There are series of examples in my textbook to decipher the chlorination selectivity.
The example was to find percentage of monohlorderviatives of n-pentane after free radical halogenation giving products 1,2,and 3-chloropentane, (assuming relative reactivity of all H-atoms to be same).
The amount of each isomers formed depends on its rate the formation, therefore, the ratio of amounts (i.e., relative amounts) is equal to the ratio of rates.
- Relative amount of products=(Total number of equivalent Hydrogen)*(Relative Reactivity)
In 1-Chloropentane, there are two 1°Carbon so six 1°H atoms. After this I cannot understand how to get 4 and 2 H-equivlents for followup products.
First, I tried to do it the same way I did for n-butane (six 1°H and four 2°H equivalents for chlorbutane and 2-chlorobutane, respectively.) But my answer was too wrong, I searched some sites to find how to find what I was missing?
It seems to me, at least my initial approach was right, H-equivalents means 1°/2°/3° H-atoms and the rest is to go by formula to get the answer? I tried again, but I am still no able to understand how are 4 and 2 equivalents for the rest of product?
Do H-equivalents means something else and what key component am I missing?
Here is one of the solution (without assuming same reactivity for H equivalents):