1. http://assiduous-atom.tripod.com/ (slides can be accessed here)
CHM247 Tutorial #2
(slides can be accessed here)

2. Question #1 – Degree of Unsaturation
“Calculate the degree of unsaturation in the following molecules”

3. Question #1 – Degree of Unsaturation
“Calculate the degree of unsaturation in the following molecules”

4. Question #1 – Degree of Unsaturation

5. Question #1 – Degree of Unsaturation
“Calculate the degree of unsaturation in the following molecules”

6. Solving Spectra
Find the D.O.U. provided you have the molecular formula
Try to pick out functional groups from the IR
Use the rest of the data, and your intuition to come up with “fragments” – then piece those fragments together
(hopefully) Useful Advice:
Count the # of H’s and C’s in the 1H and 13C NMR. Do they match the #’s in the molecular formula? If not, this could be an indication of the unknown possessing symmetry
As a general rule: a system of the form RCH2-X or –X-CH3, where X is O, N, Cl. . .etc. The chemical shift of the H’s will be ~to the electronegativity of X. i.e. for CH3F, the 1H NMR will give a peak at around 4ppm
Here are some common signals:

7. Question #2 (a) C11H14O IR: 1725, 1505, 1610 cm-1
“Draw the structures of the following unknown compounds, using the formulae and spectroscopic information provided.”
Typical for aldehydes
C11H14O IR: 1725, 1505, 1610 cm-1
1H NMR: δ 9.7 (singlet 1H), 7.5 (multiplet, 5H), 1.9 (quartet, 2H), 1.5 (singlet, 3H), 0.9 (triplet, 3H)
Aromatic region
1 – find D.O.U.
2 – look for possible functional groups in the IR
1725cm-1  ketone or aldehyde
1505, 1610cm-1  alkene or aromatic ring
3 – start to piece together fragments and fit them with the NMR data
Often a quartet 2H and triplet 3H in a spectrum is signature for a –CH2CH3 fragment

8. Question #2 (a) C11H14O IR: 1725, 1505, 1610 cm-1
1H NMR: δ 9.7 (singlet 1H), 7.5 (multiplet, 5H), 1.9 (quartet, 2H), 1.5 (singlet, 3H), 0.9 (triplet, 3H)
Fragments:
singlet 1H, 9.7ppm
triplet 3H/ quartet 2H
. . . and there’s only one
carbon left
multiplet, 5H ~7-8ppm
probably a monosubstituted
aromatic ring

9. Question #2 (a) C11H14O IR: 1725, 1505, 1610 cm-1
1H NMR: δ 9.7 (singlet 1H), 7.5 (multiplet, 5H), 1.9 (quartet, 2H), 1.5 (singlet, 3H), 0.9 (triplet, 3H)

10. Question #2 (b) 3 equivalent H’s = -CH3 b) C5H10O3 IR: 1741 cm-1
1H NMR: δ 4.3 (triplet, 2H), 3.6 (triplet, 2H), 3.2 (singlet, 3H), 2.1 (singlet, 3H)
3 equivalent H’s = -CH3
Looks like –CH2CH2-
1 – find D.O.U.
2 – look for possible functional groups in the IR
1741cm-1  C=O, but too high a wavenumber to be a ketone (~ ) or an aldehyde (~ )  this is probably an Ester (consult IR table)
3 – start to piece together fragments and fit them with the NMR data
. . . All you have left over is -O-

11. Question #2 (b) -OCH2CH2O- b) C5H10O3 IR: 1741 cm-1
1H NMR: δ 4.3 (triplet, 2H), 3.6 (triplet, 2H), 3.2 (singlet, 3H), 2.1 (singlet, 3H)
-OCH3
Now, trying to put the pieces together
Leads to 4 possibilities. . .
Also – if you look it up, H’s on Carbon adjacent to carbonyls give peaks ~2-2.5ppm
This would give a
Different NMR coupling pattern than the one observed
Same here
Best Proposed Structure

12. Question #2 (c) Ratio = 2:3  2 –CH2- ‘s and 2 –CH3’s
that don’t couple to
each other?
c) C7H10N2 IR: 2230 cm-1
1H NMR: 2.3 (singlet, 4H), 1.1 (singlet, 6H)
1 – find D.O.U.
2 – look for possible functional groups in the IR
2230cm-1  could be an Alkyne or a Nitrile
Left over  C2N2, which could be = x2
3 – start to piece together fragments and fit them with the NMR data
 note there are 2 peaks for 10 H’s !! This means there is probably
a high degree of symmetry in the molecule!
For 2 methyl groups to be the same, they either have to be both on the same carbon, or the molecule must have a mirror plane in the center

13. Question #2 (d) d) C9H11Cl IR: 1505, 1610 cm-1
1H NMR: δ 7.2 (doublet, 2H), 7.1 (doublet, 2H), 3.1 (septet, 1H), 1.3 (doublet, 6H)
Usually Isopropyl
1 – find D.O.U.
2 – look for possible functional groups in the IR
1505, 1610cm-1  could be an Alkene, or an Aromatic Ring
3 – start to piece together fragments and fit them with the NMR data
. . . And what’s left over?? 

14. Question #3
“Study the spectra below and deduce the structure of Compound A”
A : C6H10O
 D.O.U. = 2
~2800cm-1, aldehyde C-H
These 2 must be coupling
~9.4ppm
Aldehyde C-H
~1640cm-1
Alkene
Alkene region – CH
Next to a CH3
~1695cm-1
Conjugated aldehyde
Typical –CH2CH3