Concept explainers
Interpretation:
The strucutral formula of the given compound K has to be decucted with the help of the given
Concept Introduction:
The
Chemical shift: The NMR spectrum of any compound is taken with reference to a standard compound called reference compound. Generally, tetramethylsilane (TMS) is taken as the reference compound. The methyl protons of TMS are equivalent and produces only one sharp peak at the rightmost end of the scale.
13C NMR Spectroscopy: This type of NMR splitting of signals tells us numbers of hydrogens atoms are attached to each carbon. The triangle rule (n+1) C. The chemical shift explains the different hybridization
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Chapter 21 Solutions
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_forwardThe 1H-NMR spectrum of compound R, C6H14O, consists of two signals: d 1.1 (doublet) and d 3.6 (septet) in the ratio 6:1. Propose a structural formula for compound R consistent with this informationarrow_forwardPropose a structure given the 1H and 13C NMR spectra of the unknown compound. Assign chemical shifts to corresponding hydrogen and carbon atoms Molecular Formula: C5H10O3arrow_forward
- The 'H NMR spectrum of compound A (C3H100) has four signals: a multiplet at 8 = 7.25-7.32 ppm (5 H), a singlet at d = 5.17 ppm (1 H), a quartet at d = 4.98 ppm (1 H), and a doublet at ô = 1.49 ppm (3 H). There are 6 signals in its 13C NMR spectrum. The IR spectrum has a broad absorption in the -3200 cm-1 region. Compound A reacts with KMNO4 in a basic solution followed by acidification to give compound B with the molecular formula C7H6O2. Draw structures for compounds A and B.arrow_forwardUse the following spectroscopic data to work out the structure of an organic compound: Molecular formula: C3H7NO2, IR (cm): 3001, 2987, 1553 and 1387 cm*1; 1H NMR (5): 1.0 (t, 3H), 2.1 (m, 2H), and 4.4 (t, 2H); 13C NMR (5): 10, 21, and 78. MS: 89 (M+), 43 (Base peak), 41, and 29arrow_forwardThymol (molecular formula C10H14O) is the major component of the oil of thyme. Thymol shows IR absorptions at 3500–3200, 3150–2850, 1621, and 1585 cm−1. The 1H NMR spectrum of thymol is given below. Propose a possible structure for thymol.arrow_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_forwardWhat is the structure of the compound with the formula C8H10O has six signals in its 13C-NMR spectrum. The IR spectrum shows signals at 3400 and 2900 cm-1, the 1H-NMR spectrum is: 1.19 ppm (3H triplet) 2.58 ppm (2H quartet) 4.65 ppm (1H broad singlet) 6.07 ppm (2H doublet) 7.05 ppm (2H doublet)arrow_forwardCompound A has molecular formula C5H10O. It shows three signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.1 ppm, a singlet of integral 3 at 2.14 ppm, and a quintet of integral 1 at 2.58 ppm. Suggest a structure for A and explain your reasoning.arrow_forward
- A hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?arrow_forwardCompound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.arrow_forwardFollowing are 1H-NMR spectra for compounds B (C6H12O2) and C (C6H10O). Upon warming in dilute acid, compound B is converted to compound C. Deduce the structural formulas for compounds B and C.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning