Chapter 20, Problem 20.45P

Interpretation Introduction

(a)

Interpretation:

The product with detailed mechanism for the reaction between $\text{m-ethylbenzoyl chloride}$ and ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ is to be drawn.

Concept introduction:

An acid chloride can be converted to corresponding ketone by reacting it with lithium dialkylcuprate $\left({\text{R}}_{\text{2}}\text{CuLi}\right)$. The only alkyl group from $\left({\text{R}}_{\text{2}}\text{CuLi}\right)$ acts as a nucleophile and adds to acid derivative, and undergoes elimination of the chloride ion to produce the ketone.

Answer to Problem 20.45P

The product with detailed mechanism for the reaction between $\text{m-ethylbenzoyl chloride}$ and ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ is:

Explanation of Solution

The equation for the reaction of $\text{m-ethylbenzoyl chloride}$ with ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ is:

$\text{m-ethylbenzoyl chloride}$ is the acid chloride on reaction with ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ undergoes nucleophilic addition-elimination reaction and forms a ketone product. In the first step, the ethyl group from ${{\text{(CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{)}}_{\text{2}}\text{CuLi}$ acts as a nucleophile and adds to carbonyl carbon of $\text{m-ethylbenzoyl chloride}$ and then by delocalization of oxygen lone pair removes the leaving group chloride ion and produces the corresponding ketone. 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 $\left({\text{R}}_{\text{2}}\text{CuLi}\right)$ with acid chloride.

Interpretation Introduction

(b)

Interpretation:

The product with detailed mechanism for the reaction between $\text{m-ethylbenzoyl chloride}$ and ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$ is to be drawn.

Concept introduction:

An acid chloride can be converted to an aldehyde by reacting it with ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$. ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$ is a source of hydride ion which acts as a nucleophile and reduces acid derivative to aldehyde by elimination of chloride ion.

Answer to Problem 20.45P

The product with detailed mechanism for the reaction between $\text{m-ethylbenzoyl chloride}$ and ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$ is:

Explanation of Solution

The equation for the reaction of $\text{m-ethylbenzoyl chloride}$ with ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$ is:

$\text{m-ethylbenzoyl chloride}$ is the acid chloride, which, on reaction with ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$ at ${\text{-78}}^{\text{o}}\text{C}$ undergoes nucleophilic addition-elimination reaction and forms a aldehyde product. In the first step, the hydride ion from ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$ acts as a nucleophile and adds to carbonyl carbon of $\text{m-ethylbenzoyl chloride}$ and then by delocalization of oxygen lone pair removes the leaving group chloride ion and produces the corresponding aldehyde. 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 ${\text{LiAlH(O-t-Bu)}}_{\text{3}}$ with acid chloride.

Interpretation Introduction

(c)

Interpretation:

The product with detailed mechanism for the reaction between $\text{m-ethylbenzoyl chloride}$ and $\text{NaBH}\text{,}\text{ EtOH}$ is to be drawn.

Concept introduction:

An acid chloride can be converted to a primary alcohol by reacting with $\text{NaBH}\text{,}\text{ EtOH}$. $\text{NaBH}\text{,}\text{ EtOH}$ is a source of hydride ion which acts as a nucleophile and reduces acid derivative to aldehyde by elimination of chloride ion and then to primary alcohol.

Answer to Problem 20.45P

The product with detailed mechanism for the reaction between $\text{m-ethylbenzoyl chloride}$ and $\text{NaBH}\text{,}\text{ EtOH}$ is:

Explanation of Solution

The equation for the reaction of $\text{m-ethylbenzoyl chloride}$ with $\text{NaBH}\text{,}\text{ EtOH}$ is:

$\text{m-ethylbenzoyl chloride}$ is the acid chloride, which, on reaction with $\text{NaBH}\text{,}\text{ EtOH}$ undergoes nucleophilic addition-elimination reaction and forms an aldehyde product which on further reduction produces primary alcohol. In the first step, the hydride ion from ${\text{NaBH}}_4$ acts as a nucleophile and adds to carbonyl carbon of $\text{m-ethylbenzoyl chloride}$ and then by delocalization of oxygen lone pair removes the leaving group chloride ion and produces corresponding aldehyde. In the next step, another hydride ion adds to aldehyde carbonyl and reduces it to primary 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 $\text{NaBH}\text{,}\text{ EtOH}$ with acid chloride.

Interpretation Introduction

(d)

Interpretation:

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

Concept introduction:

An acid chloride 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’s where the ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{}^{-}$ acts as a nucleophile and adds to carbonyl group of acid chloride and reduces to ketone by elimination of chloride ion. This ketone is then reduced to tertiary alcohol followed by protonation using ${\text{H}}^{+}$.

Answer to Problem 20.45P

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

Explanation of Solution

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

$\text{m-ethylbenzoyl chloride}$ is the acid chloride, 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 $\text{m-ethylbenzoyl chloride}$ and then by delocalization of oxygen lone pair removes the leaving group chloride 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 acid chloride.