Chemistry 211 Laboratory
Fall 2013
Paper Unknown
Final Reports
Complete pp. 1-4 of the attached report form for each of your Group Paper
Unknowns and submit the report to your laboratory instructor by the end of your
laboratory period during the week of December 3 - 6. See Model Report pp. 5-8.
This report form is available on the Experiment 7 page of the course website. The
report may be completed by hand or by word processor using the electronic copy
from the website. While your reports must be clearly and neatly presented, wordprocessed text and electronically produced figures are not required.
Remove this cover page before submitting your report.
CHEM 211-13
Final Group Paper Unknown Report
Group Name
Group Members:
Unknown #
Append your Group Paper Unknown data handout sheets.
Summary of Results
Fill in the tables below with data from group paper unknown spectra (See attached model
report)
Molecular Mass:
Melting Point:
˚C
Boiling Point:
MS Molecular ion Data:
M+ mass
M+ Intensity
M+1 Intensity
M+2 Intensity
Does the compound contain chlorine or bromine? How do you know? Cite data.
Does the compound contain a nitrogen atom? How do you know? Cite data.
Approximate number of carbon atoms in the molecule.
From MS M+1 peak:
Solubility tests:
Water:
1 M HCl
1 M NaOH
Is your compound acidic, basic or neutral? How do you know? Cite data.
Interpretable IR Data
(List useful peaks, you need not list all peaks in each region)
Region
Structural
 (cm-1
-1
(cm )
Feature
Region 1:3600-3200
Region 2:3100-2500
Region 3:2300-2000
Region 4:1750-1630
Region 5:1630-1400
Region 6:1350-1000
˚C
CHEM 211-13
Final Group Paper Unknown Report
2
Interpretable Mass Spectral Data
(List the largest peaks from the mass table and any others that suggest recognizable structures)
m/e Ion
M+ - Ion
Structural
Information*
M+
M+
* Provide a structure rather than a molecular formula (e.g. CH3CH2CH2+ rather than C3H7+)
1H
Chemical Shift
(
Multiplicity
NMR
Relative Areas
13C
Chemical Shift (
Interpretation*
NMR
Interpretation*
CHEM 211-13
Final Group Paper Unknown Report
3
Discussion of the Process for Determining the Unknown’s Structure
Proposed Compound Name
Proposed Compound Structure
Discussion
Your discussion must explain the sequence by which the structure was devised from the data.
Indicate how parts of the structure were identified and the logic used to finally “put the pieces
together” to create a complete structure. The order in which your data is introduced will depend
upon the logic used in devising your structure. Your discussion should be thorough but concise,
300 words or less. (See model, p. 7)
CHEM 211-13
Final Group Paper Unknown Report
4
Spectral Assignment Diagrams
Use spectral assignment diagrams similar to those provided in the attached model report (p. 8) to
illustrate how your structure accounts for the IR, MS and NMR data summarized in your data
tables. Be sure to note any significant data that cannot be explained from your structure.
IR
MS
1H-NMR
13C-NMR
CHEM 211-13
Final Group Paper Unknown Report
5
Model Report for a Final Group Paper Unknown
Summary of Results:
Molecular Mass:
163
Melting Point:
˚C
Boiling Point:
222
˚C
MS Molecular ion Data:
M+ mass
163
M+ Intensity 34.1
M+1 Intensity
4.4
M+2 Intensity
0
Does the compound contain chlorine or bromine? How do you know? Cite data.
Molecule does not contain chlorine or bromine => There is no significant M + 2+ peak
in MS.
Does the compound contain a nitrogen atom? How do you know? Cite data.
Molecule contains an odd number of nitrogen atoms => M+ is odd
Approximate number of carbon atoms in the molecule.
From MS M+1 peak:
~12
Solubility tests:
Water:
Insoluble
1 M HCl
Soluble
1 M NaOH
Insoluble
Is your compound acidic, basic or neutral? How do you know? Cite data.
Molecule is basic – more soluble in H-Cl than in water indicates that it reacts with acid => basic
IR Data
Region
Structural
 (cm-1
(cm-1)
Interpretation
3600-3200
No absorption
NO N-H or O-H
3100
Aromatic C-H
3100-2500
2950
Sat’d C-H
2300-2000
No absorption
No triple bonds
1750-1630
No absorption
No C=O
1610
Aromatic C-C
1630-1400
1530
Aromatic C-C
1210
C-N or C-O
1350-1000
1170
C-N or C-O
CHEM 211-13
Final Group Paper Unknown Report
6
Mass Spectral Data
(List the largest peaks from the mass table and any others that suggest recognizable structures)
m/e Ion
M+ - Ion
163
M+
Structural
Interpretation*
M+
148
M-15
Loss of CH3
119
M-44
?
91
M-72
77
M-86
CH2
* Provide a structure rather than a molecular formula (e.g. CH3 CH2 CH2+ rather than C3H7+)
1H
Chemical Shift
(
1.2
2.2
2.6
3.3
7.2
NMR
Multiplicity
Relative Areas
Interpretation
doublet
singlet
doublet
multiplet
multiplet
3
6
2
1
5
CH3 next to CH
2 equivalent CH3's attached to N
CH2 adjacent to C=O or aromatic ring and C-H
CH attached to O or to N & other func. group
C-H Aromatic w/5 H’s – singly substitute
aromatic ring
13C
Chemical Shift (
20
32
41
48
112
114
121
145
NMR
Interpretation
sp3 C
sp3 C
sp3 C
sp3 C attached to N or O
aromatic C
aromatic C
aromatic C
aromatic C, most sub’d.
CHEM 211-13
Final Group Paper Unknown Report
7
Proposed Compound Name
N,N-dimethyl-1-phenyl-2-propanamine
Proposed Compound Structure
CH3
CH2
CH3
C
N
CH3
H
Discussion
Your discussion must explain the sequence by which the structure was devised from the data.
Indicate how parts of the structure were identified and the logic used to finally “put the pieces
together” to create a complete structure .The order in which your data is introduced will depend
upon the logic used in devising your structure.
H-NMR (7.2  A=5) & 13C-NMR (112,114,121 &145 ) suggest phenyl group (C6H5). This
structure is supported by IR 3100 (aromatic C-H), & 1610->1470 cm-1 (aromatic C-C) and
MS-77 m/e. MS 91 & 92 m/e suggest that phenyl has an adjacent CH2. That leaves 163 – 91 =
72 mass. Odd M+ indicates a nitrogen atom. That leaves 72-14= 58 mass. 1H-NMR (2.2 
A=6, sing.) suggests 2 equivalent CH3 groups attached to an atom with no H’s. 1H-NMR (2.2 
A=3, doublet => CH3 next to C-H, 1.2  A=2, doublet => CH2 next to C-H and 3.3  A=1,
multiplet => CH next to the CH3 & CH2 gives a CH3CHCH2 structure. IR: no 3300 cm-1 for NH, so N doesn’t have any attached H’s. So pieces seem to be:
1
CH3
CH2
?
?
CH2
C
?
N
?
CH3
?
H
Mass = 42
Mass = 91
CH3
?
?
Mass = 30
Mass = 14
Total mass = 177
Comparison of the total mass above (177) with the M+ mass (163) indicates that there must be
some overlap among the pieces identified (excess mass 177-163 = 14). All carbon atoms have at
least 1 H attached, so the two methyls with NMR singlets must be attached to the nitrogen and
the chemical shifts of the methyls (2.2 ) support that attachment. That change leaves three
pieces.
CH3
CH2
Mass = 91
?
?
CH2
C
CH3
?
N
?
CH3
H
Mass = 42
Mass = 44
Total mass = 177
The Mass 91 and Mass 44 pieces have only one available attachment site (bond) and the mass 42
piece has two bonding sites. So the mass 91 and 44 pieces must be attached to the mass 42
piece. The mass overlap of 14 can be removed if the two CH2 groups are the same. (mass 14
each) The result of these changes gives the final structure.
CH3
CH2 C
H
CH3
N
CH3
CHEM 211-13
Final Group Paper Unknown Report
8
Spectral Assignment Diagrams
(See data in tables on the previous pages)
Use spectral assignment diagrams similar to those provided in the attached model report to
illustrate how your structure accounts for the IR, MS and NMR data summarized in your data
tables. Be sure to note any significant data that cannot be explained from your structure.
IR
C-H aromatic
3100 cm-1
CH3
CH2 C
CH3
N
C-N
1210 & 1170 cm-1
CH3
H
C-C aromatic
1610->1470 cm-1 sat'd C-H
2950 cm-1
MS
m/e
CH3
CH3
CH2
77
C
H
N
148
CH3 (M-15)
91
119
(M-44)
1H
2.6 A=2 doublet
H
H
H
13C
NMR
121 
1.2 A=3 doublet
CH3
CH3
CH2 C
N
CH3
H
H
H
3.3 A=1 Mult.
7.2 A=5 sing.
114 
2.2 A=6 sing.
41 
145 
121 
32 
20 
48 
112 
114 
NMR
N
32 