Organic Chemistry Study Guide and Solutions
Organic Chemistry Study Guide and Solutions
6th Edition
ISBN: 9781936221868
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 23, Problem 23.64AP
Interpretation Introduction

Interpretation:

The identification of compounds A, B and C are to be stated.

Concept introduction:

The molecule or compound which looses one or more electrons is said to be oxidized.

Amines are the organic compounds that are formed by replacement of hydrogen from ammonia. The formation of diazonium salt from aromatic amines takes place using sodium nitrite and hydrochloric acid at low temperatures. This process is known as diazotization.

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Liquid ammonia (anhydrous NH3(0)) is often used as a solvent. Like water, ammonia undergoes autoionization by the transfer of a proton: 2NH3 = NH4+ + NH2" K = 1x10-33 at -50°C Since ammonia is better proton acceptor than water, the ionization of acids is relatively enhanced in liquid ammonia. For example, acetic acid is a strong acid in liquid ammonia. Estimate [NH4+] (mol/L) in pure (i.e. neutral) liquid ammonia at -50°C. mol/L Submit Answer Tries 0/5 Calculate [NH4+] (mol/L) in a 7.07x10-5 M solution of acetic acid in liquid ammonia at -50°C. mol/L Submit Answer Tries 0/5 Estimate [NH2] (mol/L) in a 7.07x10-5 M solution of acetic acid in liquid ammonia at -50°C.
Answer ALL parts of this question. An aspirin tablet dissolved in sodium hydroxide solution gave sodium salicylate. The sodium salicylate gives a yellow colour and absorbs at 295 nm, which gave an estimated concentration of sodium salicylate of 0.00015 mol dm-³. CO₂H NaOH (aq) CO₂Na OH sodium salicylate Aspirin (a) Explain the formation of sodium salicylate using a curly arrow mechanism. (b) The solution of sodium salicylate was diluted to 50 cm³. How many moles are in the 50 cm³? (c) There were initially 250 cm³ of the original aspirin solution, so the total number of moles of sodium salicylate in the original solution is 250 times that calculated in part (b). Calculate the mass of aspirin in the tablet. (MW of aspirin = 180 gmol-¹).
11.(a) 10.06 cm3 of the stock 0.100 mol dm–3 solution of NaOH was required to fully neutralise 49.5 mg of a diprotic acid (H2A). Write a balanced equation for this reaction and hence determine the molar mass (Mr) of the acid.                                                                                                                 (b) Determine the pH of 0.100 mol dm–3 NaOH at 298 K [pKw = 14.0].

Chapter 23 Solutions

Organic Chemistry Study Guide and Solutions

Ch. 23 - Prob. 23.11PCh. 23 - Prob. 23.12PCh. 23 - Prob. 23.13PCh. 23 - Prob. 23.14PCh. 23 - Prob. 23.15PCh. 23 - Prob. 23.16PCh. 23 - Prob. 23.17PCh. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - Prob. 23.27PCh. 23 - Prob. 23.28PCh. 23 - Prob. 23.29PCh. 23 - Prob. 23.30PCh. 23 - Prob. 23.31PCh. 23 - Prob. 23.32PCh. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Prob. 23.35PCh. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Prob. 23.38PCh. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44APCh. 23 - Prob. 23.45APCh. 23 - Prob. 23.46APCh. 23 - Prob. 23.47APCh. 23 - Prob. 23.48APCh. 23 - Prob. 23.49APCh. 23 - Prob. 23.50APCh. 23 - Prob. 23.51APCh. 23 - Prob. 23.52APCh. 23 - Prob. 23.53APCh. 23 - Prob. 23.54APCh. 23 - Prob. 23.55APCh. 23 - Prob. 23.56APCh. 23 - Prob. 23.57APCh. 23 - Prob. 23.58APCh. 23 - Prob. 23.59APCh. 23 - Prob. 23.60APCh. 23 - Prob. 23.61APCh. 23 - Prob. 23.62APCh. 23 - Prob. 23.63APCh. 23 - Prob. 23.64APCh. 23 - Prob. 23.65APCh. 23 - Prob. 23.66APCh. 23 - Prob. 23.67APCh. 23 - Prob. 23.68APCh. 23 - Prob. 23.69APCh. 23 - Prob. 23.70APCh. 23 - Prob. 23.71APCh. 23 - Prob. 23.72APCh. 23 - Prob. 23.73APCh. 23 - Prob. 23.74APCh. 23 - Prob. 23.75APCh. 23 - Prob. 23.76APCh. 23 - Prob. 23.77APCh. 23 - Prob. 23.78APCh. 23 - Prob. 23.79APCh. 23 - Prob. 23.80APCh. 23 - Prob. 23.81AP
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