What are primary hydrogen and secondary hydrogen?
Regioselectivity of halogenation in the presence of secondary and tertiary carbon-hydrogen bondsEven with propane there are two different places at which the propane molecule can become a radical. In the propane molecule, primary and secondary hydrogen atoms are found on carbon atoms. If the alkane is branched, there are even tertiary hydrogen atoms on carbon atoms. The radical substitution is a reaction that is built up from several steps. The more stable the alkane radical that is formed in the first step, the lower the energy barrier between the starting materials and products and the more this alkane radical reacts in contrast to the other alkane radicals. This effect decreases as the temperature of the reaction mixture increases, i.e. at high temperatures the differences no longer have such a strong impact.
The Stability of the radical increases from a primary to a secondary to a tertiary radical. If there is no alkyl group, the charged carbon atom is virtually unprotected and this radical is therefore quite unstable. If there are one, two, three or four alkyl groups in the vicinity of the atom, the radical becomes more and more stable because the charge is more protected. The protection comes about by the -Orbital of the neighboring alkyl residue overlaps the area of residence of the lone electron and this electron thus gets a larger area of residence. Thus the produc
The is also decisive for the distribution of the products between primary, secondary and tertiary haloalkanes Reactivity of the halogen used in each case. If its reactivity is high, the stability of the radical no longer plays a role, since it reacts quickly anyway. If the reactivity is very low, then the stability is to be assigned a high priority. The effect of the halogen used can be read from the following table:
|Fluorine (at 25 ^ C)||1||1,2||1,4|
|Chlorine (at 25 ^ C)||1||4||5|
|Bromine (at 98 ^ C)||1||250||6300|
Last is the statistical probability decisive for the distribution of the products. We have already seen with methane that the methane molecule has four different hydrogen atoms, all of which can react with the same probability. In principle there would be four different products, which, however, can all be converted into one another by rotation and are therefore the same. In the case of higher alkanes there are also different hydrogen atoms which, when substituted, result in the same haloalkanes. But there are also different hydrogen atoms, which result in different haloalkanes. The hydrogen atoms whose substitution results in the same haloalkane must be counted for each haloalkane product and placed in relation to one another.
Example: We want to do a bromination. Let us consider the following molecule:
We find 15 primary hydrogen atoms, 2 secondary and 1 tertiary hydrogen atoms on the molecule. A carbon atom does not have any hydrogen atoms at all, it only has bonds to carbon atoms.
With the primary hydrogen atoms, there are now two different options for the product molecule: The six hydrogen atoms on the left react to form the following molecule (1):
If we look at all of the hydrogen atoms, we will see that all the primary hydrogen atoms on the left are equivalent. They react to one and the same molecule, since we can turn all C-C bonds at will.
On the right side, nine primary hydrogen atoms react to form the following molecule (2). So together we have 15 primary hydrogen atoms:
The two secondary hydrogen atoms react to form the following molecule (3):
One tertiary hydrogen atom reacts to form the following molecule (4):
From the table for the regioselectivity of the halogens we can now read that for bromine in a reaction at 98 ^ C the following distribution between primary: secondary: tertiary applies:
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