Question

10-1Copyright © 2016 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.1.Chlorination of pentane gives a mixture of isomers having the molecular formula C5H11Cl. The percentage of 1-chloropentane is 22%. Assuming the secondary hydrogens in pentane are equally reactive to monochlorination, what is the percentage of 3-chloropentane in the mixture?

Answer 1

Answer:

26%

Explanation:

Chlorination is a reaction that substitutes an atom of hydrogen for an atom of chlorine in a hydrocarbon. Pentane has 5 carbons and 12 hydrogens, and the chlorination can happen at the carbons 1,2 or 3. If it happens at carbon 4, the structure will be the same as carbon 2, and if it happens at carbon 5, will be the same as carbon 1.

Then, the percentage of 2-chloropentane and 3-chloropentane is 100 - 22 = 78%. As stated above, 4 hydrogens can be substituted to form 2-chloropentane (two at carbon 2, and two at carbon 4), and only two for 3-chloropentane. So, the percentage of secondary hydrogens to form the structure wanted is:

2/6 = 1/3

Thus, the percentage of it is:

(1/3)*78% = 26%