Chapter 22, Problem 22.50P

Draw the product formed when phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ is treated with each reagent.

a. ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

b. ${\text{H}}_{\text{2}}{\text{O, }}^{-}\text{OH}$

c. $\text{[1]}{\text{CH}}_{\text{3}}\text{MgBr;}\text{[2]}{\text{H}}_{\text{2}}\text{O}$

d. $\text{[1]}{\text{CH}}_{\text{3}}{\text{CH}}_2\text{Li; [2]}{\text{H}}_{\text{2}}\text{O}$

e. $\text{[1]}\text{DIBAL-H; [2]}{\text{H}}_{\text{2}}\text{O}$

f. $\text{[1]}{\text{LiAlH}}_4\text{; [2]}{\text{H}}_{\text{2}}\text{O}$

Interpretation Introduction

(a)

Interpretation: The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is to be drawn.

Concept introduction: Hydrolysis of nitriles in acidic medium converts them to the corresponding carboxylic acids. The reaction involves two parts; that is nucleophilic addition followed by nucleophilic acyl substitution.

Answer to Problem 22.50P

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is

Explanation of Solution

The given compound is a nitrile.

Hydrolysis of nitriles in acidic medium converts them to the corresponding carboxylic acids. The reaction involves two parts; that is nucleophilic addition followed by nucleophilic acyl substitution. Phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ reacts with ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ to form phenylacetic acid as depicted below.

Figure 1

Conclusion

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is drawn in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with ${\text{H}}_{\text{2}}{\text{O, HO}}^{-}$ is to be drawn.

Concept introduction: Phenylacetonitrile undergo hydrolysis in basic medium to form corresponding carboxylate ion. The mechanism of the reaction involves two parts. The first part is conversion of nitrile to a ${\text{1}}^{\text{ο}}$ amide. The second part is hydrolysis to convert ${\text{1}}^{\text{ο}}$ amide into carboxylate anion.

Answer to Problem 22.50P

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with ${\text{H}}_{\text{2}}{\text{O, HO}}^{-}$ is

Explanation of Solution

The given compound is a nitrile.

Hydrolysis of nitriles in basic medium converts them to the corresponding carboxylate anion. Nitriles react with ${\text{H}}_{\text{2}}{\text{O, HO}}^{-}$ to form corresponding carboxylate ions. Phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ reacts with ${\text{H}}_{\text{2}}{\text{O, HO}}^{-}$ to form phenylcarboxylate ion as depicted below.

Figure 2

Conclusion

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with ${\text{H}}_{\text{2}}{\text{O, HO}}^{-}$ is drawn in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{CH}}_{\text{3}}\text{MgBr; [2]}{\text{H}}_{\text{2}}\text{O}$ is to be drawn.

Concept introduction: The Grignard reagents are organometallic compounds having the general formula $\text{RMgX}$ where $\text{R}$ is an alkyl or aryl group and $\text{X}$ is a halogen atom. Nitriles react with both Grignard reagent, and organolithium reagents, followed by hydrolysis to yield ketones. The reaction utilizes ${\text{R}}^{-}$ nucleophile to yield desired product.

Answer to Problem 22.50P

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{CH}}_{\text{3}}\text{MgBr; [2]}{\text{H}}_{\text{2}}\text{O}$ is

Explanation of Solution

The given compound is a nitrile.

Nitriles react with both Grignard reagent, and organolithium reagents, followed by hydrolysis to yield ketones. The reaction utilizes ${\text{R}}^{-}$ nucleophile to yield desired product.

Phenylacetonitrile undergoes Grignard reaction with $\text{[1]}{\text{CH}}_{\text{3}}\text{MgBr;}\text{[2]}{\text{H}}_{\text{2}}\text{O}$ to form $1$-phenylpropan-$2$-one as depicted below.

Figure 3

Conclusion

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{CH}}_{\text{3}}\text{MgBr; [2]}{\text{H}}_{\text{2}}\text{O}$ is drawn in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{CH}}_{\text{3}}{\text{CH}}_2\text{Li;}\text{[2]}{\text{H}}_{\text{2}}\text{O}$ is to be drawn.

Concept introduction: Organolithium reagents are organometallic compounds having the general formula $\text{RLi}$ where $\text{R}$ can be an alkyl or aryl group. They have a carbon-lithium bond and yield similar product as Grignard reaction.

Answer to Problem 22.50P

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{CH}}_{\text{3}}{\text{CH}}_2\text{Li;}\text{[2]}{\text{H}}_{\text{2}}\text{O}$ is drawn in Figure 4.

Explanation of Solution

The given compound is a nitrile.

Nitriles react with both Grignard reagent, and organolithium reagents, followed by hydrolysis to yield ketones. The reaction utilizes ${\text{R}}^{-}$ nucleophile to yield desired product.

The treatment of phenylacetonitrile with $\text{[1]}{\text{CH}}_{\text{3}}{\text{CH}}_2\text{Li;}\text{[2]}{\text{H}}_{\text{2}}\text{O}$ leads to the formation of $1$-phenyl-$2$-butanone as depicted below.

Figure 4

Conclusion

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{CH}}_{\text{3}}{\text{CH}}_2\text{Li;}\text{[2]}{\text{H}}_{\text{2}}\text{O}$ is drawn in Figure 4.

Interpretation Introduction

Interpretation: The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}\text{DIBAL-H; [2]}{\text{H}}_{\text{2}}\text{O}$ is to be drawn.

(e)

Concept introduction: $\text{DIBAL-H}$ is a selective and a milder reducing agent than ${\text{LiAlH}}_{\text{4}}$. Nitrile compounds are converted to corresponding aldehydes in the presence of milder reducing agents like $\text{DIBAL-H}$, followed by hydrolysis.

Answer to Problem 22.50P

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}\text{DIBAL-H; [2]}{\text{H}}_{\text{2}}\text{O}$ is

Explanation of Solution

Diisobutylaluminiumhydride $\left(\text{DIBAL-H}\right)$ is a selective reducing agent. It readily converts nitriles compounds to corresponding aldehydes. Phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ reacts with $\text{[1]}\text{DIBAL-H; [2]}{\text{H}}_{\text{2}}\text{O}$ to form phenylacetic acid as depicted below.

Figure 5

Conclusion

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}\text{DIBAL-H; [2]}{\text{H}}_{\text{2}}\text{O}$ is drawn in Figure 5.

Interpretation Introduction

(f)

Interpretation: The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{LiAlH}}_4\text{; [2]}{\text{H}}_{\text{2}}\text{O}$ is to be drawn.

Concept introduction: ${\text{LiAlH}}_4$ is a strong reducing agent. The reaction of a nitrile with $\text{[1]}{\text{LiAlH}}_4\text{; [2]}{\text{H}}_{\text{2}}\text{O}$ adds two equivalent of ${\text{H}}_{\text{2}}$ across the $\text{C}\equiv \text{N}$ bond to yield primary amines.

Answer to Problem 22.50P

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{LiAlH}}_4\text{; [2]}{\text{H}}_{\text{2}}\text{O}$ is

Explanation of Solution

Nitriles react with $\text{[1]}{\text{LiAlH}}_4\text{; [2]}{\text{H}}_{\text{2}}\text{O}$ to form amines. The overall reaction adds two equivalent of ${\text{H}}_{\text{2}}$ across the $\text{C}\equiv \text{N}$ bond to yield primary amine. Phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ reacts with $\text{[1]}{\text{LiAlH}}_4\text{; [2]}{\text{H}}_{\text{2}}\text{O}$ to form $2$-phenyl ethylamine as depicted below.

Figure 6

Conclusion

The product formed from the treatment of phenylacetonitrile $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{CH}}_{\text{2}}\text{CN}\right)$ with $\text{[1]}{\text{LiAlH}}_4\text{; [2]}{\text{H}}_{\text{2}}\text{O}$ is drawn in Figure 6.