Solution Found!
Identify each compound in Figure 15.37 from its molecular
Chapter 15, Problem 43P(choose chapter or problem)
Identify each compound in Figure \(15.37\) from its molecular formula and its \({ }^{13} C\ N M R\)
Spectrum.
a. \(\mathrm{C}_{11} \mathrm{H}_{22} \mathrm{O}\)
b. \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}\)
c. \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}\)
d. \(\mathrm{C}_{6} \mathrm{H}_{12}\)
Figure 15.37
The \({ }^{13} C\ N M R\) spectra for Problem \(43\) .
Equation Transcription:
Text Transcription:
15.37
^13 C NMR
C_11 H_22 O
C_8 H_9 Br
C_6 H_10 O
C_6 H_12
43
Questions & Answers
QUESTION:
Identify each compound in Figure \(15.37\) from its molecular formula and its \({ }^{13} C\ N M R\)
Spectrum.
a. \(\mathrm{C}_{11} \mathrm{H}_{22} \mathrm{O}\)
b. \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}\)
c. \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}\)
d. \(\mathrm{C}_{6} \mathrm{H}_{12}\)
Figure 15.37
The \({ }^{13} C\ N M R\) spectra for Problem \(43\) .
Equation Transcription:
Text Transcription:
15.37
^13 C NMR
C_11 H_22 O
C_8 H_9 Br
C_6 H_10 O
C_6 H_12
43
ANSWER:
Solution 43P
(a)
The given molecular formula is. So, the double bond equivalence can be calculated as follows:
It indicates that there is a double bond or one ring.
The signal at 210 ppm is for a carbonyl carbon. There are ten other carbons in the compound and five other signals. The possible structure is as follows:
(b)
The given molecular formula is
