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For each alkane,1. Draw all the possible monochlorinated
Chapter 5, Problem 42SP(choose chapter or problem)
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
Questions & Answers
QUESTION:
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
ANSWER:
Solution 42SP
Step 1
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.