(a)
Interpretation:
The mole percent of the each given compound in the mixture is to be stated.
Concept introduction:
NMR spectroscopy is a technique used to determine a unique structure of the compounds. It identifies the carbon-hydrogen bonding of an organic compound. A hydrogen atom is known as a proton in the NMR spectroscopy.
The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. The more the shielded proton lesser will be its chemical shift value and the corresponding signal will be produced at the right-hand side or lower frequency region or vice versa.
(b)
Interpretation:
Among 1 mole percent CH3I impurity in (CH3)3C−Br, and 1 mole percent (CH3)3C−Br impurity in CH3I, the one which is more easily detected by NMR is to be stated with an explanation.
Concept introduction:
NMR spectroscopy is a technique used to determine a unique structure of the compounds. It identifies the carbon-hydrogen bonding of an organic compound. A hydrogen atom is known as a proton in the NMR spectroscopy.
The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. The more the shielded proton lesser will be its chemical shift value and the corresponding signal will be produced at the right-hand side or lower frequency region or vice versa.
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Organic Chemistry
- Shown below is the major resonance structure for a molecule. Draw the second best resonance structure of the molecule. Include all non-zero formal charges. H. H. +N=C H H H Cl: Click and drag to start drawing a structure. : ? g B S olo Ar B Karrow_forwardDon't used hand raitingarrow_forwardS Shown below is the major resonance structure for a molecule. Draw the second best resonance structure of the molecule. Include all non-zero formal charges. H H = HIN: H C. :0 H /\ H H Click and drag to start drawing a structure. ×arrow_forward
- Part II. two unbranched ketone have molecular formulla (C8H100). El-ms showed that both of them have a molecular ion peak at m/2 =128. However ketone (A) has a fragment peak at m/2 = 99 and 72 while ketone (B) snowed a fragment peak at m/2 = 113 and 58. 9) Propose the most plausible structures for both ketones b) Explain how you arrived at your conclusion by drawing the Structures of the distinguishing fragments for each ketone, including their fragmentation mechanisms.arrow_forwardPart V. Draw the structure of compound tecla using the IR spectrum Cobtained from the compound in KBr pellet) and the mass spectrum as shown below. The mass spectrum of compound Tesla showed strong mt peak at 71. TRANSMITTANCE LOD Relative Intensity 100 MS-NW-1539 40 20 80 T 44 55 10 15 20 25 30 35 40 45 50 55 60 65 70 75 m/z D 4000 3000 2000 1500 1000 500 HAVENUMBERI-11arrow_forwardTechnetium is the first element in the periodic chart that does not have any stable isotopes. Technetium-99m is an especially interesting and valuable isotope as it emits a gamma ray with a half life ideally suited for medical tests. It would seem that the decay of technetium should fit the treatment above with the result In(c/c) = -kt. The table below includes data from the two sites: http://dailymed.nlm.nih.gov/dailymed/druginfo.cfm?id=7130 http://wiki.medpedia.com/Clinical: Neutrospec_(Technetium_(99m Tc)_fanolesomab). a. b. C. Graph the fraction (c/c.) on the vertical axis versus the time on the horizontal axis. Also graph In(c/c.) on the vertical axis versus time on the horizontal axis. When half of the original amount of starting material has hours fraction remaining disappeared, c/c = ½ and the equation In(c/c.) = -kt becomes In(0.5) = -kt1/2 where t₁₂ is the half life (the time for half of the material to decay away). Determine the slope of your In(c/c.) vs t graph and…arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning