Chapter 27, Problem 27.71AP

Interpretation Introduction

(a)

Interpretation:

The synthesis of given compound from the given starting material is to be stated.

Concept Introduction:

Amino group reacts faster with the alkyl halides than the carboxylic group. The reaction of carboxylic acid can be performed only by protecting the amine group. Fmoc compound is used as the protecting agent for amine group. Amine and acid chloride react to form amide bond.

Answer to Problem 27.71AP

The synthesis of given compound from the given starting material is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 1

The amine group of the $p-\text{aminobenzoic}\text{acid}$ is protected by reacting it with Fmoc. The protected compound will then undergo esterification reaction. The carboxylic acid group reacts with the ethyl chloride to form ethyl ester of Fmoc protected $p\text{-aminobenzoic}\text{acid}$. Then deprotection of amino group is performed. Amino group will then react with benzoyl chloride to form product. The complete synthesis is shown below.

Figure 2

Conclusion

The complete synthesis of the given compound is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The synthesis of given compound from the given starting material is to be stated.

Concept Introduction:

Sodium borohydride is a strong reducing agent. The hydride ion of sodium borohydride acts as the nucleophile and gets attached to the carbonyl carbon. $\text{PCC}$ is a strong oxidizing agent, it converts the alcoholic group to aldehyde group.

Answer to Problem 27.71AP

The synthesis of given compound from the given starting material is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 3

The benzoic acid reacts with $\text{NaB}{\text{D}}_{\text{4}}$ to form deuteriated benzyl alcohol. Then deuteriated alcohol undergo oxidation reaction to form deuteriated benzaldehyde. The deuteriated benzaldehyde reacts with ammonium chloride and sodium cyanide to form deuteriated $\alpha -$ amino nitrile, which undergo hydrolysis to form desired product. The complete synthesis is shown below.

Figure 4

Conclusion

The complete synthesis of the given compound is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation:

The synthesis of given compound from the given starting material is to be stated.

Concept Introduction:

Aldehydes reacts with sodium cyanide to form nitrile compound. The nitrile compound reacts with ammonium chloride to form the $\alpha -$ amino nitrile compounds. The $\alpha -$ amino nitrile compound upon hydrolysis form $\alpha -$ amino carboxylic acid compounds.

Answer to Problem 27.71AP

The synthesis of given compound from the given starting material is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 5

The deuteriated acetaldehyde reacts with ammonium chloride and sodium cyanide to form deuteriated $\alpha -$ amino nitrile, which undergo hydrolysis to form desired product. The complete synthesis is shown below.

Figure 6

Conclusion

The synthesis of the given compound is shown in Figure 6.

Interpretation Introduction

(d)

Interpretation:

The synthesis of given compound from the given starting material is to be stated.

Concept Introduction:

Amino group reacts faster with the alkyl halides than the carboxylic group. The reaction of carboxylic acid can be performed only by protecting the amine group. Fmoc compound is used as the protecting agent for amine group. Amine and acid chloride reacts to form amide bond.

Answer to Problem 27.71AP

The synthesis of given compound from the given starting material is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 7

The cesium salt of Fmoc-proline is reacted with the resin linker. The amine is deprotecteed in pyrrolidine. Fmoc-alanine reacts with the prolin resin in presence of $\text{DCC}$ to form Fmoc-ala-pro-resin linkage. It will undergo deprotection to form alanine-proline resin.

Figure 8

The alanine-proline resin reacts with the Boc protected lysin to form the $\text{L-Lys-L-Ala-L-Pro}$ resin. Then the hydrolysis takes place in presence of trifluoroacetic acid to remove resin. The complete synthesis is shown below.

Figure 9

Figure 10

Conclusion

The synthesis of the given compound is shown in Figure 8 and Figure 9 and Figure 10.

Interpretation Introduction

(e)

Interpretation:

The synthesis of given compound from the given starting material is to be stated.

Concept Introduction:

Aldehydes reacts with sodium cyanide to form nitrile compound. The nitrile compound reacts with ammonium chloride to form the $\alpha -$ amino nitrile compounds. The $\alpha -$ amino nitrile compound upon hydrolysis form $\alpha -$ amino carboxylic acid compounds.

Answer to Problem 27.71AP

The complete synthesis of given compound is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 11

The compound $\text{ethyl-2-oxalate}$ reacts with sodium cyanide and ammonium chloride to form $\alpha -$ amino nitrile product. the cyanide ion acts as nucleophile, attacks the carbonyl carbon of the aldehyde group. Then the $\alpha -$ amino nitrile product formed reacts with acetyl chloride to form desired product.

Figure 12

Conclusion

The synthesis of the given compound is shown in Figure 12.

Interpretation Introduction

(f)

Interpretation:

The synthesis of given compound from the given starting material is to be stated.

Concept Introduction:

The alkene reacts with bromine to form dibromo alkane compound. The dibromo alkane compound is presence of sodium ethoxide and ethanol forms diene compound. Diene reacts with hydrogen bromide to form bromo alkene.

Answer to Problem 27.71AP

The complete synthesis of given compound is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 13

The compound cyclopentene undergo bromination reaction to form $\text{1, 2-dibromocyclopentane}$. The compound formed undergo $\beta -$ elimination in presence of sodium ethoxide to give $\text{cyclopenta-1, 3-diene}$, which undergo $1\text{, 4-conjugate}\text{addition}$ to form $\text{3-bromocyclopent-1-ene}$. The synthesis steps are shown below.

Figure 14

The given nitrile compound upon reaction with sodium ethoxide forms anion. The formed anion reacts with $\text{3-bromocyclopent-1-ene}$ to form amide derivative, which undergoes hydrolysis to form the desired product. The synthesis steps are shown below.

Figure 15

Conclusion

The complete synthesis of given compound is shown in Figure 14 and Figure 15.

Interpretation Introduction

(g)

Interpretation:

The synthesis of given compound from the given starting material is to be stated.

Concept Introduction:

The reaction of carboxylic acid and amines to form amide is not possible. Amines are very basic so it abstracts acidic proton from the carboxylic acid. The reaction is performed in the presence of $\text{DCC}$. The coupling reaction of acid and amine in the presence of $\text{DCC}$ forms amide.

Answer to Problem 27.71AP

The complete synthesis of given compound is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 16

The reaction between terephthalic acid and $\text{1, 4-diaminobenzene}$ reacts in presence of $\text{DCC}$ to form $\text{p-aramid}$ polymer. The carboxylate ion is formed which reacts with $\text{DCC}$. Then addition of amino group to carboxylic acid takes place which results in amide bond formation. In the last step amide compound is formed by removal of dicyclohexyl urea. The synthesis of given compound is shown below.

Figure 17

Conclusion

The synthesis of given compound is shown in Figure 17.