Organic Chemistry
Organic Chemistry
3rd Edition
ISBN: 9781119338352
Author: Klein
Publisher: WILEY
Question
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Chapter 3, Problem 44PP

(a)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  1.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

(b)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  2.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

(c)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  3.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

(d)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  4.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

(e)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  5.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

(f)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  6.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

(g)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  7.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

(h)

Interpretation Introduction

Interpretation:

The more acidic compound from each given pair of compounds should be identified in each of the given anions.

Concept introduction:

Acidic strength in the molecule depends on the stability of the Conjugate anion.

The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.

In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.

The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is Organic Chemistry, Chapter 3, Problem 44PP , additional homework tip  8.

Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.

To find: the more acidic compound from each given pair of compounds.

Blurred answer

Chapter 3 Solutions

Organic Chemistry

Ch. 3.3 - Prob. 8ATSCh. 3.3 - Prob. 9ATSCh. 3.3 - Prob. 4LTSCh. 3.3 - Prob. 10PTSCh. 3.3 - Prob. 11ATSCh. 3.3 - Prob. 12CCCh. 3.4 - Prob. 5LTSCh. 3.4 - Prob. 13PTSCh. 3.4 - Prob. 14ATSCh. 3.4 - Prob. 6LTSCh. 3.4 - Prob. 15PTSCh. 3.4 - Prob. 16ATSCh. 3.4 - Prob. 17ATSCh. 3.4 - Prob. 7LTSCh. 3.4 - Prob. 18PTSCh. 3.4 - Prob. 19PTSCh. 3.4 - Prob. 20ATSCh. 3.4 - Prob. 8LTSCh. 3.4 - Prob. 21PTSCh. 3.4 - Prob. 22ATSCh. 3.4 - Prob. 9LTSCh. 3.4 - Prob. 23PTSCh. 3.4 - Prob. 24PTSCh. 3.4 - Prob. 25ATSCh. 3.4 - Prob. 26ATSCh. 3.5 - Prob. 10LTSCh. 3.5 - Prob. 27PTSCh. 3.5 - The development of chemical sensors that can...Ch. 3.5 - Determine whether H2O would be a suitable reagent...Ch. 3.5 - Prob. 29PTSCh. 3.5 - Prob. 30ATSCh. 3.7 - Prob. 31CCCh. 3.9 - Prob. 12LTSCh. 3.9 - Prob. 32PTSCh. 3.9 - Prob. 33ATSCh. 3 - Prob. 34PPCh. 3 - Prob. 35PPCh. 3 - Prob. 36PPCh. 3 - Prob. 37PPCh. 3 - Prob. 38PPCh. 3 - Prob. 39PPCh. 3 - Prob. 40PPCh. 3 - Prob. 41PPCh. 3 - Prob. 42PPCh. 3 - Prob. 43PPCh. 3 - Prob. 44PPCh. 3 - Prob. 45PPCh. 3 - Prob. 46PPCh. 3 - Prob. 47PPCh. 3 - Prob. 48PPCh. 3 - Prob. 49IPCh. 3 - Prob. 50IPCh. 3 - Prob. 51IPCh. 3 - Prob. 52IPCh. 3 - Prob. 53IPCh. 3 - Prob. 54IPCh. 3 - Prob. 55IPCh. 3 - Prob. 56IPCh. 3 - Prob. 57IPCh. 3 - Prob. 58IPCh. 3 - Prob. 59IPCh. 3 - Prob. 60IPCh. 3 - Prob. 61IPCh. 3 - Prob. 62IPCh. 3 - Prob. 63IPCh. 3 - Prob. 64IPCh. 3 - The bengamides are a series of natural products...Ch. 3 - Prob. 66IPCh. 3 - Prob. 67IPCh. 3 - Prob. 68IPCh. 3 - Prob. 69IPCh. 3 - Prob. 70CPCh. 3 - Prob. 71CPCh. 3 - Prob. 72CPCh. 3 - Prob. 73CP
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