-ethoxy-4-methylpent-2-ene}$ as a product by the removal of hydrogen bromide. The species, ${\text{CH}}_{\text{3}}{\text{CH}}_2{\text{O}}^{-}$ acts as a nucleophile and attacks at the carbon adjacent to the double bond. As a result, $\text{Br}$ leaves as a leaving group. Therefore, the product is formed by the removal of $\text{HBr}$ which is shown in Figure 8.

Conclusion

The complete reaction between $\text{trans-1-bromo-4-methylpent-2-ene}$ and ethanol is shown in Figure 8.

Interpretation Introduction

(e)

Interpretation:

The reaction of a given compound with NBS is to be completed.

Concept introduction:

Cyclic alkenes on reaction with N-Bromosuccinimide (NBS) forms allyl bromide, that is, bromine is substituted at the allylic position. NBS is a rich source of free radical of $\text{Br}$. The mechanism followed by NBS is a radical substitution reaction.

Answer to Problem 17.45AP

The complete reaction of a given compound with NBS is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 9

The complete reaction is shown below.

Figure 10

The NBS is used for allylic bromination that is it substitutes $\text{Br}$ at allylic positions. The compound A contains two allylic positions. Therefore, two different products are formed in the presence of $\text{AIBN}$ and ${\text{CCl}}_{\text{4}}$.

Therefore, the products formed are shown in Figure 10.

Conclusion

The complete reaction of given compound with NBS is shown in Figure 10.

Interpretation Introduction

(f)

Interpretation:

The reaction between $\text{p-cymene}$ and NBS is shown is to be completed.

Concept introduction:

Cyclic alkenes on reaction with N-Bromosuccinimide (NBS) forms allyl bromide, that is, bromine is substituted at the allylic position. NBS is a rich source of free radical of $\text{Br}$. The mechanism followed by NBS is a radical substitution reaction.

Answer to Problem 17.45AP

The complete reaction between $\text{p-cymene}$ and NBS is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 11

The complete reaction is shown below.

Figure 12

The molecular formula of the product is ${\text{C}}_{\text{10}}{\text{H}}_{\text{13}}\text{Br}$. This reaction is a free radical substitution reaction. The stable free radical is formed at benzylic position.

Due to the addition of NBS, benzylic bromination of $\text{p-cymene}$ takes places which leads to the formation of $\text{1-(2-bromopropan-2-yl)-4-methylbenzene}$.

Therefore, the product formed is shown in Figure 12.

Conclusion

The complete reaction between $\text{p-cymene}$ and NBS is shown in Figure 12.

Interpretation Introduction

(g)

Interpretation:

The complete reaction in which $\text{1, 4-dihydronaphthalene}$ reacts with ${\text{Br}}_{\text{2}}$ followed by treatment with $\text{KOH}$ and ethanol is to be completed.

Concept introduction:

The elimination reaction of alkyl halide, $\text{β}$-elimination, involves removal of the halogen atom and a hydrogen atom from the adjacent carbon atom, which leads to the formation of the alkene.

Answer to Problem 17.45AP

The complete reaction in which $\text{1, 4-dihydronaphthalene}$ reacts with ${\text{Br}}_{\text{2}}$ followed by treatment with $\text{KOH}$ and ethanol is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 13

The complete reaction is shown below.

Figure 14

The addition of ${\text{Br}}_{\text{2}}$ takes place at the double bond of $\text{1, 4-dihydronaphthalene}$. Furthermore, the product formed undergoes $\text{β}$-elimination reaction takes place in presence of $\text{KOH}$ and ethanol which leads to the formation of naphthalene as a product.

Therefore, the product formed is shown in Figure 14.

Conclusion

The complete reaction in which $\text{1, 4-dihydronaphthalene}$ reacts with ${\text{Br}}_{\text{2}}$ followed by treatment with $\text{KOH}$ and ethanol is shown in Figure 14.

Interpretation Introduction

(h)

Interpretation:

The reaction between $\text{buta-1, 3-dienylbenzene}$ and $\text{HBr}$ is to be completed.

Concept introduction:

When an alkene reacts with water in the acidic medium the reaction follows Markovnikov rule which states that the negative part of the reagent attacks the carbon with the least number of hydrogen atoms attached to it.

Answer to Problem 17.45AP

The complete reaction between $\text{buta-1, 3-dienylbenzene}$ and $\text{HBr}$ is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 15

The complete reaction is shown below.

Figure 16

The compound, $\text{buta-1, 3-dienylbenzene}$, on treatment with $\text{HBr}$ gives two products. The product which follows Markovnikov rule will be obtained as a major product. Therefore compound J is obtained as a major product as ${\text{Br}}^{-}$ gets attached to that carbon which contains less number of the hydrogen atom.

Therefore, the products formed are shown in Figure 16.

Conclusion

The complete reaction between $\text{buta-1, 3-dienylbenzene}$ and $\text{HBr}$ is shown in Figure 16.

Interpretation Introduction

(i)

Interpretation:

The reaction between $\text{5-methyl-1, 2, 3, 4-tetrahydronaphthalene}$ and potassium permanganate is to be completed.

Concept introduction:

The substance which gets easily reduced is termed as a strong oxidizing agent. It is also defined as the substances which oxidize other substances by accepting their electrons. Examples of strong oxidizing agents are potassium permanganate, hydrogen peroxide and many more.

Answer to Problem 17.45AP

The complete reaction between $\text{5-methyl-1, 2, 3, 4-tetrahydronaphthalene}$ and potassium permanganate is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 17

The complete reaction is shown below.

Figure 18

The compound, ${\text{KMnO}}_{\text{4}}$ is a strong oxidizing agent. Therefore, it will convert the alkyl side chain of $\text{5-methyl-1, 2, 3, 4-tetrahydronaphthalene}$ into carboxylic acid group. As a result, $\text{cyclohexane-1, 2, 3-tricarboxylic acid}$ is obtained as a product.

Therefore, the product formed is shown in Figure 18.

Conclusion

The complete reaction between $\text{5-methyl-1, 2, 3, 4-tetrahydronaphthalene}$ and potassium permanganate is shown in Figure 18.