Chapter 20, Problem 20.7P

Draw the products formed when ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ is treated with each reagent: (a) ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$; (b) ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$; (c) ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)\text{, Pd-C}$; (d) ${\text{H}}_{\text{2}}\left(\text{excess}\right)\text{, Pd-C}$; (e) ${\text{NaBH}}_{\text{4}}\left(\text{excess}\right){\text{in CH}}_{\text{3}}\text{OH}$; (f) ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$.

Interpretation Introduction

(a)

Interpretation: The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$ is to be drawn.

Concept introduction: Treatment of carbonyl compounds with ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$ yields alcohols. It is a type of reduction reaction. The reduction of ketone by ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$ yields secondary alcohol.

Answer to Problem 20.7P

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$ is,

Explanation of Solution

The reduction of ketone by ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$ yields secondary alcohol. The double bond is not reduced by it. Hence, the product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$ is shown below.

Figure 1

Conclusion

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{2}}\text{O}$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is to be drawn.

Concept introduction: Treatment of carbonyl compounds with ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields alcohols. It is a type of reduction reaction. The reduction of an aldehyde by ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields primary alcohol and the reduction of ketone by ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yield secondary alcohol.

Answer to Problem 20.7P

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is,

Explanation of Solution

The reduction of ketone by ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields secondary alcohol. The double bond is not reduced by it. Hence, the product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is shown below.

Figure 2

Conclusion

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)\text{, Pd-C}$ is to be drawn.

Concept introduction: Treatment of carbonyl compounds with ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)\text{, Pd-C}$ yields alcohols. It is a type of reduction reaction. ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)$ gas reduces the double bond selectively.

Answer to Problem 20.7P

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)\text{, Pd-C}$ is,

Explanation of Solution

Treatment of carbonyl compounds with ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)\text{, Pd-C}$ yields alcohols. ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)$ gas reduces the double bond. Hence, the product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)\text{, Pd-C}$ is shown below.

Figure 3

Conclusion

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{1equiv}\right)\text{, Pd-C}$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{excess}\right)\text{, Pd-C}$ is to be drawn.

Concept introduction: Treatment of carbonyl compounds with ${\text{H}}_{\text{2}}\left(\text{excess}\right)\text{, Pd-C}$ yields alcohols. It is a type of reduction reaction. ${\text{H}}_{\text{2}}\left(\text{excess}\right)$ gas present in the palladium charcoal catalyst reduces double bond as well as keto group.

Answer to Problem 20.7P

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{excess}\right)\text{, Pd-C}$ is,

Explanation of Solution

Treatment of carbonyl compounds with ${\text{H}}_{\text{2}}\left(\text{excess}\right)\text{, Pd-C}$ yields alcohols. Hence, the product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{excess}\right)\text{, Pd-C}$ is shown below.

Figure 4

Conclusion

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\left(\text{excess}\right)\text{, Pd-C}$ is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation: The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}\left(\text{excess}\right){\text{in CH}}_{\text{3}}\text{OH}$ is to be drawn.

Concept introduction: Treatment of carbonyl compounds with ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields alcohols. It is a type of reduction reaction. The reduction of an aldehyde by ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields primary alcohol and the reduction of ketone by ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yield secondary alcohol.

Answer to Problem 20.7P

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}\left(\text{excess}\right){\text{in CH}}_{\text{3}}\text{OH}$ is,

Explanation of Solution

The reduction of ketone by ${\text{NaBH}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields secondary alcohol. The double bond is not reduced by it Hence, the product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}\left(\text{excess}\right){\text{in CH}}_{\text{3}}\text{OH}$ is shown below.

Figure 5

Conclusion

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBH}}_{\text{4}}\left(\text{excess}\right){\text{in CH}}_{\text{3}}\text{OH}$ is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation: The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is to be drawn.

Concept introduction: Treatment of carbonyl compounds with ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields alcohols. It is a type of reduction reaction. The reduction of an aldehyde by ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields primary alcohol and the reduction of ketone by ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yield secondary alcohol.

Answer to Problem 20.7P

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is,

Explanation of Solution

The reduction of ketone by ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ yields secondary alcohol. The double bond is not reduced by it. Hence, the product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is shown below.

Figure 6

Conclusion

The product formed by the treatment of ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CH}={\text{CH}}_{\text{2}}$ with ${\text{NaBD}}_{\text{4}}{\text{ in CH}}_{\text{3}}\text{OH}$ is shown in Figure 6.