Dolbier HW Solutions 578

Dolbier HW Solutions 578 - m/z for C 14 H 8 Cl 4 316 35 Cl...

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( c ) The number of octachlorobiphenyls will be equal to the number of dichlorobiphenyls (12). In both cases we are dealing with a situation in which eight of the ten substituents of the biphenyl system are the same and considering how the remaining two may be arranged. In the dichloro- biphenyls described in part ( b ), eight substituents are hydrogen and two are chlorine; in the octachlorobiphenyls, eight substituents are chlorine and two are hydrogen. ( d ) The number of nonachloro isomers (nine chlorines, one hydrogen) must equal the number of monochloro isomers (one chlorine, nine hydrogens). There are therefore three nonachloro de- rivatives of biphenyl. 23.29 The principal isotopes of chlorine are 35 Cl and 37 Cl. A cluster of f ve peaks indicates that dichlorodiphenyldichloroethane (DDE) contains four chlorines.
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Unformatted text preview: m/z for C 14 H 8 Cl 4 316 35 Cl 35 Cl 35 Cl 35 Cl 318 35 Cl 35 Cl 35 Cl 37 Cl 320 35 Cl 35 Cl 37 Cl 37 Cl 322 35 Cl 37 Cl 37 Cl 37 Cl 324 37 Cl 37 Cl 37 Cl 37 Cl The peak at m / z 316 therefore corresponds to a compound C 14 H 8 Cl 4 in which all four chlorines are 35 Cl. The respective molecular formulas indicate that DDE is the dehydrochlorination product of dichlorodiphenyltrichloroethane (DDT). The structure of DDT was given in the statement of the problem. This permits the structure of DDE to be assigned. SELF-TEST PART A A-1. Give the product(s) obtained from each of the following reactions: ( a ) ? (two products) Cl CF 3 KNH 2 NH 3 DDE (only reasonable dehydrochlorination product of DDT) Cl Cl C C Cl Cl 2 HCl C 14 H 9 Cl 5 DDT C 14 H 8 Cl 4 DDE 672 ARYL HALIDES...
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This note was uploaded on 10/03/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.

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