The radicals, in order of theirs decreasing stability, are to be listed. Concept introduction: A molecule that contains at least one unpaired electron is known as a radical. The relative stability of radicals is the same as that of carbocations. The stability of radicals is as follows: 3 ° > 2 ° > 1 ° > methyl

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 10, Problem 1PP

Interpretation Introduction

Interpretation:

The radicals, in order of theirs decreasing stability, are to be listed.

Concept introduction:

A molecule that contains at least one unpaired electron is known as a radical.

The relative stability of radicals is the same as that of carbocations.

The stability of radicals is as follows:

3°>2°>1°>methyl

Expert Solution & Answer

Answer to Problem 1PP

Solution:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 1

Explanation of Solution

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 2

The radical is a methyl radical. So, it is the least stable.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 3

The carbon containing the unpaired electron is primary. Thus, the radical is a primary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 4

The carbon containing the unpaired electron is tertiary. Thus, the radical is a tertiary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 5

The carbon containing the unpaired electron is secondary. Thus, the radical is a secondary radical.

Therefore, the tertiary radical is the most stable and the methyl radical is the least stable.

Hence, the decreasing order of the stability of the radical is as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 6

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

(c) The following data have been obtained for the hydrolysis of sucrose, C12H22O11, to glucose, C6H12O6, and fructose C6H12O6, in acidic solution: C12H22O11 + H2O → C6H12O6 + C6H12O6 [sucrose]/mol dm³ t/min 0 0.316 14 0.300 39 0.274 60 0.256 80 0.238 110 0.211 (i) Graphically prove the order of the reaction and determine the rate constant of the reaction. (ii) Determine the half-life, t½ for the hydrolysis of sucrose.

(III) adsorbent (b) Adsorption of the hexacyanoferrate (III) ion, [Fe(CN)6] ³, on y-Al2O3 from aqueous solution was examined. The adsorption was modelled using a modified Langmuir isotherm, yielding the following values of Kat pH = 6.5: (ii) T/K 10-10 K 280 2.505 295 1.819 310 1.364 325 1.050 Determine the enthalpy of adsorption, AadsHⓇ. If the reported value of entropy of adsorption, Aads Se = 146 J K-1 mol-1 under the above conditions, determine Aads Gº.

with full details solution please

Answer 2

Question

Chapter 10, Problem 1PP

Interpretation Introduction

Interpretation:

The radicals, in order of theirs decreasing stability, are to be listed.

Concept introduction:

A molecule that contains at least one unpaired electron is known as a radical.

The relative stability of radicals is the same as that of carbocations.

The stability of radicals is as follows:

3°>2°>1°>methyl

Expert Solution & Answer

Answer to Problem 1PP

Solution:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 1

Explanation of Solution

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 2

The radical is a methyl radical. So, it is the least stable.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 3

The carbon containing the unpaired electron is primary. Thus, the radical is a primary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 4

The carbon containing the unpaired electron is tertiary. Thus, the radical is a tertiary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 5

The carbon containing the unpaired electron is secondary. Thus, the radical is a secondary radical.

Therefore, the tertiary radical is the most stable and the methyl radical is the least stable.

Hence, the decreasing order of the stability of the radical is as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 6

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 10, Problem 1PP

Interpretation Introduction

Interpretation:

The radicals, in order of theirs decreasing stability, are to be listed.

Concept introduction:

A molecule that contains at least one unpaired electron is known as a radical.

The relative stability of radicals is the same as that of carbocations.

The stability of radicals is as follows:

3°>2°>1°>methyl

Expert Solution & Answer

Answer to Problem 1PP

Solution:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 1

Explanation of Solution

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 2

The radical is a methyl radical. So, it is the least stable.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 3

The carbon containing the unpaired electron is primary. Thus, the radical is a primary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 4

The carbon containing the unpaired electron is tertiary. Thus, the radical is a tertiary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 5

The carbon containing the unpaired electron is secondary. Thus, the radical is a secondary radical.

Therefore, the tertiary radical is the most stable and the methyl radical is the least stable.

Hence, the decreasing order of the stability of the radical is as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 6

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 10, Problem 1PP

Interpretation Introduction

Interpretation:

The radicals, in order of theirs decreasing stability, are to be listed.

Concept introduction:

A molecule that contains at least one unpaired electron is known as a radical.

The relative stability of radicals is the same as that of carbocations.

The stability of radicals is as follows:

3°>2°>1°>methyl

Expert Solution & Answer

Answer to Problem 1PP

Solution:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 1

Explanation of Solution

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 2

The radical is a methyl radical. So, it is the least stable.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 3

The carbon containing the unpaired electron is primary. Thus, the radical is a primary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 4

The carbon containing the unpaired electron is tertiary. Thus, the radical is a tertiary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 5

The carbon containing the unpaired electron is secondary. Thus, the radical is a secondary radical.

Therefore, the tertiary radical is the most stable and the methyl radical is the least stable.

Hence, the decreasing order of the stability of the radical is as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 6

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 10, Problem 1PP

Interpretation Introduction

Interpretation:

The radicals, in order of theirs decreasing stability, are to be listed.

Concept introduction:

A molecule that contains at least one unpaired electron is known as a radical.

The relative stability of radicals is the same as that of carbocations.

The stability of radicals is as follows:

3°>2°>1°>methyl

Expert Solution & Answer

Answer to Problem 1PP

Solution:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 1

Explanation of Solution

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 2

The radical is a methyl radical. So, it is the least stable.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 3

The carbon containing the unpaired electron is primary. Thus, the radical is a primary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 4

The carbon containing the unpaired electron is tertiary. Thus, the radical is a tertiary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 5

The carbon containing the unpaired electron is secondary. Thus, the radical is a secondary radical.

Therefore, the tertiary radical is the most stable and the methyl radical is the least stable.

Hence, the decreasing order of the stability of the radical is as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 6

Answer 6

Chapter 10, Problem 1PP

Interpretation Introduction

Interpretation:

The radicals, in order of theirs decreasing stability, are to be listed.

Concept introduction:

A molecule that contains at least one unpaired electron is known as a radical.

The relative stability of radicals is the same as that of carbocations.

The stability of radicals is as follows:

3°>2°>1°>methyl

Expert Solution & Answer

Answer to Problem 1PP

Solution:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 1

Explanation of Solution

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 2

The radical is a methyl radical. So, it is the least stable.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 3

The carbon containing the unpaired electron is primary. Thus, the radical is a primary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 4

The carbon containing the unpaired electron is tertiary. Thus, the radical is a tertiary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 5

The carbon containing the unpaired electron is secondary. Thus, the radical is a secondary radical.

Therefore, the tertiary radical is the most stable and the methyl radical is the least stable.

Hence, the decreasing order of the stability of the radical is as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 6

Answer 7

Chapter 10, Problem 1PP

Interpretation Introduction

Interpretation:

The radicals, in order of theirs decreasing stability, are to be listed.

Concept introduction:

A molecule that contains at least one unpaired electron is known as a radical.

The relative stability of radicals is the same as that of carbocations.

The stability of radicals is as follows:

3°>2°>1°>methyl

Answer 8

Expert Solution & Answer

Answer to Problem 1PP

Solution:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 1

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 2

The radical is a methyl radical. So, it is the least stable.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 3

The carbon containing the unpaired electron is primary. Thus, the radical is a primary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 4

The carbon containing the unpaired electron is tertiary. Thus, the radical is a tertiary radical.

The radical is given as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 5

The carbon containing the unpaired electron is secondary. Thus, the radical is a secondary radical.

Therefore, the tertiary radical is the most stable and the methyl radical is the least stable.

Hence, the decreasing order of the stability of the radical is as follows:

Organic Chemistry, Chapter 10, Problem 1PP , additional homework tip 6

Answer 12

Ch. 10 - Prob. 1PP Ch. 10 - Prob. 2PP Ch. 10 - Practice Problem 10.3 How would the molecular ion...Ch. 10 - Prob. 4PP Ch. 10 - Prob. 5PP Ch. 10 - Prob. 6PP Ch. 10 - Practice Problem 10.7 Chlorination reactions of...Ch. 10 - Prob. 8PP Ch. 10 - Prob. 9PP Ch. 10 - Prob. 10PP

Concept explainers

Answer to Problem 1PP

Explanation of Solution

Want to see more full solutions like this?

Chapter 10 Solutions