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Arrange \(\mathrm{CCl}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{NH}_{2}\) (a), \(\mathrm{CCl}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{NH}_{2}\) (b) and \(\mathrm{CCl}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (c)in order of decreasing basic nature. (a) \(C>B>A\) (b) \(C>A>B\) (c) \(\mathrm{A}>\mathrm{C}>\mathrm{B}\) (d) \(\mathrm{B}>\mathrm{A}>\mathrm{C}\)

Short Answer

Expert verified
The order of decreasing basicity is A > B > C, hence option (c) is correct.

Step by step solution

01

Understanding Basicity

The basic nature in organic compounds like amines is determined by the availability of the lone electron pair on the nitrogen atom to accept protons. Electron-withdrawing groups reduce basicity because they pull electron density away from the nitrogen, making it less likely to donate its lone pair.
02

Analyzing the Effects of Trichloromethyl Groups

In each of the given compounds, the presence of a trichloromethyl group (CCl₃) is crucial. It is a strong electron-withdrawing group, which affects the availability of the nitrogen's lone pair. The more CCl₃ groups there are between the nitrogen and the rest of the molecule, the more electron-withdrawing effect there is.
03

Comparing Compounds by Electron-withdrawing Effects

The compounds are: - (a) CCl₃(CH₂)₃NH₂, where the CCl₃ group is furthest from the amine. - (b) CCl₃(CH₂)₂NH₂, where the CCl₃ group is moderately close to the amine. - (c) CCl₃CH₂NH₂, where the CCl₃ group is closest to the amine. The order of electron-withdrawing effect on the nitrogen is then: c > b > a.
04

Arranging by Basicity

Given that more electron-withdrawing power means less basicity, arrange them in decreasing order of basicity: - (a) CCl₃(CH₂)₃NH₂ (least affected by CCl₃, most basic) - (b) CCl₃(CH₂)₂NH₂ - (c) CCl₃CH₂NH₂ (most affected by CCl₃, least basic)
05

Choosing the Correct Answer

According to the decreasing basicity order, we have: A > B > C. Thus, the correct option is (c) A > C > B.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Basicity of Amines
Amines are organic compounds that contain a nitrogen atom with a lone pair of electrons. The basicity of amines, or how readily they can accept protons, is a crucial concept in organic chemistry. This ability is primarily determined by the availability of their lone electron pair.

Nitrogen in amines can act as a Lewis base when it donates its lone pair to bond with a proton. The more available this lone pair is, the stronger the amine as a base. Factors influencing the availability of the lone pair include the surrounding chemical environment, particularly the presence of electron-donating or electron-withdrawing groups that can affect electron density around the nitrogen atom.

Understanding these concepts helps predict how different amines will react in chemical processes and how this can be manipulated in synthetic chemistry applications.
Electron-withdrawing Groups
Electron-withdrawing groups (EWGs) are atoms or groups of atoms that are highly electronegative. These groups pull electron density towards themselves and away from the rest of the molecule.

In the context of amines, if these groups are located near the nitrogen, they can significantly reduce the basicity of the amine. This is because they decrease the lone pair's availability on nitrogen by delocalizing its electron density.

A common impact of EWGs is observed in the acidity, basicity, and overall reactivity of the compound. When an EWG is close to the reactive center of a molecule, it can decrease the strength of bases while increasing the strength of acids. This occurs due to a shift in electron density, making the lone pair on nitrogen less likely to bind with protons.
Trichloromethyl Group
The trichloromethyl group, represented as \( \text{CCl}_3 \), is a powerful electron-withdrawing group. Its high electronegativity comes from the three chlorine atoms which are highly electronegative, pulling electrons toward themselves.

In molecules where a trichloromethyl group is present, it exerts a strong electron-withdrawing effect, particularly if it is close to the functional groups within the molecule, such as an amine group.

When \( \text{CCl}_3 \) is close to an amine, it decreases the electron density on the nitrogen atom. This effect reduces the availability of the lone pair on nitrogen, thereby decreasing the basicity of the amine. The closer the \( \text{CCl}_3 \) group is, the more significant the effect, making it an important structural feature to consider in synthesis and reactions.
Lone Pair Availability
The availability of a nitrogen's lone pair of electrons in amines is a fundamental concept that dictates their chemical behavior. How easily a nitrogen can use its lone pair to form a bond with a proton or other electrophiles determines an amine's basicity.

Several factors influence this availability, including the electron density around the nitrogen, which can be affected by nearby substituents.
  • Electron-donating groups increase availability by pushing more electron density towards the nitrogen.
  • Electron-withdrawing groups decrease availability by pulling electron density away.

When approaching problems involving amines, it's essential to look at the whole molecular structure. Understanding how various substituents, such as the trichloromethyl group, impact the lone pair of electrons allows for predicting the amine behavior in chemical reactions. By analyzing these aspects, chemists can manipulate reaction conditions and molecular design for desired outcomes in synthesis and functionalization.

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Most popular questions from this chapter

Which of the following reaction does not involve a carbocation as intermediate? (a) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \mathrm{AlBr}_{3}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{Br}_{2} \longrightarrow \mathrm{BrCH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{HBr} \mathrm{H}^{+}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{H}_{2} \mathrm{O}\) (d) Both (b) and (c)

Arrange the following compounds in order of increasing dipole moment. \(\mathrm{Cl}\) Clc1cccc(Cl)c1Cl Clc1ccc(Cl)cc1 Clc1ccc(Cl)c(Cl)c1 (I) (II) (III) ClC=C=C=CCl Clc1cc(Cl)cc(Cl)c1 (IV) (V) (a) \(\mathrm{II}=\mathrm{IV}<\mathrm{V}<\mathrm{I}<\mathrm{III}\) (b) \(\mathrm{IV}<\mathrm{III}<\mathrm{I}\) (c) \(\mathrm{II}<\mathrm{IV}<\mathrm{III}<\mathrm{I}<\mathrm{V}\) (d) \(\mathrm{V}=\mathrm{II}=\mathrm{IV}\)

Which of the following species on photolysis does give a carbene? (a) CC(C)=O (b) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}\) (c) \(\mathrm{CCl}_{4}\) (d) \(\mathrm{CHCl}_{3}\)

Which of the following molecules has all the effects: inductive, mesomeric and Baker Nathan effect? (a) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{C}-\mathrm{CH}_{3}\)

The stability of the following carbocations decreases in the order. (1) [C]1CCCCC1 (2) [C+]1CC1 (3) [C+]1C=CC=C1 (4) [C+]1C=CC=CC=C1 (a) \(3>2>4>1\) (b) \(4>2>3>1\) (c) \(4>3>2>1\) (d) \(4>2>1>3\)

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