Spectroscopy Discussion

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Problem 1

A hydrocarbon having a molecular mass = 112 could have the formula C9H4 (divide 112 by 12) or C8H16. Since we are told the compound contains oxygen, possible formulas include C7H12O, C6H8O2 & C5H4O3. The nmr data suggests a high degree of symmetry (one 1H nmr signal & two carbon signals). The infrared absorption at 1717 cm-1 is a sure sign that a C=O function is present. It would be difficult to write a C7H12O structure that satisfies these features. However, for C6H8O2 1,4-cyclohexanedione agrees with all the evidence. In the case of C5H4O3, 1,2,3-cyclopentanetrione comes close, but would have three 13C nmr signals.

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Problem 2

The only hydrocarbon having a molecular mass = 90 amu has the formula C7H6. Since the compound contains oxygen we must consider C6H2O, C4H10O2, C3H6O3 & C2H2O4 as possible formulas. Of these C4H10O2 is the most probable, since the m/z=45 base ion peak would result from cleaving the molecule in half. From this formula we conclude the molecule has no rings or double bonds. The infrared spectrum shows no absorption for a hydroxyl group, so the oxygen must be present in ether functions, as indicated by the strong absorption from 1105 to 1125 cm-1. The 1H nmr signals also suggest that all hydrogen atoms are on carbons bound to an oxygen. The symmetry suggested by the simple nmr spectra is best accounted for by 1,2-dimethoxyethane.

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Problem 3

The three molecular ion peaks at intervals of 2 amu are a sure indication that two chlorine atoms are present in this compound. Two bromines are not possible (mass = 158, 160 & 162), and a combination of bromine and chlorine would give peaks having a different intensity ratio. Also, the fragment ions at 111 & 113 amu have the correct intensity ratio for one chlorine. If we subtract the mass of two chlorines from the molecular ion mass, we are left with 76 amu for the carbon structure. This is most likely C6H4, and together with the nmr data suggests a benzene ring. Once more, the simplicity of the nmr spectra indicates para-substitution of the ring. The ortho and meta-dichlorobenzenes would have three & four 13C nmr signals respectively.

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