Organic Chemistry Study Guide and Solutions
6th Edition
ISBN: 9781936221868
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 27, Problem 27.74AP
Interpretation Introduction
Interpretation:
The mechanism of formation of diketopiperazine from artificial sweetener aspartame, on storage for extended period of time, is to be stated.
Concept introduction:
In large molecules, when an electrophile and a nucleophile are present in vicinity, the molecule undergoes intramolecular rearrangement reaction to form other compounds. The electrophile takes the electron from the atom having high electron density and gets negatively charged. The atom which shares its electron gets positively charged as a result. The atoms get rearranged to form new molecules as byproduct.
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Account for the following:(i) Aniline does not give Friedel-Crafts reaction.(ii) Ethylamine is soluble in water whereas aniline is not.(iii) pKb of methylamine is less than that of aniline.
22.64
Give reasons :(a) Aniline is a weaker base than cyclohexyl amine.(b) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
Chapter 27 Solutions
Organic Chemistry Study Guide and Solutions
Ch. 27 - Prob. 27.1PCh. 27 - Prob. 27.2PCh. 27 - Prob. 27.3PCh. 27 - Prob. 27.4PCh. 27 - Prob. 27.5PCh. 27 - Prob. 27.6PCh. 27 - Prob. 27.8PCh. 27 - Prob. 27.9PCh. 27 - Prob. 27.10PCh. 27 - Prob. 27.11P
Ch. 27 - Prob. 27.12PCh. 27 - Prob. 27.13PCh. 27 - Prob. 27.14PCh. 27 - Prob. 27.15PCh. 27 - Prob. 27.16PCh. 27 - Prob. 27.17PCh. 27 - Prob. 27.18PCh. 27 - Prob. 27.19PCh. 27 - Prob. 27.20PCh. 27 - Prob. 27.21PCh. 27 - Prob. 27.22PCh. 27 - Prob. 27.23PCh. 27 - Prob. 27.24PCh. 27 - Prob. 27.25PCh. 27 - Prob. 27.26PCh. 27 - Prob. 27.27PCh. 27 - Prob. 27.28PCh. 27 - Prob. 27.29PCh. 27 - Prob. 27.30PCh. 27 - Prob. 27.31PCh. 27 - Prob. 27.32PCh. 27 - Prob. 27.33PCh. 27 - Prob. 27.34PCh. 27 - Prob. 27.35PCh. 27 - Prob. 27.36PCh. 27 - Prob. 27.37PCh. 27 - Prob. 27.38PCh. 27 - Prob. 27.39PCh. 27 - Prob. 27.40PCh. 27 - Prob. 27.41PCh. 27 - Prob. 27.42PCh. 27 - Prob. 27.43APCh. 27 - Prob. 27.44APCh. 27 - Prob. 27.45APCh. 27 - Prob. 27.46APCh. 27 - Prob. 27.47APCh. 27 - Prob. 27.48APCh. 27 - Prob. 27.49APCh. 27 - Prob. 27.50APCh. 27 - Prob. 27.51APCh. 27 - Prob. 27.52APCh. 27 - Prob. 27.53APCh. 27 - Prob. 27.54APCh. 27 - Prob. 27.55APCh. 27 - Prob. 27.56APCh. 27 - Prob. 27.57APCh. 27 - Prob. 27.58APCh. 27 - Prob. 27.59APCh. 27 - Prob. 27.60APCh. 27 - Prob. 27.61APCh. 27 - Prob. 27.62APCh. 27 - Prob. 27.63APCh. 27 - Prob. 27.64APCh. 27 - Prob. 27.65APCh. 27 - Prob. 27.66APCh. 27 - Prob. 27.67APCh. 27 - Prob. 27.68APCh. 27 - Prob. 27.69APCh. 27 - Prob. 27.70APCh. 27 - Prob. 27.71APCh. 27 - Prob. 27.72APCh. 27 - Prob. 27.73APCh. 27 - Prob. 27.74APCh. 27 - Prob. 27.75APCh. 27 - Prob. 27.76APCh. 27 - Prob. 27.77APCh. 27 - Prob. 27.78APCh. 27 - Prob. 27.79APCh. 27 - Prob. 27.80APCh. 27 - Prob. 27.81APCh. 27 - Prob. 27.82APCh. 27 - Prob. 27.83AP
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- (ii) Describe the key molecular structure features of the pharmacologically active 2- oxazolidinone group of antibacterial agents.arrow_forward24.82 When the compound shown here is heated, ethene gas is evolved and a product with the formula C14H3O2 is formed. The 'H NMR and 13C NMR spectra of C14H8O2 are shown below. (There are two signals >150 ppm in the 13C NMR spectrum. Recall that the 13C NMR signal at 77 ppm is from the CDCI3 solvent.) (a) Draw the structure of C,14H3O2. (b) Draw the mechanism that accounts for its formation. (c) What is the main driving force that favors the products of this reaction? ? C14H3O2 C14H8O2 9. 8 7 6 4 3 2 1 250 200 150 100 50 Chemical shift (ppm) Chemical shift (ppm)arrow_forward2. (a) This problem deals with the acidity of amines, not their basicity (or the acidity of their conjugate acids). While cyclohexylamine is a normal amine and is an extremely weak acid (pKa ~36), aniline (Ph-NH2) is somewhat more acidic, with a pKa of ~28. Explain this observation by comparing both ơ- and Tt-effects on the acidity of these two amines. (b) Similar to phenols, substituted derivatives of aniline have explainable shifts in pKa. Explain the following pKas based on the o- and t-effects of the -OCH3 substituent. o-Methoxyaniline: 26 m-Methoxyaniline: 26 p-Methoxyaniline: 29 Here are the pKas (for the N-H bond, not O-H) for the hydroxyanilines: o-Hydroxyaniline: 33 m-Hydroxyaniline: 29 p-Hydroxyaniline: 30 (c) Briefly explain why all three of these pKas are above 28. (d) Explain why the o- isomer is the highest. (e) Make predictions for the pKas of the three cyano-anilines, and provide explanations. Use actual numbers, but only the trend and your explanations will be graded.…arrow_forward
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