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1H NMR中d=8.15ppm和d=7.31ppm附近分别有两个双重峰，这是苯环相邻两个氢的耦合产生的，且两者个数相等，都是12个H. d=10.19ppm为伯胺H的化学位移，在d=7.31ppm 和7.10ppm 是苯环上H的位移，各有一个双重峰，表明苯环上对位的硝基全部被取代了。 |
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铁杆木虫 (著名写手)
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Carena: 金币+1, 感谢应助 2013-01-12 21:55:08
qq156437683: 金币+8, ★有帮助 2013-01-23 10:45:14
Carena: 金币+1, 感谢应助 2013-01-12 21:55:08
qq156437683: 金币+8, ★有帮助 2013-01-23 10:45:14
| In 1H NMR there exists two double peaks with the same H of 12 produced by coupling of two adjacent H on benzene at diameter of 8.15 and 7.31 ppm. Diameter of 10.19 ppm is chemical shift of H of primary amine. There exist two double peaks that belong to H on benzene at diameter of 7.31 and 7.10 ppm，indicating that contrapuntal nitro on benzene is replaced. |
2楼2013-01-10 20:21:42
nwsuafliu
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Carena: 金币+1, 感谢应助 2013-01-12 21:55:16
qq156437683: 金币+7, 翻译EPI+1, ★有帮助 2013-01-23 10:45:07
Carena: 金币+1, 感谢应助 2013-01-12 21:55:16
qq156437683: 金币+7, 翻译EPI+1, ★有帮助 2013-01-23 10:45:07
| 1H NMR spectrum showed two doublets around d=8.15ppm and d=7.31ppm, respectively. They were yielded from coupling of two neighboring H on benzene, both of which with 12 H. d=10.19 ppm indicates the chemical shift of H on primary amine, whereas d=7.31 ppm and 7.10ppm is the H shift on benzene. The presence of doublet at each shift indicated that contrapuntal nitro on benzene is substituted. |
3楼2013-01-10 21:05:50