Chapter 23, Problem 23.60AP

Interpretation Introduction

(a)

Interpretation:

The steps required to obtain the pure samples of the given compound from the starting material are to be stated.

Concept introduction:

The rearrangement in which acyl azides $\left(\text{RCO}-{\text{N}}_{\text{3}}\right)$ undergoes a reaction to form the amines is known as Curtius rearrangement. The acyl azide is heated in benzene or toluene, inert solvent, which results in a loss of nitrogen to form isocyanate $\left(\text{R}-\text{N}=\text{C}=\text{O}\right)$. The acid or base hydrolysis converts the isocynate products into amines.

Answer to Problem 23.60AP

The reaction route for the synthesis of the given product is shown below.

Explanation of Solution

The separation of enantiomers occurs in three steps.

In the first step, the two enantiomers, $(\pm )\text{-2-}$ phenylbutanoic acid are reacted with $\left(R\right)\text{-}$ form of $\text{1-}$ phenylpropane $\text{-1-}$ amine to give the diasteromeric salts. Then the diasteromeric salts are separated by the process of fractional crystallization with the help of suitable solvent and the precipitates of isomers are obtained. At last, the precipitates get separated when treated with dilute $\text{HCl}$.

The separation of the two enantiomers is shown below.

Figure 1

The formation of the given product follows the curtius rearrangement. The reaction follows a concerted mechanism, a reaction in which the bond formation and bond dissociation occurs in a single step.

The $\left(R\right)\text{-}$ form of $\text{2-}$ phenylbutanoic acid obtained by the separation method is heated with ${\text{NaN}}_{\text{3}}$ in presence on thionyl chloride to form the isocyanate. Then the reaction of the isocynate with $\text{MeOH}$ results in the formation of the desired product, $\left(R\right)\text{-}$ methyl $\text{1-}$ phenylpropylcarbamate.

The reaction for the formation of $\left(R\right)\text{-}$ methyl $\text{1-}$ phenylpropylcarbamate is shown below.

Figure 2

Conclusion

The desired product, $\left(R\right)\text{-}$ methyl $\text{1-}$ phenylpropylcarbamate is formed by the curtius reacrrangent as shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The steps required to obtain the pure samples of the given compound from the starting material are to be stated.

Concept introduction:

An ester is formed by the reaction of carboxylic with alcohol in the presence of an acid catalyst. This reaction is known as Esterification.

Ester is the functional group which is formed from the reaction of an alcohol and a carboxylic acid. Esters end with a suffix, $-oate$. They are always present in the middle of the carbon chain. The carbon chain which is not bonded to the carbon of the ester group is named first, and then the carbon chain which includes the ester group $-\text{COO}-$ is named.

Answer to Problem 23.60AP

The reaction route for the synthesis of the given product is shown below.

Explanation of Solution

The $\left(S\right)\text{-}$ isomer of the $\text{2-}$ phenylbutanoic acid is obtained by the enantiomer separation steps. The $\left(S\right)\text{-}$$\text{2-}$ phenylbutanoic acid on treatment with ethanol gives the desired ester, $\left(S\right)\text{-}$ ethyl $2\text{-}$ phenylbutanoate.

The reaction of the formation of $\left(S\right)\text{-}$ ethyl $2\text{-}$ phenylbutanoate is shown below.

Figure 3

Conclusion

The formation of $\left(S\right)\text{-}$ ethyl $2\text{-}$ phenylbutanoate follows the esterification reaction and the reaction is shown in Figure 3.

Interpretation Introduction

(c)

Interpretation:

The steps required to obtain the pure samples of the given compound from the starting material are to be stated.

Concept introduction:

Hofmann rearrangement reaction is the conversion reaction of the primary amide to a primary amine with the loss of one carbon chain. The reaction requires basic conditions. The reaction of sodium hydroxide with bromine forms sodium hypobromite as an intermediate in situ.

Answer to Problem 23.60AP

The reaction route for the synthesis of the given product is shown below.

Explanation of Solution

The isomer, $\left(R\right)\text{-}$$2\text{-}$ phenylbutanoic acid reacts with ammonia and the amide group is introduced followed by the reaction with ${\text{Br}}_{\text{2}}$ and $\text{KOH}$ to give the amine with the loss of one carbon chain. Then, the reaction of amine with phosgene $\left({\text{COCl}}_{\text{2}}\right)$ forms the desired product.

The reaction of the formation of $\left(R,R\right)\text{-}$$1,3\text{-}$ bis $\left(\text{1-phenylpropyl}\right)$ urea is shown below.

Figure 4

Conclusion

The formation of $\left(R,R\right)\text{-}$$1,3\text{-}$ bis $\left(\text{1-phenylpropyl}\right)$ urea is shown in Figure 4.

Interpretation Introduction

(d)

Interpretation:

The steps required to obtain the pure samples of the given compound from the starting material are to be stated.

Concept introduction:

Hofmann rearrangement reaction is the conversion reaction of the primary amide to a primary amine with the loss of one carbon chain. The reaction requires basic conditions. The reaction of sodium hydroxide with bromine forms sodium hypobromite as an intermediate in situ.

Answer to Problem 23.60AP

The reaction route for the synthesis of the given product is shown below.

Explanation of Solution

The isomer, $\left(+-\right)\text{-}$$2\text{-}$ phenylbutanoic acid reacts with ammonia and the amide group is introduced followed by the reaction with ${\text{Br}}_{\text{2}}$ and $\text{KOH}$ to give the amine with the loss of one carbon chain. Then, the reaction of amine with phosgene $\left({\text{COCl}}_{\text{2}}\right)$ forms the desired product.

The reaction of the formation of $\left(meso\right)\text{-}$$1,3\text{-}$ bis $\left(\text{1-phenylpropyl}\right)$ urea is shown below.

Figure 5

Conclusion

The formation of $\left(meso\right)\text{-}$$1,3\text{-}$ bis $\left(\text{1-phenylpropyl}\right)$ urea is shown in Figure 5.

Interpretation Introduction

(e)

Interpretation:

The steps required to obtain the pure samples of the given compound from the starting material are to be stated.

Concept introduction:

Hofmann rearrangement reaction is the conversion reaction of the primary amide to a primary amine with the loss of one carbon chain. The reaction requires basic conditions. The reaction of sodium hydroxide with bromine forms sodium hypobromite as an intermediate in situ.

Answer to Problem 23.60AP

The reaction route for the synthesis of the given product is shown below.

Explanation of Solution

The isomer, $\left(R\right)\text{-}$$2\text{-}$ phenylbutanoic acid reacts with ammonia and the amide group is introduced followed by the reaction with ${\text{Br}}_{\text{2}}$ and $\text{KOH}$ to give the amine. The amine formed is reacted with $\text{4-}$ chlorobenzoic acid in presence of the Buchwald catalyst and a base to get the desired product,

The reaction of the formation of $\left(R\right)\text{-}$$\text{4-}$$\left(\text{1-phenylpropylamino}\right)$ benzoic acid is shown below.

Figure 6

Conclusion

The formation of $\left(R\right)\text{-}$$\text{4-}$$\left(\text{1-phenylpropylamino}\right)$ benzoic acid follows Hofmann rearrangement reaction as shown in Figure 6.