Chemistry For Today: General, Organic, And Biochemistry, Loose-leaf Version
Chemistry For Today: General, Organic, And Biochemistry, Loose-leaf Version
9th Edition
ISBN: 9781305968707
Author: Spencer L. Seager
Publisher: Brooks Cole
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Chapter 16, Problem 16.17E
Interpretation Introduction

Interpretation:

The reason as to why the boiling points of amines are lower than those of corresponding alcohols is to be stated.

Concept introduction:

Amines are a class of organic compounds which are derived from ammonia. They are formed by replacing one or more hydrogen in ammonia by organic molecules. Amines usually show hydrogen bonding. Alcohol is an organic compound containing a hydroxyl group (−OH). Hydrogen bonding forms between a hydrogen atom covalently bonded to a highly electronegative element and another electronegative atom. It is usually shown by the hydrogen covalently bonded to oxygen, nitrogen or fluorine.

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Instructions: Complete the questions in the space provided. Show all your work 1. You are trying to determine the rate law expression for a reaction that you are completing at 25°C. You measure the initial reaction rate and the starting concentrations of the reactions for 4 trials. BrO³¯ (aq) + 5Br¯ (aq) + 6H* (aq) → 3Br₂ (l) + 3H2O (l) Initial rate Trial [BrO3] [H*] [Br] (mol/L) (mol/L) | (mol/L) (mol/L.s) 1 0.10 0.10 0.10 8.0 2 0.20 0.10 0.10 16 3 0.10 0.20 0.10 16 4 0.10 0.10 0.20 32 a. Based on the above data what is the rate law expression? b. Solve for the value of k (make sure to include proper units) 2. The proposed reaction mechanism is as follows: i. ii. BrО¸¯ (aq) + H+ (aq) → HBrO3 (aq) HBrO³ (aq) + H* (aq) → H₂BrO3* (aq) iii. H₂BrO³* (aq) + Br¯ (aq) → Br₂O₂ (aq) + H2O (l) [Fast] [Medium] [Slow] iv. Br₂O₂ (aq) + 4H*(aq) + 4Br‍(aq) → 3Br₂ (l) + H2O (l) [Fast] Evaluate the validity of this proposed reaction. Justify your answer.
е. Д CH3 D*, D20
C. NaOMe, Br Br

Chapter 16 Solutions

Chemistry For Today: General, Organic, And Biochemistry, Loose-leaf Version

Ch. 16 - Prob. 16.1ECh. 16 - Prob. 16.2ECh. 16 - Prob. 16.3ECh. 16 - Prob. 16.4ECh. 16 - Prob. 16.5ECh. 16 - Prob. 16.6ECh. 16 - Prob. 16.7ECh. 16 - Prob. 16.8ECh. 16 - Give each of the following amines an IUPAC name:...Ch. 16 - Give each of the following amines an IUPAC name....
Ch. 16 - Prob. 16.11ECh. 16 - Prob. 16.12ECh. 16 - Prob. 16.13ECh. 16 - Prob. 16.14ECh. 16 - Prob. 16.15ECh. 16 - Prob. 16.16ECh. 16 - Prob. 16.17ECh. 16 - Prob. 16.18ECh. 16 - Prob. 16.19ECh. 16 - Draw diagrams similar to Figure 16.1 to illustrate...Ch. 16 - Prob. 16.21ECh. 16 - Prob. 16.22ECh. 16 - Prob. 16.23ECh. 16 - Prob. 16.24ECh. 16 - Prob. 16.25ECh. 16 - Prob. 16.26ECh. 16 - Prob. 16.27ECh. 16 - Prob. 16.28ECh. 16 - Prob. 16.29ECh. 16 - Prob. 16.30ECh. 16 - Prob. 16.31ECh. 16 - Prob. 16.32ECh. 16 - Prob. 16.33ECh. 16 - Describe the general structure of a neuron.Ch. 16 - Name the two amino acids that are starting...Ch. 16 - Prob. 16.36ECh. 16 - Prob. 16.37ECh. 16 - Prob. 16.38ECh. 16 - Prob. 16.39ECh. 16 - Prob. 16.40ECh. 16 - Prob. 16.41ECh. 16 - Prob. 16.42ECh. 16 - Why are alkaloids weakly basic?Ch. 16 - Prob. 16.44ECh. 16 - Prob. 16.45ECh. 16 - Prob. 16.46ECh. 16 - Prob. 16.47ECh. 16 - Prob. 16.48ECh. 16 - Prob. 16.49ECh. 16 - Prob. 16.50ECh. 16 - Prob. 16.51ECh. 16 - Complete the following reactions: a. b.Ch. 16 - Complete the following reactions: a. b.Ch. 16 - Prob. 16.54ECh. 16 - What are the products of the acid hydrolysis of...Ch. 16 - Prob. 16.56ECh. 16 - Prob. 16.57ECh. 16 - Prob. 16.58ECh. 16 - Prob. 16.59ECh. 16 - Prob. 16.60ECh. 16 - Prob. 16.61ECh. 16 - Prob. 16.62ECh. 16 - Prob. 16.63ECh. 16 - Prob. 16.64ECh. 16 - Prob. 16.65ECh. 16 - Prob. 16.66ECh. 16 - Prob. 16.67ECh. 16 - Prob. 16.68ECh. 16 - Prob. 16.69ECh. 16 - Prob. 16.70ECh. 16 - Prob. 16.71ECh. 16 - The stimulant in coffee is: a. tannic acid b....Ch. 16 - What are the most likely products of a reaction...
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