Chapter 20, Problem 20.38P

Draw the products formed when pentanal $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CHO}\right)$ is treated with each reagent. With some reagents, no reaction occurs.

a. ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$

b. ${\text{LiAlH}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$

c. ${\text{H}}_2\left(1\text{equiv}\right)$, $\text{Pd}-\text{C}$

d. $\text{PCC}$

e. ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$

f. ${\text{Ag}}_{\text{2}}\text{O},{\text{NH}}_{\text{4}}\text{OH}$

g. $\left[\text{1}\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$

h. $\left[\text{1}\right]{\text{C}}_6{\text{H}}_5\text{Li}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$

i. $\left[\text{1}\right]\left({\text{CH}}_3\right)\text{CuLi}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$

j. $\text{HC}\equiv \text{CNa}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$

k. ${\text{H}}_{\text{3}}\text{CC}\equiv \text{CLi}$, ${\text{H}}_{\text{2}}\text{O}$

l. The [product in (a), then TBDMS-Cl, imidazole.

Interpretation Introduction

(a)

Interpretation: The product formed by the reaction of pentanal with ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$ is to be predicted.

Concept introduction: The substitution reaction involves the replacement of one functional group by other functional group. In nucleophilic substitution an electron rich species attack the species that is deficient in electrons. The electrophile and the leaving group together form a substrate. The nucleophile attacks over the substrate and there occurs the removal of leaving group from the substrate.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$ is shown below.

Figure 1

Explanation of Solution

The reaction that shows the formation of product when pentanal reacts with ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$ is shown below.

Figure 2

In the given reaction, sodium borohydride is used as a reducing agent. Sodium borohydride is used to for the reduction of carbonyl compounds to alcohols.

Conclusion

The product formed by the reaction of pentanal with ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product formed by the reaction of pentanal with ${\text{LiAlH}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: The substitution reaction involves the replacement of one functional group by other functional group. In nucleophilic substitution an electron rich species attack the species that is deficient in electrons. The electrophile and the leaving group together form a substrate. The nucleophile attacks over the substrate and there occurs the removal of leaving group from the substrate.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with ${\text{LiAlH}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 1

Explanation of Solution

The reaction that shows the formation of product when pentanal reacts with ${\text{LiAlH}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 3

In the given reaction, ${\text{LiAlH}}_{\text{4}}$ is used as a reducing agent. Lithium aluminium hydride is used to for the reduction of carbonyl compounds to alcohols.

Conclusion

The product formed by the reaction of pentanal with ${\text{LiAlH}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is shown in Figure 1.

Interpretation Introduction

(c)

Interpretation: The product formed by the reaction of pentanal with the ${\text{H}}_2\left(1\text{equiv}\right)$, $\text{Pd}-\text{C}$ is to be predicted.

Concept introduction: Addition of hydrogen takes place across a double bond of alkenes to form alkanes. This process is known as hydrogenation. It takes place in the presence of catalysts such as nickel, platinum or palladium.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with the ${\text{H}}_2\left(1\text{equiv}\right)$, $\text{Pd}-\text{C}$ is shown below.

Figure 1

Explanation of Solution

The reaction that shows the formation of product when $\text{α},\text{β}-$ unsaturated ketone reacts with ${\text{H}}_2\left(1\text{equiv}\right)$, $\text{Pd}-\text{C}$ is shown below.

Figure 4

In the given reaction, hydrogenation of alkene takes place in the presence of palladium.

Conclusion

The product formed by the reaction of pentanal with the ${\text{H}}_2\left(1\text{equiv}\right)$, $\text{Pd}-\text{C}$ is shown in Figure 1.

Interpretation Introduction

(d)

Interpretation: The product formed by the reaction of pentanal with $\text{PCC}$ is to be predicted.

Concept introduction: $\text{PCC}$ is an oxidizing reagent and is used for the conversion of alcohols to carbonyl compounds.

Answer to Problem 20.38P

No product is formed by the reaction of pentanal with $\text{PCC}$.

Explanation of Solution

The product formed by the reaction of pentanal with $\text{PCC}$ is shown below.

Figure 5

Conclusion

No product is formed by the reaction of pentanal with $\text{PCC}$.

Interpretation Introduction

(e)

Interpretation: The product formed by the reaction of pentanal with the ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Organometallic reagents like $\text{RLi,}\text{RMgX}$ and ${\text{R}}_{\text{2}}\text{CuLi}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. Epoxide converts into alcohol on treatment with organometallic reagents.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with the ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 6

Explanation of Solution

The product formed by the reaction of pentanal with the ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 7

Conclusion

The product formed by the reaction of pentanal with the ${\text{Na}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ${\text{H}}_{\text{2}}\text{O}$ is shown in Figure 6.

Interpretation Introduction

(f)

Interpretation: The product formed by the reaction of pentanal with ${\text{Ag}}_{\text{2}}\text{O},{\text{NH}}_{\text{4}}\text{OH}$ is to be predicted.

Concept introduction:

Answer to Problem 20.38P

The product formed by the reaction of pentanal with ${\text{Ag}}_{\text{2}}\text{O},{\text{NH}}_{\text{4}}\text{OH}$ is shown below.

Figure 6

Explanation of Solution

The product formed by the reaction of pentanal with ${\text{Ag}}_{\text{2}}\text{O},{\text{NH}}_{\text{4}}\text{OH}$ is shown below.

Figure 8

Conclusion

The product formed by the reaction of pentanal with ${\text{Ag}}_{\text{2}}\text{O},{\text{NH}}_{\text{4}}\text{OH}$ is shown in Figure 6.

Interpretation Introduction

(g)

Interpretation: The product formed by the reaction of pentanal with the $\left[\text{1}\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Grignard reagent is prepared by the reaction of alkyl or aryl bromide with magnesium metal in the presence of ether. The reaction of Grignard reagent with an aldehyde/ketone followed by hydrolysis yields an alcohol.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with the $\left[\text{1}\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 9

Explanation of Solution

The reaction that shows the formation of product when $\text{α},\text{β}-$ unsaturated ketone reacts with $\left[\text{1}\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 10

In the given reaction, Grignard reagent is used. This reagent attacks on the carbonyl compound to form alkoxides. In the next step, protonation with water takes place to give the final product.

Conclusion

The product formed by the reaction of pentanal with the $\left[\text{1}\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 9.

Interpretation Introduction

(h)

Interpretation: The product formed by the reaction of pentanal with $\left[\text{1}\right]{\text{C}}_6{\text{H}}_5\text{Li}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Organometallic reagents like $\text{RLi,}\text{RMgX}$ and ${\text{R}}_{\text{2}}\text{CuLi}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. Epoxide converts into alcohol on treatment with organometallic reagents.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with $\left[\text{1}\right]{\text{C}}_6{\text{H}}_5\text{Li}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 11

Explanation of Solution

The reaction that shows the formation of product when $\text{α},\text{β}-$ unsaturated ketone reacts with $\left[\text{1}\right]{\left({\text{CH}}_{\text{2}}-\text{CH}\right)}_{\text{2}}\text{CuLi}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 12

Conclusion

The product formed by the reaction of pentanal with $\left[\text{1}\right]{\text{C}}_6{\text{H}}_5\text{Li}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 11.

Interpretation Introduction

(i)

Interpretation: The product formed by the reaction of pentanal with $\left[\text{1}\right]\left({\text{CH}}_3\right)\text{CuLi}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Organometallic reagents like $\text{RLi,}\text{RMgX}$ and ${\text{R}}_{\text{2}}\text{CuLi}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. Epoxide converts into alcohol on treatment with organometallic reagents.

Answer to Problem 20.38P

No product formed by the reaction of pentanal with $\left[\text{1}\right]\left({\text{CH}}_3\right)\text{CuLi}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Explanation of Solution

The product formed by the reaction of pentanal with $\left[\text{1}\right]\left({\text{CH}}_3\right)\text{CuLi}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 13

Aldehydes do not react with organo copper reagent because these reagents are less reactive and cannot attack to the electrophilic centre of the aldehyde group.

Interpretation Introduction

(j)

Interpretation: The product formed by the reaction of pentanal with $\text{HC}\equiv \text{CNa}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Organometallic reagents like $\text{RLi,}\text{RMgX}$ and ${\text{R}}_{\text{2}}\text{CuLi}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. Epoxide converts into alcohol on treatment with organometallic reagents.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with $\text{HC}\equiv \text{CNa}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 14

Explanation of Solution

The product formed by the reaction of pentanal with $\text{HC}\equiv \text{CNa}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 15

Interpretation Introduction

(k)

Interpretation: The product formed by the reaction of pentanal with ${\text{H}}_{\text{3}}\text{CC}\equiv \text{CLi}$, ${\text{H}}_{\text{2}}\text{O}$ is to be predicted.

Concept introduction: Organometallic reagents like $\text{RLi,}\text{RMgX}$ and ${\text{R}}_{\text{2}}\text{CuLi}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. Epoxide converts into alcohol on treatment with organometallic reagents.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with ${\text{H}}_{\text{3}}\text{CC}\equiv \text{CLi}$, ${\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 16

Explanation of Solution

The reaction that shows the formation of product when $\text{α},\text{β}-$ unsaturated ketone reacts with $\left[\text{1}\right]{\left({\text{CH}}_{\text{2}}-\text{CH}\right)}_{\text{2}}\text{CuLi}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown below.

Figure 17

Interpretation Introduction

(l)

Interpretation: The product formed by the reaction of pentanal with ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$ and then $\text{TBDMS}-\text{Cl}$ is to be predicted.

Concept introduction: Organometallic reagents like $\text{RLi,}\text{RMgX}$ and ${\text{R}}_{\text{2}}\text{CuLi}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. Epoxide converts into alcohol on treatment with organometallic reagents.

Answer to Problem 20.38P

The product formed by the reaction of pentanal with ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$ and then $\text{TBDMS}-\text{Cl}$ is shown below.

Figure 18

Explanation of Solution

The product formed by the reaction of pentanal with ${\text{NaBH}}_{\text{4}}$, ${\text{CH}}_{\text{3}}\text{OH}$ and then $\text{TBDMS}-\text{Cl}$ is shown below.

Figure 19

In the given reaction, sodium borohydride is used as a reducing agent. Sodium borohydride is used to for the reduction of carbonyl compounds to alcohols.

Conclusion

The product formed by the reaction of pentanal with the $\left[\text{1}\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$, $\left[\text{2}\right]{\text{H}}_{\text{2}}\text{O}$ is shown in Figure 9.