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c O C CH3
c Because the hydrogens marked by a and the hydrogens marked by b are separated by
more than three sigma bonds, they will also not exhibit spin-spin coupling. Thus we will
have three singlet peaks (following our assumptions). Predicting the relative locations of
the peaks was not discussed in the text, but they should be in a 3:2:5 relative area ratio
(a:b:c). A sketch of the idealized NMR spectrum is: c
CH3 C a
CH3 CHAP. 14
588 COVALENT BONDING: ORBITALS - EVEN-NUMBER PROBLEMS The different groups of equivalent hydrogen atoms are labeled a and b. Since all H
atoms are separated by more than three sigma bonds, we should have no spin-spin
coupling. Again, you do not have the information to predict where the two peaks should
be in the idealized NMR spectrum, but they should have relative areas of 9:3 (or 3:1). a b c.
ba cH H c CH2CH3
c The three different groups of equivalent hydrogen atoms are labeled a, b, and c. The H
atoms marked c will not exhibit spin-spin coupling, but the H atoms marked a and b will.
Since the H atoms marked a in the –CH3 groups neighbor –CH2 groups, a triplet line
pattern will result with intensities of 1:2:1. The H atoms marked b in the – CH2 groups
neighbor –CH3 groups, so a quartet peak should result with intensities of 1:3:3:1. The
relative area ratios of the three different H atoms (a:b:c) should be 9:6:3 (or 3:2:1). c b a CHAP. 14
60. COVALENT BONDING: ORBITALS - EVEN-NUMBER PROBLEMS C6H12 NMR spectrum:
Because only one peak is present, all the hydrogen atoms are equivalent in their
environment. Knowing this, then the process is trial and error to come up with the
correct structure for C6 H12. Following the “organic rules” in Exercise 14.70 and knowing
a double bond is present, the only possibility to explain the NMR pattern is: C H3
C H3 C C C H3 C H3
Here, all the H atoms have equivalent neighboring atoms and there would be no spin-spin
coupling since all H atoms are separated by more than three sigma bonds.
We have a quartet...
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