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Solutions for Infrared Spectroscopy and Mass Spectrometry
6 continued
(b) The strong absorption at 1691 cm–1 is unmistakably a conjugated C=O. The smaller peak at 1626 cm–1 
indicates a C=C, probably conjugated with the C=O. The two peaks at 2712 cm–1 and at 2814 cm–1 
represent H—C=O confirming that this is an aldehyde. Unsaturated =C—H above 3000 cm–1 ; saturated 
C—H below 3000 cm–1 .
(c) The strong peak at 1650 cm–1 is C=C, probably conjugated with C=O as it is unusually strong. The 
1703 cm–1 peak appears to be a conjugated C=O, undeniably a carboxylic acid because of the strong, broad 
O—H absorption from 2400-3400 cm–1 . The unsaturated =C—H is obscured by the strong O—H; 
saturated C—H below 3000 cm–1 .
(d) The C=O absorption at 1742 cm–1 coupled with C—O at 1220 cm–1 suggest an ester but 1742 cm–1 is 
too high to be conjugated. The small peak at 1604 cm–1 , peaks above 3000 cm–1, and peaks in the 600-800 
cm–1 region indicate a benzene ring. It has both unsaturated C—H above 3000 cm–1 and saturated C—H 
below 3000 cm–1 .
(a) The M and M+2 peaks of equal intensity identify the presence of bromine. The mass of M (156) minus 
the weight of the lighter isotope of bromine (79) gives the mass of the rest of the molecule: 156 − 79 = 77. 
The C6H5 (phenyl) group weighs 77; this compound is bromobenzene, C6H5Br .
(b) The m/z 127 peak shows that iodine is present. The molecular ion minus iodine gives the remainder of 
the molecule: 156 − 127 = 29 . The C2H5 (ethyl) group weighs 29; this compound is iodoethane, C2H5I .
(c) The M and M+2 peaks have relative intensities of about 3:1, a sure sign of chlorine. The mass of M 
minus the mass of the lighter isotope of chlorine gives the mass of the remainder of the molecule: 90 − 35 
= 55. A fragment of mass 55 is not one of the common alkyl groups (15, 29, 43, 57, etc., increasing in 
increments of 14 mass units (CH2)), so the presence of unsaturation and/or an atom like oxygen must be 
considered. In addition to the chlorine atom, mass 55 could be C4H7 or C3H3O. Possible molecular 
formulas are C4H7Cl or C3H3ClO.
(d) The odd-mass molecular ion indicates the presence of an odd number of nitrogen atoms (always begin 
by assuming one nitrogen). The rest of the molecule must be: 115 − 14 = 101; this is most likely C7H17. 
C7H17N is the correct formula of a molecule with no elements of unsaturation. The seven carbons probably 
include alkyl groups like ethyl or propyl or isopropyl. A hint about spectra with odd-mass molecular ions: 
Look for fragment peaks with even masses to confirm that the odd-mass peak is the parent and not just a 
fragment from an invisible molecular ion.
7
mass 29
radicals
not detected
m/z 57
+
57
CH3 CH
CH3
CH2 CH2 CH3 CH2 CH3CH CH2
CH3
CH3
8 Recall that radicals are not detected in mass spectrometry; only positively charged ions are detected.
The fragment giving m/z 57 is a primary carbocation, less stable than the more abundant secondary 
carbocations of m/z 43 and 71.
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