Organic Chemistry
4th Edition
ISBN: 9780073402772
Author: Janice G. Smith
Publisher: MCG
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Chapter 25, Problem 25.85P
Treatment of compound D with LiAlH4 followed by H2O forms compound E. D shows a molecular ion in its mass spectrum at
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Treatment 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.
Treatment of compound D with LIAIH, followed by H,O 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. E shows a
molecular ion in its mass spectrum at m/z = 75 and IR absorptions at 3636 and 3600 – 3200 cm'.
Propose structures for D and E from these data and the given 'H NMR spectra.
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Part 1 out of 2
1Η ΝMR of D
2 H
2 H
1 H
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Indicate two basic differences that exist between the spectra of 1H y 13C in NMR.
Chapter 25 Solutions
Organic Chemistry
Ch. 25 - Prob. 25.1PCh. 25 - Draw the structure of a compound of molecular...Ch. 25 - Prob. 25.3PCh. 25 - Prob. 25.4PCh. 25 - Prob. 25.5PCh. 25 - Arrange the compounds in order of increasing...Ch. 25 - Prob. 25.7PCh. 25 - Prob. 25.8PCh. 25 - Problem 25.7
Draw the product of each...Ch. 25 - Prob. 25.10P
Ch. 25 - Prob. 25.11PCh. 25 - Prob. 25.12PCh. 25 - Prob. 25.13PCh. 25 - Prob. 25.14PCh. 25 - Prob. 25.15PCh. 25 - Prob. 25.16PCh. 25 - Prob. 25.17PCh. 25 - Prob. 25.18PCh. 25 - Draw the products of each acidbase reaction....Ch. 25 - Prob. 25.20PCh. 25 - Prob. 25.21PCh. 25 - Prob. 25.22PCh. 25 - Prob. 25.23PCh. 25 - Problem 25.22
Which nitrogen atom in each compound...Ch. 25 - Which N atom in each drug is more basic? a. b. c.Ch. 25 - Prob. 25.26PCh. 25 - Prob. 25.27PCh. 25 - Prob. 25.28PCh. 25 - Prob. 25.29PCh. 25 - Prob. 25.30PCh. 25 - Problem 25.28
Draw the major product formed in...Ch. 25 - Prob. 25.32PCh. 25 - Prob. 25.33PCh. 25 - Problem 25.31
Devise a synthesis of each compound...Ch. 25 - Prob. 25.35PCh. 25 - Problem 25.33
What starting materials are needed...Ch. 25 - Problem 25.34
(a) What two components are needed...Ch. 25 - Prob. 25.38PCh. 25 - Prob. 25.39PCh. 25 - Prob. 25.40PCh. 25 - 25.37 Varenicline (trade name Chantix) is a drug...Ch. 25 - Prob. 25.42PCh. 25 - Prob. 25.43PCh. 25 - Prob. 25.44PCh. 25 - Prob. 25.45PCh. 25 - Prob. 25.46PCh. 25 - Prob. 25.47PCh. 25 - Prob. 25.48PCh. 25 - Decide which N atom in each molecule is most basic...Ch. 25 - 25.44 Rank the nitrogen atoms in each compound in...Ch. 25 - 25.45 Explain why nitroaniline is a stronger base...Ch. 25 - 25.46 Explain the observed difference in the...Ch. 25 - 25.47 Why is pyrrole more acidic than...Ch. 25 - How would you prepare 3-phenyl-1-propanamine...Ch. 25 - Prob. 25.55PCh. 25 - Prob. 25.56PCh. 25 - Prob. 25.57PCh. 25 - Prob. 25.58PCh. 25 - 25.51 How would you separate toluene , benzoic...Ch. 25 - 25.52 Draw the products formed when methylaniline ...Ch. 25 - Prob. 25.61PCh. 25 - Prob. 25.62PCh. 25 - Prob. 25.63PCh. 25 - Prob. 25.64PCh. 25 - Prob. 25.65PCh. 25 - Prob. 25.66PCh. 25 - Prob. 25.67PCh. 25 - Prob. 25.68PCh. 25 - 25.60 A chiral amine A having the configuration...Ch. 25 - 25.61 Draw a stepwise mechanism for each...Ch. 25 - 25.62 Draw a stepwise mechanism for the following...Ch. 25 - Prob. 25.72PCh. 25 - Tertiary (3) aromatic amines react with NaNO2 and...Ch. 25 - Prob. 25.74PCh. 25 - Devise a synthesis of each compound from benzene....Ch. 25 - Prob. 25.76PCh. 25 - Prob. 25.77PCh. 25 - Prob. 25.78PCh. 25 - Prob. 25.79PCh. 25 - Synthesize each compound from benzene. Use a...Ch. 25 - Prob. 25.81PCh. 25 - Devise a synthesis of each compound from benzene,...Ch. 25 - Prob. 25.83PCh. 25 - 25.72 Three isomeric compounds A, B, and C, all...Ch. 25 - 25.73 Treatment of compound D with LiAlH4 followed...Ch. 25 - Prob. 25.86PCh. 25 - 25.75 Rank the following compounds in order of...Ch. 25 - Prob. 25.88PCh. 25 - Prob. 25.89PCh. 25 - Prob. 25.90P
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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_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 hydrocarbon, compound B, has molecular formula C6H6, and gave an NMR spectrum with two signals: delta 6.55 pm and delta 3.84 pm with peak ratio of 2:1. When warmed in pyridine for three hr, compound B quantitatively converts to benzene. Mild hydrogenation of B yielded another compound C with mass spectrum of m/z 82. Infrared spectrum showed no double bonds; NMR spectrum showed one broad peak at delta 2.34 ppm. With this information, address the following questions. a) How many rings are in compound C? b) How many rings are probably in B? How many double bonds are in B? c) Can you suggest a structure for compounds B and C? d) In the NMR spectrum of B, the up-field signal was a quintet, and the down field signal was a triplet. How must you account for these splitting patterns?arrow_forward
- A 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_forwardHow could 1H NMR spectroscopy be used to distinguish among isomers A, B, and C?arrow_forwardPropose a structure for a compound that displays the given spectroscopic data. The molecular ion in the mass spectrum appears at m/z = 116. IR: 1710 (s) and 3000 (s, broad) cm¹. ¹H NMR: 8 = 0.94 (t, J = 7.0 Hz, 6H), 1.59 (m, 4H), 2.36 (quintet, J = 7.0 Hz, 1H) and 12.04 (broad s, 1H) ppm. 13C NMR: 8 = 11.7, 24.7, 48.7, and 183 ppm. Deduce the structure. Select 11 ||| ||| Q Search Draw C H O • Templates More Question Source: Vollhardt 7e - Organic Chemistry: Structure And Function P W Erase Publisher: W.H. Freeman ENG 4: 4/18arrow_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_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_forwardTreatment 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.arrow_forward
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