Concept explainers
Reaction of
following signals in its
We will learn about this reaction in Chapter 22.
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Organic Chemistry
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- The 13C-NMR spectrum of 3-methyl-2-butanol shows signals at 17.88 (CH3), 18.16 (CH3), 20.01 (CH3), 35.04 (carbon-3), and 72.75 (carbon-2). Account for the fact that each methyl group in this molecule gives a different signal.arrow_forwardWhen the 1HNMR spectrum of an alcohol is run in dimethylsulfoxide (DMSO) solvent rather than in chloroform, exchange of the Ο-H proton is slow and spin-spin splitting is seen between the Ο-H proton and C-H protons on the adjacent carbon. What spin multiplicities would you expect for the hydroxyl protons in the following alcohols? (a) 2-Methyl-2-propanol (b) Cyclohexanol (c) Ethanol (d) 2-Propanol (e) Cholesterol (f) 1-Methylcyclohexanolarrow_forwardDeduce the structure of compound C.arrow_forward
- Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.arrow_forwardA compound (L) with the molecular formula C9H10 reacts with bromine and gives an IR absorption spectrum that includes the following absorption peaks: 3035 cm ¹(m), 3020 cm ¹(m), 2925 cm ¹(m), 2853 cm ¹(w), 1640 cm ¹1(m), 990 cm ¹(s), 915 cm ¹(s), 740 cm ¹(s), 695 cm ¹(s). The ¹H NMR spectrum of L consists of: Doublet 3.1 ppm (2H) Multiplet 5.1 ppm Multiplet 7.1 (5H) ppm Multiplet 4.8 ppm Multiplet 5.8 ppm The UV spectrum shows a maximum at 255 nm. Propose a structure for compound L. Edit Drawing harrow_forwardThe 1H-NMR spectrum of compound B,C7H14O , consists of the following signals: δ 0.9 (t, 6H), 1.6 (sextet, 4H), and 2.4 (t, 4H). Draw the structural formula of compound B.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_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_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_forwardGive and explain the analysis of H-NMR and C-NMR spectra of the followingarrow_forwardThe 13C NMR spectrum of 1-bromo-3-chloropropane contains peaks at δ 30, δ 35, and δ 43. Assign these signals to the appropriate carbons.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning