Molecular formulas of organic compounds: the nitrogen rule and degree of unsaturation

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 wui the fir>! chemisr 111 write the mtrlcc- ulilr li,rniuin.; d organic wmpounds in the imm nith whirh we are i;unil:iir tudny. For 3 "iocn ctmpo~md the correct dc- scripl~on O 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 currentmicro- analvtical method which in turn has been continuallv im-

are 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 ehullios- copic 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 sDec- tromctry is r d a " rt.pliising r,, wme extenr 14tmenful annl\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 s double 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.

Addition: p + p = p Multiplication: p X p = p p + q = q P x q = P q + q = p q x q = q

( 1) 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.

(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.

CH3.CH1-CH~.LH3 0 ~ C U " ~ a [ L ~ ~ o [sH6 [ p H 8 c t ~ " 8

t l ; 5 8 M ; 6 b M ; l l b 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.

626 Journal of Chemical Education

placed by a bridging oxygen atom to give, for example, 1,2- epoxycyclohexane, 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.

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.

t r 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.

I t noteworthy that this also applies for the isotopic species containing 37C1 and slBr, since these isotopes are also of odd mass number.

(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 the 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.

( P ) Molecules with C, H, 0, S, N, and X

Comparing the previous example and remembering that H and X have the same valency, i t is now necessary to consider

Figure 2. The four ways of innoducing a nitrogen atom into cyclohexane change 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).

Table 1. Parlty of M Deduced from that of N

Parity of the number of Parity of Example N H + X X H M number

even even 6a even

OW 6b odd even ) odd

6c odd (odd OW even M

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+dl t , then thr mt,ltrulcrc,nmlnannvdd t~umlrrr <,i nitrogen atoms. It is interesting to consider in the same way the case of a radical. Starting from the general case of an alkyl radical CnH2,+,, the same successive reasoning can be applied as was used for a molecule. The conclusion reached is the inverse result of that obtained for a molecule; that is, the mass and the number of nitrogen atoms in a radical are of opposlte parity.

All the above results are summarized in Table 2. It should be remembered that in a molecule all the electrons are aired while a radical, by definition, possesses an unpaired electron. (The particular case of hiradicals will not be considered here.)

Table 2. Relative Parlty of A and N for Even- or Odd-Numbered Electron Neutral Soecles

Neutral Mass Number of Number of Species Number A N atoms electrons E

21 even

Radical1 Odd =I odd

Volume 60 Number 8 August 1983 627

In fact. the ahove is onlv a first steD toward the Nitroeen Rule, which he:tig n rule d m a s s spectrometry, mu.;t include t he c a ~ e of pu,itwe ions. The . ~ l ~ o w results will therefigre nuw la. extended I , ) includ<. the cn i of mori~tihsrgcd positive ions. \The rat her ran. casr t.f doulJ\, charz6:d i m s will 1101 he considered here.)

(2) The Nitrogen Rule If a molecule is bombarded by an electron beam in a mass

spectrometer, it can lose an electron thereby giving the molecular ion which is a positive ion with an odd number of electrons. This, in turn, can undergo various fragmentations either by simple rupture of a bond or by rearrangement or ring opening reactions. As with all chemical reactions such fragmentation must occur with overall couservation of electronic charge and number of electrons: -

As a result of this discussion of the nitrogen rule an interesting question arises: why is there only a nitrogen rule and not a similar rule for another element? This question will now be considered in detail.

(3) Nitrogen: A Special Case The p-block group of elements of the periodic table, to

which nitrogen belongs, contain all the nonmetals most of which form covalent hydrides. Many of these elements are found in covalent molecules of organic compounds. In Table 4 each p-block element is listed with its symbol, its value of Z, its value of A for the most stable isotope (and hence the most abundant naturally occurring one) and, in brackets, its most frequently encountered valency. It can be seen that there are three columns of odd-numbered mass numbers which al-

Molecule prima O + Ion Radical primA ' - + /

+e- ~ o l e c u l A 7 Ion Radical l

The molecular ion has the same characteristics, in terms of mass number rnle and numher of nitrogen atoms, as the starting molecule. The sole difference is that it now has an odd numher of electrons. Thus, rnle and N have the same parity for an ion with an odd number of electrons.

In order to deduce the relative parities of rnle and N for an ion with an even number of electrons, consider the fragmentation:

Ion R a d a i - a+ + R ~ ~ J *

I t has been shown ahove that rnle and N have the same parity for the ion radical and opposite parity for the radical. The addition laws stated earlier require, therefore, that m/e and N are also of opposite parity for an ion with an even number of electrons.

These results can he summarized as in Tahle 3, which represents the most complete form of the nitrogen rule.

Table 3. Relative Parity of m/e and N for Even- or Odd- Numbered Electron Positively charged Specles: The Nltrogen Rule

Charged Mass Number Number of Number of Species m/e N atoms electrons E

~anl i even l Odd 2: I Odd odd

Ion]+ I ,we, even

'l'hc mu.! cotnm<m statement or the nitrogen r u b is as fr,l- luws. ..An ion hs\inz ti l l irdd numlwr or'elcctnmu lsuch as rhc molecular ion) will be observed at an even mass numher mle, except when it contains an odd number of nitrogen atoms." This statement, in fact, covers only the first half of Tahle 3. Furthermore by equating rnle with A, Tables 2 and 3 can be combined to give the most general form of the nitrogen rule:

~ o l e c u q ~ a n d l o n - ~ a d a : A and N have the same parity

~ a d a and a +: A and N have opposite parity

The following two examples, in which the fragmentations have been unambiguouslv demonstrated hv mass svectrom- etry, serve to illustrate these concepts ( ~ i & 3 and 4).


ternate with two columns containing elements of even valency and even mass number. In other words, all the elements in Table 4 have a valency and mass number of equal parity with a single exception, nitrogen. Thus, nitrogen is the only ele-

CH~.CH;I' w e s29 (odd1

n e ~ t l a l fragnenf E w e n -CUT A : 1 5 Loddl le

CI+.CHz.CH3 e- n = ~ & t w m ~ CH~.CH~.CH?:

E e v e n m 0 1 0 ~ ~ I a r ion m/e = 4 4 leven1

E odd

CH~.CH~-CH:* or

C H ~ - C H . C H ~ +

w e : &3 (odd1

E even

Figure 3. Illustration of the nitrogen rule fw fragmentation of a molecule with an even number of nitrogen atoms (N = 0) (E = number of elechonst.

~ o I ~ T Y I ~ I i b n

w e ; 135 (odd1 , E odd

-HcN''

SH~CO' M b2 Ierenl * ~ N H ? : ---- c ~ H ~ : - IE=teven' m/e ;93 l o d d l M = z " O d d l mm=66levenl

E odd E odd

M e = 6 5 lodd l

E even

w e = 51 l a d d l

E e v e n E odd

Figure 4. Illustration of the nitrogen rule fw hagmentation 01 a molecule with an add number of nitrogen atoms (N = 1) (E = number of electrons)

ment found in organic molecules which has a ualency and mass number of opposite parity. I t is this fact that explains the singular role of this element and the existence of the ni-

l trogen rule.

This special case of nitrogen is comparable to another parity rule which concerns the nuclear spin I. The nuclear spin I depends on the parity of A and Z. If A is an odd numher, then the nuclear spin is an odd multiple of % (38.8% of the stahle natural nuclei). If A is an even numher, two cases are distinguished according to the parity of Z. If A and Z are both even, then I = O (59.4% of the stable natural nuclei). If A is even and Z odd, then the s ~ i n I is a non-zero whole number (1.8% of the stahle natural niclei). Of the 273 stahle natural nuclei only five have non-zero whole numbered spins, namely:

Even in this small eroun nitroeen again distinguishes itself. Table 5 lists the nudearspins and relative natural isotopic ahundances of these five elements. Each of them can he seen to exist as a mixture of two stahle isotopes, one of which is found in much hieher natural abundance. In each case the - nion: abundant isotupe is that with a halt-spin w i ~ h t h ~ e x - ceprion ui nnrorcn- 14 which has spin I = I and has a natural abundance of 6.64%. Thus, ';N ;gain occupies an unique position in that it is the nucleus of whole numbered spin and is the most abundant natural isotope.

In general the nitrogen rule allows the chemist to deduce the numher of nitrogen atoms in an organic compound from its molecular weight. This helps establish the molecular formula. Furthermore. once that formula is known. additional

~~ ~ ~ ~, ~ ~ ~ ~ ~ ~ ~ ~ - - - ~ . . - information can be deduced; namely, the number of (rings + double bonds) in the molecule. This problem will now he considered in detail.

Table 5. Natural Abundances of the Five Elements Havlng Isotopes with Whole-Numbered Spins

Nuclear-Spin Natural Abundance I S O ~ O D ~ S I %

Degree ol Unsaturation: Determination of the Number of (Rings + Double Bonds)

In rhii w t i m a numher umhirh is rhnrncteriiri; oi 81 mu- Icrul,~r furmula u,ill h. d i a ~ u ~ ~ r d . Since this inumber has been given various names in the literature, a few definitions will first he made.

molecule. This number, therefore, represents the total number of sites of multiple bonding in the molecule, with nodistinction being made between double and triple bonds. The unsaturation number IUN) of a molecule reoresents the

test (decolorization of bromine water by unsaturated compounds) and the Baeyer test (decolorization of a dilute potassium per- manganate solution under acidic conditions by unsaturated compounds) are qualitative tests for multiple earhon-carbon bonds. At the quantitative level, UN can be determined by ti- trating a known quantity of the organic compound being studied with bromine. Since 2Brz is consumed by a triple bond and 1Br2 by a double bond, it can he said that a triple bond is equivalent to two double bands. This ean be compared with the iodine index, or the HiiBL index, which is used by biochemists todescribe the unsaturation richness of an oil or grease (iodine index = quantity of iodine in grams consumed by 100 g of oil). . The number of irinm +double bonds) ( R + DB) or the degree

since creation of a ring from an aliphatic molecule requires the loss of two hydrogen atoms, as is the case for a double bond.

For the molecule shown U S , U N , and DU can be calculated:

US = 2 (double bonds) + 1 (triple bond) = 3 UN = 2 (double bonds) + 2 X 1 (triple bond) = 4, and DU = 2 (cycles) + 2 (double hands) + 2 X 1 (triple bond) = 6.

The determination of US and UN requires a knowledge of the molecular structure. On the other hand, DU can he straightforwardly calculated from the molecular formula, with no knowledge whatsoever of molecular structure. The DU is, therefore, characteristic of a given formula and a precious, although of course limited, piece of information. For example

' Since they are the most frequently used. the expressions "number of (rings + double bondsy' and "degree of unsaturation" will be used. interchangeably, throughout this work even though they are not strictly correct. While it is much less cumbersome to speak of degrees of un- Saturation it should be mentioned that, in principle, the word unsaturation implies multiple bonding so that decalin, for example, which has an unsaturation number of zero. has a deoree of unsaturation of two. which " m ght be confGsmg to some S m lar y, the numner of (r ngs T oounle oonos) s not strictly correct slnce acetylene has a numoer of (R T DB, eq.d to two Tne terms unsat-rat on noex ( 101 an0 numoor of sm?s of unsaturation" ( 11) are also encountered. The most correct expressions are "lack of hydrogen index" (10) or "number of double bond equivalents" (5).


a formula with D U = 2 could correspond to a molecule possessing either 2 rings, or 1 ring and 1 douhle hond, or 2 douhle honds and no rings, or 1 triple hond and no rings.

High values of DU are often an indication of the presence of an aromatic system in the molecule, since benzene itself has D U = 4 (1 ring and 3 douhle bonds).

For a given molecular formula, DU can he calculated by either of two methods which will he discussed separately below:

either an "empirical" method which will he developed stepwise. or by using a mathematical formula which will he estah- lished.

( 1) Determination of DUby an Empirical Method Using Successive Modifications to the Molecular Formula (a) Molecules with C and H: Hydrocarbons

Saturated aliphatic hydrocarbons have a DU equal to zero. TheDU of any given hydrocarhon is the numher of molecules of hydrogen Hz required to transform it into a saturated aliphatic hydrocarhon possessing the same numher of carbon atoms. Thus, a douhle hond or a ring correspond to a DU = 1 and a triple hond to a D U = 2. Thus, to calculateDUfor the hydrocarhon CsHlo i t is necessary to compare it to the corresponding saturated aliphatic hydrocarbon of formula C,Hz,+z where n = 5 , namely C5H12, and 12H - 10H = 2H = lHz. Thus, the formula C5H10 has D U = 1 corresponding to the presence of either a ring or a douhle bond. Similarly, naph- thalene, C1oHa when compared to the alkane CloHsz gives 22H - 8H = 14H = 7H2 and thus have DU = 7 corresponding to 2 rings and 5 douhle honds.

(h) Molecules with 0 and S

If the molecule under study contains an atom of oxygen or sulfur (or any other divalent atom) they are simply removed from the formula and one then proceeds as for a hydrocarhon. Indeed. it is easilv seen that an alcohol such as cvclohexanol. or an ether, such'as oxepane, has the same DU as the corresponding hydrocarhon after removal of the oxygen; in this case, cyclohexane. Similarly, removal of the oxygen atom from cyclohexanone (without adding any hydrogen atoms) leads to a formula with D U = 2, corresponding to one ring and one douhle hond (Fig. 5).

c 7 H l l C6H12 '7"lk c6H1b C7Ht2 nu-1 OU=l OU=l OU.2 OU=2

Figure 5. Removal of the oxygen atom from cyclohexanal, oxepane, and cyclohexanone does not affect the DU.

Note also that the same result is ohtained by replacing the oxygen atom with a methylene CH2, group. However, simply removing the oxygen is more convenient, and this method will he used for divalent atoms hereinafter. Thus, for example, to determine the DU corresponding to the formula C8H1202 the two oxygens are removed, giving C8H12, which is then compared with C8H1s tO give (18H - 12H = 6H = 3H2) DU = 3.

(c) Molecules with X = Halogen

Since H and X have the same valency, if the formula contains one or several halogen atoms they are simply replaced by hydrogen atoms. For example, to determine DU for the formula C&C14 the divaleut sulfur atoms are first removed,

giving CsC4 and the chlorine atoms are then replaced by hydrogens, giving CfiH4. which is in turn compared with C S H I ~ to give (14H - 4H = 10H = 5H2) DU = 5 for the formula CfiSeC14, indicating a highly unsaturated structure.

(d) Molecules with N or Other Trivalent Atoms

The conversion of a nitrogen containing compound to its carbon homoloeue can he represented schematically as shown - in Figure 6.

Fioure 6. The chanoe from a nitmaen containincl com~ound to its carbon hc- *- ~ .. .

rno ogbe of me $?me DU can oe elleLle0 ether of SUIJEILI ng the N ~8th Cn or ov rernuvlng hrl Ifom me molecu ar form.la

Thus, if the formula contains one or more nitrogen (or other trivalent) atoms, they are either substituted with CH in the formula or removed hv subtractina the reauired numher of NH groups. The two pbssibilities aFe equivaient although the latter becomes imoossihle when the formula contains more nitrogen atoms than hydrogens. For this reason replacement of each N with CH, which always is possible, will he used here.

For example the formula C6H13N becomes CfiH14 which is an alkane with D U = 0, so that molecules of formula C5HlsN are saturated with aliphatic compounds. Similarly, the formula CeNa, for which it is impossible to subtract NHgroups, becomes C10H4 which is then compared to C10H22, to give (22H - 4H = 18H = 9Hz) DU = 9. Tetracyauoethylene has the formula CsN4 and DU = 9.

(e) General Procedure

For any given molecular formula, all the divalent atoms (0, S, . . .) are first removed, then each halogen X is replaced by a hydrogen. Each trivalent atom (N, P, . . .) is then removed and replaced by a CH. The numher of hydrogen atoms in the resulting, hydrocarhon formula is then compared with that in the saturated aliphatic hydrocarhon with the same numher of carbon atoms, from which D U = (the numher of H in the alkane-number of H in the derived formula)/2.

~ x a m ~ l e : Given the formula C7H,jCION3: -3N

C ~ H G C I O N ~ ~ C ~ H ~ C I N ~ 2 C ~ H ~ N Z + + ClOHl0 +H +3CH

which is compared with C10H22 to give, 22H - 10H = 12H = 6H2, DU = 6.

Should a negative or fractional value for DU he ohtained, then the starting formula was incorrect since the combination of the given numher of atoms of each element of known valency is not compatible with a plausiblemolecular structure. The ways of recognizing such impossible molecular formulas will now he discussed.

(f) Cases of Impossible Formulas

An impossible molecular formula can he recognized easily on the basis of a few simple considerations. For example, it has already been shown that for a molecule containing no trivalent atoms the number of monovalent atoms (H + X) is an even numher. For example, the formula C I I H ~ ~ C ~ Z O ~ S is impossible because it contains an uneven numher of monovalent atoms.


If a molecule contains trivalent atoms, then the numher of monovalent atoms (H + X) must have the same parity as the number of trivalent atoms (see Table 1). Thus, the formula C~H&IZON~ is impossible since it contalns an even number of monovalent atoms and an odd numher of trivalent atoms.

While the numher of plurivalent atoms (restricting the discussion to atoms of valency 2, 3, or 4 ) has no upper limit, the number of monovalent atoms, on the other hand, is re- stricted by the number of trivalent and tetravalent atoms. For instance, in an alkane possessing n carbon atoms the maximum number of hydrogens is 2 n + 2. The introduction of trivalent atoms such as nitrogen into the formula of an alkane, whereby each NH2 replaces an H atom allows only one extra hydrogen atom for each nitrogen introduced. Thus, a molecule containing nC and p N can have a maximum of 2 n + 2 + p hydrogen atoms.

This can be generalized for a formula

containing a monovalent atoms, I, p divalent atoms 11, y trivalent atoms 111, and 6 tetravalent atoms IV. Then the only possible formulas are those in which a and y have the same parity and with

Rearranging eqn. (1) gives

in which a and y must have the same oaritv. which means that . . h - n 2 T j 2 T I musl b t 11 whdr nun1ln.r. In u1heru8ords. the mI\. ~xmible icmuulas are tlmic ir,r wh:nh thr uunnritv 6 - al2; 712 + 1 is a non-negative integer. It will be shown below that this quantity is none other than the DU for the formula I, IIB 111, IVa. Thus, the simplest method of recognizing whether a molecular formula is possible or not is to calculate its DU which must be a whole number.

(2) Mathematical Calculation of DU

(a) Euler's Formula

The mathematical method of calculating DU is based on the Euler formula. The numher, f , of faces (or regions) of any graph constituting p points and h sides (or lines) is given by the Euler formula:

If=k--p+ll (3)

Thus, for the graph shown: k = 11 sides (or lines), p = 7 p o i n t s , a n d f = k - p + 1 = 1 1 - 7 + 1 = 5 .

The graph possesses 5 faces or regions. A light-hearted demonstration of the Euler formula (12)

is given in the appendix.

(h) Application of the Euler Formula to the Calculation of DU for an Organic Compound

Consider the general formula for an organic compound:

1- Il0 lva vt VIr

containing

a monovalent atoms I f l divalent atoms I1 y trivalent atoms I11 6 tetravalent atoms IV r pentavalent atoms V

and f hexavalent atoms VI '

The total numher of atoms, corresponding to the points of the graph, is

p = u + B + y + 6 + c + f

The total numher of bonds, which correspond to the sides, or lines, of the graph, is obtained as follows. In order that each bond between two atoms is not counted twice, the valency of each atom is divided by two. Each valency is then multiplied by the numher of atoms of that valency and the sum made over all valencies present, to give

k = a X 1 / 2 + f l X 2 / 2 + y X 3 / 2 + 6 X 4 / 2 + c X 5 / 2 + ~ X 6 / 2

The numher of rings plus double bonds, or the DU, corresponds to the number f of faces or regions. Using eqn. (3) and the above values of p and h gives

For instance, the isopentenyl pyrophosphate molecule in which phosphorus is pentavalent:

CHS I

0 0 11 11 formula: Hl2O7C5P1

CHcC=CH-CH,-0-P-0-P-OH DU = -6 + 5 + 3 + 1 = 3 I I from esn. (4) OH OH

This occurs similarly for the ethyl sulfate molecule which contains hexavalent sulfur.

I t is particularly important to be able to calculate DU for the formula of a compound of unknown structure. For this it will he assumed initially that all elements are in their lowest valency state; for example, two for S and three for P. Under these conditions the valencies found in organic chemistry are rarely greater than four and so f = f = 0 in eqn. (4) which re- duces to

for the formula I, IIB 111, IVa Equation (I) was in fact a simplified and less general version

of eqn. (5) m26+2+y (1)

In fact, the equality a = 26 + 2 + y corresponds to a saturated comwound. that is DU = 0. The ineaualitv a < 26 + 2 t 7 vorreqxmd.. t ( , :I , Y ~ I I I ~ C , U O ~ wh1,41 i 3 u~s:~tur,~tt .d .und $,I

r\.clir. 111 this raic rhe hall dilleren8.c i1rrat.t.n thv ml!nlx r u i monovalent atoms (H + X) in the given formula and the maximum numher of monovalent atoms (26 + 2 + y) in the corresponding saturated compound gives the DU according to eqn. (5).

If the compound contains only the elements C, H, 0 , and N, as is often the case in organic chemistry, then for the general formula C,H,N,O, eqn. (5) gives


I DU = x - y / 2 + 212 + 1 (6)

Thus, the formula CllHllNOe corresponds to a DU = 11 - 5.5 + 0.5 + 1 = 7. When using eqn. (6) to calculate DU and if other elements are present, then they must be added to the number of atoms of C, H, 0, and N to which they correspond in valency, to give the more general eqn. (5). Thus, the numher of atoms of silicon is added to that of carbon, halogens to hydrogen, and phosphorus to nitrogen.

For example the formula C4SizH&1Nz0 corresponds to a D U = 6 - 5 + l + l = 3 .

In the case of a saturated polycyclic compound the DU immediately gives the number of cycles. For example, cuhane CsHs with DU = 5 is a pentacycle:

pentacyelo [4.2.0.02,5.03,8.04,7] octane

The DU of two compounds can also he compared, such as hexahelicene (C26H16, 6 rings and 13 double bonds) and co- ronene (Cz4H12, 7 rings and 12 double bonds), each having the same DU = 19. In general, two hydrocarbons have the same DU if the difference of the number of hydrogens is twice the difference of the number of carbons.

The above results are applicable to neutral molecules. Some special cases of the application of eqn. (6) will now he considered, notably for the cases of radicals and positive ions.

(c) Special Cases

(a) Elements in Higher Valency States

In applying eqn. (6) to cases where elements exist with valencies higher than their lowest normal state, such as pentavalent phosphorus and hexavalent sulfur, double bonds involvine these elements must he ienored. If one takes the molecules of isopentenyl pyrophosphate and of ethyl sulfate and writes the multiule bonds to ohosohorus and sulfur as . . dative (donor-accept&) bonds then the DU is in agreement with that calculated according to eqn. (6) .

CH, 0 0

t CHI-L-w-O-Lo--H I I OH OH

Formula: CrH1,P20, DU=5-6+l+ l=f romeun . (6 )

0,

t CB-CH,-0-S-OH

1 0

Formula: CIHsOIS D U = 2 - 3 + 1 = O h o m e q n . ( 6 )

+// 0

cH,-N'O or cH,-iv

l o ' 0 - Formula CHsNOl

D U = 1 - 1.5 + 0.5 + 1 = 1 from eun. (6)

Equation (6) similarly gives a correct result for nitro- methane

(6) Radicals and Positive Ions

The results described above for molecules are eauallv valid . "

for odd-electron ions, such as the molecular ion found in mass soectrometrv. However in the case of a radical. which lacks a-hydrogen relative to the closest correct molecular structure, the calculation of DU using eqn. (6) will give a value 0.5 in excess of the true value, due to the negative sign of y/2 in eqn. (6). For example, the radical has DU = 19 - 7.5 + 1 = 12.5, whereas the correct value is 12 as can he seen for the tri~henylmethvl radical which has this formula.

Even-electron positive cations resulting from fragmentation under electron impact also lack one hydrogen relative to the closest correct molecular structure. Thus, again the DU calculated from eqn. (6) will he greater-the real value by 0.5. For instance, the tropylium cation C7H71 +, has DU = 7 - 3.5 -5 while the real value is 4. Similarly, the henzoyl cation C7Hs0)+ has DU = 5.5 while the real value is 5.

Although this is true for gas phase fragmentations in mass spectrometry this is not necessarily the case in solution, since even-electron cations may result from protonation of a molecule. In this case there is a hydrogen atom in excess and calculation of DU using eqn. (6) will give a value inferior to the

real value by 0.5. Thus, care must he taken in calculating DU for even-electron ions.

Conclusion In summarizing, it can be seen that the nitrogen rule can be

of considerable aid in the determination of molecular formulas. The oaritv of the number of monovalent atoms in a formula is imporlant in that it governs the parity of the molecular weieht. M. which in turn allows the detection of the presence ofanodd number of nitrogen atoms in the molecule of a given organic com~ound. A knowledee of the DU of a formula also- gives vaiuahle indicationsuas to molecular structure and often the extent of conjugation. Finally, i t has been seen that in calculating DU, only the valency of each atom is important and in this respect nitrogen acts as any other trivalent atom. In contrast, for the parity of M only nitrogen plays a unique role even to the exclusion of other trivalent atoms. The importance of DU is exemplified by the recent work of McLafferty (13) which describes attempts to determine DU for an organic compound directly from its mass spectrum.

Acknowledgment The author wishes to thank warmly Drs. J. L. Aubagnac,

J. P. Billiot, and A. Fruchier of the Univers i~ des Sciences et Techniques du Languedoc at Montpellier for their advice and suggestions.

APPENDIX: Demonstration of the Euler Formula In order to demonstrate the Euler formula (12) the eranh - .

Ilviny stutlit.tl will I w rl~n4derrd lo represtlnt fields (the faces # w rt.:wtl:, bun lw~d by il svstenl iddykes (the sides ur lines) uith ( l i t exterior region Iwing cuvtwd with witter. It is then ini;~ui~~ed rnn t tl~ed\.kv.i are destngyed unr after theother until all the fields are uider water ( ~ i & 7).

To do this it is not necessary to destrov all the dvkes: ob- viously those with water on both sides canbe retained. 1f only those dykes are destroyed which have water on one side, then one extra field is submerged each time a dyke is destroyed. Since in the beginning only the outer region is under water, there are f fields to be submerged. Hence, after destroying the appropriate walls one after the other until all the fields are submerged, f dykes will have been destroyed.

If one now considers the system of dykes which remain intact, it will he shown that it is possible to go from any point to another one without crossing the water by following the intact dvkes. This. of course. was uossihle before heeinnine to destroy the dykes, since thk water lay only on the exterior. Now suppose that a dyke AB (Fig. 8) is destroyed and that as a result the system becomes separated into two completely isolated islands. The destruction of the dvke AB makes it impossible to get from A to B without enterLg the water. The water com~letelv surrounds each of the two svstems. Thus. there must Ihdw 1 1 r . t ~ water im h t h sides 01, the dyke AH be- for( it., destruc~i<,n, which ron~radicts the rule that surh dvkes are not to he destroyed.

Furthermore, to get from any point, P, to any other one, Q, there is only one path possible without entering the water.

Figure 7. (left! A collection 01 fields bordered by a system of dykes, tha extwior region being covered with water. (-- intact dyke. - - - - destroyed dyke!.

Figure 8. (center) Destruction ol the dyke AB makes it impassible to get from A to B without entering the water.

Figure 9. (right! To get from a given point, P, to another, 0, there is only one possible route without entering the water.


If this were not the case and there were two nossible oaths Literature Cited

from (Fig. then the intact dykes (1) L3eynon.J. H.,"MassSpeetrometry and its ApplicationstoOrganic Chemistry,"El- two nathwavs would form an enclosure which would not be ~ o . ~ ~ . A."~.o.~m." >man -" "" ..., submerged, which contradicts the requirement that all fields (2) ~i??ann. .. K., . ~ . ~ . " M ~ B spectrametry, organic chemical ~ppiieations," M C C ~ ~ W ~ H ~ U ,

be submerged. Thus, for a given starting point P, once all the fields are

submerged there is only one dry path between P and each of the other points. On each such pathway there is a final dyke crossed in order to reach the target point and this dyke is uniquely determined for that point. There is thus a relation- ship between the points and the intact dykes since there is an intact dyke for each target point. The starting point P not being a target point, the number of non-destroyed dykes equals p - 1. Since there are f dykes destroyed and p - 1 dykes intact, the total number of dykes, k , is thus: k = f + ( p - 1) from which

New York. LYLZ. (31 McLafferty, F. W.. "Interp~efstim of M w Spectra: An Intmduction," W. A. Benjamin

he.. New York, 1966. (1) Browning, D. R., "Spectroscopy," McCm-Hill, London. 1969. (51 Hill, H. C.. "lntroduetion to Mass Spechometry," Heydon Ltd., London, 1972. (61 Leszla, P., and Stang, P. J., "Spedrosmpie Orgsnique," Hermann Ed., Paris, 1972. (71 Silveratein, R. M., Bssslcr, C. C., and Marrill, T. C.. "SpeNometric Identification of

Organic Compounds? 3rd ed., Wiley, New Ymk. 1974. (8) Longevialle. P., "Principe de la sppctrombhie de masse des subatsnaes organiques."

Massan, Paris. 1981. (9) Cram, D. J., and Hammond, G. S., '"Organic Chemistry." McGraw-Hill, New York,

59. erds, J. H., Cram,D. L a n d Hammond. G. S.,"El&ments de Chimie Orgsnique," (10) Rich

Mecraw-Hill, Montrbal, 1968. (11) Allinger. N.L.,Cava,M.P.,DeJongh,D.C.,Johnmn,C.R.,Lebel.N.A.andStevens,

C. L., "Organic Chemistry: 2nd ed., Worth Pvhlishers Inc., New York. 1976. (12) Radmacher, H. and Tueplik, 0.. "The Enjoyment of Mathematics? Princeton Uni-

versity Press, Primeton, NJ, 1957. This baok hes been translated inta French and I f = k - p + l 1 Geman. (3' (13) Dayringer, H. E., and McLafferty, F. W., 078. Mass Spectrom~try, 12.63 (19771.


Molecular formulas of organic compounds: the nitrogen rule and degree of unsaturation

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Transcript of Molecular formulas of organic compounds: the nitrogen rule and degree of unsaturation