Molecular Formulas of Organic Compounds The Nitrogen Rule and Degree of Unsaturation Valdo Pellegrin Ecole Nationaie Superieure de Chimie, 8 Rue de I'Ecole Normale, F-34075 Montpellier Cedex, France In 1 X l 3 Gerhardr w u i the fir>! chemisr 111 write the mtrlcculilr li,rniuin.; d organic wmpounds in the imm nith whirh we are i;unil:iir tudny. For 3 "iocn ctmpo~mdthe correct dcscripl~onO i it.; mthrular timnuln requires the deterniina~iun. by elemental analysis, of the percentages of its constituent elements as well as its molecular weight M. Various qualitative and quantitative methods of specific analysis of each of the most common elements were developed between 1811 and 1831,principally by Liebig, Dumas, Thenard, Gay-Lnssac, and Berzelius. Since that time the actual principles of elemental analysis have not changed, although technical developments led to theintroduction in 1912 by Pregl of the currentmicroanalvtical method which in turn has been continuallv imare more numerous and based on a varietyof principles. For low molecular weight compounds (for example M < 600) the methods of Dumas and Meyer allow the determination of vapor densities, d , relative to the air for volatile compounds, which thus permits the calculation of M by Avogadro's law (1824). For non-volatile substances, cryoscopic and ehullioscopic methods based on Raoult's law (1880) in general give sufficiently precise measures of M. The discovery of isotopes and the advent of mass spectrometry between 1910 and 1920 were important historical events in chemistry. Indeed, once the masses and natural abundances of all the isotopes of each element became known, mass spectrometry was able to tackle the problem of molecular weight determinations of organic compounds. Since the early sixties it has become a separate organic spectroscopic method for the study of molecular fragmentations under electron imoact. As a result of technical develooments. mass sDectromctry is r d a " rt.pliising r,, wme extenr 14tmenfulannl\ies for the labor;ltur\, detrrmtnntinn of mdecular imnuias. At the same time that mass spectrometers began to appear in research laboratories, teaching manuals treating this method arrived in the libraries (e.g., references (1-8)).' In discussing the determination of molecular formulas and molecular weights of organic compounds, many of the works have, following McLafferty (31,introduced a rule, the nitrogen rule, and a formula, the formula to calculate the number of rings plus double bonds, without justifying them. Most general organic chemistry texts discuss the degree of unsaturation (DU)(9,10,11) of a compound and describe the procedure for calculating it, thereby obtaining the same result as the formula, whose origin remains unknown. The oresent article. which is onrelv didactic. wishes to demons'trate the nitrogen rule andthe fgrmula forcalculating the number of rings ~ l u double s bonds of anv common orzanic compound. Parity of the Molecular Weights of Organic Compounds and the Nitrogen Rule I t should he recalled that, by definition, the molecular weight M of a compound containing elements which possess several isotopes is calculated by using the most abundant isotope present for each element. Also the following elementary rules of addition and multiplication of even, p, and odd, q , whole numbers will be frequently assumed in what follows.
626
Journal of Chemical Education
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p p =p p+q=q q+q=p
Addition:
Multiplication:
p X p =p Pxq=P qxq=q
Parity of M I t will he shown that all organic compounds possessing some or all of the elements C, H, 0 , S, N, and X have a molecular weight which is an even number except those possessing an uneven number of nitrogen atoms. ( 1)
(a) Molecules with C and H: Hydrocarbons (C = 12, H = 1) A saturated aliphatic hydrocarbon having the formula C,H2,+2 has a molecular weight M = 12n + 2n 2, which will be an even nnmher regardless of the value of the whole number n, since both 12n and 2n 2 (the nnmher of hydrogen atoms) must be even numbers. For a given saturated aliphatic hydrocarbon, the introduction of a double bond, triple bond, or ring into the molecule will always decrease the number of hydrogen atoms by an even number; 2 in the case of a ring or double bond and 4 for a triple bond. Thus, the total number of hydrogen atoms in every hydrocarbon (saturated or not, cyclic or not) is always an even number as is M.
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CH3.CH1-CH~.LH3
[ L ~ ~ o tl;58
0 [sH6
M;6b
~
C [pH8
M;llb
a U
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~
ct~"8 N ~ ? ? 8
(b) Molecules with 0 and S ( 0 = 16, S = 32) In compounds which contain oxygen or sulfur or which possess any other common divalent atom, the total number of hydrogen atoms is always an even number. In fact, there are four ways of introducing an oxygen atom into a hydrocarbon structure such as is shown below for cyclohexane (Fig. 1). The oxygen atom can be inserted into either a C-H bond, togive cyclohexanol, or a C-C bond, togive oxepane. Neither of these two operations modifies the total number of hydrogen atoms. Alternatively, two hydrogen atoms from two different carbon atoms of the ring can bere-
Figure 1. The four ways of introducing an oxygen atom ima cyclohexane do not affect either the parity of the number of hydrogens or the parily of M.
placed by a bridging oxygen atom to give, for example, 1,2epoxycyclohexane, or indeed two hydrogens from a single carbon can he replaced with a doubly bonded oxygen, to give cyclohexanone. These two latter operations reduce the number of hydrogens by 2, an even numher, compared to the starting cyclohexane. Thus, compounds containing oxygen or sulfur will always also have an even molecular weight.
(c) Molecules with X = Halogen ( F = 19, Cl = 35, Br = 79,I = 127)
Given that halogen atoms, X, are, like H atoms, monovalent, the total numher of atoms X H of an organic compound containing C, H, 0,S, and X will always he an even number. Thus, the numher of atoms of X and the numher of H will always have the same parity.
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Figure 2. The four ways of innoducing a nitrogen atom into cyclohexanechange both the parity of the number of hydrogens and the parity of M. not only the parity of H hut also the parity of the numher of (H X). Four cases can be distinguished systematically (Table 1).
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Table 1.
Parlty of M Deduced from that of N
Parity of the number of Since the mass numhers A, of all the halogens are odd numhers, if the numher of atoms X is even then nA, is even as well as the numher of H and therefore M. If n is odd, then nA, and the numher of H are both odd and hence M is rvrn. tr can be seen thrrefure that thr rniolevttlar wei#hts of all mgani,: nmpounds pussessiny some o r all of the elimwnts (.'. H. 0. S.and X are ewn numbers. It noteworthy that this also applies for the isotopic species containing 37C1 and slBr, since these isotopes are also of odd mass number.
N even
H+X
X
even
even OW even
odd
odd
(odd
OW
Parity of H
Example number
M
even
)
6a 6b 6c M
odd
(d) Molecules with N ( N = 14) For molecules possessing nitrogen atoms, two cases will he considered separately according to whether the molecule contains halogen atoms. (0)Molecules with
C, H, 0, S, and N
Sitice multiples of 12.16.32, and 14 are alwavs even numbers, the parity of M will depend entirely on t h e parity of the numher of atoms of H. For compounds containing nitrogen atoms, the numher of N and the numher of H have the same parity.
This can he understood by noting that again there are four ways of introducing a nitrogen atom into a hydrocarbon structure such as cyclohexane (Fig. 2). Summarizing: If the number of N is even, the number of H is also even and M will be even. If the number of N is odd, the number of H is also odd and Mwill be odd.
This leads to the conclusion that for any given organic molecule possessing some or all of the elements C, H, 0,S, N, and X: If the number of N is even, then M is an even number. If the number of N is odd, then M is an odd number. In other words, the number of nitrogen atoms in an organic comoound has the same oaritv . . as its molecular weieht. i h nl~ernati\rl\., i t JU t q a n i v cmqnmnd hnsan odd-numucred ~nzslec~hr wc+dlt, then thr mt,ltrulcrc,nmlnannvdd t~umlrrr
