# Lecture Topics: I. IR spectroscopy

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1 IR and Mass Spectrometry Reading: Wade chapter 12, sections Study Problems: 12-15, 12-16, 12-23, Key Concepts and Skills: Given an IR spectrum, identify the reliable characteristic peaks Explain why some peaks are strong and some are weak or absent.. Identify functional groups by IR spectra. determine molecular weights and molecular structures from mass spectra Lecture Topics: I. IR spectroscopy Absorption spectroscopy involves the measurement of the amount of light absorbed by a compound as a function of its wavelength. Infrared spectroscopy observes the vibrations of bonds and provides evidence for the presence of certain functional groups within a structure. Specifically, the infrared region of the electromagnetic spectrum has radiation energetic enough to cause the stretching and bending of bonds in molecules. Note: infrared light does not cause electronic transitions to take place. Excitation of electrons in pi bonds is observed in UV-visible spectroscopy; excitation of electrons in sigma bonds require far UV and X-ray photons, which are much more energetic than infrared photons Remember: E=hν Where frequency ν= c/λ In an IR spectrum, we are measuring % transmission of light versus wavenumbers, ν, which is the reciprocal of wavelength (note that wavenumbers are proportional to frequency): ν (cm -1 )= 1/λ The wavenumber range of an IR spectrum is 4000 cm -1 to 600cm -1 or 2,500 nm to 16,000 nm. Note that the higher the wavenumber, the shorter the wavelength and the greater the energy of the photon causing the bond vibration sample I o I hν 100% % transmittance 0% increasing energy of photons 4000 ν (cm -1 ) common functional groups fingerprint region

2 The frequency of the stretching vibration depends on the masses of the atoms making up the bond and also the stiffness (strength) of the bond eavier atoms vibrate more slowly than lighter ones, and thus the frequency of stretching vibration decreases with increasing atomic weight: bond stretching frequency C 3000 cm -1 C D 2100 cm -1 C C 1200 cm -1 Thus, less energetic photons are required to stretch a C-C bond than a C- bond Stronger bonds are stiffer and thus require more force (requiring more energetic photons) to stretch and compress those bonds; the frequency of light required to stretch a bond increases with bond energy: bond stretching frequency C C 1200 cm -1 C=C 1660 cm -1 C C 2200 cm -1 C N 1200 cm -1 C=N 1650 cm -1 C N 2200 cm -1 C 1100 cm -1 C= 1700 cm -1 The simple stretching vibrations (in the 1600 cm -1 to 3500 cm -1 region) are the most characteristic and predictable, and absorptions in this region are used to identify functional groups in molecules. The region of an IR spectrum below 1600 cm -1 is called the fingerprint region and contains a unique set of absorptions (yet less predictable) for a given molecule; this region can be used for identifying unknown substances by comparison with spectra of known compounds. IR active bonds/functional groups For bonds to be IR active (that is, to absorb a photon of infrared light), they must possess a dipole moment, and furthermore, bond vibrations must change the dipole moment of the molecule. The electromagnetic field of an IR photon must be able to interact with a bond in order for absorption of energy to occur. Because one of the components of an electromagnetic wave is a rapidly reversing electric field, there can only be an interaction with a bond when the bond has a dipole moment. Bonds which have little or no dipole moment give rise to extremely weak (or absent) absorptions in infrared spectra. This is seen most frequently with internal alkynes:

3 3 C 3 C 2 C C 3 C 3 IR inactive 3 C IR active ydrocarbons For hydrocarbons, we can most readily observe C=C (1660 cm -1 ) and C C (2200 cm -1 ); the C-C stretch is usually buried in the fingerprint region (1200 cm -1 ) and so is not diagnostic. Conjugation generally lowers the absorption frequency, since electron delocalization in conjugated systems (dienes, trienes, aromatic rings) leads to weaker bonds: 1600 cm cm -1 For the C- stretching frequency, we find a rough correlation with the strength of the chemical bond: Csp- bonds are stronger (more tightly held electrons) the Csp 2 - bonds which are stronger than Csp 3 - bonds. bond stretching frequency C C 2900 cm cm cm -1 Again, from IR spectra we get an idea of bond strengths: compare C bonds (100 kcal/mol) at 2900 cm -1 to C-C bonds (80 kcal/mol) at 1200 cm -1. Alcohols and Amines The bond stretch in the IR spectrum is one of the most recognizable absorptions. It occurs at around 3300 cm -1 and is broad due to hydrogen bonding. Note that the C- stretch is buried in the fingerprint region ( cm -1 )

4 R 3300cm -1 broad cm -1 broad - bonding in acids RN 3300 cm -1 broad, with two spikes for 1 amines one spike for 2 amines Amines also show weaker absorptions in this region (3300 cm -1 ), with a sharp spike for each N bond present. Note that tertiary amines (R 3 N) do not give rise to N stretching absorptions, since they have no N s. Carbonyl compounds The carbonyl C= of ketones, aldehydes, and carboxylic acids have absorption frequencies around 1710 cm -1. The aldehyde C- absorbs at cm -1. Ester C= s have slightly higher frequencies of absorption at 1735 cm -1 ; strained ring ketones such as cyclobutanone also show higher frequency C= absoprtions (1785 cm -1 ). In contrast, C= bonds in conjugated ketones, aldehydes and carboxylic acids (which because of resonance have weaker C= bonds) absorb at lower frequencies ( cm -1 ). Amide carbonyls, which also display strong resonance involving the nitrogen lone pair, absorb at 1640 cm cm cm cm cm cm cm cm -1 R 1735 cm -1 N 2 - N cm cm -1 amides also show N spikes, 3500 cm -1 Note: the greater the dipole moment of a bond, the stronger the absorption observed. Consider the following trends

5 Summary Bond stretching frequency C-N 1200 cm -1 C=N 1600 cm -1 C N 2200 cm -1 strong C C 2200 cm -1 weak or absent: cm -1 C=C, C=, C=N cm -1 C C, C N cm -1 C-, -, N- Utility: IR spectroscopy indicates the functional groups in an organic compound Can confirm the identity of a compound by comparison with a known sample, taking note of the fingerprint region Says nothing about the connectivity of the carbon skeleton of the molecule or stereochemistry. 2. Mass spectrometry Provides the molecular weight and valuable information about the molecular formula using a very small amount of sample. Not absorption spectroscopy! In mass spectrometry, molecules are ionized in a vcuum, the ions are sorted according to their masses, and the abundance of each mass is recorded In electron- impact (EI-MS), electrons are used to bombard molecules, resulting in ejection of two electrons to form the molecular ion (M + ). The molecular ion may undergo fragmentation to produce multiple lower-mass ions. M + e - ==> [M] + 2e - [M] fragmentation to other ions 100% base peak-largest in spectrum assigned 100% abundance abundance M + - usually highest m/z value in spectrum fragments m/z (mass to charge ratio)

6 The ions are separated by magnetic deflection passing through a magnetic field; the curvature of the ion s path depends on the mass/charge ratio (m/z); the vast majority of ions are +1 (only positively charged fragments are detected by the mass spectrometer), so the ions are curved on the basis of their mass. The fragmentation products are often easily identifiable. Since the molecular ion is a radical cation, it fragments to produce radicals and cations; only the cation fragments (with odd mass numbers) are detected by the spectrometer. In general cation and radical stabilities help to explain the mass spectra of linear and branched alkanes; generally the more stable cation is produced in a fragmentation reaction. For allylic and benzylic systems, the allylic and benzylic cations are easily recognized fragments; for alcohols, splitting off of water (-18) is usually observed, producing another radical cation (even number), which can further split to form allylic cations (odd number) + C 3 odd number not detected odd mass number + C 3 not detected more likely odd mass number + more stable than methyl radical not detected + C 3 C 2 allylic cation- odd mass number + C 3 delocalized bezylic cation odd mass number not detected odd mass number even mass number + C 3 C 2

7 GC-MS A GC-MS is anistrument which first passes the sample through a gas chromatography (GC) column, allowing for separation of the compounds in a mixture prior to injection into the mass spectrometer. Determination of molecular formula. igh resolution mass spectrometry detects particle mass to 1 in 20,000 and provides an exact mass for the molecular ion. ne can then compare the exact mass with masses calculated from various molecular formulas. For example, if high-resolution analysis gives an exact mass of , the following information allows us to conclude that out molecular formula is C 2 4 : mass 12 C N C 3 8 C 2 4 C 2 CN 2 4 Calculated mass: eavy isotope peaks Further information can be gleaned from analysis of heavy isotope peaks (M+1, M+2), whose abundance depends on the natural abundance of the isotope. The presence of Br, Cl, I, S, and N can all be distinguished based on peaks close to M + : Summary Compounds containing have Br M+2 as large as M + Cl M+2 1/3 as large as M + I a peak at 127 for I + N an odd number for M + S M+2 4% of M + Mass spectrometry allows us to determine the molecular weight of a substance; highresolution mass spectrometry allows us accurately to determine the molecular formula of a substance Analysis of heavy isotope peaks in the mass spectrum (M+1, M+2) signals the presence of I, Cl, Br, and S. An odd molecular weight signals the presence of an odd number of nitrogens in a molecule (trivalent) Additional Problems for practice:

8 1. match the following IR spectra with the compounds listed below: 1,3-cyclohexadiene Diphenylacetylene 1-octene 2-pentene 2. match the following IR spectra with the compounds listed below butyl acetate butyramide isobutyl amine lauric acid 3. Match the following IR spectra with the compounds listed below C 4. The mass spectrum of 2-butenal shows a peak at m/z 69 that is 28.9% as intense as the base peak. Propose at least one fragmentation route to account for this peak, and explain why this fragment would be reasonably stable 5. The exact mass of a compound determined by high-resolution mass spectra is Given the following table, what is the molecular formula? Given the spectrum below, what is the structure of the compound?

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