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).

enter image description here

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)

enter image description here

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): enter image description here

  • 1
    $\begingroup$ C1 and C5 have 6H, C2 and C4 have 4H, C3 has 2H. With simplifying assumption of the same reactivity, these numbers are ratio of products. Sure, relative reactivity of primary and secondary C has to be considered. The picture says secondary carbons are 3.8 times more reactive. $\endgroup$
    – Poutnik
    Mar 16 at 11:31
  • $\begingroup$ There are 12 H-atoms of three different types in pentane. They are equally reactive, so the distribution of products will be the same as the proportion of each type of H-atom in the molecule. (You are not chlorinating 1 chloropentane). So what is the problem? The first calculation looks right. $\endgroup$
    – matt_black
    Mar 16 at 11:33
  • $\begingroup$ One thing to point out that you don't have tertiary carbon in pentane. Other than methyl groups at the ends, other three carbons are secondery. That's why they were considered extra 3.4 times reactive (reactivity of $\ce{R2CH^.}$ vs $\ce{RCH2^.}$). $\endgroup$ Mar 16 at 16:05

1 Answer 1


A normal alkane $\ce{C_nH_{2n+2}}$ has two sorts of Carbon atoms : $1$) terminal $\ce{C}$ atoms, which hold $3$ $\ce{H}$ atoms, and $2$) internal $\ce{C}$ atoms, holding only $2$ $\ce{H}$ atoms. Removing one $\ce{H}$ atom from anywhere in the molecule produces n-alkyl radicals $\ce{C_nH_{2n+1}}$. And these radicals are primary $1$-alkyl if the $\ce{H}$ atom was attached to the terminal position, and secondary alkyl if the $\ce{H}$ atom comes from internal $\ce{C}$ atoms.

The first alkane with both terminal and internal $\ce{C}$ atoms is propane $\ce{C3H8}$ with $n=3$. In propane $\ce{C3H8}$, the first and the last $\ce{C}$ atoms are terminal atoms. Removing one of the $3$ $\ce{H}$ atoms par terminal $\ce{C}$ atoms produce $1$-propyl radicals. Each terminal $\ce{C}$ atom can produce $1$-propyl radicals. Removing one of the two internal $\ce{H}$ atoms produce $2$-propyl radicals. In total removing one $\ce{H}$ of the eight $\ce{H}$ atoms in propane can produce six different $1$-propyl radicals and two different $2$-propyl radicals.

For butane $\ce{C_4H_{10}}$, there are two terminal $\ce{C}$ atoms and two internal $\ce{C}$atoms. Removing one $\ce{H}$ atom among the $10$ available can produce $2· 3 = 6$ different $1$-butyl radicals (the same number of $1$-alkyl radicals as for propane). But removing one of the $4$ $\ce{H}$ atom attached to one of the two internal $\ce{C}$ atoms can produce $4$ different $2$-butyl radicals. In total, removing one $\ce{H}$ of the ten $\ce{H}$ atoms in butane can produce six different $1$-butyl radicals and four different $2$-butyl radicals.

Now I think you can discover yourself why pentane $\ce{C_5H_{12}}$ can be chlorinated to produce $12$ possibilities of chloropentane : six $1$-chloropentane, four $2$-chloropentane, and two $3$-chloropentane

  • $\begingroup$ Thank you! 6,4,2 possibilities for respective chlorinated products. Chlorination selectivity has more to with possibility of that derivative existing than 1°/2°/3° of hydrogen. That approach worked in n-butane because of more straighforward structure but not in n-pentane or more complicated compounds. The book did not clarify in details. Your explanation was very helpful. $\endgroup$
    – Avani
    Mar 17 at 4:24

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