Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th
7th Edition
ISBN: 9781305081086
Author: STOKER, H. Stephen
Publisher: Brooks Cole
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 17, Problem 17.74EP

(a)

Interpretation Introduction

Interpretation:

“Parent” amine can be regenerated from the given salt or not has to be indicated.

Concept Introduction:

Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom.  This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.

Neutralization reaction is the one that takes place between an acid and a base to give salt as product.  As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product.  The salt formed is an amine salt.  Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor.  In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.

When a strong base is added to the amine salt, the parent amine can be obtained.  This is a reverse reaction of the amine salt formation reaction.  These reactions can be represented as shown below,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 17, Problem 17.74EP , additional homework tip  1

Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.

(b)

Interpretation Introduction

Interpretation:

“Parent” amine can be regenerated from the given salt or not has to be indicated.

Concept Introduction:

Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom.  This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.

Neutralization reaction is the one that takes place between an acid and a base to give salt as product.  As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product.  The salt formed is an amine salt.  Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor.  In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.

When a strong base is added to the amine salt, the parent amine can be obtained.  This is a reverse reaction of the amine salt formation reaction.  These reactions can be represented as shown below,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 17, Problem 17.74EP , additional homework tip  2

Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.

(c)

Interpretation Introduction

Interpretation:

“Parent” amine can be regenerated from the given salt or not has to be indicated.

Concept Introduction:

Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom.  This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.

Neutralization reaction is the one that takes place between an acid and a base to give salt as product.  As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product.  The salt formed is an amine salt.  Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor.  In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.

When a strong base is added to the amine salt, the parent amine can be obtained.  This is a reverse reaction of the amine salt formation reaction.  These reactions can be represented as shown below,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 17, Problem 17.74EP , additional homework tip  3

Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.

(d)

Interpretation Introduction

Interpretation:

“Parent” amine can be regenerated from the given salt or not has to be indicated.

Concept Introduction:

Quaternary ammonium salt is the one that has four carbon atoms attached to the nitrogen atom.  This is formed by the reaction of tertiary amine with alkyl halide in presence of a strong base.

Neutralization reaction is the one that takes place between an acid and a base to give salt as product.  As amines are bases due to the amino group in it, the reaction with inorganic acid or carboxylic acid gives salt as product.  The salt formed is an amine salt.  Proton is donated from the acid to the nitrogen atom which acts as a proton acceptor.  In simple words, it can be said that in an amine‑acid reaction, the acid loses a hydrogen ion and amine gains a hydrogen ion.

When a strong base is added to the amine salt, the parent amine can be obtained.  This is a reverse reaction of the amine salt formation reaction.  These reactions can be represented as shown below,

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th, Chapter 17, Problem 17.74EP , additional homework tip  4

Quaternary ammonium salt does not give the “parent” amine when treated with a strong base as there is no possibility of deprotonation to take place.

Blurred answer
Students have asked these similar questions
Please help me with # 4 and 5. Thanks in advance!
A small artisanal cheesemaker is testing the acidity of their milk before it coagulates. During fermentation, bacteria produce lactic acid (K₁ = 1.4 x 104), a weak acid that helps to curdle the milk and develop flavor. The cheesemaker has measured that the developing mixture contains lactic acid at an initial concentration of 0.025 M. Your task is to calculate the pH of this mixture and determine whether it meets the required acidity for proper cheese development. To achieve the best flavor, texture and reduce/control microbial growth, the pH range needs to be between pH 4.6 and 5.0. Assumptions: Lactic acid is a monoprotic acid H H :0:0: H-C-C H :0: O-H Figure 1: Lewis Structure for Lactic Acid For simplicity, you can use the generic formula HA to represent the acid You can assume lactic acid dissociation is in water as milk is mostly water. Temperature is 25°C 1. Write the K, expression for the dissociation of lactic acid in the space provided. Do not forget to include state symbols.…
Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. :0: :0 H. 0:0 :0: :6: S: :0: Select to Edit Arrows ::0 Select to Edit Arrows H :0: H :CI: Rotation Select to Edit Arrows H. < :0: :0: :0: S:

Chapter 17 Solutions

Study Guide with Selected Solutions for Stoker's General, Organic, and Biological Chemistry, 7th

Ch. 17.4 - Prob. 1QQCh. 17.4 - Prob. 2QQCh. 17.5 - Prob. 1QQCh. 17.5 - Prob. 2QQCh. 17.5 - Prob. 3QQCh. 17.6 - Prob. 1QQCh. 17.6 - Prob. 2QQCh. 17.6 - Prob. 3QQCh. 17.7 - Prob. 1QQCh. 17.7 - Prob. 2QQCh. 17.7 - Prob. 3QQCh. 17.8 - Prob. 1QQCh. 17.8 - Prob. 2QQCh. 17.8 - Prob. 3QQCh. 17.8 - Prob. 4QQCh. 17.9 - Prob. 1QQCh. 17.9 - Prob. 2QQCh. 17.10 - Prob. 1QQCh. 17.10 - Prob. 2QQCh. 17.10 - Prob. 3QQCh. 17.10 - Prob. 4QQCh. 17.11 - Prob. 1QQCh. 17.11 - Prob. 2QQCh. 17.11 - Prob. 3QQCh. 17.12 - Prob. 1QQCh. 17.12 - Prob. 2QQCh. 17.12 - Prob. 3QQCh. 17.12 - Prob. 4QQCh. 17.13 - Prob. 1QQCh. 17.13 - Prob. 2QQCh. 17.13 - Prob. 3QQCh. 17.13 - Prob. 4QQCh. 17.14 - Prob. 1QQCh. 17.14 - Prob. 2QQCh. 17.14 - Prob. 3QQCh. 17.15 - Prob. 1QQCh. 17.15 - Prob. 2QQCh. 17.16 - Prob. 1QQCh. 17.16 - Prob. 2QQCh. 17.16 - Prob. 3QQCh. 17.17 - Prob. 1QQCh. 17.17 - Prob. 2QQCh. 17.17 - Prob. 3QQCh. 17.18 - Prob. 1QQCh. 17.18 - Prob. 2QQCh. 17.18 - Prob. 3QQCh. 17.19 - Prob. 1QQCh. 17.19 - Prob. 2QQCh. 17.19 - Prob. 3QQCh. 17.19 - Prob. 4QQCh. 17 - Prob. 17.1EPCh. 17 - Prob. 17.2EPCh. 17 - Prob. 17.3EPCh. 17 - Prob. 17.4EPCh. 17 - Prob. 17.5EPCh. 17 - Prob. 17.6EPCh. 17 - Indicate whether or not each of the following...Ch. 17 - Indicate whether or not each of the following...Ch. 17 - Prob. 17.9EPCh. 17 - Prob. 17.10EPCh. 17 - Prob. 17.11EPCh. 17 - Prob. 17.12EPCh. 17 - Prob. 17.13EPCh. 17 - Prob. 17.14EPCh. 17 - Prob. 17.15EPCh. 17 - Assign a common name to each of the following...Ch. 17 - Prob. 17.17EPCh. 17 - Prob. 17.18EPCh. 17 - Prob. 17.19EPCh. 17 - Prob. 17.20EPCh. 17 - Prob. 17.21EPCh. 17 - Prob. 17.22EPCh. 17 - Prob. 17.23EPCh. 17 - Prob. 17.24EPCh. 17 - Prob. 17.25EPCh. 17 - Prob. 17.26EPCh. 17 - Prob. 17.27EPCh. 17 - Prob. 17.28EPCh. 17 - Prob. 17.29EPCh. 17 - Prob. 17.30EPCh. 17 - Prob. 17.31EPCh. 17 - Prob. 17.32EPCh. 17 - Prob. 17.33EPCh. 17 - Prob. 17.34EPCh. 17 - Determine the maximum number of hydrogen bonds...Ch. 17 - Prob. 17.36EPCh. 17 - Prob. 17.37EPCh. 17 - Prob. 17.38EPCh. 17 - Prob. 17.39EPCh. 17 - Prob. 17.40EPCh. 17 - Prob. 17.41EPCh. 17 - Prob. 17.42EPCh. 17 - Prob. 17.43EPCh. 17 - Prob. 17.44EPCh. 17 - Prob. 17.45EPCh. 17 - Prob. 17.46EPCh. 17 - Prob. 17.47EPCh. 17 - Prob. 17.48EPCh. 17 - Prob. 17.49EPCh. 17 - Prob. 17.50EPCh. 17 - Prob. 17.51EPCh. 17 - Prob. 17.52EPCh. 17 - Prob. 17.53EPCh. 17 - Prob. 17.54EPCh. 17 - Prob. 17.55EPCh. 17 - Prob. 17.56EPCh. 17 - Prob. 17.57EPCh. 17 - Prob. 17.58EPCh. 17 - Prob. 17.59EPCh. 17 - Prob. 17.60EPCh. 17 - Prob. 17.61EPCh. 17 - Prob. 17.62EPCh. 17 - Prob. 17.63EPCh. 17 - Prob. 17.64EPCh. 17 - Prob. 17.65EPCh. 17 - Prob. 17.66EPCh. 17 - Prob. 17.67EPCh. 17 - Prob. 17.68EPCh. 17 - Prob. 17.69EPCh. 17 - Prob. 17.70EPCh. 17 - Prob. 17.71EPCh. 17 - Prob. 17.72EPCh. 17 - Prob. 17.73EPCh. 17 - Prob. 17.74EPCh. 17 - Prob. 17.75EPCh. 17 - Prob. 17.76EPCh. 17 - Prob. 17.77EPCh. 17 - Prob. 17.78EPCh. 17 - Prob. 17.79EPCh. 17 - Prob. 17.80EPCh. 17 - Prob. 17.81EPCh. 17 - Prob. 17.82EPCh. 17 - Prob. 17.83EPCh. 17 - Prob. 17.84EPCh. 17 - Prob. 17.85EPCh. 17 - Prob. 17.86EPCh. 17 - Prob. 17.87EPCh. 17 - Prob. 17.88EPCh. 17 - Prob. 17.89EPCh. 17 - Prob. 17.90EPCh. 17 - Prob. 17.91EPCh. 17 - Prob. 17.92EPCh. 17 - Prob. 17.93EPCh. 17 - Prob. 17.94EPCh. 17 - Prob. 17.95EPCh. 17 - Prob. 17.96EPCh. 17 - Prob. 17.97EPCh. 17 - Prob. 17.98EPCh. 17 - Indicate whether or not each of the following...Ch. 17 - Indicate whether or not each of the following...Ch. 17 - Classify each of the following amides as...Ch. 17 - Classify each of the following amides as...Ch. 17 - Classify each of the amides in Problem 17-101 as a...Ch. 17 - Prob. 17.104EPCh. 17 - Prob. 17.105EPCh. 17 - Prob. 17.106EPCh. 17 - Prob. 17.107EPCh. 17 - Prob. 17.108EPCh. 17 - Prob. 17.109EPCh. 17 - Prob. 17.110EPCh. 17 - Assign an IUPAC name to each of the following...Ch. 17 - Assign an IUPAC name to each of the following...Ch. 17 - Prob. 17.113EPCh. 17 - Prob. 17.114EPCh. 17 - Prob. 17.115EPCh. 17 - Prob. 17.116EPCh. 17 - Prob. 17.117EPCh. 17 - Prob. 17.118EPCh. 17 - Prob. 17.119EPCh. 17 - What is the relationship between the acronym DEET...Ch. 17 - Prob. 17.121EPCh. 17 - Prob. 17.122EPCh. 17 - Prob. 17.123EPCh. 17 - Prob. 17.124EPCh. 17 - Prob. 17.125EPCh. 17 - Prob. 17.126EPCh. 17 - Prob. 17.127EPCh. 17 - Prob. 17.128EPCh. 17 - Prob. 17.129EPCh. 17 - Prob. 17.130EPCh. 17 - Prob. 17.131EPCh. 17 - Prob. 17.132EPCh. 17 - Prob. 17.133EPCh. 17 - Prob. 17.134EPCh. 17 - Prob. 17.135EPCh. 17 - Prob. 17.136EPCh. 17 - Prob. 17.137EPCh. 17 - Prob. 17.138EPCh. 17 - Prob. 17.139EPCh. 17 - Prob. 17.140EPCh. 17 - Prob. 17.141EPCh. 17 - Prob. 17.142EPCh. 17 - Prob. 17.143EPCh. 17 - Prob. 17.144EPCh. 17 - Prob. 17.145EPCh. 17 - Prob. 17.146EPCh. 17 - Prob. 17.147EPCh. 17 - Prob. 17.148EPCh. 17 - Prob. 17.149EPCh. 17 - Prob. 17.150EPCh. 17 - Prob. 17.151EPCh. 17 - Prob. 17.152EPCh. 17 - Prob. 17.153EPCh. 17 - Prob. 17.154EP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Text book image
Organic And Biological Chemistry
Chemistry
ISBN:9781305081079
Author:STOKER, H. Stephen (howard Stephen)
Publisher:Cengage Learning,
Text book image
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Cengage Learning
Text book image
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning
Nomenclature: Crash Course Chemistry #44; Author: CrashCourse;https://www.youtube.com/watch?v=U7wavimfNFE;License: Standard YouTube License, CC-BY