Chapter 23, Problem 23.65AP

Interpretation Introduction

(a)

Interpretation:

The reaction of ammonia with nitrous acid is to be completed. Also, the major product formed in the corresponding reaction is to be stated.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia with a substituent. It may be alkyl or aryl group. The formation of diazonium salt from aromatic amines takes place using sodium nitrite and hydrohalide at low temperatures. This process is known as diazotization. Aryl diazonium salts undergo a variety of specific substitution reactions in which the nucleophilic Z group replaces (a very good leaving group) to form corresponding products.

Answer to Problem 23.65AP

The reaction of ammonia with nitrous acid is shown below.

The major product formed is the nitrogen gas.

Explanation of Solution

When ammonia reacts with nitrous acid, diazotization occurs with the formation of ammonium nitrite intermediate which dissociates and liberates nitrogen gas along with the formation of water.

The complete reaction is shown below.

Figure 1

Conclusion

The reaction of ammonia with nitrous acid is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

The reaction between $\text{7a-methylhexahydro-1}H\text{-pyrrolizine}$ and methyl iodide and then with silver oxide and heat is to be completed. Also, the major product for the corresponding reaction is to be stated.

Concept introduction:

The conversion of amines into alkenes can be achieved by Hoffmann elimination reaction. Amines have poor leaving groups. They react with excess of alkyl halides to form ammonium salts. These ammonium salts undergo $\text{β}$-elimination reaction to form alkenes. The $\text{β}$-elimination refers to removal of proton from the $\text{β}$-carbon atom by the base.

Answer to Problem 23.65AP

The complete reaction between $\text{7a-methylhexahydro-1}H\text{-pyrrolizine}$ and methyl iodide and then with silver oxide and heat is shown below.

The major product will be compound $B$ which contains a stable five-membered cyclic ring.

Explanation of Solution

Hofmann elimination reaction takes place when $\text{7a-methylhexahydro-1}H\text{-pyrrolizine}$ reacts with methyl iodide and then with silver oxide. The intermediate, quaternary ammonium hydroxide formed in this reaction, consists of two $\text{β-}$ carbon atoms bearing hydrogen atom. So, two alkenes are formed by anti-elimination. The complete reaction is shown below.

Figure 2

Conclusion

The complete reaction between $\text{7a-methylhexahydro-1}H\text{-pyrrolizine}$ and methyl iodide and then with silver oxide and heat is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

The reaction between $\text{cyclopropane-1, 2, 3-tricarbonyl}\text{trichloride}$ with dimethyl amine followed by reduction with ${\text{LiAlH}}_{\text{4}}$ in water and then with methyl iodide and silver oxide is to be stated. Also, the major product is to be stated.

Concept introduction:

The conversion of amines into alkenes can be achieved by Hoffmann elimination reaction. Amines have poor leaving groups. They react with excess of alkyl halides to form ammonium salts. These ammonium salts undergo $\text{β}$-elimination reaction to form alkenes. The $\text{β}$-elimination refers to removal of proton from the $\text{β}$-carbon atom by the base.

Answer to Problem 23.65AP

The reaction between $\text{cyclopropane-1, 2, 3-tricarbonyl}\text{trichloride}$ with dimethyl amine followed by reduction with ${\text{LiAlH}}_{\text{4}}$ in water and then with methyl iodide and silver oxide is shown below.

The major product formed is $\text{1, 2, 3-trimethylenecyclopropane}$.

Explanation of Solution

When $\text{cyclopropane-1, 2, 3-tricarbonyl}\text{trichloride}$ reacts with dimethyl amine, triamide is formed which undergoes reduction on reaction with ${\text{LiAlH}}_{\text{4}}$ to form amine. The amine formed undergoes Hofmann elimination reaction with methyl iodide and silver oxide to generate an alkene which is an isomer of benzene. The complete reaction is shown below.

Figure 3

Conclusion

The reaction between $\text{cyclopropane-1, 2, 3-tricarbonyl}\text{trichloride}$ with dimethyl amine followed by reduction with ${\text{LiAlH}}_{\text{4}}$ in water and then with methyl iodide and silver oxide is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

The reaction of phenol with hydrochloric acid and sodium nitrite is to be completed. Also, the major product of the corresponding reaction is to be stated.

Concept introduction:

Electrophilic aromatic substitution reaction involves the substitution of an electrophile on an aromatic ring. The $\text{π-}$ bonds of an aromatic ring makes it electron rich which attracts an electron deficient electrophile towards itself and a substitution reaction takes place.

Answer to Problem 23.65AP

The reaction of phenol with hydrochloric acid and sodium nitrite is shown below.

The major product formed is $\text{4-nitrosophenol}$.

Explanation of Solution

The compounds $\text{HCl}$ and $\text{NaOH}$ reacts to generate nitrous acid. Electrophilic substitution reaction occurs between phenol and nitrous acid to give two niroso substituted products. The complete reaction is shown below.

Figure 4

Conclusion

The reaction of phenol with hydrochloric acid and sodium nitrite is shown in Figure 4.

Interpretation Introduction

(e)

Interpretation:

The reaction between $\text{4-nitroaniline}$ with nitrous acid is to be completed. The major product in the corresponding reaction is to be stated.

Concept introduction:

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia with a substituent. It may be alkyl or aryl group. The formation of diazonium salt from aromatic amines takes place using sodium nitrite and hydrohalide at low temperatures. This process is known as diazotization. Aryl diazonium salts undergo a variety of specific substitution reactions in which the nucleophilic Z group replaces (a very good leaving group) to form corresponding products.

Answer to Problem 23.65AP

The reaction between $\text{4-nitroaniline}$ with nitrous acid is shown below.

The major product in the corresponding reaction is $\text{1, 4-dinitrobenzene}$.

Explanation of Solution

When $\text{4-nitroaniline}$ reacts with ${\text{HNO}}_2$, diazotization reaction takes place and the amine group is converted into diazonium salt. The diazonium salt formed further reacts with ${\text{CuNO}}_{\text{2}}$ to form $\text{1, 4-dinitrobenzene}$. The complete reaction is shown below.

Figure 5

Conclusion

The reaction between $\text{4-nitroaniline}$ with nitrous acid is shown in Figure 5.

Interpretation Introduction

(f)

Interpretation:

The reaction between butylamine with oxirane is to be completed. The major product of the corresponding reaction is to be stated.

Concept introduction:

Nucleophilic substitution reaction is the reaction in which a nucleophile attacks the electrophilic center and a substituted product is formed. It takes place by the generation of an electrophilic intermediate.

Answer to Problem 23.65AP

The reaction between butylamine with oxirane is shown below.

Explanation of Solution

Oxirane is a symmetrical molecule. Therefore, the nucleophile can attack at any carbon. Nucleophilic substitution reaction takes place when butyl amine reacts with oxirane. The lone pair of electrons attacks the carbon atom of the oxirane which opens the ring. Since, the oxirane is in excess, so, the amine will react with oxirane till it’s all the hydrogen groups are replaced by the oxirane to yield a quaternary ammonium derivative as shown below.

Figure 6

Conclusion

The reaction between butylamine with oxirane is show in Figure 6.

Interpretation Introduction

(g)

Interpretation:

The reaction of $\text{2, 2-dimethyl}\text{}\text{oxirane}$ with diethylamine is to be completed.

Concept introduction:

Nucleophilic substitution reaction is the reaction in which a nucleophile attacks the electrophilic center and a substituted product is formed. It takes place by the generation of an electrophilic intermediate.

Answer to Problem 23.65AP

The reaction of $\text{2, 2-dimethyl}\text{}\text{oxirane}$ with diethylamine is shown below.

Explanation of Solution

The reaction between $\text{2, 2-dimethyl}\text{}\text{oxirane}$ and diethylamine occurs by ${\text{S}}_{\text{N}}\text{1}$ reaction mechanism. Therefore, substitution of nucleophile takes place at the most hindered site since it will form a tertiary cation as an intermediate. The complete reaction is shown below.

Figure 7

Conclusion

The reaction between $\text{2, 2-dimethyl}\text{}\text{oxirane}$ and diethylamine is shown in Figure 7.

Interpretation Introduction

(h)

Interpretation:

The reaction between phthalimide and $\text{KOH}$ and $\text{2-methyloxirane}$ is to be completed.

Concept introduction:

Gabriel synthesis is a reaction in which a primary amine is formed when a phthalimide anion is alkylated and then undergoes hydrolysis. It is the best method to prepare primary aliphatic amines. Aromatic amines cannot be prepared by this reaction.

Answer to Problem 23.65AP

The reaction between phthalimide and $\text{KOH}$ and $\text{2-methyloxirane}$ is shown below.

Explanation of Solution

Gabriel phthalimide reaction occurs when phthalimide reacts with $\text{KOH}$ to yield phthalimide anion which gets alkylated on reaction with $\text{2-methyloxirane}$ and on hydrolysis yields $\text{1-aminopropan-2-ol}$. The complete reaction is shown below.

Figure 8

Conclusion

The reaction between phthalimide and $\text{KOH}$ and $\text{2-methyloxirane}$ is shown in Figure 8.

Interpretation Introduction

(i)

Interpretation:

The reaction between $\text{2, 3-diphenyloxirane}$ and piperidine is to be completed. Also, the major product is to be stated.

Concept introduction:

Nucleophilic substitution reaction is the reaction in which a nucleophile attacks the electrophilic center and a substituted product is formed. It takes place by the generation of an electrophilic intermediate.

Answer to Problem 23.65AP

The nucleophilic substitution reaction between $\text{2, 3-diphenyloxirane}$ and piperidine is shown below.

Two enantiomeric products will be formed in equal proportions because the substrate is symmetrical, so, the attack of nucleophile is feasible equally at both the carbon sites in substituted oxirane.

Explanation of Solution

Nucleophilic substitution reaction takes place when $\text{2, 3-diphenyloxirane}$ reacts with piperidine. Since $\text{2, 3-diphenyloxirane}$ is a symmetrical compound, the nucelophile will attack at both the carbons of $\text{2, 3-diphenyloxirane}$ to form two enantiomeric products in equal amounts. The complete reaction is shown below.

Figure 9

Conclusion

The reaction between $\text{2, 3-diphenyloxirane}$ and piperidine is shown in Figure 9.

Interpretation Introduction

(j)

Interpretation:

The reaction between $\text{2, 3-dibromopropene}$ and diethylamine is to be completed.

Concept introduction:

Nucleophilic substitution reaction is the reaction in which a nucleophile attacks the electrophilic center and a substituted product is formed. It takes place by the generation of an electrophilic intermediate.

Answer to Problem 23.65AP

The reaction between $\text{2, 3-dibromopropene}$ and diethylamine is shown below.

Explanation of Solution

The nucleophilic substitution reaction takes place when diethyl amine reacts with $\text{2, 3-dibromopropene}$. The nucleophile attacks on the alkyl bromide and not the alkenyl bromide because alkenyl bromide is already electron rich due to the presence of double bonds. So, electron-electron repulsion will take place between nucleophile and alkenyl bromide. The complete reaction is shown below.

Figure 10

Conclusion

The reaction between $\text{2, 3-dibromopropene}$ and diethylamine is shown in Figure 10.

Interpretation Introduction

(k)

Interpretation:

The reaction of nitrobenzene with ${\text{H}}_{\text{2}}\text{,}\text{Pd/C}$ and butanal is to be completed.

Concept introduction:

Catalytic hydrogenation is a reduction process of addition of hydrogen atoms in an alkene or an alkyne by using a metal catalyst. The number of bonds is reduced between carbon atoms. Amines can be prepared from nitrobenzene by reduction with ${\text{H}}_{\text{2}}\text{,}\text{Pd/C}$, $\text{Sn/HCl}$, or $\text{Fe/HCl}$. When amines are reacted with aldehydes, imines are formed.

Answer to Problem 23.65AP

The reaction of nitrobenzene with ${\text{H}}_{\text{2}}\text{,}\text{Pd/C}$ and butanal is shown below.

Explanation of Solution

When nitrobenzene is catalytically hydrogenated with ${\text{H}}_{\text{2}}\text{,}\text{Pd/C}$, the amine is formed which reacts with butanal to form an imine. Imine on reduction with ${\text{H}}_{\text{2}}\text{,}\text{Pd/C}$ forms a disubstituted amine. The reaction is shown below.

Figure 11

Conclusion

The reaction of nitrobenzene with ${\text{H}}_{\text{2}}\text{,}\text{Pd/C}$ and butanal is shown in Figure 11.