Organic Chemistry
Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter SRP, Problem 1P

Arrange the compounds of each of the following series in order of increasing acidity:

Chapter SRP, Problem 1P, 1.	Arrange the compounds of each of the following series in order of increasing acidity:

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Interpretation Introduction

Interpretation:

The compounds of each of the given series are to be arranged in order of increasing acidity.

Concept introduction:

Acidity depends on the pKa of the conjugate acid. If the value of the pKa is higher, the conjugate acid is more stableand the conjugate base is more unstable.

Resonance structures are the structures in which two or more possible electron structures are drawn. In the resonance structure, the position of the atoms is the same but position of the electrons is different.

Resonance causes delocalization of electron pairs, which increases the stability of the base.

Answer to Problem 1P

Solution:

(a)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  1

(b)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  2

(c)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  3

(d)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  4

(e)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  5

Explanation of Solution

a)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  6

The α-hydrogen atoms of carbonyl groups are acidic. The acidity arises from the electron withdrawing effect of the carbonyl and resonance stabilization of its conjugate base. The electron donating effect of Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  7 groups tends to destabilize anions. In diketone, there is an active methylene, adjacent to two carbonyl groups. This indicates more resonance stabilization. The charge of anion can be delocalized to both oxygen atoms. The hydroxyl proton in carboxylic acid is an α-proton. On comparing the acidity of carboxylic acids and alcohols, alcohol is less acidic than carboxylic acid.

So, the increasing order of the acidity is as follows:

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  8

b)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  9

Phenol is more acidic than cyclohexanol because the resonance stabilization in both is different.

In the case of cyclohexane carboxylic acid, negative charge is shared between two different oxygen atoms making it more stabilized than phenoxide. Hence, the removal of proton from cyclohexane carboxylic acid is easier than phenol, making it more acidic than phenol.

So, the increasing order of the acidity is as follows:

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  10

c)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  11

In the case of carboxylic acids, electron substituents increase acidity by inductive electron donation. The electron-donating tert-butyl group destabilizes the conjugate base of benzoic acid, making it less acidic.

So, the increasing order of the acidity is as follows:

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  12

d)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  13

The electron-withdrawing chloro groups increase the acidity of carboxylic acid by increasing the stability of the carboxylate ion. Hence, the carboxylic acid with more chloro groups is more acidic.

So, the increasing order of the acidity is as follows:

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  14

e)

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  15

The lone pair electron in aniline is localized on the nitrogen atom, whereas onbenzamide, it is delocalized between oxygen and nitrogen via resonance. Therefore, benzamide is more acidic than aniline.

So, the increasing order of the acidity is as follows:

Organic Chemistry, Chapter SRP, Problem 1P , additional homework tip  16

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