l15e

l15e - CH 203 O R G A N I C C H E M I S T R Y I NMR...

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NMR spectroscopy: Problem set © Bruno I. Rubio 1 CH 203 O R G A N I C C H E M I S T R Y I NMR spectroscopy: Problem set The following tables are helpful. Type of hydrogen ! [ppm] Type of hydrogen ! [ppm] H O variable X H (X = I, Cl, Br) 2.2–4.2 H N variable H 2.2 H R 0.8–1.6 O H 3.2–3.8 H 1.6 H C C 4–7 O H 2.0–2.2 H 6.5–8.5 H 2–3 H O 8–10 N H 2.2–2.8 O O H 10–14
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NMR spectroscopy: Problem set © Bruno I. Rubio 2 Stretch " [cm –1 ] alcohol O—H 3650–3200 N—H 3500–3300 carboxylic acid O—H 3300–2500 Csp–H 3300 benzene or alkene Csp 2 –H 3100 Csp 3 –H 2950 aldehyde Csp 2 –H 2700 C # C 2260–2100 C # N 2260–2220 C=O 1870–1650 alkene C=C 1680–1600 benzene C=C 1600 (and sometimes 1500) Problem Prepare a sketch of the NMR spectrum of each compound. Your sketch should (1) have the correct number of signals; (2) demonstrate that you can make a good guess about the approximate chemical shift of each signal; (3) show the correct splitting of each signal. Make the size of the signal roughly correspond to the relative number of hydrogens producing that signal. (a) H 3 C OCH 3 O (b) H 3 C OH O CH 3 (c) H 3 C O O CH 3 (d) H 3 C O O CH 3 CH 3 (e) O O CH 3 CH 3 H 3 C (f) OCH 3 O H 3 C H 3 C CH 3 (g) O 2 N O CH 3 Answer The actual NMR spectrum of each compound is presented; compare it to your sketch.
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NMR spectroscopy: Problem set © Bruno I. Rubio 3 (a) ! 1.15, 3H, t ( J = 7 Hz); ! 2.32, 2H, q ( J = 7 Hz); ! 3.67, 3H, s (b) ! 1.20, 6H, d ( J = 7 Hz); ! 2.58, 1H, sep ( J = 7 Hz); ! 11.88, 1H, s (c) ! 1.14, 3H, t ( J = 7 Hz); ! 1.26, 3H, t ( J = 7 Hz); ! 2.32, 2H, q ( J = 7 Hz); ! 4.13, 2H, q ( J = 7 Hz)
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NMR spectroscopy: Problem set © Bruno I. Rubio 4 (d) ! 1.23, 6H, d ( J = 7 Hz); ! 2.02, 3H, s; ! 4.99, 1H, sep ( J = 7 Hz) (e) ! 1.12, 3H, t ( J = 7 Hz); ! 1.23, 6H, d ( J = 7 Hz); ! 2.28, 2H, q ( J = 7 Hz); ! 5.00, 1H, sep ( J = 7 Hz) (f) ! 1.20, 9H, s; ! 3.66, 3H, s (g) ! 1.46, 3H, t ( J = 7 Hz); ! 4.13, 2H, q ( J = 7 Hz); ! 6.94, 2H, d ( J = 8.4 Hz); ! 8.18, 2H, d ( J = 8.4 Hz)
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NMR spectroscopy: Problem set © Bruno I. Rubio 5 Integration The number of hydrogens responsible for a signal is given by that signal’s integral. On a spectrum an integral appears as a stair step whose height is proportional to the number of hydrogens that produce each signal. Example
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l15e - CH 203 O R G A N I C C H E M I S T R Y I NMR...

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