Degree of Unsaturation (DOU) is also known as Double Bond Equivalent. If the molecular formula is given, DoU can be calculated as follows
DoU = 2C+2+N−X−H
2
C is the number of carbons
N is the number of nitrogens
X is the number of halogens (F, Cl, Br, I)
H is the number of hydrogens
Molecular formula C4H6Cl2
Therefore sites of unsaturation = [(2*4)+2-6-2]/2 = 1
1H NMR data:
NMR peak value (ppm) |
Multiplicity |
Protons |
Indication |
2.18 |
singlet |
3H |
CH3 attached to quaternary carbon |
4.16 |
Doublet |
2H |
CH2 attached to CH and chlorine |
5.71 |
triplet |
1H |
CH attached to CH2 and alkene |
Therefore, structure of the molecule is
10. (8 points) For each of the following molecules predict the "C and 'H NMR spectra....
3. Propose structures for compounds that fit the following 1H NMR data: a. C.HgCl2: 8 1.60 (doublet, 3H), 2.15 (multiplet, 2H), 3.72 (triplet, 2H), and 4.27 (multiplet, 1H) b. C«H,Br: 8 1.1 (doublet, 6H), 1.9 (multiplet, 1H), and 3.4 (doublet, 2H) c. C-H140:8 0.9 (triplet, 6H), 1.6 (sextet, 4H), and 2.4 (triplet, 4H) d. CsH1002: 8 1.2 (doublet, 6H), 2.0 (singlet, 3H) and 5.0 (septet, 1H)
3. Propose structures for compounds that fit the following 1H NMR data: a. C4H2Cl2: 8 1.60 (doublet, 3H), 2.15 (multiplet, 2H), 3.72 (triplet, 2H), and 4.27 (multiplet, 1H) b. C4H,Br: 8 1.1 (doublet, 6H), 1.9 (multiplet, 1H), and 3.4 (doublet, 2H) c. C-H140:8 0.9 (triplet, 6H), 1.6 (sextet, 4H), and 2.4 (triplet, 4H) d. C5H1002:8 1.2 (doublet, 6H), 2.0 (singlet, 3H) and 5.0 (septet, 1H)
3. Propose structures for compounds that fit the following 1H NMR data: a. C.H.Cl2: 8 1.60 (doublet, 3H), 2.15 (multiplet, 2H), 3.72 (triplet, 2H), and 4.27 (multiplet, 1H) b. C.H.Br: 8 1.1 (doublet, 6H), 1.9 (multiplet, 1H), and 3.4 (doublet, 2H) c. CH 40:8 0.9 (triplet, 6H), 1.6 (sextet, 4H), and 2.4 (triplet, 4H) d. CsH1002: 8 1.2 (doublet, 6H), 2.0 (singlet, 3H) and 5.0 (septet, 1H)
Propose structures that fit the following formula and NMR data. (a) C3H100 singlet at 2.10 8 (3H) doublet at 0.958 (6H) multiplet at 2.43 8 (1H) (b) C10H14 singlet at 1.30 8 (9 H) singlet (broad) at 7.308 (5H) (c) C9H11Br quintet at 2.15 8 (2H) triplet at 2.758 (2H) triplet at 3.38 8 (2H) singlet at 7.22 8 (51) IR spectrum: strong peak near 1720 cm (d) C18H14 O2 Singlet at 2.20 8 (3H) Singlet at 5.08 8 (1H)...
IL (20 points) The 'H NMR, "C NMR, Mass Spectra and IR spectra for a mystery compound with the formula CsH20 are shown below. Please draw the structure for the compound in the box below. Note: The numbers on top of the 'H NMR peaks are the number of protons associated with that peak. A chart for 'H NMR, IR and C NMR spectra shift values are on the following pages. FTIR 13C NMR Zoom Out 'H NMR Zoom Out...
For the protons labeled Ha and Hb in the structure below, predict the characteristics of their signals in the H NMR spectrum: the approximate chemical shift, the splitting pattern, and the integration value of their signals. Approximate Integration Splitting chemical shift value O 1H H NMR signal O 1 ppm O singlet for H O doublet O 2H 2 ppm O 3H O 3-4 ppm O triplet Br O 4H O 5-6 ppm O quartet O 5H O 7-8 ppm...
2. Use the 'H NMR and IR data to determine the structure of the following compounds and name them. Compound A Molecular formula: CroH IR absorptions at N/A H NMR data: 1.3 (singlet, 9H), 7.0 to 7.5 (multiplet, 5H) ppm Compound B Molecular formula: CHO IR absorptions at 1H NMR data: 1735-1745, 1050 cm 0.93 (doublet, 6H), 1.52 (multiplet, 2H), 1.69 (multiplet, 1H), 2.04 (singlet, 3H), and 4.10 (triplet,2H) ppm Compound C Molecular ion: IR absorptions at 1710 cm 1H...
Propose structures for compounds that fit the following 'H NMR data. CH,Br 2H quintet at 2.48, J=6 Hz 4H triplet at 3.5 8, J= 6 Hz 2. C.H.O 2H quartet at 2.5 8, 3H Singlet at 2.0 8, 3H triplet at 1.08 1. C,H40 6H triplet at 0.9 8, J = 7 Hz 4H sextet at 1.6 8, J= 7 Hz 4H triplet at 2.48, J= 7 Hz CH4 6H doublet at 1.28, J= 7 Hz 3H singlet at 2.38...
just the circled ones only 8.10 Suggest structures consistent with the following 'H NMR data: a. Сно triplet (1.0 ppm, 3H) singlet (2.1 ppm, 3H) quartet (2.4 ppm, 2H) b. С Н.о, triplet (1.2 ppm, 3H) singlet (2.1 ppm, 3H) quartet (4.1 ppm, 2H) усно singlet (2.4 ppm, 3H) multiplet (7.5 ppm, 5H) singlet (1.6 ppm, 6H) singlet (3.1 ppm, 2H) multiplet (7.3 ppm, 5H) 1. Сно, triplet (1.2 ppm, 3H) singlet (2.4 ppm, 3H quartet (4.2 ppm, 2H) ....
4. Consider the following two diastercomers: Both compounds generate 'H NMR spectra with the following signals: A multiplet between 7.20-7.63 ppm with an integration of 5H Two separate doublets between 5.70-6.50 ppm, each with an integration of 1H A singlet around 1.30 ppm with an integration of 9H Despite the similarities, how could you differentiate between the H NMR spectra of these isomers? 5. Assign each set of protons to their appropriate signals in the 'H NMR spectrum shown below....