Haloallenes: Chiral Compounds without Chiral Carbon Atoms

Four questions regarding the structure, stereochemistry, and symmetry of haloallenes. Keywords (Audience):. Upper-Division Undergraduate. Keywords ...
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Haloallenes: Chiral Compounds without Chiral Carbon Atoms

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lgor Novak

National University of Singapore Singapore 0511 The followina auestion can be useful for advanced undergraduate or biginning postgraduate physical chemistry courses in Svmmetrv and Grouo Theorv and Molecular ~ ~ e c t r o s c o~each&s ~< of under&aduate" organic chemistrv course (stereochemistrv) mav find parts two and three of it useful too, especially because the answer can be arrived a t most easily by the use of molecular models. Question (11 Using W, X,

(2) The requirement for chirality is that no S. axis exists. The point groups of the static structures contain no S,, axes. Furthermore, restricted rotation ensures that rearraneement does not result in time-averaeed structure with an S, axis. (3) In order to calculate probability ( p = nllnz) one must know the number of chiral haloallenes (nl) and the total number of haloallenes (nz). ForDU symmetry CE,l) = 5 which leaves 4 after subtracting the allene itself. Thus, there are four molecules of that symmetry: FzCCCFz,ClZCCCCIz,Br2CCCBrz,IzCCCIz For Cz Symmetry C(5,2)= 10. For Cz, symmetry C(5,2)= 10. .For C. symmetry general formulas are: X2CCCXY and XzCCCYW.There are two distinct permutations among X, Y

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substituents in X.S!CCXY and three ~ermutationsof X. Y.. W substrtuentr in X,CCCY\V. Taking thrsc prnnmntions into nrrount onrgcts 1C*5;1t7 3C(5;J, = 50 halosllencs belmgmg to the C, point group. For C1 symmetry general formulas can be XYCCCXW and XYCCCWZ. XYCCCXW gives [using C(5,3) = 10 and similarly multiplying by three1 a total of 30. XYCCCWZ can he calculated from C(5,4)= 5 and multiplying by three to take combinations: XYCCCWZ. XWCCCYZ. and XZCCCYW. The total for C1 symmetry is then 3C(5,3)+ 3C(5,4)= 45. Adding the relevant numbers and remembering that bath Cz and C , molecules are c h i d we get 55. We must finally multiply by two because each c h i d molecule exists as a pair of enantiomers to get nl = 110. Adding all haloallenes gives the grand total nz = 174 with probability p=110/174=0.632

Y,Z to represent four different substi-

tuents attached to terminal carbons, draw generic structures of all possible mono-, di-, tri-, and tetra-haloallenes, then specify point group (symmetry) of each. Don't forget that haloallenes may possess one to three hydrogens and halogen atoms should include F, C1, Br, and I. (2) Explain why haloallenes can be chiral without possessing chiral carbon centers? (3) Calculate the probability that you will select a chiral haloallene molecule (in a single try) from a collection of all possible haloallenes! (4) Suggest ways of determining the symmetry of an unknown haloallene using methods of o~ticals ~ e c t r o s- c- o ~ ~ ! Hints: The numberof different combinations (C) and permutations (PI ofn things takenp a t a time are given by: C(np) = n!/Ip!(n-p)!land P(n,p)= n(n-l)(n-2)..(n-p+1) Acceptable Solution

(1) The drawing is

(63.2%).

(4) We have five possible symmetries: Du, CZ., C,, CZ, and C,. Measurements of the anele of rotation of plane DOlarized light will differentiate C; and Cz symmetkes frbm the remainine three:. onlv " C, and CI svmmetries will eive non-zero an&. The altfrnaiive meihdd for recoLmizinf C , and C?svmmctries is circular dichroism rCT) soectroscoov. An attempt to observe a pure microwave (I& spect&n ') can be used to establish whether a molecule has a permanent dipole moment (Cz,, C,) or not (DM).In ordertodistinmish structures further, one can derive svmmetries of vibrational normal modes i n haloallenis and compare Raman and IR vibrational spectrum of the unknown molecule: C ~ V 6a1 + la2 + 4bi

Cs

Cn C1

10a'+ 5a" 8a+ 7b 15a

+ 4b2

(all modes except az Raman & IR active) (all modes IR & Raman active) (all modes IR & Raman active) (all modes IR & Raman active)

If the molecule belongs to Cz, it will exhibit one fewer fundamental band in IR than in Raman spectrum while C, molecule will have the same number of fundamentals in IR and Raman. To distinguish between Cz and C1 one can use polarized Raman scattering to find out the number of totally symmetric modes in the spectrum (depolarization ratio); they differ in Cz and C1 symmetries. Volume 71 Number 7 July 1994

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