Why is oxygen (or other elements) isn't present in the formulae for calculation of the value of degree of unsaturation $(\mathrm{DU})$ as it usually more common than nitrogen and halogens.

And whats so special about nitrogen and halogens to be in the formulae

$$\mathrm{DU} = C + 1 - \frac{H + X - N}{2},$$

where $C,$ $H,$ $N,$ $X$ are the numbers of carbon, hydrogen, nitrogen and halogen atoms present in the compound, respectively?

  • 2
    $\begingroup$ Try to derive DU=f(C,H,X,N,O) for various heterogenous componds. How does DU change, if -CH2- is replaced by -CO-, or C-H by C-OH, or C-C by C-O-C ? Do the same for other elements. $\endgroup$
    – Poutnik
    Jul 10 '20 at 13:12
  • 3
    $\begingroup$ Forget the formula! Try the methods here under Degree of Unsaturation. ursula.chem.yale.edu/~chem220/chem220js/… $\endgroup$
    – user55119
    Jul 10 '20 at 13:39
  • $\begingroup$ Hint:Think about how many bonds that atom can form and make sure you so not change the DBI $\endgroup$ Jul 10 '20 at 13:43

So here is the method I learned years ago without memorizing an awkward formula. The concept is to reduce the molecular formula to a hydrocarbon and compare it to the most saturated hydrocarbon bearing the same number of carbon atoms. Consider the infamous free-base, hydroxychloroquine, $\ce{C18H26ClN3O}$.

The following atom replacements may be done in any order. Ignore divalent atoms ($\ce{O,S}$). The formula is now $\ce{C18H26ClN3}$. Replace all halogens with an equal number of hydrogens $\Rightarrow \ce{C18H27N3}$. Substitute $\ce{CH}$ for each nitrogen or trivalent atom $\Rightarrow \ce{C21H30}$. The most saturated $\ce{C21}$ hydrocarbon is $\ce{C21H44}$. Now do the math: $\ce{C21H44}$ - $\ce{C21H30}$ = $\ce{H14}$. Now divide 14 by 2. The Degree of Unsaturation is 7.

As to the question as to why oxygen is dropped from the equation, if the oxygen in hydroxychloroquine weren't there, the degree of unsaturation would still be the same. For example, the Degrees of Unsaturation of tetrahydrofuran and cyclobutane are the same, D.U. = 1.

  • $\begingroup$ The question was about why you would not need to consider Oxygen. Not on how to find DU, $\endgroup$ Jul 10 '20 at 15:24
  • $\begingroup$ Thats a nice way for DU calculation. $\endgroup$ Jul 10 '20 at 16:04
  • $\begingroup$ If you think so, you might want to consider accepting it. ;) $\endgroup$
    – user55119
    Jul 10 '20 at 19:51

The degree of unsaturation is defined as the index of hydrogen deficiency (IHD) that determines the total number of rings and π bonds. It means the removal of two hydrogen atoms from a molecule is equal to one added $\mathrm{DU}.$

$$\text{rings} + \pi~\text{bonds} = C - \frac{H}{2} - \frac{X}{2} + \frac{N}{2} + 1$$

If you add a halogen to a molecule, you need to remove a hydrogen atom (decreased half-one $\mathrm{DU}).$ If you do the same for a nitrogen atom, you need to remove a hydrogen atom from the molecule and add two others to the nitrogen (for amines as a saturated substituent), so you have one more hydrogen (increased half-one $\mathrm{DU}).$ As a result, the number of halogen and nitrogen atoms must be considered in the equation, but about other atoms such as oxygen $(\ce{C-OH})$ and sulfur $(\ce{C-SH}),$ there is no change in the total number of hydrogen atoms when adding them as saturated substituents to a molecule.

In the case of the unsaturated substituents such as imines and carbonyls, it's clear that the total number of hydrogen atoms decreases, so the $\mathrm{DU}$ increases.

  • $\begingroup$ Please note that Chemistry.SE has a native support for MathJax, so there is no need to insert equations as images. Feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk
    Jul 10 '20 at 14:36

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