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When exactly 1 mole of methane is mixed with exactly 1

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

Solution for problem 47SP 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 47SP

When exactly 1 mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a chlorination reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane, as well as unreacted methane.

(a) Explain how a mixture is formed from this stoichiometric mixture of reactants, and propose mechanisms for the formation of these compounds from chloromethane.

(b) How would you run this reaction to get a good conversion of methane to CH3Cl? Of methane to CCl4?

Step-by-Step Solution:

Answer :

Step 1 </p>

When one mole of methane is reacted with the chlorine it gives hydrochloric acid and methyl radical and the process goes on giving the di ,tri tetra methane molecules .

As CH3Cl is produced it can compete with CH4 for available Cl radical generating CH2Cl2.

This can generate CHCl3 .

Step 2</p>

Step 3 of 3

Chapter 4, Problem 47SP is Solved
Textbook: Organic Chemistry
Edition: 8
Author: L.G. Wade Jr
ISBN: 9780321768414

Organic Chemistry was written by and is associated to the ISBN: 9780321768414. This textbook survival guide was created for the textbook: Organic Chemistry, edition: 8. This full solution covers the following key subjects: methane, mixture, Mole, reaction, tetrachloromethane. This expansive textbook survival guide covers 25 chapters, and 1336 solutions. The answer to “When exactly 1 mole of methane is mixed with exactly 1 mole of chlorine and light is shone on the mixture, a chlorination reaction occurs. The products are found to contain substantial amounts of di-, tri-, and tetrachloromethane, as well as unreacted methane.(a) Explain how a mixture is formed from this stoichiometric mixture of reactants, and propose mechanisms for the formation of these compounds from chloromethane.(b) How would you run this reaction to get a good conversion of methane to CH3Cl? Of methane to CCl4?” is broken down into a number of easy to follow steps, and 85 words. Since the solution to 47SP from 4 chapter was answered, more than 317 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 47SP from chapter: 4 was answered by , our top Chemistry solution expert on 05/06/17, 06:41PM.

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