Question

Arrange the following in the increasing order of their basic character. Give reason:

Answer 1

Step 1: Understanding the Factors Affecting Basicity The basicity of amines is primarily determined by the electron density on the nitrogen atom. The more electron-donating the substituents, the more basic the amine will be, as they increase the availability of the lone pair on nitrogen for protonation. Conversely, electron-withdrawing groups decrease the electron density on nitrogen and thus reduce its basicity.
Electron-donating groups (such as \(–NH_2\) and \(–CH_3\)) increase the basicity because they increase the electron density on nitrogen.
Electron-withdrawing groups (such as \(–NO_2\) and –CN) decrease the basicity because they pull electron density away from nitrogen, making the lone pair less available to accept a proton.

Step 2: Analyzing the Substituents Let's break down the substituents attached to the nitrogen atoms in the given compounds:
\(–NH_2\) is an amine group, which is an electron-donating group. Thus, it will increase the electron density on nitrogen, making the amine more basic.
\(–NO_2\) is a strong electron-withdrawing group. It pulls electron density away from the nitrogen, thus decreasing the availability of the nitrogen's lone pair and reducing the basicity.
\(–CH_3\) is an electron-donating group, but not as strongly as \(–NH_2\) . Therefore, it increases basicity, though to a lesser extent compared to \(–NH_2\) .

Step 3: Arranging in Order Now, let's arrange the compounds in the order of increasing basicity, considering the strength of electron-donating or electron-withdrawing effects: \(–NO_2\) (electron-withdrawing) will have the lowest basicity.
\(–NO_2\) (with \(–NO_2\) at the para position) is less basic than NH₂ due to the electron-withdrawing nature of \(–NO_2\).
\(–NH_2\) will be more basic than both \(–NO_2\) and \(–CH_3\), as it is a strong electron-donating group. \(–CH_3\) is a mild electron-donating group, so it will be more basic than \(–NO_2\) but less basic than \(–NH_2\). Thus, the increasing order of basicity is: \[ \text{NO}_2 < \text{NO}_2< \text{NH}_2< \text{NH}_2 < \text{CH}_3 \]

Answer 2

Step 1: Understanding the Concept of Basicity Basicity of amines is defined as the tendency of the nitrogen atom to donate its lone pair of electrons to a proton. The basicity is higher when the lone pair of electrons on nitrogen is more available for protonation.
In aniline (C₆H₅NH₂), the lone pair of electrons on the nitrogen atom can interact with the benzene ring via resonance, making the nitrogen’s lone pair less available for protonation. This reduces the electron density on nitrogen, making aniline a weaker base than methylamine (CH₃NH₂), where the methyl group is an electron-donating group, increasing the electron density on nitrogen and making its lone pair more available for protonation.
Step 2: Resonance Effect in Aniline In aniline, the nitrogen's lone pair of electrons can be delocalized into the aromatic ring, creating a resonance structure where the nitrogen’s lone pair is not as available for protonation. This is called the resonance effect, and it reduces the basicity of aniline compared to methylamine.
Step 3: Methylamine and the Inductive Effect Methylamine, on the other hand, has a methyl group (–CH₃) attached to nitrogen. The methyl group is an electron-donating group via the inductive effect, which pushes electron density towards the nitrogen, making the nitrogen’s lone pair more available to accept a proton, thereby increasing its basicity. Thus, the basicity of methylamine is greater than that of aniline, and consequently, the pKₐ of aniline is more than that of methylamine.

Answer 3

Step 1: Understanding the Effect of Alkyl Groups on Basicity The basicity of amines in aqueous solutions increases with the number of alkyl groups attached to the nitrogen atom. Alkyl groups are electron-donating groups, and they increase the electron density on nitrogen, making the lone pair more available to accept a proton. 
Trimethylamine \((CH_3)_3N\) has three methyl groups attached to nitrogen, making it highly electron-donating and thus the most basic.
Dimethylamine \((CH_3)_2\)NH has two methyl groups, so it is also basic, but not as basic as trimethylamine.
Methylamine \((CH_3NH_2)\) has one methyl group, which donates less electron density than two or three groups.
Ammonia \((NH_3)\) has no alkyl groups, so it is the least basic.

Step 2: Arranging in Order of Basicity The increasing order of basicity is: 

\[(NH_3) < (CH_3NH_2) < (CH_3)_2NH < (CH_3)_3N\]

Answer 4

Step 1: The Ammonolysis Process Ammonolysis of alkyl halides involves reacting an alkyl halide with ammonia (NH₃), where the nucleophilic nitrogen from ammonia attacks the electrophilic carbon in the alkyl halide, displacing the halide ion and forming a primary amine. However, this reaction suffers from a major drawback: the product amine can further react with excess alkyl halide, leading to the formation of secondary and tertiary amines.
Step 2: Multiple Reactions The alkyl groups attached to the nitrogen of the amines formed during ammonolysis can make the reaction progress further to form secondary and tertiary amines. This leads to a mixture of amines, rather than a pure primary amine. For example, if methylamine (CH₃NH₂) is formed in the first step, it can then react with another molecule of alkyl halide to form dimethylamine (CH₃NHCH₃), and further to trimethylamine (CH₃N(CH₃)₂).
Step 3: Alternative Methods To obtain pure primary amines, other methods, such as reductive amination or Gabriel synthesis, are preferred as they avoid the formation of secondary and tertiary amines.