Chapter 13 Organic Compounds Containing Nitrogen (Amines)

1. Discuss the factors affecting the basicity of amines in detail. Include all structural, electronic, and solvent effects.

Answer:
The basicity of amines depends on the following factors:

1. Inductive Effect (+I Effect):

• Alkyl groups donate electron density to nitrogen via the inductive effect, increasing the availability of the lone pair for protonation.
• Example: The order of basicity in gaseous phase is:
  Tertiary amines > Secondary amines > Primary amines > Ammonia.

2. Resonance Effect:

• In aromatic amines (e.g., aniline), the lone pair on nitrogen is delocalized into the benzene ring through resonance, reducing its availability for protonation.
• Aromatic amines are less basic than aliphatic amines.

3. Steric Hindrance:

• Tertiary amines are sterically hindered due to bulky alkyl groups, which reduce their basicity, especially in aqueous solutions. This explains why the order of basicity in water is:
  Secondary amines > Primary amines > Tertiary amines.

4. Solvation Effect:

• In aqueous solutions, the cation formed after protonation is stabilized by hydrogen bonding with water.
• Primary amines form the strongest hydrogen bonds, making them more basic in water than tertiary amines.

Example for Comparison:

• Ammonia: Basicity increases in the order:
  Ammonia < Primary Amine < Secondary Amine (aqueous phase).
• Aniline vs. Aliphatic Amines: Aniline’s basicity is lower due to resonance delocalization.

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2. Describe the mechanism of the Gabriel phthalimide synthesis and explain why this method is used for preparing primary amines but not secondary or tertiary amines.

Answer:

Mechanism:

1. Formation of Potassium Phthalimide:
   Phthalimide reacts with KOH to form potassium phthalimide:

   $$\text{C}_6\text{H}_4(\text{CO})_2\text{NH} + KOH \rightarrow \text{C}_6\text{H}_4(\text{CO})_2\text{N}K + H_2O$$

2. Nucleophilic Substitution:
   Potassium phthalimide reacts with an alkyl halide ($R-X$ ) via $S_N2$ mechanism to form N-alkylphthalimide:

   $$\text{C}_6\text{H}_4(\text{CO})_2\text{N}K + R-X \rightarrow \text{C}_6\text{H}_4(\text{CO})_2\text{N}R + KX$$

3. Hydrolysis:
   The N-alkylphthalimide is hydrolyzed with an aqueous base or acid to yield a primary amine:

   $$\text{C}_6\text{H}_4(\text{CO})_2\text{N}R + 2H_2O \rightarrow R-NH_2 + \text{C}_6\text{H}_4(\text{COOH})_2$$

Why only Primary Amines?

• The reaction involves nucleophilic substitution ($S_N2$ ), which requires an unhindered substrate.
• Secondary or tertiary alkyl halides do not undergo this reaction efficiently due to steric hindrance and competition from elimination reactions.

Significance:

Gabriel synthesis is preferred for preparing pure primary amines without contamination from secondary or tertiary amines.

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3. Explain the diazotization reaction of primary aromatic amines. Discuss its mechanism and importance in organic synthesis.

Answer:

Reaction:

Primary aromatic amines react with nitrous acid (generated in situ from NaNO₂ and HCl) at 0–5°C to form diazonium salts.

General Equation:

$$\text{C}_6\text{H}_5\text{NH}_2 + HNO_2 + HCl \rightarrow \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- + 2H_2O$$

Mechanism:

1. Formation of Nitrous Acid:

   $$NaNO_2 + HCl \rightarrow HNO_2 + NaCl$$

2. Protonation of HNO₂:

   $$HNO_2 + H^+ \rightarrow NO^+ + H_2O$$

3. Electrophilic Attack on Aniline:
   The $NO^+$ electrophile reacts with the lone pair of electrons on nitrogen in aniline to form a diazonium ion.

Importance in Organic Synthesis:

• Diazonium salts are versatile intermediates used for:
  • Substitution reactions (e.g., Sandmeyer reaction, Gattermann reaction).
  • Coupling reactions to form azo dyes.
  • Preparing phenols, halides, and other aromatic derivatives.

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4. Design a numerical problem involving amines and explain the solution in detail.

Problem: Calculate the pH of a 0.1 M aqueous solution of methylamine ($CH_3NH_2$ ). Given $K_b = 4.4 \times 10^{-4}$ .

Solution:

1. Dissociation Equation:

   $$CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-$$

2. Expression for $K_b$ :

   $$K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$$

3. Assumption: Let $[OH^-] = x$ . Initially, $[CH_3NH_2] = 0.1$ . At equilibrium:

   $$K_b = \frac{x^2}{0.1 - x}$$

   Since $x \ll 0.1$ :

   $$K_b = \frac{x^2}{0.1}$$

4. Calculate $x$ :

   $$x^2 = K_b \times 0.1 = (4.4 \times 10^{-4}) \times 0.1 = 4.4 \times 10^{-5}$$ $$x = \sqrt{4.4 \times 10^{-5}} = 6.63 \times 10^{-3} \, \text{M}$$

5. Find pH:

   $$[OH^-] = 6.63 \times 10^{-3} \, \text{M}$$ $$pOH = -\log(6.63 \times 10^{-3}) = 2.18$$ $$pH = 14 - 2.18 = 11.82$$

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5. Compare the reactivity of aliphatic and aromatic amines towards electrophilic substitution and explain the influence of substituents.

Answer:

1. Aliphatic Amines:

• React readily due to high electron density on nitrogen.
• Example: Reaction with alkyl halides to form higher amines.

2. Aromatic Amines:

• Substituents on the aromatic ring influence reactivity:
  • Electron-donating groups (e.g., -CH₃, -OH): Increase reactivity by activating the ring.
  • Electron-withdrawing groups (e.g., -NO₂, -COOH): Decrease reactivity by deactivating the ring.

Electrophilic Substitution in Aniline:

• Directs electrophiles to ortho and para positions due to the activating effect of the amino group.
• Example: Bromination of aniline gives 2,4,6-tribromoaniline.

6. Illustrate and explain the Hinsberg test for the distinction of primary, secondary, and tertiary amines.

Answer:

Principle:

The Hinsberg test utilizes benzene sulfonyl chloride ($C_6H_5SO_2Cl$ ) as a reagent to distinguish between primary, secondary, and tertiary amines based on their solubility and reactivity.

Procedure and Observations:

1. Primary Amines:

   • React with $C_6H_5SO_2Cl$ to form a sulfonamide.
   • The sulfonamide is soluble in an alkaline medium due to the formation of a salt.

   Reaction:

   $$R-NH_2 + C_6H_5SO_2Cl \rightarrow R-NH-SO_2C_6H_5 + HCl$$

2. Secondary Amines:

   • React with $C_6H_5SO_2Cl$ to form a sulfonamide, but it is insoluble in an alkaline medium.

   Reaction:

   $$R_2NH + C_6H_5SO_2Cl \rightarrow R_2N-SO_2C_6H_5 + HCl$$

3. Tertiary Amines:

   • Do not react with $C_6H_5SO_2Cl$ . Hence, no sulfonamide is formed, and the amine remains insoluble in the reaction mixture.

Conclusion:

• Soluble sulfonamide indicates a primary amine.
• Insoluble sulfonamide indicates a secondary amine.
• No reaction indicates a tertiary amine.

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7. Explain the preparation of aromatic primary amines by reduction of nitro compounds with equations. Why is this method preferred for large-scale production?

Answer:

Reaction:

The reduction of nitro compounds to primary amines is achieved using various reducing agents such as tin and HCl, iron and HCl, or catalytic hydrogenation.

General Equation:

$$R-NO_2 + 6[H] \rightarrow R-NH_2 + 2H_2O$$

Example (Aniline Preparation):

$$C_6H_5NO_2 + 6[H] \xrightarrow{\text{Sn/HCl or Fe/HCl}} C_6H_5NH_2 + 2H_2O$$

Catalytic Hydrogenation:

$$C_6H_5NO_2 + 3H_2 \xrightarrow{\text{Ni/Pt}} C_6H_5NH_2 + 2H_2O$$

Advantages for Large-Scale Production:

1. High yield and purity of the product.
2. Simple process involving readily available reducing agents.
3. Minimal side reactions, particularly in catalytic hydrogenation.

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8. Discuss the synthesis, properties, and applications of diazonium salts in organic chemistry.

Answer:

Synthesis:

Diazonium salts are prepared by diazotization of primary aromatic amines with nitrous acid at 0–5°C.

Equation:

$$C_6H_5NH_2 + HNO_2 + HCl \rightarrow C_6H_5N_2^+Cl^- + 2H_2O$$

Properties:

1. Highly reactive and unstable above 5°C.
2. Soluble in water and decompose to nitrogen gas upon heating.

Applications:

1. Substitution Reactions:

   • Conversion to halides, phenols, or cyanides via Sandmeyer or Gattermann reactions.
   $$C_6H_5N_2^+Cl^- \xrightarrow{\text{CuX}} C_6H_5X + N_2$$

2. Coupling Reactions:

   • Formation of azo dyes, e.g., methyl orange.
   $$C_6H_5N_2^+Cl^- + C_6H_5OH \rightarrow C_6H_5-N=N-C_6H_4OH$$

3. Industrial Use:

   • Azo dyes for textiles and indicators in titrations.

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9. Compare the boiling points of methylamine, ethylamine, and aniline. Justify the differences based on molecular structure and hydrogen bonding.

Answer:

Comparison:

• Methylamine: Lower boiling point (~6°C) due to weaker intermolecular hydrogen bonding and smaller molecular mass.
• Ethylamine: Slightly higher boiling point (~16°C) as it has a larger molecular mass and stronger dispersion forces.
• Aniline: Highest boiling point (~184°C) due to stronger hydrogen bonding and higher molecular mass.

Explanation:

• Hydrogen Bonding: Amines can form hydrogen bonds with each other, but the strength varies depending on the alkyl or aryl group attached to nitrogen.
• Molecular Size: Larger molecules have stronger van der Waals forces, increasing boiling points.
• Aniline: The benzene ring provides additional dispersion forces, further raising the boiling point.

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10. A 0.2 M solution of ethylamine ($C_2H_5NH_2$ ) has a pH of 11.87. Calculate $K_b$ of ethylamine.

Answer:

Given Data:

• $C_2H_5NH_2$ concentration = 0.2 M
• $pH = 11.87$

Step 1: Calculate $pOH$ :

$$pOH = 14 - pH = 14 - 11.87 = 2.13$$

Step 2: Calculate $[OH^-]$ :

$$[OH^-] = 10^{-\text{pOH}} = 10^{-2.13} \approx 7.41 \times 10^{-3} \, \text{M}$$

Step 3: Use $K_b$ Expression:

$$K_b = \frac{[OH^-]^2}{[C_2H_5NH_2]}$$ $$K_b = \frac{(7.41 \times 10^{-3})^2}{0.2} = \frac{5.49 \times 10^{-5}}{0.2} = 2.745 \times 10^{-4}$$

Final Answer:

$$K_b = 2.75 \times 10^{-4}$$

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11. Discuss the reactivity of alkyl halides with ammonia in the presence of excess ammonia. Explain the reaction mechanism.

Answer:

Reaction:

Alkyl halides react with excess ammonia to form primary amines.

Mechanism:

1. Nucleophilic Substitution:
   Ammonia attacks the alkyl halide, displacing the halide ion:

   $$RX + NH_3 \rightarrow RNH_3^+ + X^-$$

2. Deprotonation:
   The protonated amine loses a proton to form a primary amine:

   $$RNH_3^+ + NH_3 \rightarrow RNH_2 + NH_4^+$$

3. Importance of Excess Ammonia:
   Excess ammonia minimizes the formation of secondary and tertiary amines by preventing further substitution of the primary amine.

Applications:

This method is commonly used to prepare pure primary amines.

12. Explain why aniline is less basic than ethylamine, both experimentally and theoretically.

Answer:

Experimentally:

• The basicity of a compound is determined by its ability to donate a lone pair of electrons.
• The $pK_b$ value for aniline ($C_6H_5NH_2$ ) is higher (~9.4) than ethylamine ($C_2H_5NH_2$ ), indicating lower basicity.

Theoretical Explanation:

1. Resonance Effect:

   • In aniline, the lone pair of electrons on nitrogen delocalizes into the benzene ring through resonance, reducing its availability for protonation.
   • In ethylamine, no such delocalization occurs, so the lone pair is fully available.

2. Inductive Effect:

   • The ethyl group in ethylamine has a +I (electron-donating) effect, increasing the electron density on nitrogen and enhancing basicity.
   • The benzene ring in aniline has a weak -I effect, which reduces the electron density on nitrogen.

3. Hybridization of Nitrogen:

   • In both, nitrogen is $sp^3$ -hybridized, but resonance in aniline weakens the basic character.

Conclusion:

Ethylamine is more basic due to the absence of resonance and the presence of a +I effect.

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13. Write the complete mechanism of Gabriel phthalimide synthesis for the preparation of primary amines. Why is this method unsuitable for aryl amines?

Answer:

Steps in the Mechanism:

1. Formation of Potassium Phthalimide:

   $$C_6H_4(CO)_2NH + KOH \rightarrow C_6H_4(CO)_2N^-K^+ + H_2O$$

2. Alkylation:
   The nucleophilic substitution of the phthalimide anion with an alkyl halide forms an N-alkylphthalimide:

   $$C_6H_4(CO)_2N^- + RX \rightarrow C_6H_4(CO)_2NR + KX$$

3. Hydrolysis:
   Acidic or basic hydrolysis of the alkylphthalimide yields a primary amine:

   $$C_6H_4(CO)_2NR + 2H_2O + HCl \rightarrow RNH_2 + C_6H_4(CO)_2H$$

Why Unsuitable for Aryl Amines?

1. Aryl halides do not undergo nucleophilic substitution easily due to resonance stabilization of the C–X bond.
2. The strong $sp^2$ -hybridized carbon in aryl halides resists attack by nucleophiles.

Conclusion:

The method is ideal for alkyl but not aryl amines.

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14. Discuss the effect of substituents on the basicity of amines with examples.

Answer:

Types of Substituents:

1. Electron-Donating Groups (+I or +M):

   • Increase the electron density on nitrogen, enhancing basicity.
   • Example: $C_6H_5NH_2$ with –CH$_3$ is more basic than aniline.

2. Electron-Withdrawing Groups (-I or -M):

   • Decrease the electron density on nitrogen, reducing basicity.
   • Example: Nitroaniline ($C_6H_5NO_2$ ) is less basic than aniline.

Specific Examples:

1. p-Toluidine ($C_6H_4NH_2$ -CH$_3$ ):
   • +I effect of –CH$_3$ increases basicity.
2. p-Nitroaniline ($C_6H_4NH_2$ -NO$_2$ ):
   • -M effect of –NO$_2$ significantly reduces basicity.

Conclusion:

Substituents affect the basicity of amines by altering electron density through inductive or resonance effects.

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15. A mixture of aniline and p-nitroaniline is separated by steam distillation. Explain the principle behind this method.

Answer:

Principle of Steam Distillation:

• Steam distillation separates components of a mixture based on their volatility and immiscibility with water.
• Aniline, being volatile, distills with steam, whereas p-nitroaniline, being less volatile, remains in the distillation flask.

Why This Works:

1. Aniline:
   • Boiling point ~184°C but distills at a lower temperature with steam due to partial pressure contribution.
2. p-Nitroaniline:
   • Higher melting point and lower volatility; does not distill easily with steam.

Procedure:

1. Add water and heat the mixture.
2. Collect the aniline-water distillate, leaving p-nitroaniline behind.

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16. Explain the role of ammonia in the Hofmann bromamide reaction.

Answer:

Hofmann Bromamide Reaction:

Primary amides react with bromine and alkali to form primary amines.

General Reaction:

$$RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$$

Role of Ammonia (Intermediary Step):

1. Formation of Intermediate:

   • Ammonia is generated in situ when the amide is treated with alkali and bromine.
   • This ammonia is nucleophilic, attacking the bromo intermediate.

2. Facilitates Rearrangement:

   • Leads to the formation of an isocyanate intermediate, which hydrolyzes to form the amine.

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17. Compare the solubility of lower and higher aliphatic amines in water.

Answer:

1. Lower Amines:

   • Highly soluble due to hydrogen bonding with water molecules.
   • Example: Methylamine and ethylamine dissolve easily.

2. Higher Amines:

   • Solubility decreases as alkyl chain length increases, reducing polarity.
   • Example: Hexylamine and octylamine are nearly insoluble.

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18. Provide the reaction of tert-butylamine with nitrous acid and explain why no evolution of gas is observed.

Answer:

Reaction:

$$(C_4H_9)NH_2 + HNO_2 \rightarrow \text{No Reaction}$$

Reason:

• Tertiary amines do not react with nitrous acid due to the absence of an alpha hydrogen necessary for diazotization or evolution of $N_2$ .

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19. Design an experiment to prepare acetanilide from aniline and justify each step.

Answer:

1. Reaction with Acetic Anhydride:

   $$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$$

2. Acetanilide Formation:

   • Acetanilide is less reactive and safer for handling than aniline.

20. Compare the reduction of nitrobenzene using different reducing agents and explain the products formed.

Answer:

Reduction of Nitrobenzene ($C_6H_5NO_2$ ) Using:

1. Tin (Sn) and HCl:

   $$C_6H_5NO_2 + 6[H] \xrightarrow[]{Sn/HCl} C_6H_5NH_2 + 2H_2O$$
   • Forms aniline as the product.

2. Iron (Fe) and HCl:

   • Similar to Sn/HCl, forms aniline.
   • Advantage: Fe is less toxic and environmentally friendly.

3. Catalytic Hydrogenation:

   $$C_6H_5NO_2 + 3H_2 \xrightarrow[]{Pt/Pd} C_6H_5NH_2 + 2H_2O$$
   • Requires expensive catalysts but is cleaner.

4. Partial Reduction (Zn/Ammonium Chloride):

   $$C_6H_5NO_2 \rightarrow C_6H_5NHOH \rightarrow C_6H_5N=O$$
   • Forms phenylhydroxylamine or azoxybenzene depending on conditions.

Comparison and Applications:

• Sn/HCl and Fe/HCl are cost-effective for industrial applications.
• Catalytic hydrogenation is preferred for high-purity aniline.

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21. Why do diazonium salts decompose on heating? Write the decomposition reaction of benzene diazonium chloride.

Answer:

Reason for Decomposition:

1. Diazonium salts contain a weak $N_2^+$ bond.
2. Heating provides energy to break this bond, liberating nitrogen gas, which is highly stable.

Reaction:

$$C_6H_5N_2Cl \xrightarrow{\Delta} C_6H_5Cl + N_2$$

Observation:

• A brisk evolution of nitrogen gas is observed.
• The reaction is exothermic.

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22. Discuss the effect of substituents on the stability of diazonium salts with examples.

Answer:

1. Electron-Donating Groups (EDG):

   • Example: –OH, –NH$_2$ , –OCH$_3$ .
   • Destabilize the diazonium salt by increasing electron density on the aromatic ring, weakening the $N_2^+$ -bond.

2. Electron-Withdrawing Groups (EWG):

   • Example: –NO$_2$ , –CN, –SO$_3$ H.
   • Stabilize the diazonium salt by delocalizing the positive charge.

Conclusion:

EWG enhances stability, making diazonium salts more useful in reactions like azo dye formation.

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23. Calculate the pH of a 0.1 M methylamine solution. $K_b = 4.4 \times 10^{-4}$ .

Answer:

Step 1: Reaction Equation

$$CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-$$

Step 2: Expression for $K_b$ :

$$K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}$$

Step 3: Approximation:

$$[OH^-] = \sqrt{K_b \cdot C} = \sqrt{4.4 \times 10^{-4} \cdot 0.1}$$ $$[OH^-] = \sqrt{4.4 \times 10^{-5}} = 6.63 \times 10^{-3} \text{ M}$$

Step 4: pH Calculation:

$$pOH = -\log{[OH^-]} = -\log{(6.63 \times 10^{-3})} = 2.18$$ $$pH = 14 - 2.18 = 11.82$$

Final Answer:

The pH is 11.82.

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24. Explain why secondary amines have higher boiling points than tertiary amines but lower boiling points than primary amines.

Answer:

1. Hydrogen Bonding:

   • Primary amines: Two hydrogens available for hydrogen bonding, leading to stronger intermolecular forces.
   • Secondary amines: Only one hydrogen for hydrogen bonding.
   • Tertiary amines: No hydrogens on nitrogen, so no hydrogen bonding.

2. Intermolecular Forces:

   • Secondary amines have weaker hydrogen bonding compared to primary amines.
   • Tertiary amines rely only on van der Waals forces, leading to even lower boiling points.

Conclusion:

Boiling points: Primary > Secondary > Tertiary.

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25. Describe the Hinsberg test for distinguishing primary, secondary, and tertiary amines.

Answer:

Principle:

The Hinsberg test utilizes the differential reactivity of amines with benzene sulfonyl chloride ($C_6H_5SO_2Cl$ ).

Procedure and Observation:

1. Primary Amines:

   • React to form sulfonamides soluble in alkali: $$RNH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHR + HCl$$

2. Secondary Amines:

   • React to form sulfonamides insoluble in alkali: $$R_2NH + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NR_2 + HCl$$

3. Tertiary Amines:

   • Do not react, as they lack hydrogen for the reaction.

Conclusion:

The Hinsberg test reliably distinguishes all three types of amines based on solubility and reactivity.