Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 19, Problem 80E
Interpretation Introduction
Interpretation:
The functional group present in sulfanilamide which is an antibacterial is to be stated.
Concept introduction:
The specific moieties or groups present in a molecule which are responsible for the particular kind of
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When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M* = 163 m/z)
is formed. In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm. The 13C NMR and DEPT spectra
are provided. Draw the structure of the product as the resonance contributor lacking any formal charges.
13C NMR
DEPT 90
200
160
120
80
40
0
200
160
120
80
40
0
DEPT 135
T
200
160
120
80
40
0
Draw the unknown amide.
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Chapter 19 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 19 - Prob. 1CECh. 19 - Prob. 2CECh. 19 - Prob. 3CECh. 19 - Prob. 4CECh. 19 - Prob. 5CECh. 19 - Prob. 6CECh. 19 - Prob. 7CECh. 19 - Prob. 1KTCh. 19 - Prob. 2KTCh. 19 - Prob. 3KT
Ch. 19 - Prob. 4KTCh. 19 - Prob. 5KTCh. 19 - Prob. 6KTCh. 19 - Prob. 7KTCh. 19 - Prob. 8KTCh. 19 - Prob. 9KTCh. 19 - Prob. 10KTCh. 19 - Prob. 11KTCh. 19 - Prob. 12KTCh. 19 - Prob. 13KTCh. 19 - Prob. 14KTCh. 19 - Prob. 15KTCh. 19 - Prob. 16KTCh. 19 - Prob. 17KTCh. 19 - Prob. 18KTCh. 19 - Prob. 19KTCh. 19 - Prob. 20KTCh. 19 - Prob. 21KTCh. 19 - Prob. 22KTCh. 19 - Prob. 23KTCh. 19 - Prob. 24KTCh. 19 - Prob. 1ECh. 19 - Prob. 2ECh. 19 - Prob. 3ECh. 19 - Prob. 4ECh. 19 - Prob. 5ECh. 19 - Prob. 6ECh. 19 - Prob. 7ECh. 19 - Prob. 8ECh. 19 - Prob. 9ECh. 19 - Prob. 10ECh. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Prob. 59ECh. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65ECh. 19 - Prob. 66ECh. 19 - Prob. 67ECh. 19 - Prob. 68ECh. 19 - Prob. 69ECh. 19 - Prob. 70ECh. 19 - Prob. 71ECh. 19 - Prob. 72ECh. 19 - Prob. 73ECh. 19 - Prob. 74ECh. 19 - Prob. 75ECh. 19 - Prob. 76ECh. 19 - Prob. 77ECh. 19 - Prob. 78ECh. 19 - Prob. 79ECh. 19 - Prob. 80ECh. 19 - Prob. 81ECh. 19 - Prob. 82ECh. 19 - Prob. 83ECh. 19 - Prob. 84ECh. 19 - Prob. 85ECh. 19 - Prob. 86ECh. 19 - Prob. 1STCh. 19 - Prob. 2STCh. 19 - Prob. 3STCh. 19 - Prob. 4STCh. 19 - Prob. 5STCh. 19 - Prob. 6STCh. 19 - Prob. 7STCh. 19 - Prob. 8STCh. 19 - Prob. 9STCh. 19 - Prob. 10STCh. 19 - Prob. 11STCh. 19 - Prob. 12STCh. 19 - Prob. 13STCh. 19 - Prob. 14STCh. 19 - Prob. 15STCh. 19 - Prob. 16STCh. 19 - Prob. 17STCh. 19 - Prob. 18ST
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- Give the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.arrow_forwardFive isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compoundarrow_forward7. The three sets of replicate results below were accumulated for the analysis of the same sample. Pool these data to obtain the most efficient estimate of the mean analyte content and the standard deviation. Lead content/ppm: Set 1 Set 2 Set 3 1. 9.76 9.87 9.85 2. 9.42 9.64 9.91 3. 9.53 9.71 9.42 9.81 9.49arrow_forward
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- A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. • If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T G OH де OH This transformation can't be done in one step.arrow_forwardMacmillan Leaming Draw the major organic product of the reaction. 1. CH3CH2MgBr 2. H+ - G Select Draw Templates More H о QQarrow_forwardDraw the condensed structure of 3-hydroxy-2-butanone. Click anywhere to draw the first atom of your structure.arrow_forward
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