Following are 'H-NMR and BC-NMR spectral data for compound G (CH0). From this information, deduce the structure of compound G. 'H-NMR 13C-NMR 2.50 (t, 2H) 210.19 126.82 3.05 (t, 2H) 136.64 126.75 3.58 (s, 2H) 133.25 45.02 7.1-7.3 (m, 4H) 128.14 38.11 127.75 28.34

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Following are 'H-NMR and BC-NMR spectral data for compound G (CH0). From
this information, deduce the structure of compound G.
'H-NMR
13C-NMR
2.50 (t, 2H)
210.19 126.82
3.05 (t, 2H)
136.64 126.75
3.58 (s, 2H)
133.25
45.02
7.1-7.3 (m, 4H)
128.14
38.11
127.75
28.34
Transcribed Image Text:Following are 'H-NMR and BC-NMR spectral data for compound G (CH0). From this information, deduce the structure of compound G. 'H-NMR 13C-NMR 2.50 (t, 2H) 210.19 126.82 3.05 (t, 2H) 136.64 126.75 3.58 (s, 2H) 133.25 45.02 7.1-7.3 (m, 4H) 128.14 38.11 127.75 28.34
Expert Solution
Step 1

Calculation of double bond equivalence:

DBE=C+1-H2-X2+N2=10+1-102=6

Thus the given compound must have 6 double bonds or rings or their combination.

 

 

Step 2

Since there is a multiplet of 4 H at 7.1-7.3 ppm, thus the given compound must contain a benzene ring.

The ratio of protons is 2:2:2:4 or 1:1:1:2.

There are 6 signals of 'C' atom within the range of 136.64 to 126.82, thus the given compound must have:

Chemistry homework question answer, step 2, image 1

 

 

 

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