Chapter 20, Problem 20.30P

What reagent is needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}\text{COCl}$ into each compound?

a. b. c. d.

Interpretation Introduction

(a)

Interpretation: The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is to be predicted.

Concept introduction: Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organocuprate give ketone as a product. Alkyl group from organocuprate occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.30P

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is $\text{LiAl}{\left(\text{OBu}\right)}_{\text{3}}\text{H}$.

Explanation of Solution

The given compound is aldehyde. Acid chlorides on reaction with mild reducing agent like lithium tri-tert-butoxyaluminium hydride $\left(\text{LiAl}{\left(\text{OBu}\right)}_{\text{3}}\text{H}\right)$ converts into aldehyde as shown in Figure 1.

Figure 1

Therefore, $\text{LiAl}{\left(\text{OBu}\right)}_{\text{3}}\text{H}$ is needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound.

Conclusion

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is $\text{LiAl}{\left(\text{OBu}\right)}_{\text{3}}\text{H}$.

Interpretation Introduction

(b)

Interpretation: The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is to be predicted.

Concept introduction: Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organocuprate give ketone as a product. Alkyl group from organocuprate occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.30P

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$.

Explanation of Solution

The structure of given ketone compound shows that the hydroxyl group is attached to tertiary carbon atom. Acid chlorides on reaction with two equivalents of Grignard reagent convert into tertiary alcohol. The first equivalent of Grignard reagent remove good leaving $\text{Cl}$ group and second equivalent converts ketone to alcohol. The conversion of ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound by the use of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$ is shown in Figure 2.

Figure 2

Therefore, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$ is needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound.

Conclusion

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{MgBr}$.

Interpretation Introduction

(c)

Interpretation: The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is to be predicted.

Concept introduction: Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organocuprate give ketone as a product. Alkyl group from organocuprate occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.30P

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is ${\left({\text{CH}}_3{\text{CH}}_2\right)}_{\text{2}}\text{CuLi}$.

Explanation of Solution

The structure of given compound shows that the carbonyl carbon is attached to ethyl group. Ethyl group can occupy the position of leaving group $\text{Cl}$ when ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ is treated with ${\left({\text{CH}}_3{\text{CH}}_2\right)}_{\text{2}}\text{CuLi}$ as shown in Figure 3.

Figure 3

Therefore, ${\left({\text{CH}}_3{\text{CH}}_2\right)}_{\text{2}}\text{CuLi}$ is needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound.

Conclusion

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is ${\left({\text{CH}}_3{\text{CH}}_2\right)}_{\text{2}}\text{CuLi}$.

Interpretation Introduction

(d)

Interpretation: The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is to be predicted.

Concept introduction: Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organocuprate give ketone as a product. Alkyl group from organocuprate occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.30P

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is ${\text{LiAlH}}_4$.

Explanation of Solution

Acid chlorides on reaction with strong reducing agent like lithium aluminium hydride $\left({\text{LiAlH}}_4\right)$ converts into primary alcohol as shown in Figure 4.

Figure 4

Therefore, ${\text{LiAlH}}_4$ is needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound.

Conclusion

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_2{\text{CHCH}}_{\text{2}}\text{COCl}$ to given compound is ${\text{LiAlH}}_4$.