14C_s11_lec01_e02_key

42726 ppm multiplet 5 5h phenyl group c6h5 382 ppm

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Unformatted text preview: .26 ppm multiplet 5 5H Phenyl group C6H5 3.82 ppm doublet 1 1H CH in CHCH 3.58 ppm singlet 1 1H CH or OH 2.57 ppm pentet 1 1H CH in CH3CHCH* or CH in CH2CHCH2 *This implication is selected because elsewhere in the 1HNMR data there is a CH3CH implication (1.14 ppm) but not a CH2CH2 implication. 1.14 ppm doublet 3 3H CH3 in CH3CH or 3 x CH in CHCH 0.94 ppm singlet 9 9H 3 x C H 3 or 9 x C H Totals 20 20 H C6H5 + CH + OH + CH + CH3 + (3 x CH3) = C12H20O 13 C-NMR: 175.4 ppm (singlet) is an ester C=O. The 151.3 ppm through 121.6 ppm signals are consistent with a phenyl group. The molecule has 14 carbons whereas the 13C-NMR has 10 signals, indicating the molecule has some equivalent carbons (some symmetry). Atom check: C14H20O3 (from mass spectrum) - C12H20O (from 1H-NMR) - CO2 (ester from IR or 13C-NMR) = C. That this carbon is not accounted for in the 1H-NMR means it is not bonded to a hydrogen atom. That this carbon is not accounted for in the IR means it is not part of a functional group that appears in the 4000–1450 cm-1 range of the IR. DBE check: Five (calculated for C14H20O3) - 1 (C=O) - 4 (benzene ring) = 0. All DBE used. Pieces: Ph CH in CHCH OH CH in CH3CHCH CH3 in CH3CH 3 x CH3 O=C–O (ester) C Assembly: The 1H-NMR splitting suggests that the CH3, CH, and CH pieces join to form CH3CHCH. Ph OH CH3CHCH O=C–O (ester) C 3 x CH3 The methyl groups of 3 x CH3 are equivalent, and not coupled. This can happen only if they are part of a tert-butyl group. Ph OH CH3CHCH O=C–O (ester) (CH3)C There are several ways to assemble these pieces into a structure that is consistent with the given data, any of which was accepted for full credit. (Note the C=O stretching frequency is not consistent with the carbonyl directly bonded to the benzene ring.) The actual structure is: OH O C(CH3)3 O CH3...
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This note was uploaded on 09/06/2011 for the course CHEM 140A taught by Professor Whiteshell during the Fall '04 term at UCSD.

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