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Peroxides are often added to free-radical reactions as

Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L.G. Wade Jr ISBN: 9780321768414 33

Solution for problem 50SP Chapter 4

Organic Chemistry | 8th Edition

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Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L.G. Wade Jr

Organic Chemistry | 8th Edition

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Problem 50SP

Peroxides are often added to free-radical reactions as initiators because the oxygen–oxygen bond cleaves homolytically rather easily. For example, the bond-dissociation enthalpy of the O-O bond in hydrogen peroxide( H-O-O-H) is only 213 kJ mol (51 kcal mol). Give a mechanism for the hydrogen peroxide-initiated reaction of cyclopentane with chlorine. The BDE for HO-Cl is 210 kJ mol (50 kcal/mol).

Step-by-Step Solution:

Solution 50SP :

Step 1 :

Here, we have to propose a mechanism for the hydrogen peroxide- initiated reaction of cyclopentane with chlorine.

In hydrogen peroxide(H2O2) initiated reaction of cyclopentane(C5H10) with chlorine(Cl), two mechanisms are possible depending on whether the hydrogen peroxide radical (HO •) reacts with cyclopentane or with chlorine.

Step 2:

The 2 mechanisms are as follows :

Mechanism 1 :

 Here, hydrogen peroxide reacts with chlorine.

The initiation reaction is as follows :

1. HO_OH → 2 HO •                                        Ho = + 213 kJ/mol(+ 51 kcal/mol)

Here, hydrogen peroxide atom breaks into free radical, i.e the HO-OH bond of hydrogen peroxide breaks and reactive intermediates are generated.

2. HO • + Cl_Cl → HO_Cl + Cl •                        Ho = + 242 kJ/mol(+ 58 kcal/mol)

                                                Ho = + 242 - 210 = + 32 kJ/mol.

Here,  the Cl-Cl bond is broken and HO • reacts with it and forms HO_Cl bond. Given that the BDE(Bond Dissociation Energy) for HO-Cl is 210 kJ/mol(50 kJ/mol). Hence the change in enthalpy in the reaction is  Ho = + 242 - 210 = + 32 kJ/mol.

Here, the initiation step is exothermic process as 32 kJ/mol of energy is released from the system.

The propagation reaction is as follows :

3.

        Ho = + 242 - 210 = + 32 kJ/mol.

The above reaction is the propagation reaction. In the propagation reaction, one reactive intermediate i.e the free radical, reacts with stable molecule and forms product and another reactive intermediate is generated as it is seen in the reaction above.  

4.

Ho = + 397 - 431 = - 34 kJ/mol.

In the above propagation reaction, the reactive intermediate of the cyclopentane reacts with Cl-Cl bond and generates another reactive intermediate. The change in enthalpy is also shown above.

Step 3 of 3

Chapter 4, Problem 50SP is Solved
Textbook: Organic Chemistry
Edition: 8
Author: L.G. Wade Jr
ISBN: 9780321768414

The full step-by-step solution to problem: 50SP from chapter: 4 was answered by , our top Chemistry solution expert on 05/06/17, 06:41PM. Since the solution to 50SP from 4 chapter was answered, more than 951 students have viewed the full step-by-step answer. Organic Chemistry was written by and is associated to the ISBN: 9780321768414. This full solution covers the following key subjects: mol, Bond, oxygen, hydrogen, peroxide. This expansive textbook survival guide covers 25 chapters, and 1336 solutions. This textbook survival guide was created for the textbook: Organic Chemistry, edition: 8. The answer to “Peroxides are often added to free-radical reactions as initiators because the oxygen–oxygen bond cleaves homolytically rather easily. For example, the bond-dissociation enthalpy of the O-O bond in hydrogen peroxide( H-O-O-H) is only 213 kJ mol (51 kcal mol). Give a mechanism for the hydrogen peroxide-initiated reaction of cyclopentane with chlorine. The BDE for HO-Cl is 210 kJ mol (50 kcal/mol).” is broken down into a number of easy to follow steps, and 60 words.

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