CHM247 Tutorial #2

1. CHM247 Tutorial #2 http://assiduous-atom.tripod.com/ (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) “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 (~1730- 1680) or an aldehyde (~1720-1695)  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. Left over  C2N2, which could be = 2 – look for possible functional groups in the IR 2230cm-1 could be an Alkyne or a Nitrile 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”  D.O.U. = 2 A : C6H10O ~2800cm-1, aldehyde C-H These 2 must be coupling ~9.4ppm Aldehyde C-H ~1640cm-1 Alkene ~1695cm-1 Conjugated aldehyde Alkene region – CH Next to a CH3 Typical –CH2CH3