Organic Chemistry
Organic Chemistry
3rd Edition
ISBN: 9781119338352
Author: Klein
Publisher: WILEY
Question
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Chapter 3, Problem 46PP

(a)

Interpretation Introduction

Interpretation:

For the given set of reactions direction of equilibrium reaction and mechanism of the reactions has to be found.

Concept introduction:

According to Bronsted-Lowry concept, acid is proton donor and base is proton acceptor. Curved arrows show the reaction mechanism of the acid-base reaction. The mechanism of proton transfer always involves at least two curved arrows. Where one arrow should shows the abstraction of hydrogen by the base and other arrow shows the gain of negative charge by the acid.

The favored direction of equilibrium is predicted by comparing the stability of the acids or the bases in each side of the given reactions.

The anion involving in resonance in a molecule will have greater stability as compared to molecule’s having single anion. Because of the 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.

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 sp3<sp2sp    .

(b)

Interpretation Introduction

Interpretation:

For the given set of reactions direction of equilibrium reaction and mechanism of the reactions has to be found.

Concept introduction:

According to Bronsted-Lowry concept, acid is proton donor and base is proton acceptor. Curved arrows show the reaction mechanism of the acid-base reaction. The mechanism of proton transfer always involves at least two curved arrows. Where one arrow should shows the abstraction of hydrogen by the base and other arrow shows the gain of negative charge by the acid.

The favored direction of equilibrium is predicted by comparing the stability of the acids or the bases in each side of the given reactions.

The anion involving in resonance in a molecule will have greater stability as compared to molecule’s having single anion. Because of the 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.

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 sp3<sp2sp    .

(c)

Interpretation Introduction

Interpretation:

For the given set of reactions direction of equilibrium reaction and mechanism of the reactions has to be found.

Concept introduction:

According to Bronsted-Lowry concept, acid is proton donor and base is proton acceptor. Curved arrows show the reaction mechanism of the acid-base reaction. The mechanism of proton transfer always involves at least two curved arrows. Where one arrow should shows the abstraction of hydrogen by the base and other arrow shows the gain of negative charge by the acid.

The favored direction of equilibrium is predicted by comparing the stability of the acids or the bases in each side of the given reactions.

The anion involving in resonance in a molecule will have greater stability as compared to molecule’s having single anion. Because of the 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.

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 sp3<sp2sp    .

(d)

Interpretation Introduction

Interpretation:

For the given set of reactions direction of equilibrium reaction and mechanism of the reactions has to be found.

Concept introduction:

According to Bronsted-Lowry concept, acid is proton donor and base is proton acceptor. Curved arrows show the reaction mechanism of the acid-base reaction. The mechanism of proton transfer always involves at least two curved arrows. Where one arrow should shows the abstraction of hydrogen by the base and other arrow shows the gain of negative charge by the acid.

The favored direction of equilibrium is predicted by comparing the stability of the acids or the bases in each side of the given reactions.

The anion involving in resonance in a molecule will have greater stability as compared to molecule’s having single anion. Because of the 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.

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 sp3<sp2sp    .

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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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