Chapter 20, Problem 20.36P

Interpretation Introduction

(a)

Interpretation: The products formed by the treatment of ${\text{NaBH}}_{\text{4}}\text{,}{\text{CH}}_{\text{3}}\text{OH}$ with compound A and B are to be drawn.

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. Ketones on reaction with Grignard reagent converts into tertiary alcohol whereas aldehydes give primary and secondary alcohol.

Answer to Problem 20.36P

The products formed by the treatment of ${\text{NaBH}}_{\text{4}}\text{,}{\text{CH}}_{\text{3}}\text{OH}$ with compound A is,

Figure 1

Compound B do not react with ${\text{NaBH}}_{\text{4}}\text{,}{\text{CH}}_{\text{3}}\text{OH}$.

Explanation of Solution

The ball and stick model of compound A and B shows that black balls represent carbon atom, white balls represent hydrogen atom and red balls represent oxygen atom. Therefore, the structure of A and B is,

Figure 2

Ketones on reaction with ${\text{NaBH}}_{\text{4}}$ form secondary alcohol however, esters do not reaction with ${\text{NaBH}}_{\text{4}}$. Therefore, the products formed by the treatment of ${\text{NaBH}}_{\text{4}}\text{,}{\text{CH}}_{\text{3}}\text{OH}$ with compound A is,

Figure 3

Conclusion

The products formed by the treatment of ${\text{NaBH}}_{\text{4}}\text{,}{\text{CH}}_{\text{3}}\text{OH}$ with compound A is shown in Figure 3. Compound B do not reaction with ${\text{NaBH}}_{\text{4}}\text{,}{\text{CH}}_{\text{3}}\text{OH}$.

Interpretation Introduction

(b)

Interpretation: The products formed by the treatment of ${\text{LiAlH}}_{\text{4}}\text{,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are to be drawn.

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. Ketones on reaction with Grignard reagent converts into tertiary alcohol whereas aldehydes give primary and secondary alcohol.

Answer to Problem 20.36P

The products formed by the treatment of ${\text{LiAlH}}_{\text{4}}\text{,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are,

Figure 4

Explanation of Solution

The structure of A and B is shown in Figure 2.

Ketones on reaction with ${\text{LiAlH}}_{\text{4}}$ form secondary alcohol and esters form primary alcohol. Therefore, the products formed by the treatment of ${\text{LiAlH}}_{\text{4}}\text{,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are,

Figure 5

Conclusion

The products formed by the treatment of ${\text{LiAlH}}_{\text{4}}\text{,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are shown in Figure 5.

Interpretation Introduction

(c)

Interpretation: The products formed by the treatment of ${\text{CH}}_3\text{MgBr,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are to be drawn.

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. Ketones on reaction with Grignard reagent converts into tertiary alcohol whereas aldehydes give primary and secondary alcohol.

Answer to Problem 20.36P

The products formed by the treatment of ${\text{CH}}_3\text{MgBr,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are,

Figure 6

Explanation of Solution

The structure of A and B is shown in Figure 2.

Ketones and esters on reaction with Grignard reagent form tertiary alcohol. Therefore, the products formed by the treatment of ${\text{CH}}_3\text{MgBr,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are,

Figure 7

Conclusion

The products formed by the treatment of ${\text{CH}}_3\text{MgBr,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are shown in Figure 7.

Interpretation Introduction

(d)

Interpretation: The products formed by the treatment of ${\text{C}}_6{\text{H}}_5\text{Li,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are to be drawn.

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. Ketones on reaction with Grignard reagent converts into tertiary alcohol whereas aldehydes give primary and secondary alcohol.

Answer to Problem 20.36P

The products formed by the treatment of ${\text{C}}_6{\text{H}}_5\text{Li,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are,

Figure 8

Explanation of Solution

The structure of A and B is shown in Figure 2.

Ketones and esters on reaction with organolithium reagent form tertiary alcohol. Therefore, the products formed by the treatment of ${\text{C}}_6{\text{H}}_5\text{Li,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are,

Figure 9

Conclusion

The products formed by the treatment of ${\text{C}}_6{\text{H}}_5\text{Li,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are shown in Figure 9.

Interpretation Introduction

(e)

Interpretation: The products formed by the treatment of ${\text{Na}}_2{\text{Cr}}_2{\text{O}}_7\text{,}{\text{H}}_2{\text{SO}}_4\text{,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B are to be drawn.

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. Ketones on reaction with Grignard reagent converts into tertiary alcohol whereas aldehydes give primary and secondary alcohol.

Answer to Problem 20.36P

Compound A and B do not reaction with ${\text{Na}}_2{\text{Cr}}_2{\text{O}}_7\text{,}{\text{H}}_2{\text{SO}}_4\text{,}{\text{H}}_{\text{2}}\text{O}$. Hence, no product is formed.

Explanation of Solution

The structure of A and B is shown in Figure 2.

Ketones and esters do not undergo oxidation reaction with ${\text{Na}}_2{\text{Cr}}_2{\text{O}}_7\text{,}{\text{H}}_2{\text{SO}}_4\text{,}{\text{H}}_{\text{2}}\text{O}$. Therefore, no product is formed by the treatment of ${\text{Na}}_2{\text{Cr}}_2{\text{O}}_7\text{,}{\text{H}}_2{\text{SO}}_4\text{,}{\text{H}}_{\text{2}}\text{O}$ with compound A and B.

Conclusion

Compound A and B do not reaction with ${\text{Na}}_2{\text{Cr}}_2{\text{O}}_7\text{,}{\text{H}}_2{\text{SO}}_4\text{,}{\text{H}}_{\text{2}}\text{O}$. Hence, no product is formed.