Compound C SH 3H 2H 2H 10 8 7 3 2 0 ppm Chemical shift (8) Micrometers 2.5 100 10 11 12 13 14 15 20 90 80 * 70 60 50 40 30 20 10 Compound C 4000 3600 3200 2800 2400 2000 1800 1600 1400 1200 1000 800 600 400 Wavenumber (cm-!) Transmittance (%)

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter4: Introduction To Quantum Mechanics
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Problem 10P: Use the data in Figure 4.8 to estimate the ratio of radiation intensity at 10,000 Å (infrared) to...
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Compound C shows a molecular ion at m/z 148 and other prominent peaks at m/z 105 and 77. Following are its infrared and 1H-NMR spectra.

Q.) Deduce the structural formula of compound C

Compound C
SH
3H
2H
2H
10
8
7
3
2
0 ppm
Chemical shift (8)
Micrometers
2.5
100
10
11
12
13
14 15
20
90
80
* 70
60
50
40
30
20
10
Compound C
4000
3600
3200
2800
2400
2000
1800
1600
1400
1200
1000
800
600
400
Wavenumber (cm-!)
Transmittance (%)
Transcribed Image Text:Compound C SH 3H 2H 2H 10 8 7 3 2 0 ppm Chemical shift (8) Micrometers 2.5 100 10 11 12 13 14 15 20 90 80 * 70 60 50 40 30 20 10 Compound C 4000 3600 3200 2800 2400 2000 1800 1600 1400 1200 1000 800 600 400 Wavenumber (cm-!) Transmittance (%)
Expert Solution
Step 1

Introduction-

In order to determine the structure of the compounds through NMR and IR we need to know the basic functional group peaks and IR stretching frequencies of some functional groups. 

NMR is always the best technique for identification of compound and its exact structure. Only functional group detection is done through IR spectrum.

Refer below link for NMR peak chart-

https://www.cpp.edu/~lsstarkey/courses/NMR/NMRshifts1H-general.pdf  

Refer below link for IR stretching frequencies-

https://www.cpp.edu/~psbeauchamp/pdf/spec_ir_nmr_spectra_tables.pdf

Step 2

Explanation-

If we observe the given NMR correctly, there are 5 protons in the aromatic region that is 6-8ppm indicating that our desired compound is having aromatic ring.

In the NMR there is a triplet of 2 protons at 3ppm and multiplet of 2 protons at 2ppm and a triplet of 3 protons at 1ppm. This clearly indicates that the compound is having an alkyl most probably propyl side chain.

From the IR spectrum we can come to know that there is a –C=O spectrum at 1690 that it makes us assured of presence of keto group in it.

From the molecular ion peak mass spectrum we know the mass of compound is 148.

Thus the compound is phenyl propyl ketone.

 

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