# Percentages of radical constitutional isomers [duplicate]

Is it possible to use the radical selectivity factors for chlorination based on primary, secondary, and tertiary substitutions to calculate the expected percentage of each constitutional isomer formed in the chlorination reaction? If so, how?

• The gist of this question is different and yet you could basically copy and paste the answer here to have it answered. If I vote for duplicate, the question will be hammered; I don’t want to do that. – Jan Nov 16 '17 at 9:30
• @Jan Thanks! I see expected percentages but not how they were derived from selectivity factors. – alehul Nov 16 '17 at 17:18

Your question is a bit hard to understand, you should change the wording or add more explanation. But I believe you want to know how to calculate the percentage of a particular isomer from the relative rates of chlorination. Well the calculation is as follows:

First, derive the ratio of different isomers. To find the terms of ratio, multiply the number of hydrogens which can be substitued to form the isomer, by the rate of halogenation of that type of hydrogen.

Next, calculate percentage from ratio. This is very easy arithmetic; you can do this by creating a fraction with the value of that particular isomer on the numerator and sum of all values from the ratio on the denominator. Then multiply the fraction by 100.

The ratio of reactivity of tertiary($\ce{3^\circ}$), secondary($\ce{2^\circ}$) and primary($\ce{1^\circ}$) H-atoms for chlorination is $\ce{5 : 3.8 : 1}$ at 298K. For bromination of $\ce{3^\circ}$, $\ce{2^\circ}$, $\ce{1^\circ}$ H, the ratio of reactivity is $\ce{1600:82:1}$ at 400K

Example: Monochlorination of n-butane. Here, two isomers are formed, 1-chlorobutane, and 2-chlorobutane.

Now, substitution of six $\ce{1^\circ}$ H (shown by blue box) results in 1-chlorobutane

and substitution of four $\ce{2^\circ}$ H (shown by red box) results in 2-chlorobutane

So, the ratio is 1-chlorobutane : 2-chlorobutane $\ce{=6\times1 : 4\times3.8= 6:15.2}$

Calculate percentage: 1-chlorobutane is $\ce{ \frac{6}{6+15.2}\times 100\%= 28.3\%}$

and similarly, for 2-chlorobutane, percentage is $\ce{71.7\%}$