Step 5: Draw the structure that best fits the mass spec data. Let's review the results of the previous steps. Step 1: The molecular ion is 27. Step 2: There are no Cl or Br atoms based on isotope pattern. Step 3: The molecular ion is odd, so there is one nitrogen. Now consider Step 4 as it relates to the unknown compound. Step 4: examine the largest peaks and calculate the difference from the molecular ion. Look for common alkyl fragments. The mass spec data shows only one base peak at m/z 27, and a smaller peak at 26, so there are no alkyl fragments. One nitrogen atom will have a molecular weight of 14, leaving 13 amu for the remaining unknown portion. A molecular weight of 13 amu can only correspond to one carbon atom and one nitrogen atom, giving the molecular formula of CHN.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Step 5: Draw the structure that best fits the mass spec data.
Let's review the results of the previous steps.
Step 1: The molecular ion is 27.
Step 2: There are no Cl or Br atoms based on isotope pattern.
Step 3: The molecular ion is odd, so there is one nitrogen.
Now consider Step 4 as it relates to the unknown compound.
Step 4: examine the largest peaks and calculate the difference from the molecular ion. Look for common alkyl fragments.
The mass spec data shows only one base peak at m/z 27, and a smaller peak at 26, so there are no alkyl fragments. One nitrogen
atom will have a molecular weight of 14, leaving 13 amu for the remaining unknown portion. A molecular weight of 13 amu
can only correspond to one carbon atom and one nitrogen atom, giving the molecular formula of CHN.
Deduce the structure of the unknown compound.
Select
Draw
Rings
More
Erase
H
Transcribed Image Text:Step 5: Draw the structure that best fits the mass spec data. Let's review the results of the previous steps. Step 1: The molecular ion is 27. Step 2: There are no Cl or Br atoms based on isotope pattern. Step 3: The molecular ion is odd, so there is one nitrogen. Now consider Step 4 as it relates to the unknown compound. Step 4: examine the largest peaks and calculate the difference from the molecular ion. Look for common alkyl fragments. The mass spec data shows only one base peak at m/z 27, and a smaller peak at 26, so there are no alkyl fragments. One nitrogen atom will have a molecular weight of 14, leaving 13 amu for the remaining unknown portion. A molecular weight of 13 amu can only correspond to one carbon atom and one nitrogen atom, giving the molecular formula of CHN. Deduce the structure of the unknown compound. Select Draw Rings More Erase H
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