Use of Degrees of Unsaturation in Solving Organic Structural Problems LeRoy H. Klemm University of Oregon, Eugene, OR 97403 Determination of the com~letestructural formula of a newly isolated or synthesized organic compound may involve extensive, sophisticated procedures, including chemical transformations and spectral studies. I n general, for each new compound produced in chemical research one obtains elemental analysis and, thereby, a molecular formula for the substance. Acommon structural problem encountered in organic chemistry textbooks, homework, and examinations provides the molecular formula of the compound considered. In this communication I show that valuable structural information is easily calculated from the molecular formula and that this can help greatly in a n initial mental assessment of structural features present in the compound. The calculation provides a numerical quantity variously called the degree(s) of unsaturation (DU) (I), in this article. elements of unsaturation (2).units of unsaturation (3). index ol'hydrogen deficiency (41;sum of double bonds and rines tSOI)AIO .5,. or sinlolv a rinr or unnaturatlon 16.. ~ h i ~ r e s e n t a t i oofnDU diffeis c o n s ~ e r a h l yin these references. In SODAR one must consider a triple bond a s the equivalent of two double bonds. In ref 6 the term unsaturation refers to a double or triple bond and must be distinguished from degrees of unsaturation. In refs 4 and 6 the authors consider hydrocarbons only and, thereby, limit the usefulness of the DU concept. References 3 and 5 extend the calculation to molecules containing halogen or oxygen, whereas refs 1and 2 include nitrogen atoms a s well. None of these references considers organic molecules with other elements, particularly sulfur, silicon, and phosphorus, which are also presented here. Extension to other elements is also possible. Surprisingly, none of the numerous other recent or historically well-known organic chemistry textbooks examined even mentions the DU concept. Calculations for a Hydrocarbon First. consider what demees of unsaturation mean in a hydrocarbon. The stand:~;d of reference ri.e., I)U = 0)is a saturated molecule or alkane, C,Hz, .2. The /)(I fnr another hydrocarbon (ZJcontaining n carbon atoms per molecule equals the number of moleculrit of II, that must, at least in theory, be added to % t o i h t n ~ nthe alkane, or alternatively must be removed from the alkane to product: 2.If Z has the atmolerular fonnuln C.,.H,.. then %is short two hvdroeen " oms or one Hz molecule for a n alkane, and DU = 1.Z could have either one ring structure or one double bond. If Z, instead, has the formula C,H2,.2 then its DU = 2, and Z could have one of these possible structural features: two rings, two double bonds, one ring plus one double bond, or one triple bond. Benzene, CsH6, for example, is compared with hexane, CsHla, as shown below.
-
C6H14
- C6Hs Hs or 4M. Therefore,DU= 4
This result is consistent with the usual structural formula 0 which indicates three double bonds plus one ring. In fact, when one has a hydrocarbon with DU 2 4 one should be on the alert for the possible presence of a n aromatic ring. A simple structural problem was posed in the first chapter of a n older textbook (71,which, unfortunately, does not introduce DU to the reader. Application of DU should assist in solving the problem, a s follows. A compound shown by analysis to have the formula C9Hlo gives benzoic acid on vigorous oxidation. What are the four pos-
sible structures for this compound?
- CgHio Hlo or DU = 5 H
benzoic acid There must he 5-4=1
degree of unsaturation in a side chain. C9Hl0- C6H5 = C3H5
for the side chain, which can contain one ring or one double bond. In other words, it can he
Without use of DU, the student may tend to overlook the possibility of cyclopropylhenzene a s a n answer. The fourth of these side chains can exist a s cis and trans isomers, so five structures are, indeed, possible. However, such isomerization is not introduced in the book until later. Calculations for Molecules with Heteroatoms To handle molecules containing elements other than carbon and hydrogen, first calculate the molecular formula of the hydrocarbon, called the equivalent hydrocarbon (EH), which would have the same DU as the compound being considered. This is accomplished by a series of simple rules applied consecutively in any order to the molecular formula, a s follows. 1. Replace all halogen atoms (i.e., monocovalent atoms) by an equal number of hydrogen atoms in the formula. 2. Delete oxygen and sulfur (i.e., dicovalent atoms) from the formula. 3. Delete all nitrogen atoms (i.e.,tricavalent atoms) fmm the formula as NH units. Do likewise for phosphorus as PH units. 4. Replace silicon atoms (i.e.,tetracovalent atoms)by carbon atoms, or alternatively delete all silicon atoms as SiH2units.
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Then compare the resultant EH with the corresponding alkane, a s previously, and calculate DU for the original compound.
From rule
Examples
From rule 2
Compound A Compound A h a s the formula C7H5Br02S. Replacing bromine with hydrogen gives C7H5Br02SoC7H602S where the symbol o means is equivalent to. Deleting oxygen and sulfur gives
o C2,H3,N203PS2Si C20H37ClIN203PS2Si
C20H39N203PS2Si o C,,H3&PSi CznH39NzPSi o C20H36Si From rule 4 (replacing silicon by C) C2,H3,Si o C21H36= EH
One can now propose numerous possible structures for A. Three examples are given below.
Rationalization for DU Values Now consider the reasons why calculating DU actually works correctly. First we look a t cases, illustrated by chlorine, oxygen, nitro0 CH2, t gen, and silicon, where only one heteroatom ,CH--eC-C=-S-Br (i.e., a n atom other than carbon or hydrogen) CH2 is present in the functional group, as given in examples 1-9.
n
77
yo
1
H~C-C=C-C-C-~HZ I I Br H
,4.
"
Count the DU in each structure a s a check on the answer. Sulfoxide, sulfone, and sulfonyl groups are written with semipolar (also called dative or coordinate covalent) bonds, rather than a s S-0 double bonds, a s given in some textbooks. This allows consistency with the DU concept, follows the structural preference of some chemists (including this author), and avoids the problem of worrying about the oxidation state of the sulfur. I n fact, the middle structure shows a peroxide, in which the oxygen h a s the unusual oxidation state of -1. Oxidation states can be ignored i n the use of DU. Also, semipolar bonds do not count a s degrees of unsaturation. Cases involving nitrogen in various oxidation states are shown in the next example. Compound B Draw five possible structures for compound B, molecular formula C6HgNO2. Solution C5HgN02o C5H,N o C5H8 = EH C5H,, - C6H8= H4or 2H2 Hence,
DU=2 Possible structures with nitrogen in different functional groups and various oxidation states are the following.
fl
fl
CH3C-yCH2CCkb H
HG:? H
H
F
CHPC-CEN I CHzOH
0 0
CHFCH(CH~)~-N// \o
11 OH H-C-CH-C-H I
C H ~ - ~ H ~
Compound C Calculate the DU for compound C of molecular formula C20H37C11N203PS2Si.
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Journal of Chemical Education
In 1the monocovalent halogen atom has merely replaced a monocovalent bydrogen atom from a hydrocarbon. Thus, one obtains the equivalent hydrocarbon by reversal of this process, and without any change in DU. Three situations arise for dicovaleut oxygen. I t may occupy a position between two carbon atoms (as in 2) or between a carbon and a bydrogen ( a s i n 3). Alternatively, i t may be doubly bonded to a carbon atom (as in 4). I n 2 and 3 one visualizes simply extruding the oxygen atom to leave a hydrocarbon (propane here) behind. The presence of the oxygen atom has not changed DU a t all. In 4 (DU = 1) it is apparent that the oxygen atom has replaced two hydrogen atoms from a parent hydrocarbon. For 5 extrusion of a n NH unit is analogous to removal of oxygen in 2, whereas removal of NH from 6 is analogous to removal of oxygen from 4. For the nitrile 7 we visualize tautomerization of the molecule (no loss of hydrogen atoms) to 8. Then we can remove NH to IeaveDU = 2. In 9 replacement of silicon by carbon or extrusion of SiH2 does not change the value of DU. Some cases where more than one beteroatom is present in a functional group may be visualized a s illustrated below.
112uopui-e substances form a crystalline molecular cam. pound or complex. Two examples are anthracene.1,3,5-trinitrobenzene, Cl4Hl0*C6H3N3Oe,mp 164",a stoiehiometric complex (11) C1,H2,.4.4CH,N,S, p-di-t-butylbenzene*thioureainclusion complex,a nonstoichiametric,crystalline complex (12,131
. y\l
2 rings + 1 double bond
For calculation of DU one should treat each component separately. Experimentally, the worker is prepared for these cases inasmuch a s the two components of the complex are deliberately mixed to obtain a crystalline product. 112uopure substances react chemically to form a salt. Examples are ammonium trifluoroacetate (8)
Limitations on DUCalculation The author has encountered five situations in which a calculation of DU may give a useless, doubtful, or erroneous result, a s indicated here, unless one recognizes and circumvents the problem. The compound does not h u e a n EH. Examples are HN03, HzSO4, CS, NH4C1, CzN2, and CO. Calculations of DU should not be made from the molecular formulas in such trivial cases. An example of the difficulty that arises is illustrated by comparison of CO and Con. One might be inclined to write a n EH a s CHo in both cases. This would give DU = 2 for both compounds-wrong for CO, but correct for COz. The molecular formula as deriued from elemental analysis may include soluent of crystallization or adherence.
C2H4F3N02 or CF3C02-NH,' and N-methylpyridinium iodide (commonly called pyridine methiodide), C&IN or [C5H5NCH31+I-,mp 117" (14). Using the molecular formulas directly, one calculates DU values of 0 and 3. However, from the known structural formulas,
we see that the correct DU values are 1and 4. The erroneous DU values arise because nitrogen is tetracovalent in these salts, rather than tricovalent a s assumed in Rule 3. One can circumvent this problem by considering these salts as molecular compounds CF3CO2H.NH3 and CSH~N-CH~I and treating them a s in the preceding sec-
Three examples include
.
oxalie acid dihydrate, Structural Studies on an Unknown Amine C2H204.2Hz0,mp 106" (8) .a heterocyclic disulfone disulThe following tests and observations were made on liquid amine A. Complete the third column with fonamide with one-half male of the structural inferencethat you make from each numbered test observation. (The blank third column 2-butanone of crystallization, has been filled in with the expected answers.)
.
propargyl phenylpropiolate, a Test or Reaction liquid which distills in vacuo 1. Elemental analysis on A with retention of water as ClzH802.1/2H20,bp 68.5" (0.5 2. Hinsberg test on A bp mm) or CIzHsO2~1/4H2O, lZ09(0.8 mm) (10) 3. Polarimetry on A In all of these cases one obtains a meaningful DU calculation 4. Vigorous oxidation of A only by ignoring the solvent pre- (boiling NaOH-KMnOeHzO) sent. Thus, for oxalic acid the molecule CzHzO4corresponds to 5. Treatment of A with CHd EH = C2H2 and DU = 2, correct values. If the molecular formula 6. Treatment of A with is taken as C2H606 instead, one Ph-N=C=S concludes erroneously that DU = 7. Write structural formulas for 0. In a n experimental situation A, 6, and C. be alert to the possible retention of solvent in the sample being analyzed. When the presence of solvent is suspected, check the 8. Write a balanced equation for s a m p l e by spectral o r o t h e r reaction 5. means to ascertain the appropriate molecular formula for calculation of DU.
Result or Observation
Structural Inference
1. Mol. formula of A is CgH13N.
1. For A. D U = 4 .
2. White ppt forms.
2. A is a secondary amine.
It is insoluble in H z 0 or in aqueous HCI. 3. A is dextrorotatory.
3. A contains a chiral carbon atom.
4. After reaction, the addition of 4. For B, DU= 5. HCI gives white solid B, C7H602. is probably C6HsC02H.
5. Compound C, CioHtdN forms.
5. A forms a monomethiodide,
6. A reacts to give a crystalline solid.
6. A is a primary or secondary amine.
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tion. Again the formation of such salts is usually deliberate and, hence, the problem can be anticipated. The substance is a free radical For simplicity we will consider only the case where the free radical results from removal of a monocovalent atom, particularly a hydrogen atom, from a parent stable molecule to leave a stable free radical. The formation of the free radical galvinoxyl, C 2 g H 4 1 0 2 (8, by abstraction of a hydrogen atom from its precursor, C29H4202,will illustrate the point. The precursor has DU = 9. For galvinoxyl itself the EH is C9.Hx. "" -.and a calculated DU = 9.5. Findine a nonintegral value for DU should immediately alert a itudent to the ~ossibilitvthat some covalencv rule does not a ~ n l v to the molecule" under consideration, or that a n error h a s arisen somewhere. Complete Solution of a Structural Determination Problem To illustrate the use of DU plus other information in the complete solution of a structural determination problem I present here a typical sort of question taken from a n hour examination during the third term of a three-term (i.e.. one academic yearlorganic chemistry lecture course a t the Universitv of Oreeon. The students will have iust completed the textbook chapter on amines and will have conducted a n experiment (during the previous term) on the identification of a n unknown liquid amine by a combination of measurement of boiline point, chemical tests, and determination of melting i n one or more crystalline derivatives. For the laboratory- experiment the student . was provided with a sheet of names of 30 possible amines with their boiling points, as well a s melting points on some of their crystalline derivatives. The stockroom personnel then handed out coded samples, one to each student a t random. The examination question follows. (See the table.) The total question was worth 30 points: 9 points for part 7 and 3 points for each of the other parts. Because the examination was given to a class of 150-250 students in a room where students sit in adiacent seats. two other modifications of this question were made for'^ as C8HllN (a orimam amine) and A a s ClnHlrN(a tertiarv amine). Then examination papers were distributed in 1-2-3-1-2 ... or-
-
A"
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Journal of Chemical Education
der. An average grade on this question was 20-24 points with 10-15% of the class obtaining full credit. There were always some who failed to recognize the presence of a benzene ring and others who tried to write a structure for A directly, based only on its molecular formula. I n honors sections of organic chemistry the author would introduce svectral data ('H and 13C NMR. mass.. IR., and sometimes even UV)as part of the experimental portion. The students were then provided with some sheets of IR and NMR tables. I n a short course in organic chemistry for students who do not take laboratory, I still use DU and structural problems, hut only based on information from the textbook and lectures. Conclusion The concept of degrees of unsaturation in organic chemistry is readily understood and applied by students in undergraduate organic chemistry. I t facilitates solving of nroblems in structural determination. where one annlies experimental and theoretical concepts'to the development of one (or more) rational structural formulas from a n initial molecular formula. Use of the DU concept should be made a standard part of every beginning organic chemistry course.
..
Literature Cited 1. MeMurry, J.Organic Chemistry: BrookdCole: Monterey. C& 1984: pp 12P125. 2. Wade, L. G. Organic Chemistv; Prentice-Hall: Englewood Cliffs, NJ, 1967; pp 243245
..
4. siomonr,T. iv G.oraonie Chemistry: 4th eb.: wiley: ~ e ~wo r k1988: , pp 247-248. New York, 1987;pp 168-189. Seealso 5. Carey,FA.OrgonicCh~mislry;MeGraw-Hill: Atkins, R. C.; Carey. F. A. Olgonic Chemistr/--AB~ie/Cou~m:Mecraw-Hill: New York. 1990: PO 265-266. ; J. S. Oqanic Chemistry; BmokdCole: Monterey CA. 6. Fessenden. R ~ J .Fessenden,
- a""-. .-"", ,?me.
7. Robe*, J. D.;case"0.M. C.BosicPn"ciple~~fOlgani~Ch.milry: Benjamin, New York, 1964: p 21. 6. AIdrich CofologofFina Chemicals:Aldrieh Chemical: M i h a t h e . WI, 1988-1989. 9. K1emm.L. H.; Lane, W.; Hall. E. J Hdamcycl. Chem 1981.28, 187. 10. Klemm, L. H.:Klemm, R. A.; Snnthanam, P. 8.; White, D. V J. Org. C h m . 1911.36, 2169. 11. Ref7; p875. 6thed.;Heath Lexington. MA, 12. Fieser,L. E;Williamson, K L.OrgonicErp~rimsnts; 1987; pp 146149. 13. Fieser. L. F:Fieser, M. Reogenfs for Organic Synthesis:W~lqv:New York, 1968; p >>M
14. Shrine&R. L.; Fuson, R. C.:Curtin. D. Y The Systematic Idanlifieation of Organic Compounds, 4th ed.:W h y : New York. 1956; pp 228-229.296.
