Alkanes - Stability of Radical Intermediate

Why is it that when excess propane reacts with chlorine gas under the presence of sunlight, the ratio of 1-chloropropane to 2-chloropropane is less than 3:1? This is despite there are 6 Hydrogen atoms can be substituted to give 1-chloropropane and 2 of those can be substituted to give the latter.

To examine the situation, let's focus on the propagation step of the Free Radical Substitution mechanism of the reaction between propane and chlorine.

In the propagation step, the rate determination step, is most likely the reaction between the alkane molecule and the halogen radical. This is because the bond broken is the C-H single bond; the strongest bond broken in all the steps in the mechanism.

If we see all the C-H bonds in the alkane molecule are equally strong, then there is no discrimination to which Hydrogen atom is favoured to be removed. This would have resulted in the above mentioned 3:1 ratio. But it is not true, there is a difference

From the illustration above and the definition of a radical, having substitutents (such as alkyl substituents) who are willing to donate some electron density to the carbon radical, help to make the carbon radical less electron deficient and hence more stable.

The resultant: A stable product would be more likely be produced than an unstable product. Hence, the CH₃CHCH₃ radical is more readily formed than then CH₃CH₂CH₂ radical. - (This is nature's phenomenon.)

However, as the halogen radicals are highly reactive (in fact almost all radical are reactive), probability is also an important factor, hence CH₃CH₂CH₂ radical is still formed in larger quantities since there is a higher chance of forming.

The different stability observe in the two radical intermediate implies that the E_a of the propagation step which forms the CH₃CHCH₃ radical is lower than that which is required to form the other radical.

Since in the formation of alkyl radical the enthalpy change for that reaction is different (due to different radical stability) while the same H-X bond is formed. Therefore, the C-H bond strength in the alkane molecule is definitely not identical.

In addition, there is another means to allieviate the electron deficiency of a carbon radical. We must first examine the geometry of the radical.

When the radical centre is adjacent to a carbon-carbon double bond, the following phenomenon (known as delocalisation by resonace) occurs, resulting the radical to become more stable.

Hence, the stabilisation of a radical intermediate can occurs through (i) electron donation or (ii) delocalisation - Both methods work because in their own way they help to allieviate the electron deficiency with a carbon radical has to endure.

In conclusion, this also further adds to the value that despite your Data Booklet giving you the bond energy of the C-H bond to be 410 kJ mol^-1, the C-H bonds in a molecule is definitely far from identical.

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^{Article written by Kwok YL 2009.}

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