Asked Jun 19, 2019

Predict the ratios of products that result from chlorination of isopentane (2-methylbutane). please provide an understanding for how to find the ratio


Expert Answer

Step 1

The percentage and name of monochlorinated products formed in the chlorination of 2-methylpentane are given below.


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(M,-CM-CH-CH Liopatone (2 medyliotar) CM-CICNIC t 2-dhbawo-1-mely bubare (a/.) + Chy-CH-CH-CH, tCM-ČH-C CM 1-dkbne-2-ndayl hatne 1-dlone-3-me y tatre Ca y) +Ch-CH-C-- 1-delono-3 metby bubne isy.)

Step 2

The number of hydrogen atoms present at the carbon undergoing substitution in each of the product formed in the chlorination of isopentane are as follows:


1 hydrogen atoms in 2-chloro-2-methylbutane

2 hydrogen atoms in 2-chloro-3-methylbutane

6 hydrogen atoms in 1-chloro-2-methyl butane

3 hydrogen atoms in 1-chloro-3-methylbutane


The percentage of each of the product formed in terms of substitution of one hydrogen is calculated by dividing the percentage of product formed by the number of hydrogen atoms present at substituted carbon atom as shown below.


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22o 22% for 2- chloro - 2-methylbutane 1 33% 16.5% for 2 chloro - 3 -methylbutane 2 30% 5% for 1 - chloro - 2-methylbutane 6 15 9 5 % for 1- chloro - 3 - methylbutane en

Step 3

The simplest whole number ratio of products formed in the c...


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2(22 16.5:5: 5)= 44:33:10:10


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