Chapter 9, Problem 9.31P

Interpretation Introduction

(a)

Interpretation:

The equation for the preparation of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{MgBr}$ is to be stated.

Concept introduction:

Organometallic reagents like $\text{RLi}$ and $\text{RMgX}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. They are formed when alkyl halide is treated with active metal atom in the presence of a specific solvent.

Answer to Problem 9.31P

The equation for the preparation of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{MgBr}$ is shown below.

${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{Br}+\text{Mg}\stackrel{\text{diethyl}\text{ether}}{\to }{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{MgBr}$

Explanation of Solution

A Grignard reagent is formed when alkyl halide is treated with $\text{Mg}$ in the presence of anhydrous diethyl ether. The given Grignard reagent ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{MgBr}$ is prepared from alkyl halide ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{Br}$ as shown below.

${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{Br}+\text{Mg}\stackrel{\text{diethyl}\text{ether}}{\to }{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{MgBr}$

Conclusion

The equation for the preparation of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{MgBr}$ is shown below.

${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{Br}+\text{Mg}\stackrel{\text{diethyl}\text{ether}}{\to }{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-\text{MgBr}$

Interpretation Introduction

(b)

Interpretation:

The equation for the preparation of $\text{Ph}-\text{Li}$ is to be stated.

Concept introduction:

Organometallic reagents like $\text{RLi}$ and $\text{RMgX}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. They are formed when alkyl halide is treated with active metal atom in the presence of a specific solvent.

Answer to Problem 9.31P

The equation for the preparation of $\text{Ph}-\text{Li}$ is shown below.

$\text{Ph}-\text{Br}+2\text{Li}\stackrel{\text{hexane}}{\to }\text{Ph}-\text{Li}+2\text{LiBr}$

Explanation of Solution

The organolithium compound is formed when alkyl halide is treated with two equivalents of active lithium metal in the presence of hexane. The given organolithium compound $\text{Ph}-\text{Li}$ is prepared from alkyl halide $\text{Ph}-\text{Br}$ as shown below.

$\text{Ph}-\text{Br}+2\text{Li}\stackrel{\text{hexane}}{\to }\text{Ph}-\text{Li}+2\text{LiBr}$

Conclusion

The equation for the preparation of $\text{Ph}-\text{Li}$ is shown below.

$\text{Ph}-\text{Br}+2\text{Li}\stackrel{\text{hexane}}{\to }\text{Ph}-\text{Li}+2\text{LiBr}$

Interpretation Introduction

(c)

Interpretation:

The equation for the preparation of ${\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-{\text{CH}}_2\right]}_3\text{B}$ is to be stated.

Concept introduction:

Organometallic reagents like $\text{RLi}$ and $\text{RMgX}$ are the source of ${\text{R}}^{-}$, which acts as a nucleophile. They are formed when alkyl halide is treated with active metal atom in the presence of a specific solvent.

Answer to Problem 9.31P

The equation for the preparation of ${\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-{\text{CH}}_2\right]}_3\text{B}$ is shown below.

$3{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}\underset{\text{THF}}{\overset{{\text{BH}}_{\text{3}}}{\to }}{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-{\text{CH}}_{\text{2}}\right]}_3\text{B}$

Explanation of Solution

The organoboron compound is formed when an alkene is treated with boron hydride in the presence of THF (tetrahydrofuran). The given organolithium compound ${\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-{\text{CH}}_2\right]}_3\text{B}$ is prepared from three equivalents of $2$-methylpropene as shown below.

$3{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}\underset{\text{THF}}{\overset{{\text{BH}}_{\text{3}}}{\to }}{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-{\text{CH}}_{\text{2}}\right]}_3\text{B}$

Conclusion

The equation for the preparation of ${\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-{\text{CH}}_2\right]}_3\text{B}$ is shown below.

$3{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}\underset{\text{THF}}{\overset{{\text{BH}}_{\text{3}}}{\to }}{\left[{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CH}-{\text{CH}}_{\text{2}}\right]}_3\text{B}$