EBK ORGANIC CHEMISTRY
6th Edition
ISBN: 8220103151757
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 13, Problem 13.11P
Interpretation Introduction
Interpretation:
The
Concept introduction:
Nuclear magnetic resonance is a type of spectroscopy in which number of different kind of protons present in different environment can be detected. Chemical shift is defined as the distance from the reference point. Protons present in different environment gives different absoprtion peak in
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Relative Intensity
Part VI. consider the multi-step reaction below for compounds
A, B, and C.
These compounds were subjected to mass spectrometric analysis and
the following spectra for A, B, and C was obtained.
Draw the structure of B and C and match all three compounds
to the correct spectra.
Relative Intensity
Relative Intensity
100
HS-NJ-0547
80
60
31
20
S1
84
M+
absent
10
30
40
50
60
70
80
90
100
100-
MS2016-05353CM
80-
60
40
20
135 137
S2
164 166
0-m
25
50
75
100
125
150
m/z
60
100
MS-NJ-09-43
40
20
20
80
45
S3
25
50
75
100
125
150
175
m/z
Don't used hand raiting and don't used Ai solution
Predicting the pro
Predict the major products of this organic reaction.
Explanation
Check
m
☐
+
5
1.03
Click and drag t
drawing a stru
2. (CH₂)₂S
3
2
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Chapter 13 Solutions
EBK ORGANIC CHEMISTRY
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Prob. 13.4PCh. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10P
Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - Prob. 13.24PCh. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27PCh. 13 - Prob. 13.28PCh. 13 - Prob. 13.29PCh. 13 - Prob. 13.30PCh. 13 - Prob. 13.31PCh. 13 - Prob. 13.32PCh. 13 - Prob. 13.33PCh. 13 - Prob. 13.34PCh. 13 - Prob. 13.35PCh. 13 - Prob. 13.36APCh. 13 - Prob. 13.37APCh. 13 - Prob. 13.38APCh. 13 - Prob. 13.39APCh. 13 - Prob. 13.40APCh. 13 - Prob. 13.41APCh. 13 - Prob. 13.42APCh. 13 - Prob. 13.43APCh. 13 - Prob. 13.44APCh. 13 - Prob. 13.45APCh. 13 - Prob. 13.46APCh. 13 - Prob. 13.47APCh. 13 - Prob. 13.48APCh. 13 - Prob. 13.49APCh. 13 - Prob. 13.50APCh. 13 - Prob. 13.51APCh. 13 - Prob. 13.52APCh. 13 - Prob. 13.53APCh. 13 - Prob. 13.54APCh. 13 - Prob. 13.55APCh. 13 - Prob. 13.56APCh. 13 - Prob. 13.57APCh. 13 - Prob. 13.58APCh. 13 - Prob. 13.59APCh. 13 - Prob. 13.60APCh. 13 - Prob. 13.61APCh. 13 - Prob. 13.62APCh. 13 - Prob. 13.63APCh. 13 - Prob. 13.64APCh. 13 - Prob. 13.65APCh. 13 - Prob. 13.67AP
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- starting material target If so, draw a synthesis below. If no synthesis using reagents ALEKS recognizes is possible, check the box under the drawing area. Be sure you follow the standard ALEKS rules for submitting syntheses. + More... X Explanation Check C टे Br T Add/Remove step ☐ Br Br © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacarrow_forwardDon't used hand raitingarrow_forwardRelative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/zarrow_forward
- Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100arrow_forwardPart IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained. Draw the structures of these fragments.arrow_forwardFor each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.arrow_forward
- A molecule shows peaks at 1379, 1327, 1249, 739 cm-1. Draw a diagram of the energy levels for such a molecule. Draw arrows for the possible transitions that could occur for the molecule. In the diagram imagine exciting an electron, what are its various options for getting back to the ground state? What process would promote radiation less decay? What do you expect for the lifetime of an electron in the T1 state? Why is phosphorescence emission weak in most substances? What could you do to a sample to enhance the likelihood that phosphorescence would occur over radiationless decay?arrow_forwardRank the indicated C—C bonds in increasing order of bond length. Explain as why to the difference.arrow_forwardUse IUPAC rules to name the following alkanearrow_forward
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