Question

Give the structure that corresponds to the following molecular formula and NMR spectra: CsHyo TH NMR: 80.91 (6H, doublet); 8 1.17 (6H, singlet); 8 1.48 (1H, broad singlet); 8 1.65 (1H, septet). BC NMR: 0 17.6, 26.5, 8 38.7, 8 73.2. .

Answer 1

Molewhor formalp - L6Hk a sites of unsaturation = 2(CH) -H = 2(6+1)-16 = R-ko set es of unsafuoaljon zo godicafe No double bond Ring (111) 'H HMIR - 0 0.918 (6H, doablet) indicate two CH3 Dext to CH 2CH₂ - CH UITE (GH, singlet) indicate 2CH3 next to carbon 2CH₂-C C 1488 ( 1H broad singlet) andicate -OH group W 1.656 CH septet) indicate CH next to two. CH3 -CH13 CHE BCNMR 1768 - - CH3 20.58 - - CH₂ 38.76 - CH 7328 - carbon next to oxygen

from the above data the probable structure is t ez 2 - CH -ca 2,3-dimethyl-2-butanol 2,3-dim 13 OH 73.25 H3 - CH-(- (Hz tech + CH3 17 17.68 CH - Hof UH 38 76 1.650 1.488 26.50 BCNMR 1 H NMR