碳核磁共振譜ppt課件-wenkub.com
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【正文】 ? Two isomeric hydrocarbons, A and B, of C5H10 were separated by GC and gave the following 13C data: A: 13 (q), 17 (q), 26 (q), 118 (d), and 132 (s) ppm B: 13 (q), 22 (q), 31 (t), 108 (t), and 147 (s) ppm ? A pound gives a molecular ion in its EI spectrum at m/z 80. This value is unchanged when the pound is introduced into the mass spectrometer in the presence of D2O. A high resolution measurement establishes the molecular formula as C5H6N2. The pound gives the following 13C NMR spectrum: 16 (t), 22 (t), 119 (s) ppm. Deduce its structure. ? Three pounds of molecular formula C4H8O have the following 13C spectra, and the described features in their IR spectra. ? Compound 1: IR: 1730 cm1, 13C NMR: , , , ppm. ? Compound 2: IR: 3200 (broad) cm1, 13C NMR: , , , ppm. ? Compound 3: IR: no peaks except CH and fingerprint, 13C NMR: and ppm. Suggest a structure for each pound, and then see whether your suggestions are patible with the following information. Compound 1 react with NaBH4 to give pound 4, C4H10O, IR 3200 (broad) cm1 and 13C NMR , , , and ppm. Compound 2 reacts with hydrogen over a palladium catalyst to give the same product 4, while pound 3 reacts with neither reagent. ? Three pounds, 13, of formula C4H6 have been prepared by the the routes shown below. Suggest structures of these isomers. C lN a N H2t B u O HC o m p o u n d 1 : 1H N M R : 5 . 3 5 ( 2 H , s ) , ( 1 . 0 ( 4 H , s ) 1 3C N M R : 3 d i f f e r e n t s i g n a l sC lN a N H2T H FC o m p o u n d 2 : 1H N M R : 0 . 8 4 ( 2 H , d , J = 1 H z ) , 2 . 1 3 ( 3 H , s ) , 6 . 4 ( 1 H , t , J = 1 H z ) 1 3C N M R : 4 d i f f e r e n t s i g n a l sC lL i N H2/ d i o x a n e C o m p o u n d 3 : 1H N M R : 7 . 1 8 ( 2 H , d , J = 1 H z ) , 1 . 4 6 ( 1 H , t q , J = 5 , 1 H z ) , 0 . 9 7 ( 3 H , d , J = 5 H z ) 1 3C N M R : 3 d i f f e r e n t s i g n a l sCompound 1 reacts with mchloroperbenzoic acid to give pound 4,which rearranges to pound 5 in the presence of LiI. Suggest structures for pounds 4 and 5. Compound 4 (C4H6O): IR: no strong peaks outside fingerprint 1H NMR: (2H, s), (4H, AB q)ppm Compound 5 (C4H6O): IR: 1770 cm1 1H NMR: (4H, t, J = 5 Hz)), (2H, q, J = 5 Hz)ppm ? The following pound reacts with hydroxide ion to give a product C16H12O3. Deduce the structure of the product. OOB rB rIR: 1700 cm1 13C NMR: 12 aromatic carbons (8 d, 4 s) and signals at 63 (t), 74 (t), 109 (s) and 189 (s) ppm。 碳譜的解析步驟 ? 對碳譜各譜線進(jìn)行歸屬。 ? 碳原子級數(shù)的確定 。 疊烯的中央碳原子出峰位置也大于 碳譜按化學(xué)位移值一般可分為下列三個區(qū) ， 根據(jù)這三個區(qū)域可大致歸屬譜圖中各譜線的碳原子類型 。 ? 分析化合物結(jié)構(gòu)的對稱性。 ?
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