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CHM247 Tutorial!!

CHM247 Tutorial!!. Rob Webster rwebster@chem.utoronto.ca. IR Spectroscopy. What is IR spec. and why do I care? A technique that tells us something about the structure of molecules

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CHM247 Tutorial!!

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  1. CHM247 Tutorial!! Rob Webster rwebster@chem.utoronto.ca

  2. IR Spectroscopy • What is IR spec. and why do I care? • A technique that tells us something about the structure of molecules • Chemical bonds will absorb radiation at a specific frequency, allowing hem to reach an excited state of vibrational motion • Although the energies of EMR absorbed are very specific, peaks in the spectrum are subject to line broadening from various sources • Many functional groups have characteristic peaks/peak patterns, some of which are important to know in this course Alcohols, Ketones, Aldehydes, Carboxyllic acids, Alkenes, Alkynes and Nitriles.

  3. Question #1 “State the infrared spectral changes you would expect to see on conversion of Compound X into Compound Y in each of the cases below” Alkyne Alkene In other words  What are the main differences between the IR of X and Y?

  4. Here’s a sample IR spectrum of an alkyne and one of an alkene. . .

  5. Question #1 “State the infrared spectral changes you would expect to see on conversion of Compound X into Compound Y in each of the cases below” Aldehyde + Alcohol Acid + Ketone In other words  What are the main differences between the IR of X and Y?

  6. Question #1 “State the infrared spectral changes you would expect to see on conversion of Compound X into Compound Y in each of the cases below” • Characteristic Absorbances: • X: • Aldehyde: ~1730cm-1 • Alcohol: 3400-3650cm-1 (broad) • Y: • Carboxyllic acid: 3100-2500(broad), ~1730-1700cm-1 (sharp) • Ketone: 1715cm-1 (for saturated, aliphatic) Aldehyde + Alcohol Acid + Ketone In other words  What are the main differences between the IR of X and Y?

  7. Question #1 “State the infrared spectral changes you would expect to see on conversion of Compound X into Compound Y in each of the cases below” • Characteristic Absorbances: • X: • Aldehyde: ~1730cm-1 • Nitrile: 2200-2260cm-1 Y: Alcohol: 3400-3650cm-1 (broad) Amine: two peaks (weak) 3300-3500cm-1 *peaks will likely overlap In other words  What are the main differences between the IR of X and Y?

  8. Question #2 “State the number of different hydrogen “types” in each of the following molecules and predict how many peaks you would see in the proton NMR spectrum of each one” • Different hydrogen “types” refers to H’s in different chemical environments • Look for planes of symmetry (Remember that you can rotate around the axis of single bonds) • ***Special exceptions: chiral molecules can show more complex, yet predictable 1H NMR spectra

  9. Question #2

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  12. Degree of Unsaturation • Formula is based on connectivity of atoms (i.e. carbon makes 4 bonds, oxygen 2, hydrogen/halides each make 1 bond etc.) • It tells you the total number of double bonds/rings in a molecule, but nothing else about the properties of a compound

  13. Question #3 “Compound A is achiral, has a molecular formula C5H10O2, contains two methyl groups and two tertiary functional groups. It has a broad infrared absorption band in the 3250-3600cm-1 region, and no absorption in the 1620-1800cm-1 region Therefore, A must contain at least 1 double bond, or 1 ring 3250-3600cm-1 alcohol; and 1620-1800cm-1 evidence that there is no alkene!! Therefore A must contain a ring! However, the problem also states A has two methyl groups. The formula contains 5 carbons, so 5 - 2 = 3 carbons in the ring. Also, it states that there are two tertiary functional groups  so that means two tertiary alcohols.

  14. Question #3 Meso, compound: achiral (R, S) = (S, R)

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