SolnsChap14EVENZum7e

Thus we will have three singlet peaks following our

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Unformatted text preview: . c H c H O H c CH2 b H c O C CH3 a H 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 a b b. O b CH3 C a CH3 C CH3 a 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. CH2CH3 ba cH H c CH2CH3 ba CH3CH2 ab H 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. 14 589 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. C4H10O spectrum: We have a quartet...
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