Organic Chemistry Study Guide and Solutions
6th Edition
ISBN: 9781936221868
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 12, Problem 12.11P
Interpretation Introduction
Interpretation:
The structures of five isomeric
Concept introduction:
Infrared spectroscopy deals with the infrared region ranging from
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There are several isomeric alcohols and ethers of molecular formula C5H12O. Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers.
Isomer A: δ = 0.92 (t, 7.8 Hz, 3 H), 1.20 (s, 6H), 1.49 (q, 7.8 Hz, 2H), 1.85 (s, 1H) ppm
Isomer B: δ = 1.19 (s, 9 H), 3.21 (s, 3H) ppm
The IR spectrum of a sample contains absorptions at 1750 cm-1. To what class of
organic compound does this sample most likely belong?
Alkene
Ketone
Alkane
Alcohol
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.
Chapter 12 Solutions
Organic Chemistry Study Guide and Solutions
Ch. 12 - Prob. 12.1PCh. 12 - Prob. 12.2PCh. 12 - Prob. 12.3PCh. 12 - Prob. 12.4PCh. 12 - Prob. 12.5PCh. 12 - Prob. 12.6PCh. 12 - Prob. 12.7PCh. 12 - Prob. 12.8PCh. 12 - Prob. 12.9PCh. 12 - Prob. 12.10P
Ch. 12 - Prob. 12.11PCh. 12 - Prob. 12.12PCh. 12 - Prob. 12.13PCh. 12 - Prob. 12.14PCh. 12 - Prob. 12.15PCh. 12 - Prob. 12.16PCh. 12 - Prob. 12.17PCh. 12 - Prob. 12.18PCh. 12 - Prob. 12.19PCh. 12 - Prob. 12.20PCh. 12 - Prob. 12.21PCh. 12 - Prob. 12.22PCh. 12 - Prob. 12.23APCh. 12 - Prob. 12.24APCh. 12 - Prob. 12.25APCh. 12 - Prob. 12.26APCh. 12 - Prob. 12.27APCh. 12 - Prob. 12.28APCh. 12 - Prob. 12.29APCh. 12 - Prob. 12.30APCh. 12 - Prob. 12.31APCh. 12 - Prob. 12.32APCh. 12 - Prob. 12.33APCh. 12 - Prob. 12.34APCh. 12 - Prob. 12.35APCh. 12 - Prob. 12.36APCh. 12 - Prob. 12.37APCh. 12 - Prob. 12.38APCh. 12 - Prob. 12.39APCh. 12 - Prob. 12.40APCh. 12 - Prob. 12.41APCh. 12 - Prob. 12.42APCh. 12 - Prob. 12.43APCh. 12 - Prob. 12.44AP
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- Consider the following three compounds: CH,=CH, CH,=CH-CH=CH, CH,=CH-CH=CH-CH=CH, (1) (2) (3) Compound 1 contains a simple isolated carbon-carbon double bond, but the other two have conjugated double bonds. (i) Which compound will have the largest absorption (max) Value? (ii) Give a reason for your answer in part (b)(i).arrow_forwardAs 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?arrow_forwardFollowing are the 13C and 1H spectra for one of four isomeric bromoalkanes with formula C4H9Br. Assign a structure of the isomer.arrow_forward
- Cyclohex-2-enone has two protons on its carbon-carbon double bond (labeled Ha and Hb) and two protons on the carbon adjacent to the double bond (labeled Hc). (a) If Jab = 11 Hz and Jbc = 4 Hz, sketch the splitting pattern observed for each proton on the sp2 hybridized carbons. (b) Despite the fact that Ha is located adjacent to an electron-withdrawing C=O, its absorption occurs upfield from the signal due to Hb (6.0 vs. 7.0 ppm). Offer an explanation.arrow_forwardThe mass spectrum of 2,3-dimethylpentane [(CH3)2CHCH(CH3)CH2CH3]shows fragments at m/z = 85 and 71. Propose possible structures for theions that give rise to these peaks.arrow_forwardCyclohex-2-enone has two protons on its carbon–carbon double bond (labeled Ha and Hb) and two protons on the carbon adjacent to the double bond (labeled Hc). (a) If Jab = 11 Hz and Jbc = 4 Hz, sketch the splitting pattern observed for each proton on the sp2 hybridized carbons. (b) Despite the fact that Ha is located adjacent to an electron-withdrawing C = O, its absorption occurs upeld from the signal due to Hb (6.0 vs. 7.0 ppm). Offer an explanation.arrow_forward
- (b) Distinguish between the following pairs of compounds with the help of their UV-absorption characteristics : 3,5-Dimethyl-N,N-demethyl aniline 2,6-dimethyl-N, N-dimethyl and Lang aniline CH3 -CH3 and CH3 CH3arrow_forwardN-propylbenzene, C6H5CH2CH3, contains C (sp3) -H and C (sp2) -H bonds. Its IR spectrum shows strong or medium absorptions at 3085, 3064, 3028, 2960, 2931 and 2873 cm ^ -1, as well as bands below 1600cm -1. Which statement is wrong? A.) Stretching of the C (sp3) -H bonds results in absorptions at lower wave numbers than the stretching of the C (sp2) -H bonds. B.) The absorptions at 2960, 2931 and 2873 cm ^ -1 are assigned to stretching of the C (sp3) -H bonds. C.) The absorptions at 3085, 3064 and 3028 cm ^ -1they are assigned to stretching of the C (sp2) -H bonds. D.) Each absorption can be assigned to the stretch mode of a particular bond in the propylbenzene molecule.arrow_forwardThere are several isomeric alkanes of molecular formula C6H14.Two of these exhibit the following 1H-NMR spectra. Propose a structure for each of the isomers. Isomer A: δ = 0.84 (d, 12 H), 1.39 (septet, 2H) ppm Isomer B: δ = 0.84 (t, 3 H), 0.86 (s, 9H), 1.22 (q, 2H) ppmarrow_forward
- The NMR spectrum of bromocyclohexane indicates a low field signal (1H) at δ 4.16. To room temperature, this signal is a singlet, but at -75 ° C it separates into two peaks of unequal area (but totaling one proton): δ 3.97 and δ 4.64, in ratio 4.6: 1.0. How do you explain the doubling in two peaks? According to the generalization of the previous problem, what conformation of the molecule predominates (at -75 ° C)? What percentage of the molecules does it correspond to? Solve all parts otherwise down vote and hand written solutionarrow_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_forwardReaction of (CH3)3CCHO with (C6H5)3P=C(CH3)OCH3, followed bytreatment 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 thestructure of R?arrow_forward
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