1 / 32

Ultraviolet Spectroscopy

Ultraviolet Spectroscopy. Chapter 16, Smith (Pages 595-597). What is our approach to spectroscopy?. We will learn how to get the essential information from four kinds of spectroscopy and combine that information to deduce the structure of a compound.

rendor
Download Presentation

Ultraviolet Spectroscopy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Ultraviolet Spectroscopy Chapter 16, Smith (Pages 595-597)

  2. What is our approach to spectroscopy? • We will learn how to get the essential information from four kinds of spectroscopy and combine that information to deduce the structure of a compound. • The first kind is Ultraviolet Spectroscopy (UV).

  3. What is the most important thing to remember about spectroscopy? • That “spectroscopic data” is a physical property of molecules and that properties of molecules are a direct result of the structure of those molecules. • Thus, various forms of spectroscopy will give us “data” that reveals a little bit (i.e., partial structures) about the structure of a compound.

  4. How will we use this data? • We will correlate the spectroscopic data with specific partial structures. • We will find as many partial structures as possible from the data and then combine all of the partial structures into a complete structure of the only compound whose structure could produce the spectra we interpreted.

  5. What kind of data do we get? • In general, we get numbers (units). • Specific numbers reveal specific partial structures. • For example, the number 3400 cm-1 in an IR spectrum reveals the -OH group as a partial structure. • The number 3 over a signal at d = 2.0 in a 1H NMR spectrum reveals a methyl group next to a carbonyl group.

  6. Must I memorize dozens of numbers? • No, we hope to see enough examples so that you will become so familiar with the numbers that you know them much as you know your telephone number--not because you memorized the numbers but because you use them frequently enough that you simply remember them.

  7. What kinds of numbers are important for UV? • A constant with the Greek symbol epsilon (e), which tells us how much UV light a given compound absorbs. • Epsilon (e) is the molar absorptivity or molar extinction coefficient. • The value of e can vary from zero (no absorption of UV) to over 10,000 (very strong absorption)

  8. How do we characterize various values of epsilon? • Strong (e > 1000) • Weak (e < 100) • End absorption (e can’t be determined) • No absorption (e = 0)

  9. How do we use these four classifications of UV? • We associate or correlate partial structures with each of the four categories. • The next few slides will show these correlations.

  10. Partial structures:Strong UV (e > 1000) • Conjugated p systems--(alternating double bond--single bond--double bond) give e > 1000.

  11. If e > 1000, what kind of partial structure is indicated? • The compound MUST CONTAIN a conjugated p system. • It might be aromatic (like benzene). • It might be a conjugated polyene (like 1,3-butadiene).

  12. If a compound contains a conjugated p system what is e? • e must be > 1000

  13. Problem • A compound has a strong UV and the calculation of p + r for the compound is three. • What partial structure is present in the compound?

  14. Problem • A compound (C7H8) has e = 15,000. Draw the structure of the compound.

  15. Partial structures:Weak UV (e < 100) • Aldehydes or Ketones give e < 100

  16. If e < 100, what kind of partial structure is indicated? • The compound must have an aldehyde or ketone carbonyl. • There is no heteroatom next to the carbonyl carbon (only C or H).

  17. Given an aldehyde or ketone, what is e? • e < 100

  18. Problem • A compound (C4H8O) has e < 100. Draw two possible structures for the compound.

  19. Partial structures:UV End Absorption (No e) • Esters give end absorption.

  20. IfUV = end absorption, what partialstructure is indicated? • A carbonyl group with an Oxygen atom next to the carbonyl carbon is indicated. • An ester is a carbonyl with an O but not OH (acid) next to a carbonyl.

  21. Problem • A compound (C2H4O2) shows end absorption in its UV spectrum, what is its structure?

  22. Partial structures:No UV (inactive) • For CHM 201: UV = Wavelength l = 200-400 nm, the range of a typical instrument. • The UV absorption in such compounds as ethene lies outside the above region. • All partial structures, not covered in our strong, weak or end absorption categories will be considered inactive in the UV. (A ground rule, because of 200-400 nm)

  23. How is UV absorption measured? • A sample of a given concentration (C) in moles/liter = molarity, is placed in a 1 cm wide cuvette (pathlength L), and ultraviolet light, starting at a wavelength (l) of 400 nm and scanning down to 200 nm is passed thru. • The absorbance (A) is plotted vs wavelenth (l). • A is proportional to C x L for a compound that absorbs UV light.

  24. What info is recorded by the analyst? • The plot of A vs. l will have one or more maxima or lamda max (lmax). • The A value at each lmax is recorded. • Since A is proportional to C x L, the proportionality can be made an equation by inserting a proportionality constant e. • A = eCL (e is calculated and recorded for each lmax).

  25. Absorbance • The incident UV light = 1.00 • The compound absorbs some of it, say 80%. • The absorbance A is recorded as 0.80, because the compound absorbed 80% of the incident radiation. • A has no units; it’s a decimal.

  26. A typical calculation • Data: A = 0.815, L = 1 cm; C = 5.43 x 10-5 M; and l = 275 nm; solvent = cyclohexane • A = eCL; e = A/CL • e = (0.815)/(5.43 x 10-5 M)(1 cm) = 15,000 (no units); reported as e = 15,000, lmax = 275 nm (cyclohexane) • What kind of partial structure is indicated?

  27. How does UV work? • Loosely held p electrons in a given molecular orbital are excited into a higher energy molecular orbital by UV energy.

  28. p* Orbitals • An electron moves (transitions) from one orbital to an empty (unoccupied) orbital that normally is empty. • The name of the orbital where the electron winds up is always called p*. • The name of the orbital where the electron starts out is either a p or an n (n stands for nonbonding)

  29. Strong UV vs. Weak UV • Strong UV = p to p* transition • Weak UV = n to p* transition

  30. Molecular Orbital Theory • MO theory is a complementary theory to Valence-Shell-Electron-Pair-Replusion or VSEPR theory. • Every time two atomic orbitals overlap to make a bond, they make two new orbitals—a bonding and an antibonding orbital.

  31. Molecular Orbital Theory • Electrons normally fill bonding orbitals. • The antibonding orbitals (*) are normally empty of electrons, because they are of high energy and there is no net bonding in these orbitals. • The theory is most applicable to p bonds in organic chemistry.

  32. Summary • Strong UV, e > 1000, (conj p sys) mostly aromatic (like benzene) • Weak UV, e < 100, (aldehyde or ketone) • End Absorption, no e (ester) • No UV (everything else) • e = constant (defined, no units)

More Related