Instant Notes: Analytical Chemistry

compounds or whose spectra are not present in the database. For example, the

presence of a strong peak in an IR spectrum near 1700 cm−^1 should suggest a high

probability that the sample might be a carbonyl compound.

The conditions under which each spectrum has been obtained must be taken

into consideration. For example, if the UV, IR and NMR spectra were run in

solution, what was the solvent? The instrumental parameters also need to be

considered. In MS, the type of ionization used will affect the spectrum obtained.

Sometimes the source of the analytical sample is known, and this can be a

great help in elucidating the identity of the material.

It is a worthwhile exercise to follow the same general scheme and to note

down the information that is deduced from the study of each spectrum. One

suggested scheme is given below, but the value of ‘feedback’ in checking the

deductions must not be overlooked.

(i) Empirical formula.Occasionally, if the sample has been analyzed to find

the percentage of carbon, hydrogen, nitrogen, sulfur and other elements,

and to deduce the percentage of oxygen by difference, this can be a useful

first step. If this information is not available, it may be found from the MS

if an accurate relative molecular mass has been measured.

Example: A solid sample contained C 75.5%, H 7.5%, and N 8.1% by weight.

What is the empirical formula of the sample?

Dividing by the relative atomic masses gives the ratio of numbers of

atoms, noting that there must be (100 −75.5 −7.5 −8.1) =8.9% oxygen.

C =75.5/12 =6.292

H = 7.5/1 =7.5

N = 8.1/14 =0.578

O = 8.9/16 =0.556

This corresponds (roughly) to C 11 H 13 NO, with an RMM of 175, which may

give a molecular ion in the mass spectrum.

(ii) Double bond equivalents.The presence of unsaturation in a structure

should be considered. Since a saturated hydrocarbon has the formula

CnH2n+ 2 , and since a single-bonded oxygen can be thought of as equivalent

to –CH 2 −, and a single-bonded nitrogen as –CH<, the number of double

bonds, or rings, called the double bond equivalents (DBE) for the

compound is given by:

DBE =(2n 4 + 2 +n 3 −n 1 )/2

where n 4 is the number of tetravalent atoms (e.g., carbon), n 3 is the number

of trivalent atoms (e.g., nitrogen), n 1 is the number of monovalent atoms

(e.g., hydrogen or halogen).

Therefore, for benzene, C 6 H 6 , the DBE is (14 −6)/2 =4, that is, three

double bonds and one ring.

For the example in (i) above, C 11 H 13 NO, the DBE is (24 + 1 −13)/2 =6,

which would correspond to one benzene ring (4) plus one −C=C- plus one

>C=O. Note that other spectra must be used to distinguish between a ring

and a double bond or between a −C=C- and a >C=O.

(iii) The IRspectrum gives evidence about the presence or absence of functional

groups as discussed in Topics E10 and E11. The example in (ii) above would

be solved if the infrared spectrum showed no carbonyl to be present. The

Spectrometric
identification

286 Section F – Combined techniques