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In this case, we can rule out C10H6N2 because the IR spectrum shows that there is a carbonyl
in Zone 4, thus revealing an oxygen in the formula.  C10H6N2 lacks an oxygen.  You can also
rule out C10H202 because it does not correspond with the integrals given in the 1
H‐NMR.
The integral equals 9 (2+1+6), and H2 is not a proportionate ratio.  That leaves us with
C10H180 as our formula.
C11 Possibility
154 – (12 amu per C x 11C) = 22 amu left for hydrogen, nitrogen and oxygen
O N H Formula Does it work?
0 0 34 C10H34 No, violates H‐Rule
1 0 18 C10H180 Acceptable
2 0 2 C10H202 Acceptable
0 2 6 C10H6N2 Acceptable
O N H Formula Does it work?
C11H22 and C11H8N can be ruled out because they lack an oxygen.  C11H60 can also be ruled
out because it does not correspond to the integral given to us on the 1
H‐NMR.
Therefore, our final formula is C10H18O.
The last step is to calculate the DBE: 10 – (18/2) + (0/2) + 1 = 2 (possibly two rings or 2 pi
bonds or a ring and a pi bond)

1. Now we have to analyze the IR spectrum.  We do this by looking at each zone and seeing if
there are peaks that correspond to different functional groups.
Zone 1: Alcohol O‐H: absent; no peak present
Amine/Amide N‐H: absent; no peak, no nitrogen in formula
Terminal Alkyne ≡C‐H: absent; no peak in zone 3
Zone 2: Aryl/Vinyl sp2
C‐H: absent; no peak > 3000 cm‐1
Alkyl sp3 C‐H: present; peaks < 3000 cm‐1
Aldehyde C‐H: absent, no peak at approx. 2700 cm‐1
Carboxylic Acid C‐H: absent; no broad peak
Zone 3: Alkyne C≡C: absent; no peak
Nitrile C≡N: absent; no peak, no nitrogen in the formula
Zone 4: Carbonyl C=O: present, peak at 1716 cm‐1  ketone
Zone 5: Benzene C=C: absent; no peak, not enough DBE
0 0 22 C11H22 Acceptable
1 0 6 C11H60 Acceptable
0 1 8 C11H8N Acceptable
Alkene C=C: absent; no peak
From the IR spectrum, we see that there is a ketone and alkyl sp3
C‐H bonds within the
structure.
2. Now we have to analyze the 1
H‐NMR given to us.  First, put all given information into a
table:
ppm Neighbors Integration # H Implications
2.1 ppm Singlet Integration = 2
1.9 ppm Singlet Integration = 1
1.1 ppm Singlet Integration = 6
Now we have to calculate how many hydrogen atoms are present for each signal.  When
you add up each integral value, you get a total of 9.  There are 18 hydrogen atoms (as seen
in the molecular formula), so it is a 1:2 ratio.  This means that you multiply each integral by
2 to get the number of hydrogens:
ppm Neighbors Integration # H Implications
2.1 ppm Singlet Integration = 2 2×2 = 4
1.9 ppm Singlet Integration = 1 1×2 = 2
1.1 ppm Singlet Integration = 6 6×2 = 12
Now we have to fill in the implications:
ppm Neighbors Integration # H Implications
2.1 ppm Singlet Integration = 2 2×2 = 4 2x: CH2         4x: CH
1.9 ppm Singlet Integration = 1 1×2 = 2 CH2           2x: CH
1.1 ppm Singlet Integration = 6 6×2 = 12 12x: CH              4x: CH3
6x: CH2