Rationale for 6n+2 rule for the number of pi bonds in a non-cyclic molecule

  • #1
nomadreid
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In several websites there is a rule of thumb to determine the number of pi bonds in a non-cyclic molecule called the "6n+2 rule", in which n = the number of non-hydrogen atoms in the molecules, and V = total number of valence electrons, so that the number of electrons involved in pi bonds is 6n+2-V (i.e., the number of pi bonds is (6n+2-V)/2. This rule was introduced in the article "Lewis Structures and the octet rule. An automatic procedure for writing canonical forms." by Lever, A.B.P. in the J.Chem Educ. 1972. 49(12), pages 819-821. Unfortunately, there is a paywall to this article, and none of the websites which are freely available explain why the rule (generally) works. I considered that the number 6 might have something to do with a maximum of three sigma bonds around an atom and the 2 to do with a hydrogen bond, but that didn't pan out. (Also, the examples tend to be organic molecules, but is this necessary?) In short, why does the rule work (when it works)? Thanks very much for any pointers.
 
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  • #2
I hadn't heard of this rule, but it is equivalent to the one I learnt back in the day. A saturated alkane has the formula CnH2n+2, and the number of units of unsaturation (which can include pi bonds and rings) is half the deficiency of H atoms, i.e. for a molecule CnHm it is equal to (2n+2-m)/2.

If we consider each C atom to have 4 valence electrons and each H atom 1, V = 4n+m and your formula becomes the same as mine. (I think mine is easier because you don't have to work out the number of valence electrons.)

Note that -O- or -NH- is isoelectronic with -CH2-, with 6 valence electrons, so you can use the same formula in your formulation, with n the number of non-H atoms. (In mine, you ignore O's and subtract 1H for every N; n remains the number of C atoms.)
 

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