Deduce the structures of compound I and compound II and draw the structure benea

Question

Deduce the structures of compound I and compound II and draw the structure beneath the 1H NMR spectrum of each compound. Beneath the formula, IR, and l3C NMR, indicate what this information tells you about the compound! Be sure to assign all of the hydrogens in your structure to their peaks in the 1H NMR! The peaks and coupling constants, etc. for the 1H NMR are all listed in a table as well. COMPOUND I Molecular formula = C6H14O IR: Major peaks at 3200-3400, 2930 cm-1 13C NMR: 4 peaks 1H NMR: 1.65 ppm, J = 7 Hz, 1 H 1.45 ppm, singlet, 1 H 1.15 ppm, singlet, 6 H 0.9 ppm, doublet, J = 7 Hz, 6 H Molecular formula = C6H11Br IR: Major peaks at 3030, 2960, 1650 cm-1 13C NMR: 6 peaks 1H NMR: 5.1 ppm, triplet, J = 7 Hz, 1 H 3.3 ppm, triplet, J = 7 Hz, 2 H 2.5 ppm, quartet, J = 7 Hz, 2 H 1.7 ppm, singlet, 3 H 1.6 ppm, singlet, 3H COMPOUND II Molecular formula = C6H11Br IR: Major peaks at 3030,2960, 1650 cm-1 13CNMR: 6 peaks? 1H NMR: 5.1 ppm, triplet, J = 7 Hz, 1 H 3.3 ppm, triplet, J = 7 Hz, 2 H 2.5 ppm, quartet, J = 7 Hz, 2 H 1.7 ppm, singlet, 3 H 1.6 ppm, singlet, 3H

Explanation / Answer

a) C6H14 O , no unsaturation so either alcohols or ethers , peaks at 3200-3400 cm-1 indicates alochols, 4 types C since 4 C NMr peaks , H NMR - septet at 1.65 (1H) indicates CH surrounded by 2 CH3 groups , doublet at 0.9 ppm (6H) confirm CH-(CH3)2, singlet at 1.15 (6H) indicates 2CH3 adjacent to C-OH. compound is 2 , 3 dimethyl 2 butanol - (CH3)2-C(OH)-CH-(CH3)2 b) degree of unsaturation = ( 6x2+2 -11)/2 = 1 , so i double bond, IR peaks at 1650 cm- shows C=C, triplet at 5.1 (1H) indicates CH2-CH= , singlets at 1.6 and 1.7 ppm 3 H indicates 2 CH3 adjacent to C= , hence given molecule is 2 methyl 5 bromo 2 butene. (CH3)2-C=CH-CH2-CH2Br

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