The last example was fully saturated but it is usually a help in deducing the structure of an unknown compound if, once you know the atomic composition, you immediately work out how much unsaturation there is. It may seem obvious to you that, as C4H11NO has no double bonds, then C4H9NO (losing two hydrogen atoms) must have one double bond, C4H7NO two double bonds, and so on. Well, it’s not quite as simple as that. Some possible structures for these formulae are shown below.

DOUBLE BOND EQUIVALENTS HELP IN THE SEARCH FOR A STRUCTURE

Some of these structures have the right number of double bonds (C=C and C=O), one has a triple bond, and three compounds use rings as an alternative way of ‘losing’ some hydrogen atoms. Each time you make a ring or a double bond, you have to lose two more hydrogen atoms. So double bonds (of all kinds) and rings are called double bond equiva-lents (DBEs). You can work out how many DBEs there are in a given atomic composition just by making a drawing of one possible structure for the formula (all possible structures for the same for mula have the same number of DBEs). Alternatively, you can calculate the DBEs if you wish. A saturated hydrocarbon with n carbon atoms has (2n + 2) hydrogens. Oxygen doesn’t make any difference to this: there are the same number of Hs in a saturated ether or alcohol as in a saturated hydrocarbon.

So, for a compound containing C, H, and O only, take the actual number of hydrogen atoms away from (2n + 2) and divide by two. Just to check that it works, for the unsaturated ketone C7H12O the calculation becomes:

1. Maximum number of H atoms for 7Cs: 2n + 2 = 16

 2. Subtract the actual number of H atoms (12): 16 – 12 = 4

3. Divide by 2 to give the DBEs: 4/2 = 2

Here are two more examples to illustrate the method. This unsaturated cyclic acid has: 16 – 10 = 6 divided by 2 = 3 DBEs and it has one alkene, one C O, and one ring. Correct. The aromatic ether has 16 – 8 = 8 divided by 2 gives 4 DBEs and it has three double bonds in the ring and the ring itself. Correct again. A benzene ring always gives four DBEs: three for the double bonds and one for the ring. Nitrogen makes a difference. Every nitrogen adds one extra hydrogen atom because nitrogen can make three bonds. This means that the formula becomes: subtract actual number of hydrogens from (2n + 2), add one for each nitrogen atom, and divide by two. We can try this out too. Here are some example structures of compounds with seven C atoms, one N and an assortment of unsaturation and rings.

The saturated compound has (2n + 3) Hs instead of (2n + 2). The saturated nitro compound has (2n + 2) = 16, less 15 (the actual number of Hs) plus one (the number of nitrogen atoms) = 2.

Divide this by 2 and you get 1 DBE, which is the N O bond. We leave the third and fourth examples for you to work out, but the last compound (we shall meet this later as DMAP) has: 1. Maximum number of H atoms for 7Cs: 2n + 2 = 16 2. Subtract the actual number of H atoms (10): 16 – 10 = 6 3. Add number of nitrogen’s: 6 + 2 = 8 4. Divide by 2 to give the DBEs: 8/2 = 4 There are indeed three double bonds and a ring, making four in all. Make sure that you can do these calculations without much trouble. If you have other elements too it is simpler just to draw a trial structure and find out how many DBEs there are. You may prefer this method for all compounds as it has the advantage of giving you one possible structure before you really start. One good tip is that if you have few hydrogens relative to the number of carbon atoms (at least four DBEs) then there is prob ably an aromatic ring in the compound. Knowing the number of double bond equivalents for a formula derived by high-resolution mass spectrometry is a quick short cut to generating some plausible structures. You can then rule them in or rule them out by comparing with IR and NMR data.

●Working out the DBEs for an unknown compound 1 Calculate the expected number of Hs in the saturated structure (a) For Cn there would be 2n + 2H atoms if C, H, O only.

 (b) For CnNm there would be 2n + 2 + mH atoms.

 2 Subtract the actual number of Hs and divide by 2. This gives the DBEs.

3 If there are other atoms (Cl, B, P, etc.) it is best to draw a trial structure.

 4 A DBE indicates either a ring or a double bond (a triple bond is two DBEs).

 5 A benzene ring has four DBEs (three for the double bonds and one for the ring).

6 If there are few Hs, e.g. less than the number of Cs, suspect a benzene ring.

 7 A nitro group has one DBE only