What is the correct answer to this Question?
I attached the Question and the figure
What is the correct answer to this Question? I attached the Question and the figure 6H...
hnmr spectroscopy
Analyze the spectrums and determine the structures
Please help. I’m really confused with how some of the
structures might be.
2) C,HuO 3H fer 4H 0.8 PPM 0. 1.0 1.4 1.2 2.2 1.0 1. 2.0 2.8 2.4 2.0 40 z00a 4000 3000 252e 1500 1720 2202 4) CHo 6H 1H Methylee protos adjacent to chiral centers may not be 1H 1H equlvalent 1.0 PPM 3.0 2.0 0.5 3:5 2.5 1.5 42 28 158e 4000 3500 2500 200 C-...
can someone answer the
question 3 and 4
3. Propose structures for compounds that fit the following 1H NMR data: a. C4H4Cl2: 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) 4. Predict the products of the...
V. Draw the structures of Compound 23A (6 pts), Compound 23B (6 pts) and clearly indicate your assignments of all proton resonances using the lower case letter associated with the signal (6 pts). Calculate the Unsaturation Index of each compound (2 pts) (20 points total) 1.24 3.0 1.26 2.5 e = 1.25 ppm doublet, 6H 2.0 a = 10.0 ppm 1.5 Compound 23a: C10H420 singlet, 1H 10.00 7.83 7.85 7.09 7.07 1.0 Unsat. Index = (2C+2-H-X+N)/2 = X=number of halogens...
V. Draw the structures of Compound 23A (6 pts), Compound 23B (6 pts) and clearly indicate your assignments of all proton resonances using the lower case letter associated with the signal (6 pts). Calculate the Unsaturation Index of each compound (2 pts) (20 points total) 1.24 3.0 1.26 2.5 e = 1.25 ppm doublet, 6H 2.0 a = 10.0 ppm 1.5 Compound 23a: C10H420 singlet, 1H 10.00 7.83 7.85 7.09 7.07 1.0 Unsat. Index = (2C+2-H-X+N)/2 = X=number of halogens...
need help understanjng what the expected would be molecular
formula is C4H6ClN
HNMR Spectrum chemical shift, Hz 1200 900 2400 2100 1800 1500 600 3000 2.147 2.104 absorption 0.SHE 2,578 16.01 3.678 8 - 5 6 2 4 3 chemical shift, ppm (8) Observed Chemical Shift Table F: 'HNMR Functional Group Analysis Structural Expected Functional Group Fragment Name Chemical Structural Formula Shift Functionalized Structural Fragment 3.672 CH2-CH2 2.578 CH2-CH2 2.1256 CH2-CH2-CH2N HC-CN 2005 Chemical shift, st R-H or CH can...
Question 7 What is the major product of the reaction shown below? H+ CH3-CH2-CH=CH2 + HOH A) OH OH CH3 - CH2 -CH-CH2 B) OH | CH3 - CH2 -CH-CH3 OH OH CH3 -CH-CH - CH3 D) OH CH3 - CH2 -C=CH2 Choice A Choice B Choice C Choice D
6) Identify the CHCl isomer using the following proton NMR data: doublet 81.04 (6H) A) (CH3)3CCI multiplet 8 1.95 (17) doublet 8 3.35 (2H) B) CH2CH2CH2CH2C1 C) CH2CH2CHCH ĆI D) (CH3),CHCH,CI 7) Which compound below has only three peaks in its C-NMR spectrum? a) ortho-dichlorobenzene c) para-dichlorobenzene b) meta-dichlorobenzene d) chlorobenzene 8) Which one of the following isomers of CsHis has only two peaks in its C NMR? A) CH3(CH2).CH B) CH3CHCHCHÖCHCH; CH, CHz CH, C) CH3CHCHCHCH CH3 CH3 D)...
Can
you very that i got the correct names. also #8
Name the following compounds. OH CH3 -CH-CH2 - CH3 butanol 5. =0 CH3 -CH2-C-CH2-CH2 - CH2 - CH3 Hexanoic acid 6. II Propanal 7. CH3 -CH 2 -C-H 8. CH3 CH2-CH2-CH-CH2-CHE OH 9. CH3-CH2-CH2-CH2-CH2-CH2-CH2-OH heptanol - CH₂=0 CH3 - C-CH2-CH2-CH3 Pentanone 10.
1) What would be the partial interpretation (concerning the number of chemically equivalent hydrogen atoms in a molecule) of a doublet which integrates for 1 H and is located at 11 in the 1H NMR? For this question, consider only the integration of the peak. CH3 CH 2 equivalent CH2 CH2 2 equivalent CH 2) Which of these choices best describes the interpretation of a 1H NMR spectral peak that was recorded as 1.0 (9H, s)? The underlined hydrogen atom is...
organic
V. Draw the structures of Compound 23A (6 pts), Compound 23B (6 pts) and clearly indicate your assignments of all proton resonances using the lower case letter associated with the signal (6 pts). Calculate the Unsaturation Index of each Compound (2 pts) (20 points total) 1.24 30 1 25 e = 1.25 ppm 2.5 doublet, 6H 20 a = 10.0 ppm Compound 23a: C10H2O 15 singlet, 1H 7.83 7.85 7.09 707 Unsat. Index = (2C+2-H-X+N/2= X=number of halogens d=3.0...