Question

There are seven isomers of C4H10O. Draw line diagrams of the seven isomers and then match each isomer with the following H-NMR data.

NOTE: s = singlet, d = doublet, t = triplet, m = multiplet

i , 0.95 (t, 3H); 1.52 (m, 2H); 3.30 (s, 3H); 3.40 (t, 2H)

ii , 1.15 (s, 1H); 1.29 (s, 9H)

iii. , 1.20 (t, 3H); 3.45 (quartet, 2H)

iv. , 0.90 (d, 6H); 1.78 (m, 1H); 2.45 (s, 1H); 3.30 (d, 2H)

v , 1.13 (d, 6H); 3.30 (s, 3H); 3.65 (septet, 1H)

vi , 0.95 (t, 3H); 1.50 (m, 4H); 2.20 (s, 1H); 3.70 (t, 2H)

vii , 0.92 (t, 3H); 1.18 (d, 3H); 1.45 (m, 2H); 1.80 (s, 1H); 3.75 (m, 1H)

Answer 1

All isomers of C4H100 (a) H-NMR 1.19 OH 2o OH3.65 1.19 옥 I OH - OH (b) H-NMR 1.45 (c) H-NMR 3.50 3.50 Haas 3.50 0.90 OH3.65 153 O 1.10 1.10 ' per ' PPM (d) H-NMR 1.490 (e) H-NMR 0.90 337 3.30 13 PPM PPM (f) H-NMR 0.91 (E) H-NMR.48 1.18 11.46 OH3.65 0.90 0.91 3.49 OH PPM PPM i 2, 0.95 (t, 3H); 1.52 (m, 2H); 3.30 (s, 3H); 3.40 (t, 2H) → (b) ii., 1.15 (s, 1H); 1.29 (s, 9H) → (a) iii. 2, 1.20 (t, 3H); 3.45 (quartet, 2H) → (c) iv. E, 0.90 (d, 6H); 1.78 (m, 1H); 2.45 (s, 1H); 3.30 (d, 2H) → (f) v12, 1.13 (d, 6H); 3.30 (s, 3H); 3.65 (septet, 1H) → (e) vi 2, 0.95 (t, 3H); 1.50 (m, 4H); 2.20 (s, 1H); 3.70 (t, 2H) → (d) vii 2, 0.92 (t, 3H); 1.18 (d, 3H); 1.45 (m, 2H); 1.80 (s, 1H); 3.75 (m, 1H) → (g)