2444-091911

2444-091911 - Example 6 P (88) (100) P+1 (88) (5.55) P+2...

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P (88) (100) P+1 (88) (5.55) P+2 (88) (0.35) C 5 H 12 O No rings or ! -bonds 1O: look for OH (alcohol) or C=O (aldehyde or ketone) No C=O OH An alcohol Example 6 IR: 3450, 2900, 1460, 1100-1190, 910-990, 780 NMR: 1.0 t (6H) 1.4 q (4H) 3.3. S (1H) - exchanges with D 2 O 3.4 m (1H) 2 CH 3 groups, triplet 2 CH 3 -CH 2 - units OH CH ??? 2 -CH 2 -CH x Looks like 2 CH 2 So 2 CH 3 CH 2 C 5 H 12 O - OH - CH 3 CH 2 - -CH 3 CH 2 ____________ CH ? but must be next to O because it is downfield OH a b c d Must be CH-OH
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Note. In an NMR tube - take the proton NMR. The OH shows up as a broad singlet. Add D 2 O - take the spectrum again. The H exchanges with D and disappears or is greatly diminished. O H 0 1 2 3 PPM O D 0 1 2 3 PPM Remember that in mass spectrometry, many alcohols do not show a parent (P) but rather a prominent P–18 peak (loss of water) D 2 O and Alcohols
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Formula: C 9 H 18 A hydrocarbon 1 ring or ! -bond; fits the alkene formula an alkene or cyclic alkane. Example 7 C=C An alkene
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Formula: C 9 H 18 an alkene Example 7 cont. 0 1
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This note was uploaded on 02/25/2012 for the course CHEM 244 taught by Professor Jardin,j during the Fall '08 term at UConn.

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2444-091911 - Example 6 P (88) (100) P+1 (88) (5.55) P+2...

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