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For each alkane,1. Draw all the possible monochlorinated

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

Solution for problem 42SP 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 42SP

For each alkane,

1. Draw all the possible monochlorinated derivatives.

2. Determine whether free-radical chlorination would be a good way to make any of these monochlorinated derivatives. (Will the reaction give mostly one major product?)

3. Which monobrominated derivatives could you form in good yield by free-radical bromination?

(a) cyclopentane

(b) methylcyclopentane

(c) 2-methylpentane

(d) 2,2,3,3-tetramethylbutane

Step-by-Step Solution:

Solution 42SP

Step 1 </p>

Halogenation is a substitution, where a halogen atom replaces a hydrogen atom.

(a)

1. In cyclopentane, all ten hydrogen atoms are identical, and it does not matter which hydrogen is replaced. The possible product of chlorination of cyclopentane is only one as there is one secondary hydrogen atoms in this compound. So, the possible product is as follows:

2. For this compound, chlorination would work as it forms only one product.

3. The bond-dissociation enthalpies of H-Cl (431 kJ) and H-Br (368 kJ), the HBr bond is weaker, and abstraction of a hydrogen atom by Br radical is edothermic. So bromination is much slower than chlorination. Hence, bromination on secondary carbon would not be anticipated to be a high-yielding process.

Step 2 </p>

(b)

1. Methylcyclopentane has three types of hydrogens which are primary, secondary, and tertiary hydrogen atoms. In the chlorination of methylcyclopentane, four monochlorinated products are possible.

2. Chlorination would not be a good method to make only one product from the mixture.

3. Bromination is more selective than chlorination. In the bromination of methylcyclopentane, the tertiary hydrogen atom is abstracted more often because the tertiary radical and the transition state leading to it are lower in energy than the primary and secondary radical and there transition states. Monobromination at the tertiary carbon would give high yield by free-radical bromination of methylcyclopentane.

Step 3 of 4

Chapter 4, Problem 42SP is Solved
Step 4 of 4

Textbook: Organic Chemistry
Edition: 8
Author: L.G. Wade Jr
ISBN: 9780321768414

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