Amines

Due to salt formation with AlCl , aniline becomes a Lewis base, deactivating the benzene ring.
Step 1: Role of Lewis Acid Catalyst in Friedel-Crafts Reaction - Friedel-Crafts alkylation and acylation require a Lewis acid catalyst like AlCl to generate the electrophile.

Step 2: Interaction Between Aniline and AlCl - Aniline ( ) has a lone pair on nitrogen, which interacts with AlCl , forming a salt. - This deactivates the benzene ring, making it less reactive toward electrophilic substitution.

Step 3: Conclusion Due to this salt formation, aniline does not undergo Friedel-Crafts reaction.

N,N-diethylbenzenesulphonamide does not contain any hydrogen atom attached to nitrogen, making it non-acidic and thus insoluble in alkali.
Step 1: Understanding the Nature of N,N-Diethylbenzenesulphonamide - In order for a compound to dissolve in alkali, it must have an acidic hydrogen atom that can react with OH .

Step 2: Why It Is Insoluble - Sulphonamides that contain a hydrogen attached to nitrogen are acidic and dissolve in alkali. - However, in N,N-diethylbenzenesulphonamide, both hydrogen atoms on nitrogen are replaced by ethyl groups, making it non-acidic and hence, insoluble in alkali.

Here's the explanation:

1. Transformation Objective:
The reaction requires converting propanamide to propanamine, which involves reducing the amide group (-CONH₂) to an amine group (-CH₂NH₂).

2. Reagent Analysis:

A) Excess H₂:
While catalytic hydrogenation can reduce some functional groups, it's generally ineffective for directly reducing amides to amines. It requires harsh conditions and may cause unwanted side reactions.

B) Br₂ in aqueous NaOH (Hoffmann Bromamide Degradation):
This reagent degrades amides into primary amines with one fewer carbon atom (via an isocyanate intermediate). This would convert propanamide (C₃) to ethanamine (C₂), which is not the desired product.

C) Iodine in the presence of red phosphorus:
This combination is typically used for reducing carboxylic acids to alkanes and is ineffective for amide-to-amine reduction.

D) LiAlH₄ in ether:
Lithium aluminum hydride (LiAlH₄) is a powerful reducing agent that directly converts amides to amines by replacing the carbonyl oxygen with hydrogens, making it the ideal choice for this transformation.

Conclusion:
Option (D) LiAlH₄ in ether is the only reagent that achieves the desired conversion without altering the carbon chain length.

Final Answer:
The correct answer is (D).

To arrange the given amines in increasing order of their basic character, we need to consider the electronic effects of the substituents on the aromatic ring and their positions relative to the amino group ( ).

Given Compounds:
1. Aniline ( ) 2. p-Nitroaniline ( ) 3. p-Toluidine ( )

Key Concepts:

1. Basicity of Amines: The lone pair on nitrogen determines basicity. Electron-donating groups ( ) increase electron density on nitrogen, enhancing basicity, while electron-withdrawing groups ( ) decrease it.
2. Positional Effects: The effect of substituents is stronger at the para position than at the meta position due to resonance and inductive effects.

Analysis of Given Amines:

1. p-Nitroaniline ( ): The group is strongly electron-withdrawing through both resonance and inductive effects, significantly reducing the electron density on nitrogen. This makes it the weakest base among the three.
2. Aniline ( ): The group in aniline is less basic than aliphatic amines due to resonance delocalization of the lone pair into the benzene ring. It has no additional substituents to alter its basicity further.
3. p-Toluidine ( ): The group is electron-donating (+I effect), increasing electron density on nitrogen. This makes it the strongest base among the three.

Increasing Order of Basicity:

Reason:
The group destabilizes the protonated form (ammonium ion) by withdrawing electron density, while the group stabilizes it by donating electron density. Aniline, with no substituents, has intermediate basicity.

Final Answer:
The increasing order of basic character is:

To solve the problem, we need to explain why amides are less basic than amines.

1. Understanding Basicity:
Basicity refers to a compound's ability to accept protons (H⁺). The more readily a compound can donate its lone pair of electrons to bond with a proton, the more basic it is. In general, the basicity of nitrogen-containing compounds like amines and amides depends on the electron density around the nitrogen atom, as it is the site of protonation.

2. Structure of Amines:
Amines (R-NH₂) have a nitrogen atom with a lone pair of electrons. This lone pair is readily available for protonation, making amines relatively basic. The nitrogen in amines is not involved in any resonance structures that significantly reduce its electron density, allowing it to readily accept a proton.

3. Structure of Amides:
Amides (R-CO-NH₂) contain a carbonyl group (C=O) attached to a nitrogen atom. The nitrogen in amides is involved in resonance with the carbonyl group. This resonance delocalizes the lone pair of electrons on the nitrogen, making it less available for protonation. The carbonyl oxygen pulls electron density away from the nitrogen, which significantly reduces the nitrogen's ability to accept a proton.

4. Effect of Resonance on Basicity:
In amides, the resonance between the nitrogen's lone pair and the carbonyl group weakens the nitrogen's electron density, making it less basic. In contrast, amines do not experience this resonance effect, so the nitrogen’s lone pair remains more available for protonation, making them more basic.

5. Final Answer:
Amides are less basic than amines because the nitrogen in amides is involved in resonance with the carbonyl group, which reduces the availability of the nitrogen's lone pair for protonation. In contrast, the nitrogen in amines is not involved in such resonance and retains a higher electron density, making amines more basic.

To solve the problem, we need to explain why aromatic primary amines cannot be prepared by the Gabriel Phthalimide synthesis.

1. Understanding the Gabriel Phthalimide Synthesis:
The Gabriel Phthalimide synthesis is a method used to prepare primary amines by alkylating phthalimide (a derivative of phthalic acid) with an alkyl halide in the presence of a strong base like potassium hydroxide (KOH). This results in the formation of an alkylated phthalimide, which, upon hydrolysis, yields the corresponding primary amine.

2. The Mechanism:
The Gabriel synthesis works well with alkyl halides, where the alkyl group can be easily transferred to the phthalimide anion. The reaction is typically carried out in a basic medium, which deprotonates the phthalimide, making it a good nucleophile. The alkyl halide then reacts with the phthalimide anion to form an imide intermediate, which, after hydrolysis, leads to the primary amine.

3. Issue with Aromatic Compounds:
The problem arises when trying to apply this reaction to aromatic amines (amines attached to a benzene ring). Aromatic amines are not easily alkylated using the Gabriel synthesis due to the instability of the intermediate that would form when phthalimide reacts with an aromatic alkyl halide.

4. The Challenge with Aromatic Rings:
Aromatic rings, such as benzene, have a stable π-electron cloud that resists nucleophilic attack. In addition, when trying to alkylate the phthalimide with an aromatic alkyl group (like phenyl group), the reaction is not efficient. This is due to the instability of the intermediate carbocation that would be formed in an aromatic system during the alkylation step.

5. Final Answer:
Aromatic primary amines cannot be prepared by the Gabriel Phthalimide synthesis because aromatic rings resist nucleophilic substitution reactions due to their stable electron structure, and the alkylation step does not proceed efficiently.

To solve the problem, we need to arrange the given compounds in increasing order of their boiling points.

1. Understanding the Boiling Points:
Boiling points of compounds are influenced by intermolecular forces such as hydrogen bonding, dipole-dipole interactions, and London dispersion forces. Stronger intermolecular forces usually result in higher boiling points.

2. Analyzing the Compounds:
The given compounds are:
- (CH₃)₂NH (Dimethylamine)
- CH₃CH₂NH₂ (Ethylamine)
- CH₃CH₂OH (Ethanol)

3. Boiling Points of Compounds:
- (CH₃)₂NH (Dimethylamine): This compound is a simple amine. It exhibits hydrogen bonding, but it is weaker compared to alcohols. - CH₃CH₂NH₂ (Ethylamine): This is another amine, but it has a slightly higher boiling point than dimethylamine due to its larger molecular size and stronger dipole-dipole interactions. - CH₃CH₂OH (Ethanol): This compound has a hydroxyl group (-OH), which allows for strong hydrogen bonding, leading to a higher boiling point compared to the amines.

4. Order of Boiling Points:
We can now arrange the compounds in increasing order of their boiling points based on the strength of the intermolecular forces:

Increasing Order of Boiling Points:
(CH₃)₂NH < CH₃CH₂NH₂ < CH₃CH₂OH

Final Answer:
The compounds in increasing order of their boiling points are:
(CH₃)₂NH < CH₃CH₂NH₂ < CH₃CH₂OH.