Interpretation:
The structure of the compound C in the given reactionis to be drawn and explanation of high degree of symmetry of compound D is to be suggestedby making structure.
Concept introduction:
According to
The aromatic compound contains conjugated pi bonds.
All the atoms in the ring have an unhybridized
Antiaromatic compounds contain
The numerical proportion of signals generated in
Nuclear Magnetic Resonance (NMR) is one of the most capable analytical techniques used for determining the functional groups and how the atoms are structured and arranged in a molecule.
Few elements, such as
In
Induced magnetic field consists of electricity generated from movement in a magnetic field.
Any signal’s position on the X-axis in the
The number of signals in
The area covered by the signal is proportional to the number of equivalent protons causing the signal.
The hydrogen atom on adjacent carbon atoms splits the peak into two or more peaks. One, two, and three hydrogen atoms split the peak into two, three and four peaks, which further, is referred to as doublet, triplet or quartet.
The decrease in the electron density around a proton deshields the signal downfield at a larger value of chemical shift.
The increase in electron density shields the signal upfield at a lower value of chemical shift.
Cyclopentadiene, on treatment with sodium, loses a proton to form cyclopentadienyl anion
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Organic Chemistry
- Compound I (C11H14O2) is insoluble in water, aqueous acid, and aqueous NaHCO3, but dissolves readily in 10% Na2CO3 and 10% NaOH. When these alkaline solutions are acidified with 10% HCl, compound I is recovered unchanged. Given this information and its 1H-NMR spectrum, deduce the structure of compound I.arrow_forwardPropose a structure consistent with the following spectral data for a compound C8H18O2: IR: 3350 cm-1 1H NMR: 1.24 δ (12 H, singlet); 1.56 δ (4 H, singlet); 1.95 δ (2 H, singlet)arrow_forwardTreatment of 3,4-dibromohexane with strong base leads to loss of 2 equivalents of HBr and formation of a product with formula C6H10. Three products are possible. Name each of the three, and tell how you would use 1H and 13CNMR spectroscopy to help identify them. How would you use UV spectroscopy?arrow_forward
- 24. The 1HNMR spectrum of an unknown acid has the following peaks: 12.71 ppm (1H, S); 8.04 ppm (2H, d); 7.30 ppm (2H, d); 2.41 ppm (3H, s) Which structure best fits this spectral information? Dlan OH A B علی مود C OH D OHarrow_forwardPropose a structure for D (molecular formula CgH,CIO,) consistent with the given spectroscopic data. 13C NMR signals at 30, 36, 128, 130, 133, 139, and 179 ppm 1Η ΝMR of D 4H 2H 2 H 1H 12 10 8 4 2 ppmarrow_forwardA ¹H NMR spectrum is shown for a molecule with the molecular formula of CsH10O2. Draw the structure that best fits this data. 10 pom aarrow_forward
- punqe You will characterize your product by 'H NMR spectroscopy. How many different proton environments do you expect your product compound to exhibit? How many "C environments would you expect to see in the "C NMR spectrum? OS IS TS ES YS SS 9S LS S 6S 09 19 19 99 59 99 L9 19 69 OL IL TL EL YL SL YL LL L 6L O1 I'8 X: parts per Million: Protonarrow_forwardWhen 2-bromo-3,3-dimethylbutane is treated with K+ −OC(CH3)3, a singleproduct T having molecular formula C6H12 is formed. When 3,3-dimethylbutan-2-ol is treated with H2SO4, the major product U has thesame molecular formula. Given the following 1H NMR data, what are thestructures of T and U? Explain in detail the splitting patterns observedfor the three split signals in T.1H NMR of T: 1.01 (singlet, 9 H), 4.82 (doublet of doublets, 1 H, J = 10, 1.7 Hz), 4.93 (doublet of doublets, 1 H, J = 18, 1.7 Hz), and 5.83 (doublet ofdoublets, 1 H, J = 18, 10 Hz) ppm1H NMR of U: 1.60 (singlet) ppm Additional problems on the spectroscopy of alkenes are given in Chapters A–C:Mass spectrometry: A.16b, A.20, A.23Infrared spectroscopy: B.5, B.7(A), B.12c, B17a, B.18cNuclear magnetic resonance spectroscopy: C.12a; C.15d, e; C.29d; C.32d;C.37; C.38d, f; C.43i, j; C.44; C.45; C.49d, f; C.50b; C.51c; C.55arrow_forward7. NMR spectroscopy is used to evaluate the success of the reaction. Match the protons of phenacetin with their 'H NMR signals. (The aromatic protons are assigned for you.) A H3C E H B C D 1 2 2 2 3 11 10 9 8 7 6 5 1 HSP-01-202 ppmarrow_forward
- Deduce the structure of a compound with molecular formula C5H100 that exhibits the following ¹H and ¹³C NMR spectra. IH NMR CNMR 150 10 Structure A Structure B Structure C Structure D 24 1C 100 B 2H H 20 50 10 D 311arrow_forward35ee eee 25ee zeee 1see 2H 3H зн 2H 2H 2H PPM Determine the structure of A based on these spectra. Indicate clearly the signals for each type of protons for NMR analysis. Compound A gives NMR and IR spectra has the formula C3H1403 1047,42 317.5 8'2aarrow_forward11 10 9 8 7 9 +6 5 -3 2 1 ppmarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning