1 (518)

Prévia do material em texto

510 CHAPTER 14 
 
Now, we must consider the relative abundances of each peak. The peak at m/z = 423.0647 is the base peak, because it 
has a relative abundance of 100% (it is, by definition, the most abundant ion). The peak at 424.0681 has an abundance 
of 22.5%, relative to the most abundant ion. This is consistent with an ion having 21 carbon atoms, because the natural 
abundance of 13C is 1.07%. [21 × 1.07% = 22.5%] 
The peak at 425.0605 has an abundance of 4.21%, relative to the most abundant ion. This is consistent with an ion 
having one sulfur atom, because the natural abundance of 34S is 4.21%. [1 × 4.21% = 4.21%] 
 
(b) The peak at m/z = 425.0605 is consistent with an ion in which two of the 12C atoms are replaced with two 13C 
atoms: 
 
12C19
13C2H15N2O6S¯ = 
(19 × 12.000) + (2 × 13.0034) + (15 × 1.0078) + (2 × 14.0031) + (6 × 15.9949) + (1 × 31.9721) 
= 425.0715 
 
 
14.69. 
(a) Tautomerization occurs rapidly (and is difficult to prevent) so an IR spectrum of A is essentially an IR spectrum of 
a mixture of A and B. The signal at 1740 cm-1 corresponds to the nonconjugated C=O bond of the ester group in A, 
and the signal at 1720 cm-1 corresponds to the nonconjugated C=O bond of the ketone group in A, just as we would 
expect. 
 
 
 
The C=O bond in compound B is part of a conjugated ester, so it appears at a lower wavenumber than a typical C=O 
bond of an ester. This must be the signal at 1660 cm-1. This might be initially surprising, as we might have expected a 
conjugated ester to produce a signal around 1700 cm-1. But on further inspection, we recognize that this conjugated 
system has an OH group, whose lone pairs are participating in resonance. 
 
 
 
As such, there is one additional resonance structure (highlighted above); this additional resonance structure gives the 
C=O bond additional single bond character (relative to other conjugated esters). 
Finally, the signal at 1620 cm-1 corresponds with the conjugated C=C bond in B. 
 
(b) Compound B is capable of forming an intramolecular hydrogen bonding interaction, which lowers the energy of 
that tautomer. As such, it has a more significant presence at equilibrium. 
 
 
 
There is one additional factor that favors the enol in this case, which we will discuss in Chapter 16. Specifically, we 
will see that conjugation is a source of stabilization. Compound B possesses this type of stabilization, while compound 
A does not. 
 
 
14.70. 
The two compounds are isomers (both have the molecular formula C8H9BrO) so we expect the molecular ion from each 
of them to have the same m/z value. Recall that there are two isotopes of bromine with approximately equivalent 
natural abundance (81Br and 79Br), which results in two molecular ion peaks for each compound: 
www.MyEbookNiche.eCrater.com