I've been doing some thinking about this double bond equivalent formula that's introduced in Thushan's book as a handy trick to know, I'd just like some clarification on it in order to get my head around it.
So
So, it's presented in the AN Chem book as:
 + 1) - n(H,F,Cl,Br,I) + n(N)}{2})
 + 1) - n(H)}{2})
The difference between the two being that the F, Cl, Br, I and N were not needed in the second case
http://en.wikipedia.org/wiki/Double_bond_equivalentThe Wikipedia page states that for the particular case of "molecules containing only carbon, hydrogen, monovalent halogens, nitrogen, and oxygen" as:
 - \frac{n(H)}{2} - \frac{n(X)}{2} + \frac{n(N)}{2} + 1)
Which is just exactly the same formula, just not simplified (I find this form a bit easier to remember, personal preference).
It also has a more general formula, based on valency. Just looking at it, I think I can see how we could derive our more specific case from this general formula.
I don't really prefer using this formula for most questions, since there's usually an alternative way of working it out (often longer, but I'm more comfortable doing it that way), but I do see how it's useful for double checking. I saw that it was particularly handy in that IR question, where you had to determine the structure of a given CxHyOz - with the formula allowing you to see that there would exist so many double bonds.
I have a few questions though, the first question is what concerns me most. The third and fifth question is also something that I'm concerned about.
1. Where did this formula come from? Who came up with it? How was it derived? How was it proved?I can see that if we just had alkanes (saturated hydrocarbons):
 - \frac{n(H)}{2} - \frac{n(X)}{2} + \frac{n(N)}{2} + 1)
We know that there's no saturation, no halogens, nitrogens, so:
 - \frac{n(H)}{2} + 1)
 - \frac{n(H)}{2})
 - n(H))
 = 2n(C) +2)
Which is equivalent to the general formula for alkanes:
I guess just looking at saturated hydrocarbons, we already knew that
 = 2n + 2)
- which would be the maximum number possible, since saturation would only decrease this number due to the less number of bonds that can be formed.
Less number of bonds would be why valency comes up as a factor in the general case given in the Wikipedia page.
This next bit is a bit ill-explained, but I'll have a shot at trying to get my thoughts into coherent words anyway:
Is that also why atoms such as Nitrogen and Halogens are a concern - nitrogen can form 3 bonds and halogens can form one bond. The DBE is relating the affect of saturation to numbers of atoms - where saturation affects the number of hydrogen (that wiki page also seems to place emphasis on hydrogen, which got me thinking about it from that perspective).
Nitrogen is able to form 3 bonds, which means that for every extra nitrogen, we can fit in more hydrogens. Since halogens form one bond, the number of them that we fit into a structure would also decrease with saturation (thinking in terms of hydrogens, placing a halogen would mean that we couldn't put a hydrogen in because of that).
Why doesn't oxygen have an affect on the number of hydrogens though? It can form 2 bonds, so even if we stick in an 'o' into an atom (e.g. look at an alkane and then it's corresponding alkanol), there's still a place for the hydrogen.
I have no clue, not even fairly vague ideas like above, about the
in the formula though.
2. I'm assuming that only natural numbers will be input and output of the formula. 3. How should we use this formula on our exams?Is this okay to use on VCAA exams? Should we restrict it's usage for just confirming our answers in our heads? One of the sample answers in book (AOS2, Test 2, Q2d) uses the phrase "as depicted by the formula".
4. Interpreting the output of the formulaGiven that:
implies no double bonds, no rings.
implies either 1 double bond or 1 ring
implies either 2 double bonds, 2 rings or 1 double bond and 1 ring.
and so on.
So this formula isn't too useful unless you have extra information that would allow you to eliminate the chance of rings - meaning that the number will be just the number of double bonds.
If we were given the molecular formula of a fatty acid, and asked to calculate how many double bonds it has, would this formula be a safe bet? Considering that VCAA will keep to the VCE course and that we were sure that the fatty acid didn't contain any rings, it should be mostly good to use?
e.g.
Arachidonic Acid:
It's fairly clear that this is a unsaturated fatty acid, it doesn't fit our general formula for saturated fatty acids:
 - \frac{n(H)}{2} + 1 = 20 + \frac{32}{2} + 1 = 20 - 16 + 1 = 5)
5 double bonds, which is correct:
5. Limitations/Things to watch out forI guess this relates to the previous question, where we'd have to watch out for rings etc. That Wikipedia link includes triple bonds (which contributes a value of '2' to the DBE), which I assume we don't really need to worry about.
Seeing that this formula isn't on the VCE course, it's not something that we'd be expected to use, so there's probably not too much to worry about, since there won't be any questions where we'd be having to rely on it - but it does seem handy for extra double checking and confirming that you haven't made any mistakes. I guess I just want to know if there should be anything to worry about when using it though.
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