1 Instrumental Analysis Tutorial 5

ANNOUNCEMENT FOR GROUPS having tutorial on Thursday 25 th For groups 2 and 6: Compensation for the tutorial due on Thursday 25 th October (feast day off) till take place on Saturday 20 th October, 3 rd slot in H3. For Group 1 and 7:the compensation will take place on Saturday 20 th October, second slot in the instrumental analysis lab. 2

For groups 5 and 8: a compensation for the tutorial session due on the 28 th of October (day off for feast) is scheduled on Thursday 1 st of November 1 st slot in H11. 3 ANNOUNCEMENT FOR GROUPS having tutorial on Sunday 28 th

Quiz 2 Quiz 2 will take place on Tuesday 30 th of October, from 9:15 to 9:45 Am, covering lectures 4, 5 and 6 + their tutorials. Locations are in the following slide.

Locations LocationGroup Lecture Hall H5 T 2 T 6 T 7 Biotechnology group Lecture Hall H1T 1 T 8 Exercise Room - C5.105 T 9 Exercise Room - C5.204T 3 Exercise Room - C5.206T 4 Exercise Room - C5.209T 5 Exercise Room - C5.210 T 10

6 Objectives Bond properties and Absorption trends Uses of IR spectrum Base values for absorption bands How to approach the analysis of a spectrum (what you can tell at a glance) Is a carbonyl group present? Acids, esters, amides, acid anhydrides, aldehydes, and ketones  If C=O is absent Alcohols, amines and ethers Double Bonds and/or aromatic Rings Triple Bonds Hydrocarbons Correlation chart

7 Bond properties and Absorption trends oA diatomic molecule can be considered as two vibrating masses connected by a spring (the bond). oThe frequency of vibration depends on the force constant (K) of the bond, or its stiffness, and the masses (m 1 and m 2 ) of the two bonded atoms. oWhere  is the reduced mass of the system: oK is constant that will vary from one bond to another. K triple bond = 3 K single bond K double bond = 2 K single bond K triple bond = 15 x 10 5 dyne/cmK double bond = 10 x 10 5 dyne/cm K single bond = 5 x 10 5 dyne/cm Hooke’s law

8 ONE CAN NOTICE 5 IMPORTANT OBSERVATIONS: 1.Stronger bonds have larger force constant K and will vibrate at higher frequencies than weaker bonds 2.Bonds between atoms of higher masses (larger  ) will vibrate at lower frequencies than bonds between lighter atoms. 3.Bending motions tend to be easier than stretching motions, and the force constant K is smaller. 4.Hybridization is found to affect the force constant K. bonds are stronger in the order sp > sp 2 > sp 3. The C-H stretching frequencies can explain this nicely:

9 This can be explained by the fact that As s-character increase, the electronegativity of carbon increases. So, bond is shortened and becomes stronger (larger K and higher ) 5.Resonance will also affect the strength and length of the bond, and hence its force constant K. s-character Normal ketone has its C=O stretch vibration at 1715 cm -1. Ketonic group which is conjugated with a double bond will absorb at a lower frequency near 1680 cm -1. This is because resonance lengthens the C=O bond distance and give it more single bond character. This has the effect of reducing the force constant (less than the typical value for the double bond, < 10x10 5 ), and the absorption is shifted to lower frequency.

10 Uses Of The Infrared Spectrum  Different types of bonds have different natural frequencies of vibration.  The same type of bond in two different compounds is usually in a slightly different environment.  NO TWO MOLECULES OF DIFFERENT STRUCTURE WILL HAVE EXACTLY THE SAME INFRARED ABSORPTION PATTERN OR INFRARED SPECTRUM.  Thus the infrared spectrum can be used for molecules much as a fingerprint can be used for humans. 1.the best method of infrared qualitative analysis is direct comparison of the spectrum of the analyte to spectra of standards that have been obtained under identical conditions. A compilation (library) of IR spectra of compounds, obtained with the same instrument, is normally stored as a data base and used for qualitative analysis.

Frequency (cm -1 ) =C-H 2.More important use of IR spectrum is that it gives structural information about a molecule.  The absorption of each type of bond (N  H, C  H, O  H, C  X, C=O, C  O, C  C, C=C, C C, C N, etc.) are regularly found in certain small portions of the vibrational infrared region.  It is recommended at first to establish the “broad patterns” given below. Then as a second step a “typical absorption value” can be “memorized” for each of the functional groups in this pattern. str. Olefinic aromatic C-H str. aliphatic C-H str. O= Aldehyde Fermi resonance C C N C=C C=N C=C aromatic N-H bend C  C C  O C  N Str. O  H C  H bend Fingerprint Region C-H out of plane bending Aromatic compound O  H N  H Bond str. acid chlorides anhydrides esters ketones aldehydes carboxylic acids amides C=O Regions where various common types of bonds absorb (Base values)

12

Reminder for IR solving of tutorial 5

Functional groups Ester Ether Amide

Functional groups Aldehyde Ketone Acid

Functional groups Alcohol : Acid anhydride

Functional groups Primary amine: RNH 2 Secondary amine: R 2 NH Aliphatic C-H: -C-H ( SP 3 ) Olefenic/Aromatic C-H: =C-H- (SP 2 ) Acetylenic C-H: ≡ C-H (SP)

Don’t forget that the wavenumber of vibration is calculated by Hook’s law

19 Fingerprint Region ( cm -1 )  A small difference in the structure and skeleton of a molecule result in a significant change in the distribution of absorption peaks in the fingerprint region of the IR-spectrum.  A compound can be positively identified by comparing its fingerprint region with the fingerprint region of the spectrum of known sample of the compound.  Exact interpretation of individual peaks of the spectra in this region is seldom possible because of the complexity of the spectra in this region.

20 Figures illustrate the unique character of IR spectra, especially in the fingerprint region. The two molecules differ just in the position of O-H group, yet the two spectra differ dramatically in appearance in the fingerprint region. The fingerprint region The functional group region Fingerprint region is characteristic of the compound as a whole

21 How to approach the analysis of a spectrum (what you can tell at a glance) In trying to analyze the spectrum of an unknown, you should concentrate your first efforts toward determining the presence (or absence) of a few major functional groups. The C=O, O  H, N  H, C  O, C=C, C C, C N peaks are the most noticeable and give immediate structural information if they are present.

22 Scheme for identification of different functional groups in a compound: 1.Is a carbonyl group present? The C=O group gives rise to a strong absorption in the region cm -1. The peak is often the strongest in the spectrum and of medium width. You can’t miss it.

23 2.If C=O is present, check the following types (if absent, go to 3). ACIDSis O-H also present? - broad absorption near cm -1 OH stretch

24 ESTERSis a C-O also present? - strong intensity absorption near cm -1.

25 AMIDESis N-H also present? - medium absorption near 3500 cm -1. Sometimes a double peaks, with equivalent halves (symmetric and asymmetric N-H stretching vibrations). N-H stretch (pair) C=O

26 ANHYDRIDEhave two C=O absorptions near 1810 and 1760 cm -1 Symmetric stretching Asymmetric stretching

27 CH 3 -C=O(X) strong & independent 1681 CH 3 C O NH CH 3 C O H 1715 CH 3 C O CH 3 CH 3 C O OCH CH 3 C O Cl + - : C=O weakened by resonance electron donation C=O strengthened by Inductive electron withdrawal

28 Aldehyde is aldehyde C-H present? Two weak absorptions near 2850 and 2750 cm -1 (Fermi resonance) on the right hand side of C-H absorptions. Fermi resonance can occur when the frequency of one of the ground vibrations in a molecule is close to the frequency of overtone of other oscillation of the same molecule and respective oscillations are mechanically coupled with each other. The C-H stretch is in Fermi resonance with the first overtone of the C-H bending motion of the aldehyde. The normal frequency of the C-H bending motion of an aldehyde is at 1390 cm -1. As a result of this interaction, one energy level drops to ca and the other increases to ca cm -1. Ketones the five choices have been eliminated

29 3.If C=O is absent ALCOHOLS - check for O-H PHENOLS - broad absorption near cm confirm this by finding C-O near cm -1.

30 AMINES- check for N-H - medium absorption(s) near 3500 cm -1. primary amine (two N-H peaks)

31 ETHERS check for C-O (and absence of O-H) near cm -1 Secondary amine (one N-H peak) C-O

32 4.Double Bonds and/or aromatic Rings - C=C is a weak absorption band near 1650 cm medium to strong absorptions in the region cm -1 often imply an aromatic ring - confirm the above by consulting the C-H region; aromatic and vinyl C-H occur to the left of 3000 cm -1 (aliphatic C-H occurs to the right of this value) Alkene

33 Mono- substituted (two peaks near each other at cm -1 The infrared spectrum reveals substitution patterns in benzene derivatives.

34 5.Triple Bonds - C N is a medium, sharp band absorption near 2250 cm -1 - C C is a weak but sharp absorption near 2150 cm -1 - Acetylenic C-H stretching is near 3300 cm -1.

35 6.Hydrocarbons - None of the above are found. - major absorption just above 3000 cm -1 -to the right of 3000 cm -1 -(C-H stretching for aliphatic compounds) - very simple spectrum. C-H stretching

Refreshing question How to know whether the band just to the left of the 3000cm -1 is for olefinic or aromatic C-H stretching !!! Remember the OOP bands ( Out Of Plane) appears in the fingerprint region below 1000 cm -1

37 Important IR Stretching Frequencies BondBase ValueStrength / ShapeComments 1 C=O 1715 (typical) s, "finger"Exact position depends on the environment of carbonyl 2 O-H alcohol s, brd Broad due to H bonding 3 O-H acid s, very brd 4 N-H m Can tell primary from secondary 5 C-H Just to the right of 3000 s Aliphatic, sp 3 Just to the left of 3000 s Olefinic or aromatic, sp s Acetylenic, sp w, Fork like Fermi resonance of aldehydic group 6 C-O s Also check for OH and C=O 7 C=C 1650 w alkene m-s aromatic Alkene w due to low polarity Aromatic usually in pairs 8 C≡C 2150 w, sharp Characteristic since little else around it 9 C≡N 2250 m, sharp Characteristic since little else around it