(a)
Interpretation:
The appearance of the high resolution 13C spectrum of methyl formate when the protons are not decoupled should be predicted.
Concept introduction:
13C isotope of carbon has low
(b)
Interpretation:
The appearance of the high resolution 13C spectrum of acetaldehyde when the protons are not decoupled should be predicted.
Concept introduction:
13C isotope of carbon has low natural abundance. 12C is the most abundant isotope in nature. But it is NMR inactive because the spin quantum number is zero. Since 13C is less likely to find in nature there is a very low probability of finding two 13C nuclei which are close to each other. So there is no observable spin-spin coupling between adjacent carbons in 13C NMR spectra. But there are 13C and 1H coupling which leads to large number of splitting patterns in the spectrum. To obtain simplified 13C spectrum scientists use a method called broadband decoupling. This technique avoids the C-H coupling signal, so that all carbon signals appear as singlets.
(c)
Interpretation:
The appearance of the high resolution 13C spectrum of acetone when the protons are not decoupled should be predicted.
Concept introduction:
13C isotope of carbon has low natural abundance. 12C is the most abundant isotope in nature. But it is NMR inactive because the spin quantum number is zero. Since 13C is less likely to find in nature there is a very low probability of finding two 13C nuclei which are close to each other. So there is no observable spin-spin coupling between adjacent carbons in 13C NMR spectra. But there are 13C and 1H coupling which leads to large number of splitting patterns in the spectrum. To obtain simplified 13C spectrum scientists use a method called broadband decoupling. This technique avoids the C-H coupling signal, so that all carbon signals appear as singlets.

Want to see the full answer?
Check out a sample textbook solution
- Draw the mechanism of the reaction.arrow_forward9. Draw all of the possible Monochlorination Products that would Result From the Free Radical Chlormation OF 23,4-TRIMethyl Pentane b. Calculate the To Yield For the major • Product given the Following Relative Restritus For 1° 2° and 30 Hydrogens toward Free Radical Chloration 5.0: 38 : 1 30 2° 1° C. what would be the major product in the Free Radical brominator Of the Same Molecule. Explain your Reasoning.arrow_forwardWhat is the complete reaction mechanism for the chlorination of Ethane, C2H6?arrow_forward
- A 13C NMR spectrum is shown for a molecule with the molecular formula of C6H100. Draw the structure that best fits this data. 220 200 180 160 140 120100 80 60 40 20 Drawingarrow_forwardPlease help me figure out the blan areas with step by step calculations.arrow_forwardneeding help draw all of the possible monochlorination products that would result from the free radical chlorination of 2,3,4-trimethylpentanearrow_forward
- HAND DRAWarrow_forwardBased on the 1H NMR, 13C NMR, DEPT 135 NMR and DEPT 90 NMR, provide a reasoning step and arrive at the final structure of an unknown organic compound containing 7 carbons. Dept 135 shows peak to be positive at 128.62 and 13.63 Dept 135 shows peak to be negative at 130.28, 64.32, 30.62 and 19.10. Provide assignment for the provided structurearrow_forwardO Predict the 'H NMR integration ratio for the following structure. IV I. 3 H A II. 1 H III. 2 H IV. 3 H I. 3 H B II. O H III. 2 H IV. 3 H I. 3 H C II. 2 H III. 2 Harrow_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
