Double-bond equivalents, structure

The long-range coupling of 2.2 Hz which appears in A and B, two quaternary C atoms in the NMR spectrum with appropriate shifts (5c = 76.6 and 83.0) and the two double-bond equivalents (molecular formula C TZ/oO) suggest that a CC triple bond links the two structural fragments. Hence the compound is identified as hex-3-yn-l-ol (C) in accordance with the coupling patterns. 

In this book, in order that you can concentrate your attention on the NMR spectra, we shall provide you with the molecular formula in all cases. This in turn provides you with information which can be extremely useful during the process of solving the structure if the molecule only contains C, H, N and O then you can use the molecular formula to obtain the number of so-called double bond equivalents, i.e. information on the degree of unsaturation. Though there are various formulas which can be devised to do this, we recommend the calculation using the following formula for a molecule CaHbOcNd, the number of double bond equivalents DBE is calculated as follows... 

Micaceol (12) was isolated as a light yellow amorphous solid. Its UV spectram showed maximum absorption at 246 mu indicahng the presence of a conjugated n system. The IR spectmm exhibited intense absorption bands at 3,425 (OH) and 1,601 (C = C) cm". The Chemical Ionization Mass Spectrum (CI-MS) of 12 showed molecular ion peak at m/z 399 [M-H]+. The HREIMS of this compound showed molecular ion peak at m/z 398.2392, which provided the molecular formula (calcd = 398.2360) and indicated the presence of six degrees of unsatura-hon in 12. These six double bond equivalents were accounted for by the steroidal skeletons with two double bonds incorporated in its structure. 

The number of double bond equivalents corresponds to the difference between the molecular formula and that for the saturated acyclic parent compound. Each DBE (double bond or ring) results in the subtraction of 2 hydrogens or halogens from the molecular formula of this parent structure. 

Double bond equivalents help in the search for a structure... 

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 unsat-uration there is. This is usually expressed as double bond equivalents . It may seem obvious to you that, if 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. 

Some of these structures have the right number of double bonds (C=C and C=0), one has a triple bond, and three compounds use rings as an alternative way of losing some hydrogeii 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 Equivalents (DBEs). 

Four compounds having the molecular formula C4H6O2 have the IR and - C NMR spectra given below. How many DBEs (Double Bond Equivalents) are there in C4H6O2 What are the structures of the four compounds You might again find it helpful to draw out some or all possibilities before you start. 

A monosubstituted pyridine ring and a methyl group add up to CgJT N. The atoms C and O which are missing from the empirical formula and a double-bond equivalent indicate a carbonyl group. The only structure compatible with the presence of these fragments is 3-acetylpyridine. 

---