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

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The "6n+2 rule" is a guideline for determining the number of pi bonds in non-cyclic molecules, where n represents the number of non-hydrogen atoms and V is the total number of valence electrons. This rule, introduced by A.B.P. Lever in 1972, calculates the number of pi bonds as (6n+2-V)/2. There is a lack of accessible explanations for why this rule generally holds true, leading to discussions about its origins and applicability to various organic molecules. The formula aligns with the concept of units of unsaturation in hydrocarbons, where the deficiency of hydrogen atoms also indicates the presence of pi bonds. Understanding these relationships can enhance the comprehension of molecular structures in organic chemistry.
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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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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.)
 
Thanks very much, mjc123. This is a very good new rule for me! (Well, all of them are new to me, as Chemistry is not my field). I will add it to the list that I am making of such rules. (I think I have figured out why the 6n+2 rule (usually) works, and the next one -- that appears the easiest to show -- is for hydrocarbon chains, that if X= number of carbon atoms, Y= number of hydrogen atoms, then the number of pi bonds is (2X-Y)/2 +1. )
 
Thanks to the mentor Berkeman who cleaned up my post a bit.
 
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