Question

based on the following C NMR and H NMR spectra which compound is present

4-heptanone

2-heptanone

3-hexanone

2-hexanone

cyclopentanone

2,4 dimethyl-3-pentanone

3-methyl-2-hexanone

5-methyl-3-hexanone

99.3865 41.7679 anabeth 5 Carbon Types NTRAL " t 96.0000 32.5955 opperall 1019 Beton C'NMR - Neutral 8 70.5926 W 20.7677 27.6109 77.3567 5 carbon types 22 2735 200.3641 Trol] (ppm)

$Dean, Elizabeth MM NUTRAL unknown neutral S:\m\data\ c os Sep Spee Fall 2019\Sectie fre] 14 2.4234 2.3980 2.3726 2.1150 1.$

Answer 1

Here,

protons (H) are denoted by alphabetics

and Carbons (13 C) are denoted by numerical values.

If we consider theoretically we supposed to get the NMR of the given compounds are as follows

4-haptonin many country contains: 4 types of carbon atoms for 13-C NMR 3 types of proton for H-NMR a triplet by multiplate C triplet 1, 2, 3, 4 Total four 13 C-NMR signals Total three H-NMR signals 2-heptanona : CHCH24 b c d e f & 7 types of . carbon atoms 1, 2, 3, 4, 5, 6, 7 contains: 6 types of poton a singlet b triplet e multiplet → multiples e multiplet & tripler Total : Seven 13 C-NMR signals Total = Six H-NMR signals

3-hexanone het CH₂ CH2 CH2 CH2 a b ? " ¿ á e contains : -NMR & six 13 C-NMR signals 1, 2, 3, 4, 5, 6 & five signals a = singlet tripler b= quatrate c I triplet d = multiplet e triplet 2-Hexanone cizre che & v contains: & Six. 13 C-NMR signals 1, 2, 3, 4, 5, 6 Five H-NMR signals a = singlet b= triplet e = multipler d = multiplet e = triplet

Cyclopentanone DO 2 ctha cita 3 CH₂ CH2 contains: & 13 C-NMR Two H-NMR Signals a= triplet b = multplet Three signals 2,4-dimethyl-3-pentanone et sich "(the Contains: Two H-NMR Three- 13 C-NMR signals signals 1, 2, 3 a = doublet b= multipler

et. 3 Methyl-2-hexanone CH Contains: Six-H-NMR signals geven 13 C-NMR signals 1, 2, 3, 4, 5, 6, 7 a = singlet b- multipler e = doublet d = multipler e = multiplet f = tripler 5-methyl-3-hexanone C eth Set ест, 1 cih an contains: Five six 13 C-NMR signals 1, 2, 3, 4, 5, 6 H-NMR signals a = tripler b= quatrate c = doubler d = multiplar e = doublet

But in case of 5-methyl-3-hexanone d and e proton signals under get merged ( virtual coupling) thus giving a multiplet of three protons around 1.5 ppm. Hence showing a four H-NMR signal instead of five signals.

Again in case of 13C-NMR 5 and 4 numbered carbon atoms signals are merged due to close chemical shift value ( due to high shielding) and hence the combined signal of two Carbons comes out at around 38 ppm. Hence showing a five 13C-NMR signal instead of six signals.