Concept explainers
(a)
Interpretation:
The strucutral formula for each of the given compound has to be proposed using the given NMR data.
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
(a)
Explanation of Solution
Index of Hydrogen Deficiency (IHD) calculation,
Given molecular formula F is
We calculate the
From the molecular formula
The one signal in the
There are also one
Therefore, the based on above spectral details the structure that is consistent with the all of these facts is 4-Bromo-1- ethyl ester.
(b)
Interpretation:
The strucutral formula for each of the given compound has to be proposed using the given NMR data.
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
(b)
Explanation of Solution
Index of Hydrogen Deficiency (IHD) calculation,
Given molecular formula F is
We calculate the
From the molecular formula
The one signal in the
There are also one
Therefore, the based on above spectral details the structure that is consistent with the all of these facts is 4-Bromo-1- methylacetamide.
(c)
Interpretation:
The strucutral formula for each of the given compound has to be proposed using the given NMR data.
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
(c)
Explanation of Solution
Index of Hydrogen Deficiency (IHD) calculation,
Given molecular formula F is
We calculate the
From the molecular formula
The one signal in the
There are also one
Therefore, the based on above spectral details the structure that is consistent with the all of these facts is acid substituted phenylacetamide.
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Chapter 21 Solutions
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- (a) Compound 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. (b) Compound B has molecular formula C8H6O2. The IR, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for B and explain your reasoning. (c) Compound C has molecular formula C5H8O. The IR, mass, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for C and explain your reasoning.arrow_forward(a) Compound 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. (b) Compound B has molecular formula C8H6O2. The IR, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for B and explain your reasoning. (c) Compound C has molecular formula C5H8O. The IR, mass, 1H-NMR, and 13C-NMR spectra are shown below, they are also downloadable for closer inspection by clicking the link under the spectral data. Suggest a structure for C and explain your reasoning.arrow_forwardWhen compound A (C5H12O) is treated with HBr, it forms compound B (C5H11Br). The 1H NMR spectrum of compound A has a 1H singlet, a 3H doublet, a 6H doublet, and two 1H multiplets. The 1H NMR spectrum of compound B has a 6H singlet, a 3H triplet, and a 2H quartet. Identify compounds A and B.arrow_forward
- Identify the structure of compound C (molecular formula C11H15NO2), which has an IR absorption at 1699 cm−1 and the 1H NMR spectrum shown below.arrow_forwardTreatment of compound C (molecular formula C4H8O) with C6H5MgBr, followed by H2O, affords compound D (molecular formula C10H14O). Compound D has a strong peak in its IR spectrum at 3600–3200 cm−1. The 1H NMR spectral data of C and D are given. What are the structures of C and D? Compound C signals at 1.3 (singlet, 6 H) and 2.4 (singlet, 2 H) ppm Compound D signals at 1.2 (singlet, 6 H), 1.6 (singlet, 1 H), 2.7 (singlet, 2 H), and 7.2 (multiplet, 5 H) ppmarrow_forwardA and B are isomeric dicarbonyl compounds of the molecular formula C5H&O2. The 'H NMR spectrum of A contains a singlet at 2.05 ppm and another singlet at 5.40 ppm. The 'H NMR spectrum of B contains three signals: a singlet at 2.3 ppm, a triplet at 1.10 ppm and a quartet at 2.70 ppm. Suggest structures for A and B and draw them in their respective boxes below. 1st attemptarrow_forward
- Compound H (C8H6O3) gives a precipitate when treated with hydroxylamine in aqueous ethanol and a silver mirror when treated with Tollens solution. Following is its 1H-NMR spectrum. Deduce the structure of compound H.arrow_forwardCompound 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_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
- 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_forwardA compound with molecular formula C3H8O produces a broad signal between 3200 and 3600 cm1 in its IR spectrum and produces two signals in its 1°C NMR spectrum. Draw the structure of the compound.arrow_forwardCompounds A and B are isomers having the molecular formula C4H8O3. Identify A and B on the basis of their 1H NMR spectra.Compound A: δ 1.3 (3H, triplet); 3.6 (2H, quartet); 4.1 (2H, singlet); 11.1 (1H, broad singlet)Compound B: δ 2.6 (2H, triplet); 3.4 (3H, singlet); 3.7 (2H triplet); 11.3 (1H, broad singlet)arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning