CHAPTER THIRTEEN Spectroscopy protons in the OCH,CN group the other. These two sets of protons give rise to the two peaks that we see in the NMR spectrum and can be assigned on the basis of their chem ical shifts. The protons in the OCH2CN group are connected to a carbon that bears two lectronegative substituents (O and C=N) and are less shielded than those of the CH3O group, which are attached to a carbon that bears only one electronegative atom(O). The signal for the protons in the OCH,CN group appears at 84.1 ppm; the signal corre- sponding to the CH3o protons is at 83.3 ppm. Another way to assign the peaks is by comparing their intensities. The three equiv alent protons of the CH3o group give rise to a more intense peak than the two equiva- lent protons of the OCH_CN group. This is clear by simply comparing the heights of the peaks in the spectrum. It is better, though, to compare peak areas by a process called integration. This is done electronically at the time the NMr spectrum is recorded, and the integrated areas are displayed on the computer screen or printed out. Peak areas are proportional to the number of equivalent protons responsible for that signal It is important to remember that integration of peak areas gives relative, not absolute, proton counts. Thus, a 3: 2 ratio of areas can, as in the case of CH3OCH,CN, correspond to a 3: 2 ratio of protons. But in some other compound a 3: 2 ratio of areas might correspond to a 6: 4 or 9: 6 ratio of protons PROBLEM 13.5 The 200-MHZ H NMR spectrum of 1, 4-dimethylbenzene looks exactly like that of CH3 OCH2 CN except the chemical shifts of the two peaks are 8 2.2 ppm and 87.0 ppm. Assign the peaks to the appropriate protons of 1,4- dimethylbenzene Protons are equivalent to one another and have the same chemical shift when they are in equivalent environments. Often it is an easy matter to decide, simply by inspec tion, when protons are equivalent or not. In more difficult cases, mentally replacing a proton in a molecule by a"test group"can help. We'll illustrate the procedure for a sim- ple case-the protons of propane. To see if they have the same chemical shift, replace one of the methyl protons at C-1 by chlorine, then do the same thing for a proton at C-3. Both replacements give the same molecule, 1-chloropropane. Therefore the methyl protons at C-1 are equivalent to those at C-3 CH CHCH CICHCH.CH CHCHCHCI Propane 1-Chloropropane 1-Chloropropane If the two structures produced by mental replacement of two different hydrogens in a molecule by a test group are the same, the hydrogens are chemically equivalent. Thus, the six methyl protons of propane are all chemically equivalent to one another and have he same chemical shift Replacement of either one of the methylene protons of propane generates 2-chloro- propane. Both methylene protons are equivalent. Neither of them is equivalent to any of the methyl protons The H NMR spectrum of propane contains two signals: one for the six equiva lent methyl protons, the other for the pair of equivalent methylene protons PROBLEM 13.6 How many signals would you expect to find in the H NMr spec trum of each of the following compounds? (a)1-Bromobutane (c) Butane (b)1-Butanol d)1, 4-Dibromobutane Back Forward Main MenuToc Study Guide ToC Student o MHHE Websiteprotons in the OCH2CN group the other. These two sets of protons give rise to the two peaks that we see in the NMR spectrum and can be assigned on the basis of their chem￾ical shifts. The protons in the OCH2CN group are connected to a carbon that bears two electronegative substituents (O and CPN) and are less shielded than those of the CH3O group, which are attached to a carbon that bears only one electronegative atom (O). The signal for the protons in the OCH2CN group appears at  4.1 ppm; the signal corre￾sponding to the CH3O protons is at  3.3 ppm. Another way to assign the peaks is by comparing their intensities. The three equiv￾alent protons of the CH3O group give rise to a more intense peak than the two equiva￾lent protons of the OCH2CN group. This is clear by simply comparing the heights of the peaks in the spectrum. It is better, though, to compare peak areas by a process called integration. This is done electronically at the time the NMR spectrum is recorded, and the integrated areas are displayed on the computer screen or printed out. Peak areas are proportional to the number of equivalent protons responsible for that signal. It is important to remember that integration of peak areas gives relative, not absolute, proton counts. Thus, a 3:2 ratio of areas can, as in the case of CH3OCH2CN, correspond to a 3:2 ratio of protons. But in some other compound a 3:2 ratio of areas might correspond to a 6:4 or 9:6 ratio of protons. PROBLEM 13.5 The 200-MHz 1 H NMR spectrum of 1,4-dimethylbenzene looks exactly like that of CH3OCH2CN except the chemical shifts of the two peaks are  2.2 ppm and  7.0 ppm. Assign the peaks to the appropriate protons of 1,4- dimethylbenzene. Protons are equivalent to one another and have the same chemical shift when they are in equivalent environments. Often it is an easy matter to decide, simply by inspec￾tion, when protons are equivalent or not. In more difficult cases, mentally replacing a proton in a molecule by a “test group” can help. We’ll illustrate the procedure for a sim￾ple case—the protons of propane. To see if they have the same chemical shift, replace one of the methyl protons at C-1 by chlorine, then do the same thing for a proton at C-3. Both replacements give the same molecule, 1-chloropropane. Therefore the methyl protons at C-1 are equivalent to those at C-3. If the two structures produced by mental replacement of two different hydrogens in a molecule by a test group are the same, the hydrogens are chemically equivalent. Thus, the six methyl protons of propane are all chemically equivalent to one another and have the same chemical shift. Replacement of either one of the methylene protons of propane generates 2-chloro￾propane. Both methylene protons are equivalent. Neither of them is equivalent to any of the methyl protons. The 1 H NMR spectrum of propane contains two signals: one for the six equiva￾lent methyl protons, the other for the pair of equivalent methylene protons. PROBLEM 13.6 How many signals would you expect to find in the 1 H NMR spec￾trum of each of the following compounds? (a) 1-Bromobutane (c) Butane (b) 1-Butanol (d) 1,4-Dibromobutane CH3CH2CH3 Propane ClCH2CH2CH3 1-Chloropropane CH3CH2CH2Cl 1-Chloropropane 498 CHAPTER THIRTEEN Spectroscopy Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website