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Concept explainers
Interpretation:
When the compounds A and B in the previous problem are irradiated, two stereoisomeric compounds, C and D, respectively, are obtained, each of which contains a cyclobutene ring. The structure for C and D, and their formation is to be explained. The irradiation of either A or B does not give back previtamin D2 – the statement is to be verified.
Concept introduction:
Generally, electrocyclic reactions are a pericyclic reaction which occur intramolecularly. These reactions will result in the formation of ring compounds under the influence of heat or light. Notably, in this process one new sigma bond is formed and one old π-bond is consumed. Intriguingly, the reverse ring opening electrocyclic reaction can also be possible to occur under the same reaction mechanism but in reverse manner. In-phase orbital overlap results in symmetry allowed electrocyclic reactions.
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Chapter 28 Solutions
EBK ORGANIC CHEMISTRY
- Reaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed by treatment with aqueous acid, affords R (C7H14O). R has a strong absorption in its IR spectrum at 1717 cm−1 and three singlets in its 1H NMR spectrum at 1.02 (9 H), 2.13 (3 H), and 2.33 (2 H) ppm. What is the structure of R? We will learn about this reaction in Chapter 18.arrow_forwardTreatment of isobutene [(CH3)2C = CH2] with (CH3)3CLi forms a carbanion that reacts with CH2=O to form H after water is added to the reaction mixture. H has a molecular ion in its mass spectrum at m/z = 86, and shows fragments at 71 and 68. H exhibits absorptions in its IR spectrum at 3600–3200 and 1651 cm−1, and has the 1H NMR spectrum given below. Whatis the structure of H?arrow_forwardTreatment of compound E (molecular formula C4H8O2) with excess CH3CH2MgBr yields compound F (molecular formula C6H14O) after protonation with H2O. E shows a strong absorption in its IR spectrum at 1743 cm-1. F shows a strong IR absorption at 3600–3200 cm-1. The 1H NMR spectral data of E and F are given. What are the structures of E and F?Compound E signals at 1.2 (triplet, 3 H), 2.0 (singlet, 3 H), and 4.1 (quartet, 2 H) ppmCompound F signals at 0.9 (triplet, 6 H), 1.1 (singlet, 3 H), 1.5 (quartet, 4 H), and 1.55 (singlet, 1 H) ppmarrow_forward
- Treatment of isobutene [(CH3)2C=CH2] with (CH3)3CLi forms a carbanionthat reacts with CH2=O to form H after water is added to the reactionmixture. H has a molecular ion in its mass spectrum at m/z = 86, andshows fragments at 71 and 68. H exhibits absorptions in its IR spectrumat 3600−3200 and 1651 cm−1, and has the 1H NMR spectrum given below.What is the structure of H?arrow_forwardThe treatment of isoprene [CH2=C(CH3)CH=CH2] with one equivalent ofmCPBA forms A as the major product. A gives a molecular ion at 84 in itsmass spectrum, and peaks at 2850–3150 cm−1 in its IR spectrum. The 1HNMR spectrum of A is given below. What is the structure of A?arrow_forwardCompound A (C10H14) shows prominent peaks in its mass spectrum at m/z 134 and 91. Compound B (also C10H14) shows prominent peaks at m/z 134 and 119. On vigorous oxidation with chromic acid, compound A gave benzoic acid, but compound B was nonreactive. From this information, deduce the structures of both compounds, and then draw the structure of A. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading.arrow_forward
- Propose a structure consistent with each set of data.a. Compound A:Molecular formula: C8H10OIR absorption at 3150–2850 cm–11H NMR data: 1.4 (triplet, 3 H), 3.95 (quartet, 2 H), and 6.8–7.3 (multiplet, 5 H) ppm b. Compound B:Molecular formula: C9H10O2IR absorption at 1669 cm–11H NMR data: 2.5 (singlet, 3 H), 3.8 (singlet, 3 H), 6.9 (doublet, 2 H), and 7.9 (doublet, 2 H) ppmarrow_forwardGive the structure of the major diastereoisomer formed in the two reactions below. Explain the stereochemical outcome with the aid of Newman projections/transition state diagrams.arrow_forward1) In the following reactions, give the nature and possibly the stereochemistry of A1, A2 and B and justify the proportions obtained. 4 1=3 |=1 CI ull OH KOtBu 3 DMSO H₂SO4 majority A1 A1 + 2)When the first compound is treated with MeONa in methanol, compounds A1 and B are also obtained, as well as a higher molecular weight compound C, which has a higher molecular weight. B are also obtained as well as a higher molecular weight compound C whose 1H NMR spectrum is given below. Justify the formation of C. 2 PPM equimolar mixture + isomere of A1 1=3 mobilie B A2 1=3 1=3 1 1=3 Oarrow_forward
- Treatment of compound C (molecular formula C9H12O) with PCC affords D (molecular formula C9H10O). Use the 1H NMR and IR spectra of D to determine the structures of both C and D.arrow_forward1 (a) In the following reactions, (1) LIAIH, A (2) H2O mCPBA B (i) Draw the structure of compounds A and B. (ii) For each reaction, explain the type of reaction involved. (iii) Explain the successful transformation of compound A using mass spectra.arrow_forward(a) The 'H-NMR spectrum of cyclobutanone shows two signals - signal A at 3.00 ppm and signal B at 1.95 ppm. Give the multiplicity of each signal. cyclobutanone (b) When cyclobutanone is treated with D20 and NaOD, the only signal observable in the 1H-NMR is a singlet at 2.00 ppm. Explain why this is the case. [Note: Deuterium atoms do not display signals in the TH-NMR spectrum]arrow_forward
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