Question

7. Figure out what compounds 1 and 2 are. Draw out each structure. Support your conclusions with the calculation for elemental analysis, interpretation of spectral data, and include the electron count for each compound. Use the back of the page, if necessary. 'Mo is the most abundant isotope of Mo (24%) and 35Cl is the most abundant isotope of Cl (76%). (16 pts) a) 2 MoCl3 + 6 Na[C3H5] → 21 + 6 NaCl + "Cp-dimer" b) 1 + Zn(CH3)2 → 2 + ZnCl2 Compound 1: green crystals 'H NMR (CDC13): 8 -5.0 ppm (s, 10H) Anal Calc'd: C: 40.43%; H: 3.39%; CI: 23.87% EIMS: 299 m/z (P+) Compound 2: orange-red crystals 'H NMR (CDC13): 84.07 ppm (s, 10H), 0.22 ppm (s, 6H) Anal Cale'd: C: 56.25%; H: 6.29% EIMS: 258 m/z (P+) Other important MS data: 243, 228, 163, 98 m/z

Answer 1

The structure of Compound 1 is Compound a 13 COMPOUND 1 n Elemental Analysis: - 1 "has I Molybdenum atom of mass 98 , 2 chlorine atoms of mass 35.453 ; 10 hydrogen atoms of mass 1.00784 and 10 carbon atoms of mass 12. 10.0784 XLOO = 3.37 % Total Mass = 1498 + 2x 35.453+ 10x1.00784 +10 x 12 = 299 which is the molecular mass of Compound 1. Percentage of carbon = 120 Percentage of hydrogen = 10.00 Percentage of chlorine = 70.906 460o = 23.71% 299 All these values match with the experimental data. HNMR shows a singlet peak (10 hydrogen atoms) at 5ppm This can be assigned to to hydrogen atoms of two cyclopentadienyl rings. (The hydrogen atoms are shown in red colour) EIMS In mass spectrum peak at 299 (m/2) corresponds to Molecular in peak. Electron Count. Moc GE con Mo-adiss-bés Total electron count = 6+2+10 Electron Count :- (Mochy 62% ests).] = 18ēs C2SQC- 2 es 2mCHS) - 2x5 = 10 és 2012 des us = lo es

H COMPOUND 2 (Mo ( Cs) (n5 Cghts)] Elemental Analysis: - Compound a has I Molybdenum atom of mass 98, 16 hydrogen atoms of mass 1.00784 and 1a carbon atoms of mass la. . Molecular mass of compound 2 = 1x98 +16* 1.00784 * 12 X12 = 258 % of carbon = Mass of carbon *100 = 144 xloo - 55.8% molecular mass 258 % of hydrogen = 16.125 x100 = 6.25% 258 These values match with experimental data. HNMR shows a singlet peak at 4:07 ppm which corresponds to to hydrogen atoms present in two cyclopentadienyl risgs. Peak at 0.22.ppm Corresponds to six hydrogen atons present in two methyl groups. EIMS 258 Molecular in peak. 243 → peak one methyl due to fragmentation lie after the removal of group; 258-15 = 243 228 - fragmentation peak after the removal of second methyl group; 243-15 = 228. 163 Fragmentation peak after the removal of one cyclopentadienyl nag; 228-65 = 163 98 fragmentation peak after the removal of second cyclopentadienyl sing, 163-65 = 98 Electron Count : - (Mo (CH₃), (43 - Chs) ] → Mo-4d565-bes. २ CH3 > २६s . i. Total electron Count = 6+2+10=18 2456th) - 2x5 = lo ēs.