Chapter 20, Problem 20.44P

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

a. ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{CHO}$ c. ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{C}\left(\text{OH}\right){\left({\text{C}}_6{\text{H}}_5\right)}_2$

b. ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COCH}={\text{CH}}_2$ d. ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_2\text{OH}$

Interpretation Introduction

(a)

Interpretation:

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

Concept introduction:

Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organometallic compounds give ketone or alcohol as a product. Alkyl group from organometallic compound occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.44P

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

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

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

Concept introduction:

Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organometallic compounds give ketone or alcohol as a product. Alkyl group from organometallic compound occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.44P

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

Explanation of Solution

The structure of given ketone compound shows that one ethene molecule is attached to carbonyl carbon. Acid chlorides on reaction with one equivalent of organocuprate reagent convert into ketone. The organocuprate reagent removes the good leaving $\text{Cl}$ atom and leads to the formation of ketone. The conversion of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COCl}$ to given compound by the use of ${\left({\text{H}}_{\text{2}}\text{C}=\text{CH}\right)}_2\text{CuLi}$ is shown in Figure 2.

Figure 2

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

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

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

Concept introduction:

Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organometallic compounds give ketone or alcohol as a product. Alkyl group from organometallic compound occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.44P

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COCl}$ to given compound is ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{MgBr}$.

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 ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COCl}$ to given compound by the use of two equivalents of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{MgBr}$ is shown in Figure 3.

Figure 3

Therefore, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{MgBr}$ is needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COCl}$ to given compound.

Conclusion

The reagent needed to convert ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{CHCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{COCl}$ to given compound is ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{MgBr}$.

Interpretation Introduction

(d)

Interpretation:

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

Concept introduction:

Acid chlorides, which contains good leaving group $\text{Cl}$, on reaction with organometallic compounds give ketone or alcohol as a product. Alkyl group from organometallic compound occupies the position of leaving group $\text{Cl}$.

Answer to Problem 20.44P

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

Conclusion

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