Chapter 20, Problem 20.46P

Interpretation Introduction

(a)

Interpretation:

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{3}}{\text{O}}^{+}$ is to be drawn.

Concept introduction:

An ester can be converted to a primary alcohol by reacting with $\text{LiAlH}\text{,}{\text{ H}}_{\text{3}}{\text{O}}^{+}$. $\text{LiAlH}$ is a source of hydride ion which acts as a nucleophile and reduces ester to aldehyde by elimination of methoxide ion and then reduces aldehyde to primary alcohol followed by protonation.

Answer to Problem 20.46P

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{3}}{\text{O}}^{+}$ is:

Explanation of Solution

The equation for the reaction of methyl benzoate with ${\text{LiAlH}}_{\text{4}}{\text{, then H}}_{\text{3}}{\text{O}}^{+}$ is:

Methyl benzoate is an ester, which, on reaction with ${\text{LiAlH}}_{\text{4}}{\text{/H}}_{\text{3}}{\text{O}}^{+}$ undergoes nucleophilic addition-elimination reaction and forms a benzaldehyde product which is further on reduction produces benzoyl alcohol. In the first step, the hydride ion from ${\text{LiAlH}}_4$ acts as a nucleophile and adds to carbonyl carbon of methyl benzoate and then by delocalization of oxygen lone pair removes the leaving group methoxide ion and produces benzaldehyde. In the next step, another hydride ion adds to benzaldehyde carbonyl and reduces it to primary alcohol as benzoyl alcohol. The product with detailed mechanism is as follows:

Conclusion

The product with detailed mechanism for the given reaction is drawn based on the reactivity of ${\text{LiAlH}}_{\text{4}}{\text{/H}}_{\text{3}}{\text{O}}^{+}$ with an ester.

Interpretation Introduction

(b)

Interpretation:

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{Li (excess), then H}}_{\text{3}}{\text{O}}^{+}$ is to be drawn.

Concept introduction:

An ester can be converted to a tertiary alcohol by reacting it with ${\text{RCuLi (excess), then H}}_{\text{3}}{\text{O}}^{+}$. ${\text{R}}^{-}$ is a nucleophile and adds to carbonyl group of ester and reduces to ketone by elimination of methoxide ion. This ketone further reduces to tertiary alcohol followed by protonation using ${\text{H}}_{\text{3}}{\text{O}}^{+}$.

Answer to Problem 20.46P

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{Li (excess), then H}}_{\text{3}}{\text{O}}^{+}$ is:

Explanation of Solution

The equation for the reaction of methyl benzoate with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{Li (excess), then H}}_{\text{3}}{\text{O}}^{+}$ is:

Methyl benzoate is an ester, which, on reaction with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{Li (excess), then H}}_{\text{3}}{\text{O}}^{+}$ undergoes nucleophilic addition-elimination reaction and forms a ketone product which on further reduction produces tertiary alcohol. In the first step, ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{}^{-}$ from ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{Li }$ acts as a nucleophile and adds to carbonyl carbon of methyl benzoate and then by delocalization of oxygen lone pair removes the leaving group methoxide ion and produces the corresponding ketone. In the next step, another ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{}^{-}$ adds to ketone carbonyl and reduces it to tertiary alcohol by deprotonation of negatively charged oxygen. The product with detailed mechanism is as follows:

Conclusion

The product with detailed mechanism for the given reaction is drawn based on the reactivity of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{Li (excess), then H}}_{\text{3}}{\text{O}}^{+}$ with an ester.

Interpretation Introduction

(c)

Interpretation:

Whether methyl benzoate can react with ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ is to be determined.

Concept introduction:

The reagent ${\text{R}}_{\text{2}}\text{CuLi}$ is not reactive enough to reduce an ester, thus, ester cannot undergo reaction with it.

Answer to Problem 20.46P

Mmethyl benzoate cannot react with ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ and can be indicated as:

Explanation of Solution

The equation for the reaction of methyl benzoate with ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ is:

Methyl benzoate is an ester and ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ is ${\text{R}}_{\text{2}}\text{CuLi}$ type reagent which is not so reactive, and so, cannot react with ester. Thus, there is no reaction and can be indicated as:

Conclusion

It is determined that no reaction occurs based on the reactivity of ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$.

Interpretation Introduction

(d)

Interpretation:

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{MgBr (excess), then H}}^{+}$ is to be drawn.

Concept introduction:

An ester can be converted to a tertiary alcohol by reacting it with ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{MgBr (excess), then H}}^{+}$. ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{MgBr}$ is the Grignard reagent where ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{}^{-}$ acts as a nucleophile and adds to carbonyl group of ester and reduces to ketone by elimination of methoxide ion. This ketone then reduces to tertiary alcohol followed by protonation using ${\text{H}}^{+}$.

Answer to Problem 20.46P

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{MgBr (excess), then H}}^{+}$ is:

Explanation of Solution

The equation for the reaction of methyl benzoate with ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{MgBr (excess), then H}}^{+}$ is:

Methyl benzoate is an ester, which, on reaction with ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{MgBr (excess), then H}}^{+}$ undergoes nucleophilic addition-elimination reaction and forms a ketone product which on further reduction produces tertiary alcohol. In the first step, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{}^{-}$ from ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{MgBr}$ acts as a nucleophile and adds to carbonyl carbon of methyl benzoate and then by delocalization of oxygen lone pair removes the leaving group methoxide ion and produces the corresponding ketone. In the next step, another ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{}^{-}$ adds to ketone carbonyl and reduces it to tertiary alcohol by deprotonation of negatively charged oxygen. The product with detailed mechanism is as follows:

Conclusion

The product with detailed mechanism for the given reaction is drawn based on the reactivity of ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{MgBr (excess), then H}}^{+}$ with an ester.

Interpretation Introduction

(e)

Interpretation:

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$ is to be drawn.

Concept introduction:

An ester can be reduced to aldehyde without reducing it further to alcohol using a specific reagent such as ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$. The single hydride ion from ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$ adds to carbonyl of an ester and reduces it to aldehyde in acidic condition.

Answer to Problem 20.46P

The product with detailed mechanism for the reaction between methyl benzoate and ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$ is:

Explanation of Solution

The equation for the reaction of methyl benzoate with ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$ is:

Methyl benzoate is an ester, which, on reaction with ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$ undergoes nucleophilic addition-elimination reaction and forms a aldehyde product. In the first step, the hydride ion from ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$ acts as a nucleophile and adds to carbonyl carbon of methyl benzoate to produce tetrahedral intermediate which in presence of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ gives corresponding hemiacetal. In excess of ${\text{H}}_{\text{3}}{\text{O}}^{+}$ the hemiacetal is converted to benzaldehyde. The product with detailed mechanism is as follows:

Conclusion

The product with detailed mechanism for the given reaction is drawn based on the reactivity of ${\text{DIBAH, then H}}_{\text{3}}{\text{O}}^{+}$ with an ester.