Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 15.SE, Problem 29AP
3-Chlorocyclopropene, on treatment with AgBF4, gives a precipitate of AgCl and a stable solution of a product that shows a single 1H NMR absorption at 11.04 δ. What is a likely structure for the products, and what is its relation to HĂ¼ckel’s rule?
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Treatment of 2-methylpropanenitrile [(CH3)2CHCN] with CH3CH2CH2MgBr, followed by aqueous acid, affords compound V, which has molecular formula C7H14O. V has a strong absorption in its IR spectrum at 1713 cm−1, and gives the following 1H NMR data: 0.91 (triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and 2.60 (septet, 1 H) ppm. What is the structure of V? We will learn about this reaction in Chapter 20.
Treatment of anisole (CH3OC6H5) with Cl2 and FeCl3 forms P, which has peaks in its mass spectrum at m/z = 142 (M), 144 (M + 2), 129, and 127. P has absorptions in its IR spectrum at 3096–2837 (several peaks), 1582, and 1494 cm-1. Propose possible structures for P.
As reaction of (CH3)2CO with LIC≡CH followed by H2O affords compound D, which has a molecular ion in its mass spectrum at 84 and prominent absorptions in its IR spectrum at 3600−3200, 3303, 2938, and 2120 cm−1. D shows the following 1H NMR spectral data: 1.53 (singlet, 6 H), 2.37 (singlet, 1 H), and 2.43 (singlet, 1 H) ppm. What is the structure of D?
Chapter 15 Solutions
Organic Chemistry
Ch. 15.1 - Prob. 1PCh. 15.1 - Give IUPAC names for the following compounds:Ch. 15.1 - Prob. 3PCh. 15.2 - Pyridine is a flat, hexagonal molecule with bond...Ch. 15.3 - Prob. 5PCh. 15.4 - Draw the five resonance structures of the...Ch. 15.4 - Prob. 7PCh. 15.4 - Prob. 8PCh. 15.5 - Prob. 9PCh. 15.5 - Prob. 10P
Ch. 15.6 - Prob. 11PCh. 15.6 - How many electrons does each of the four nitrogen...Ch. 15.SE - Give IUPAC names for the following substances (red...Ch. 15.SE - All-cis cyclodecapentaene is a stable molecule...Ch. 15.SE - 1, 6-Methanonaphthalene has an interesting 1H NMR...Ch. 15.SE - Prob. 16VCCh. 15.SE - Azulene, an isomer of naphthalene, has a...Ch. 15.SE - Give IUPAC names for the following compounds:Ch. 15.SE - Draw structures corresponding to the following...Ch. 15.SE - Prob. 20APCh. 15.SE - Prob. 21APCh. 15.SE - Draw and name all possible aromatic compounds with...Ch. 15.SE - Propose structures for aromatic hydrocarbons that...Ch. 15.SE - Look at the three resonance structures of...Ch. 15.SE - Prob. 25APCh. 15.SE - Prob. 26APCh. 15.SE - Look at the five resonance structures for...Ch. 15.SE - Prob. 28APCh. 15.SE - 3-Chlorocyclopropene, on treatment with AgBF4,...Ch. 15.SE - Prob. 30APCh. 15.SE - Prob. 31APCh. 15.SE - Prob. 32APCh. 15.SE - Which would you expect to be most stable,...Ch. 15.SE - How might you convert 1, 3, 5, 7-cyclononatetraene...Ch. 15.SE - Calicene, like azulene (Problem 15-17), has an...Ch. 15.SE - Pentalene is a most elusive molecule that has been...Ch. 15.SE - Prob. 37APCh. 15.SE - Prob. 38APCh. 15.SE - Compound A, C8H10, yields three substitution...Ch. 15.SE - Prob. 40APCh. 15.SE - Propose structures for compounds that fit the...Ch. 15.SE - Prob. 42APCh. 15.SE - Prob. 43APCh. 15.SE - N-Phenylsydnone, so-named because it was first...Ch. 15.SE - Prob. 45APCh. 15.SE - Prob. 46APCh. 15.SE - Prob. 47APCh. 15.SE - Propose a structure for a molecule C14H12 that has...Ch. 15.SE - The proton NMR spectrum for a compound with...Ch. 15.SE - The proton NMR spectrum of a compound with formula...Ch. 15.SE - Aromatic substitution reactions occur by addition...Ch. 15.SE - Prob. 52APCh. 15.SE - Consider the aromatic anions below and their...Ch. 15.SE - After the reaction below, the chemical shift of Ha...Ch. 15.SE - Prob. 55APCh. 15.SE - Azo dyes are the major source of artificial color...
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- 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_forwardTreatment of 2-methylpropanenitrile [(CH3)2CHCN] withCH3CH2CH2MgBr, followed by aqueous acid, affords compound V, whichhas molecular formula C7H14O. V has a strong absorption in its IRspectrum at 1713 cm−1, and gives the following 1H NMR data: 0.91(triplet, 3 H), 1.09 (doublet, 6 H), 1.6 (multiplet, 2 H), 2.43 (triplet, 2 H), and2.60 (septet, 1 H) ppm. What is the structure of V?arrow_forwardTreatment of butan-2-one (CH3COCH2CH3) with strong base followed by CH3I forms a compound Q, which gives a molecular ion in its mass spectrum at 86. The IR (> 1500 cm−1 only) and 1H NMR spectra of Q are given below. What is the structure of Q?arrow_forward
- How could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?arrow_forwardA molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm1. Provide the structures of the starting material (1 pt) and product (2 pts) NaOH H20 C5H1,Br 6H triplet 4H quintet 1H quintet 10 8 7 6. 4 3 1 HSP-06-347 ppmarrow_forwardTreatment of compound D with LiAlH4 followed by H2O forms compound E. D shows a molecular ion in its mass spectrum at m/z = 71 and IR absorptions at 3600–3200 and 2263 cm−1. E shows a molecular ion in its mass spectrum at m/z = 75 and IR absorptions at 3636 and 3600–3200 cm−1. Propose structures for D and E from these data and the given 1H NMR spectra.arrow_forward
- Reaction of butanenitrile (CH3CH2CH2CN) with methylmagnesium bromide (CH3MgBr), followed by treatment with aqueous acid, forms compound G. G has a molecular ion in its mass spectrum at m/z = 86 and a base peak at m/z = 43. G exhibits a strong absorption in its IR spectrum at 1721 cm−1 and has the 1H NMR spectrum given below. What is the structure of G?arrow_forwardA molecule of the molecular formula C5H11Br gives rise to the NMR spectrum below. When reacted with NaOH and water, it forms a product which by NMR has 2 protons 1H at 5.4 ppm and 1H at 5.5 ppm each having a 3J coupling of 17 Hz. (other protons also present) The product also has an IR stretch at 1550 cm 1. Provide the structures of the starting material and product. NaOH H,0 C;H„Br 6H triplet 4H quintet 1H quintet 10 8 Ppmarrow_forwardReaction of (CH3)2CO with LiCCH followed by H2O gives compound X, which has a molecular ion in the mass spectrum at 84. It also has prominent absorptions in the IR spectrum at 3600-3200, 3303, 2938, and 2120 cm-1. The proton NMR shows a singlet at 1.53 (6H), a singlet at 2.37 (1H) and a singlet at 2.43 (1H). What is the structure for compound X?arrow_forward
- Identify the structures of isomers A (molecular formula C9H10O).Compound A: I R peak at 1742 cm−1; 1H NMR data (ppm) at 2.15 (singlet, 3H), 3.70 (singlet, 2 H), and 7.20 (broad singlet, 5 H).arrow_forwardAn unknown compound D exhibits a strong absorption in its IR spectrumat 1692 cm−1. The mass spectrum of D shows a molecular ion at m/z =150 and a base peak at 121. The 1H NMR spectrum of D is shown below.What is the structure of D?arrow_forwardTreatment of compound C (molecular formula C9H12O) with PCC affordsD (molecular formula C9H10O). Use the 1H NMR and IR spectra of D todetermine the structures of both C and D.arrow_forward
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