Problem 47 - A Complete Resource Book in Chemistry for JEE Main

Question

Arrange the following in increasing basicity order: I. \(\mathrm{Cl}_{2} \mathrm{CHCH}_{2} \mathrm{NH}_{2}\) II. \(\mathrm{Cl}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) IV. \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (a) \(\mathrm{I}<\mathrm{III}<\mathrm{II}<\mathrm{IV}\) (b) \(\mathrm{II}<\mathrm{IV}<\mathrm{III}<\mathrm{I}\) (c) \(\mathrm{IV}<\mathrm{II}<\mathrm{I}<\mathrm{III}\) (d) \(\mathrm{I}<\mathrm{II}<\mathrm{IV}<\mathrm{III}\)

Step-by-Step Solution

Answer

The answer is option (c): IV < II < I < III.

Step 1: Understanding the Basicity

The basicity of an amine is determined by the electron-donating or electron-withdrawing nature of the groups attached to the nitrogen atom. Electron-withdrawing groups lower the basicity by making it harder for the nitrogen to donate its lone pair, while electron-donating groups increase the basicity.

Step 2: Analyzing Compound I

Compound I (\(\mathrm{Cl}_{2} \mathrm{CHCH}_{2} \mathrm{NH}_{2}\)) contains two chlorine atoms attached to the carbon chain. Chlorine is an electron-withdrawing group, which decreases the basicity of the amine.

Step 3: Analyzing Compound II

Compound II (\(\mathrm{Cl}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\)) has three chlorine atoms, increasing the electron-withdrawing effect compared to compound I, leading to even lower basicity for this compound.

Step 4: Analyzing Compound IV

Compound IV (\(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\)) has a trifluoromethyl group, which is a strong electron-withdrawing group. This makes the compound less basic compared to those with fewer or weaker electron-withdrawing groups.

Step 5: Arranging in Increasing Order of Basicity

By comparing the electron-withdrawing effects, we conclude that Compound IV (\(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\)) is the least basic due to the strong effect of the trifluoromethyl group. Compound II (\(\mathrm{Cl}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\)) is next due to three chlorines. Compound I (\(\mathrm{Cl}_{2} \mathrm{CHCH}_{2} \mathrm{NH}_{2}\)) follows due to two chlorines.

Key Concepts

Electron-withdrawing groups

Electron-withdrawing groups play a crucial role in determining the basicity of organic compounds. In simple terms, these groups attract electrons towards themselves and away from the nitrogen in amines. They can significantly alter the ability of an amine to donate its lone pair of electrons, which is the essence of basicity.

Take chlorine (C"). It is considered an electron-withdrawing group. When attached to a carbon chain, chlorine atoms will pull electron density away from the nitrogen atom. This means the nitrogen in the amine is less inclined to donate its electrons, thus reducing its basicity.

The effect becomes even more pronounced with multiple electron-withdrawing groups. For instance, in the exercise, compounds with two or three chlorine atoms showed varying degrees of basicity reduction. More chlorine means more electron withdrawal, leading to lower basicity.

Amine basicity

The basicity of an amine is a fundamental concept, reflecting the amine's ability to accept protons. This ability is influenced by the nature of substituents attached to it. In amines, the basicity is largely dependent on the availability of the lone pair of electrons on the nitrogen atom.

To understand this concept better, consider the electronic environment around the nitrogen. Electron-donating groups can enhance the availability of these electrons, often increasing basicity. In contrast, electron-withdrawing groups decrease the willingness of the nitrogen atom to share its electrons, decreasing basicity.

In the context of the exercise, the clustering of multiple electron-withdrawing groups like chlorine or trifluoromethyl groups around the nitrogen atom impacts its basicity by making it more challenging for the nitrogen to participate in proton acceptance. Therefore, the presence or absence, as well as the strength of these groups, is key to comparing the basicity of different amines.

Organic chemistry problem-solving

Tackling organic chemistry problems requires a systematic approach that combines understanding of concepts with critical analysis. Problem-solving, especially in contexts like the exercise about basicity order, starts with identifying the key components of the molecules involved.

Following these steps can help:

• Start by determining the relevant functional groups in each compound.
• Assess the nature of these groups, whether they are electron-donating or electron-withdrawing.
• Compare the effects of these groups on the compound's ability to carry out reactions, such as donating electrons or accepting protons.

By understanding and ordering the electron-withdrawing power of different groups, you can predict outcomes like basicity. In complex scenarios, focusing on comparing these effects among molecules helps achieve clarity. Therefore, practice in this careful analysis not only resolves the problem at hand but also sharpens your overall chemistry problem-solving skills.