Do the following for each problem: - Identify the major functional groups in the
ID: 1018114 • Letter: D
Question
Do the following for each problem:
- Identify the major functional groups in the IR spectra
- Identify all the peaks in the H- and C-NMR spectra
- Identify the molecular mass from the mass spectrum
- Determine the pattern of fragmentation from the mass spectrum
- Generate a molecular formula (hint: try using the rule of 13)
-Determine the index of hydrogen deficency
- Determine the structure of the unknown compound
A)
B)
11 10 9 HSP-01-105 OD 5D 4000 100 80 60 40- 20- ppm 3000 MS-NW-3158 2000 200 180 160 140 120 100 89 60 40 20 1000 500 1500 HAVENUHHERI 150 100Explanation / Answer
The formula is used in interpreting mass spectra.
Molecular formula=CnHn+r
If you have heteroatoms, you adjust the formula. For example
For O, add O and subtract CH4.
For N, add N and subtract CH2.
For 35Cl, add Cl and subtract C2H11
165/13=12r7
The rule of 13 predicts a formula of C12H12+4=C12H19.
The odd mass and the odd number of H atoms make it reasonable to add an N atom and subtract CH2:
N-15=N-(CH+H) = N-CH2
This gives a formula C11H17N.
The peak at 76 suggests C6H4, so the peak at 45 is the other fragment.
43/13=3r3, which gives a formula of C3H3+3=C3H6.
If it shows a strong peak at 1700 cm-1.
This shows the presence of an O atom.
O-16=16(CH + 2)=16 - CH4
The formula becomes C2H2NO.
The infrared spectrum and NMR Spectrum may give further information
1H-NMR:
The peak value near about = 2.5 ppm shows presence of aliphatic –CH3 group and between = 7-8 ppm shows ortho coupling pattern which showed monosubstituted aromatic ring.
13C-NMR:
IR-
The compound looks like aromatic (band 1400, 1500 and 1600 cm-1 and below 900). It may be monosubstituted benzene derivative.
The absorption peak near 1700 cm-1 shows presence of carbonyl group. (-C=O)
MF may be C8H7NO3
Hydrogen Deficiency is =(no of C+1)- No. of H/2
Subtract CH2 for a N
= 8-3=5
The structure May be,
Answer B.
The peak at 45 is the other fragment.
46/13=3r7, which gives a formula of C3H3+6=C3H10.
If it shows a broad peak at 3500 cm-1.
This shows the presence of an OH group and hence presence of O.
O-16=16(CH + 2)=16 - CH4
The formula becomes C2H6O.
The infrared spectrum and NMR Spectrum may give further information
1H-NMR:
The peak value near about = 1 ppm shows presence of aliphatic –CH3 group and between = 3-4 ppm shows pentate pattern which CH2 -CH3 group.
13C-NMR:
IR-
The compound looks like aromatic (band 3500cm-1 and shows -OH presence and near 2900-3000 shows aliphatic streching ).
MF may be C2H6O
Hydrogen Deficiency is =(no of C+1)- No. of H/2
DBU= 3-3=0
The structure May be, Ethanol.
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