Arrange the following in the decreasing order of basic strength in aqueous solution:

A. \(CH_3NH_2\)

B. \((CH_3)_3N\)

C. \((CH_3)_2NH\)

D. \(NH_3\)

E. \(C_6H_5NH_2\)

Choose the correct answer from the options given below:

C > A > B > D > E

The correct answer is option 2. C > A > B > D > E.

Let us go through a detailed explanation of the basicity of the given compounds in aqueous solution, focusing on the structure, electron-donating effects, steric hindrance, and resonance, all of which influence the ability of the amine nitrogen to accept a proton.

Factors That Influence Basicity:

Electron-donating or withdrawing groups: Alkyl groups (e.g., methyl groups, \(CH_3\)) are electron-donating via the inductive effect (+I effect). They push electron density towards the nitrogen, making it easier for nitrogen to accept a proton (\(H^+\)), thus increasing basicity. Aromatic rings (e.g., in aniline, \(C_6H_5NH_2\)) are electron-withdrawing. The lone pair of nitrogen can delocalize into the aromatic ring, reducing the electron density available for protonation, which weakens basicity.

Steric hindrance: In tertiary amines like trimethylamine \((CH_3)_3N\), the bulky methyl groups around the nitrogen create steric hindrance, making it harder for the nitrogen to interact with protons in solution, thus reducing the basicity.

Solvation in aqueous solution: In water, the protonated amines need to be stabilized by solvation. Smaller or less hindered amines are better solvated, which helps to stabilize the protonated form, increasing their basicity.

Detailed Analysis of Each Compound:

C. \( (CH_3)_2NH \) (Dimethylamine):

It is a secondary amine with two electron-donating methyl groups. The two methyl groups increase electron density on the nitrogen through the +I effect. This makes it easier for nitrogen to accept a proton, increasing its basicity. Moderate steric hindrance, but not as much as in tertiary amines. This is the most basic among the given compounds due to the strong +I effect from two methyl groups and relatively low steric hindrance.

A. \( CH_3NH_2 \) (Methylamine):

It is a primary amine with one methyl group attached to nitrogen. The single methyl group donates electron density via the +I effect, but to a lesser extent than dimethylamine. Very little steric hindrance, so it’s easily solvated in water. Methylamine is slightly less basic than dimethylamine because there’s only one methyl group donating electron density.

B. \( (CH_3)_3N \) (Trimethylamine):

A tertiary amine with three methyl groups attached to nitrogen. Although three methyl groups are present, their combined +I effect is offset by significant steric hindrance, which makes it difficult for nitrogen to accept a proton. High steric hindrance due to the bulk of the three methyl groups. This reduces its ability to accept protons and makes it harder to be solvated in water. Despite having three electron-donating groups, trimethylamine is less basic than dimethylamine and methylamine due to steric hindrance

D. \( NH_3 \) (Ammonia):

A simple amine with no alkyl groups. Ammonia has no electron-donating alkyl groups, so it has less electron density on nitrogen compared to alkyl-substituted amines. None, so it is easily solvated. Ammonia is less basic than methylamine and dimethylamine because there are no electron-donating groups to enhance the electron density on nitrogen.

E. \( C_6H_5NH_2 \) (Aniline):

An aromatic amine with a phenyl group attached to nitrogen. The lone pair on the nitrogen in aniline can delocalize into the aromatic ring, reducing the electron density on nitrogen. This makes it much less available for protonation. Minimal, but the delocalization of the lone pair significantly reduces its basicity. Aniline is the least basic among the given compounds due to the electron-withdrawing effect of the phenyl ring.

Conclusion:

Based on these factors, the decreasing order of basic strength in aqueous solution is as follows:

\(\text{C} > \text{A} > \text{B} > \text{D} > \text{E}\)

Where:

C = \((CH_3)_2NH\) (Dimethylamine) is the most basic due to strong +I effect and moderate steric hindrance.

A = \(CH_3NH_2\) (Methylamine) is next, slightly less basic than dimethylamine.

B = \((CH_3)_3N\) (Trimethylamine) is weaker due to steric hindrance.

D = \(NH_3\) (Ammonia) is weaker than alkylamines due to lack of electron-donating groups.

E = \(C_6H_5NH_2\) (Aniline) is the least basic due to electron delocalization into the phenyl ring.

Thus, option 2: C > A > B > D > E is the correct answer.