Concept explainers
Three isomeric compounds A, B, and C, all have molecular formula
Compound A: IR peak at
Compound C: IR peaks at
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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_forwardKetones undergo a reduction when treated with sodium borohydride, NaBH4. What is the structure of the compound produced by reaction of 2-butanone with NaBH4 if it has an IR absorption at 3400 cm-1 and M+=74 in the mass spectrum?arrow_forwardPropose a structure for compound X (molecular formula C6H12O2), which gives a strong peak in its IR spectrum at 1740 cm−1. The 1H NMR spectrum of X shows only two singlets, including one at 3.5 ppm. The 13C NMR spectrum is given below. Propose a structure for X.arrow_forward
- Use the 1H NMR and IR data to determine the structure of each compound.arrow_forwardIdentify the structures of isomers H and I (molecular formula C8H11N).a.Compound H: IR absorptions at 3365, 3284, 3026, 2932, 1603, and 1497 cm−1b.Compound I: IR absorptions at 3367, 3286, 3027, 2962, 1604, and 1492 cm−1arrow_forward08) The NMR spectra of the two isomeric compounds with formula C3H5ClO2 are shown in letters a and b. Low-field protons appearing in the NMR spectrum around 12.1 and 11.5 ppm, respectively, are shown highlighted. Draw the structures of the isomers.arrow_forward
- Compound P has molecular formula C5H9ClO2. Deduce the structure of P from its 1H and 13C NMR spectra.arrow_forwardA compound with molecular formula C3H8O produces a broad signal between 3200 and 3600 cm³¹ in its IR spectrum and produces two signals in its 13C NMR spectrum. Draw the structure of the compound. Draw Your Solutionarrow_forwardCompound A has molecular formula C7H7X. Its 1H-NMR spectrum shows a singlet at 2.26 ppm and two doublets, one at 6.95 ppm and one at 7.28 ppm. The singlet has an integral of three and the doublets each have an integral of two. Its 13C-NMR shows five signals. The mass spectrum of A shows a peak at m/z = 170 and another peak at m/z = 172; the relative height of the two peaks is 1:1 respectively. - Identify what atom X is, explaining your reasoning - Identify Compound A, explaining your reasoning Compound A is treated with a mixture of nitric and sulfuric acids to generate Compound B. The 1H-NMR spectrum of B shows two singlets, one at 2.52 pm and one at 8.13 ppm. The 13C-NMR spectrum of B shows five signals. The mass spectrum of B shows a peak at m/z = 260 and another peak at m/z = 262; the relative height of the two peaks is 1:1 respectively. - Identify compound B, explaining your reasoning Compound B is treated with sodium ethoxide to generate compound C. The 1H-NMR spectrum of C shows…arrow_forward
- Identify the structures of D and E, isomers of molecular formulaC6H12O2, from their IR and 1H NMR data. Signals at 1.35 and 1.60 ppm inthe 1H NMR spectrum of D and 1.90 ppm in the 1H NMR spectrum of Eare multipletsarrow_forwardWhat is the structure of the compound with the formula C6H12O2, if it has a strong IR signal near 2900 and 1740 cm-1, the 1H-NMR spectrum is: 1.12 ppm (3H triplet) 1.25 ppm (6H doublet) 2.27 ppm (2H quartet) 5.02 ppm (1H septet)arrow_forwardCompound B has molecular formula C9H12O. It shows five signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.22 ppm, a septet of integral 1 at 2.86 ppm, a singlet of integral 1 at 5.34 ppm, a doublet of integral 2 at 6.70 ppm, and a doublet of integral 2 at 7.03 ppm. The 13C-NMR spectrum of B shows six unique signals (23.9, 34.0, 115.7, 128.7, 148.9, and 157.4). Identify B and explain your reasoning.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning