(a) Interpretation: It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position. Concept introduction: The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

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Answer 1

Question

Chapter 6, Problem 6.76P

Interpretation Introduction

(a)

Interpretation:

It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position.

Concept introduction:

The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

Interpretation Introduction

(b)

Interpretation:

It is to be explained how the meta isomer of nitro phenol has the highest pKa value than ortho and para isomer.

Concept introduction:

Greater the resonance stabilization of the conjugate base, stronger is the acid. The electron withdrawing group withdraws electron from ortho and para position effectively than from the meta position by resonance effect. Thus, the electron density at ortho and para positions is reduced than that at meta.

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Answer 2

Question

Chapter 6, Problem 6.76P

Interpretation Introduction

(a)

Interpretation:

It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position.

Concept introduction:

The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

Interpretation Introduction

(b)

Interpretation:

It is to be explained how the meta isomer of nitro phenol has the highest pKa value than ortho and para isomer.

Concept introduction:

Greater the resonance stabilization of the conjugate base, stronger is the acid. The electron withdrawing group withdraws electron from ortho and para position effectively than from the meta position by resonance effect. Thus, the electron density at ortho and para positions is reduced than that at meta.

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Answer 3

Question

Chapter 6, Problem 6.76P

Interpretation Introduction

(a)

Interpretation:

It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position.

Concept introduction:

The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

Interpretation Introduction

(b)

Interpretation:

It is to be explained how the meta isomer of nitro phenol has the highest pKa value than ortho and para isomer.

Concept introduction:

Greater the resonance stabilization of the conjugate base, stronger is the acid. The electron withdrawing group withdraws electron from ortho and para position effectively than from the meta position by resonance effect. Thus, the electron density at ortho and para positions is reduced than that at meta.

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Answer 4

Question

Chapter 6, Problem 6.76P

Interpretation Introduction

(a)

Interpretation:

It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position.

Concept introduction:

The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

Interpretation Introduction

(b)

Interpretation:

It is to be explained how the meta isomer of nitro phenol has the highest pKa value than ortho and para isomer.

Concept introduction:

Greater the resonance stabilization of the conjugate base, stronger is the acid. The electron withdrawing group withdraws electron from ortho and para position effectively than from the meta position by resonance effect. Thus, the electron density at ortho and para positions is reduced than that at meta.

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Answer 5

Chapter 6, Problem 6.76P

Interpretation Introduction

(a)

Interpretation:

It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position.

Concept introduction:

The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

Interpretation Introduction

(b)

Interpretation:

It is to be explained how the meta isomer of nitro phenol has the highest pKa value than ortho and para isomer.

Concept introduction:

Greater the resonance stabilization of the conjugate base, stronger is the acid. The electron withdrawing group withdraws electron from ortho and para position effectively than from the meta position by resonance effect. Thus, the electron density at ortho and para positions is reduced than that at meta.

Answer 6

Chapter 6, Problem 6.76P

Interpretation Introduction

(a)

Interpretation:

It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position.

Concept introduction:

The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

Answer 7

Chapter 6, Problem 6.76P

Interpretation Introduction

(a)

Interpretation:

It is to be explained how the pKa value of phenol decreases when the nitro group is attached to the phenol ring at ortho, meta, and para position.

Concept introduction:

The pKa value decreases with an increase in the strength of an acid. The acid whose conjugate base is most stable is the stronger acid. The adjacent electron withdrawing group delocalizes the negative charge on the conjugate base and stabilizes it through the resonance effect. Greater the resonance stabilization, stronger is the acid. The electronegative atom adjacent to the negatively charged atom also stabilizes the conjugate base through inductive effect. The effect of resonance stabilization is greater than the inductive effect.

Answer 8

Interpretation Introduction

(b)

Interpretation:

It is to be explained how the meta isomer of nitro phenol has the highest pKa value than ortho and para isomer.

Concept introduction:

Greater the resonance stabilization of the conjugate base, stronger is the acid. The electron withdrawing group withdraws electron from ortho and para position effectively than from the meta position by resonance effect. Thus, the electron density at ortho and para positions is reduced than that at meta.

Answer 9

Interpretation Introduction

(b)

Interpretation:

It is to be explained how the meta isomer of nitro phenol has the highest pKa value than ortho and para isomer.

Concept introduction:

Greater the resonance stabilization of the conjugate base, stronger is the acid. The electron withdrawing group withdraws electron from ortho and para position effectively than from the meta position by resonance effect. Thus, the electron density at ortho and para positions is reduced than that at meta.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 6 - Prob. 6.1P Ch. 6 - Prob. 6.2P Ch. 6 - Prob. 6.3P Ch. 6 - Prob. 6.4P Ch. 6 - Prob. 6.5P Ch. 6 - Prob. 6.6P Ch. 6 - Prob. 6.7P Ch. 6 - Prob. 6.8P Ch. 6 - Prob. 6.9P Ch. 6 - Prob. 6.10P

Solution Summary: The author explains how the pKa value of phenol decreases with an increase in the strength of an acid.

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