Comments on homework Lots of “Index of Hydrogen Deficiency” Math errors Matching arrows with HR MS data & Molecular formulas Following instructions: -typed & chemdraw work -Not applying rules for fragmentation -Not providing mechanisms Poor mechanism presentations (Review arrow pushing) Some organizational problems in “presentation” of homework

Parent ion: m/z 150

1.6 Match each of the exact masses to the following mass spectra. (A–W) Note that two compounds have the same exact mass, and you will need to consider the CI mass spectrum when given: (a) 56.0264, (b) 73.0896, (c) 74.0363, (d) 89.0479, (e) 94.0535, (f) 96.0572, (g) 98.0736, (h) 100.0893, (i) 102.0678, (j) 113.0845, (k) 114.1043, (l) 116.0841, (m) 116.1206, (n) 122.0733, (o) 122.0733, (p)126.1041, (q) 138.0687, (r) 150.0041, (s) 152.0476, (t) 156.9934, (u) 161.9637, (v) 169.9735, (w) 208.0094.

(r) 150.0041 -78.9183 = 71.0858

71.0858 C5H11

C5H11Br IHD = 0 Therefore it is a pentylbromide isomer A propyl radical [M –Br] or [M -79] [M - 43] [M - 29] m/z 71 = m/z 107 = C2H4Br+ m/z 121 = C3H6Br+

Fragmentation mechanisms must be consistent with valences of atoms -radical & cation chemistry

Some helpful hints: Draw hydrogens in. Use proper arrow pushing –exactly like in mechanisms. Gas phase radical cations have plenty of energy to rearrange and fragment.

M-1 m/z 39, C3H3+ [M -17] or [M-OH] M M+1

M-1 m/z 39, C3H3+ [M -17] or [M-OH] M

Chapter 2: Infrared Spectroscopy

Duality wave/particle of the light Electromagnetic Energy: Light E = hυ h: Planck constant h = 6.626x10-34 J.s υ=c/λ Duality wave/particle of the light

Infrared Spectroscopy The vibrational IR extends from 2.5 x 10-6 m (2.5 m) to 2.5 x 10-5 m (25 m). The frequency of IR radiation is commonly expressed in wavenumbers. Wavenumber: The number of waves per centimeter, with units cm-1 Expressed in wavenumbers, the vibrational IR extends from 4000 cm-1 to 400 cm -1

Molecular vibrations Fundamental stretching and bending vibrations for a methylene group.

Molecular Vibrations Consider two covalently bonded atoms as two vibrating masses connected by a spring. The total energy is proportional to the frequency of vibration; E = hn where h is Planck’s constant. The frequency of a stretching vibration is given by an equation derived from Hooke’s law. K = a force constant, which is a measure of bond strength, m = reduced mass of the two atoms, (m1m2)/(m1 + m2), where m is the mass of the atoms in amu.

Potential energy diagrams. Curve 1, harmonic oscillator Potential energy diagrams. Curve 1, harmonic oscillator. Curve 2, anharmonic oscillator

Molecular Vibrations From this equation, we see that the peak position of a stretching vibration is proportional to the strength of the vibrating bond. is inversely proportional the masses of the atoms connected by the bond. The intensity of absorption depends primarily on the polarity of the vibrating bond.

i)   Stretching frequencies are higher than corresponding bending frequencies. (It is easier to bend a bond than to stretch or compress it.) ii)   Bonds to hydrogen have higher stretching frequencies than those to heavier atoms. iii)   Triple bonds have higher stretching frequencies than corresponding double bonds, which in turn have higher frequencies than single bonds.

Comparison Beetween Dispersion Spectrometer and FTIR To separate IR light, a grating is used. Detector Grating Slit In order to measure an IR spectrum, the dispersion Spectrometer takes several minutes. Also the detector receives only a few % of the energy of original light source. Sample To select the specified IR light, A slit is used. Light source Fixed CCM An interferogram is first made by the interferometer using IR light. FTIR In order to measure an IR spectrum, FTIR takes only a few seconds. Moreover, the detector receives up to 50% of the energy of original light source. (much larger than the dispersion spectrometer.) Detector B.S. Sample Moving CCM The interferogram is calculated and transformed into a spectrum using a Fourier Transform (FT). IR Light source

Figure 16-9 Single- and double-beam spectra of atmospheric water vapor and CO2. In the lower, single-beam trace, the absorption of atmospheric gases is apparent. The top, double-beam trace shows that the reference beam compensates nearly perfectly for this absorption and allows a stable 100% T baseline to be obtained. (From J. D. Ingle Jr. and S. R. Crouch, Spectrochemical Analysis, p. 409. Englewood Cliffs, NJ: Prentice-Hall, 1988. With permission.)

IR Source  Sources of continuous radiation  A hot material emits a continuum of radiation. Blackbody (no envelope): intensity highest near 5000 cm-1; about 100 times lower near 500 cm-1. a) Nichrome coil heated electrically to 1100oC and a black oxide film forms. Simple, robust, reliable, long lifetime.

Figure 16-4 Specpral distribution of energy from a Nernst glower operated at approximately 2200 K.

Mirror moves to create path difference between 2 beams Interferometer produces plot of intensity vs time during 1 scan (interferogram).Interference of beams occurs from fixed and moving mirrors. Mirror moves to create path difference between 2 beams Transmits 50%, reflects 50% At t=0, zero path difference between 2 beams for monochromatic radiation; constructive interference; maximum signal at detector.At a later time, path difference = /2; destructive interference; minimum signal. Later on, …. …maximum signal. So interferogram is a cosine wave.

Interference is a superpositioning of waves FTIR seminar Relationship between light source spectrum and the signal output from interferometer Light source spectrum I Signal output from interference wave Az Monochromatic light Dichroic light Continuous spectrum light Wavenumber u Time t S I SAz Wavenumber u Time t I(t) b (u) u Time t Wavenumber All intensities are standardized.

Centreburst grows when more frequencies are present One frequency Three frequencies Many frequencies

Peak width When band is broader, interferogram wings decay faster: more frequencies give more chance of cancellation

Relation between spectrum and interferogram. 3 interferogram parameters important: A, P, T

Fourier Transform Time axis by FFT Wavenumber Single strength SB 4000 Wavenumber[cm-1] 400 Optical path difference [x] (Interferogram) (Single beam spectrum) Time axis by FFT Wavenumber

End of scan from (i) white light source or (ii) centreburst Schematic of steps in spectrum collection End of scan from (i) white light source or (ii) centreburst Background spectrum

This only comes from aliasing and can be prevented. 5. Resolution Advantages of FT-instrument over dispersive one Fellgett advantage All frequencies are measured simultaneously. Typical scan times are only a few seconds. 2. Jacquinot advantage The energy throughput is higher for any resolution, giving a higher signal:noise ratio. 3. Connes advantage The laser wavelength is used as a reference for the calculation of band positions, and is precise. 4. Stray light This only comes from aliasing and can be prevented. 5. Resolution This is constant for the whole spectral range 6. Robustness FT instruments only have 1 moving part

Instrument scanning Signal: noise ratio, S/N  (measurement time)0.5 S/N  (no. of scans)0.5 How many scans do I need to reduce the noise in 1 scan by a factor of 4? Often 1 scan of sample is ratiod against 1 scan of (empty) background. In the ranges 70-35%T and 0-35%T normally the ratio of background:sample scans is increased to 1:2 and 1:4 respectively. When the energy throughput is reduced by a factor of x for the sample spectrum, x times more scans are required.

Resolution

Sample preparation Methods Transmission Solids: KBr Pellet Liquids: NaCl Plates Quick press KBr pellet press

Instrumental Setup: Attenuated Total Reflectance (ATR) Technique IR radiation passes through (IRE-internal reflection element) crystal and hits sample at a 45 degree angle IRE made of high refractive index material (zinc selenide, diamond, germanium Incident radiation penetrates into sample (~1 micrometer) where it can be absorbed Attenuated radiation is reflected

Sample preparation Methods ATR Liquids and solids loaded directly onto crystal Arm Applies pressure to solids for uniform contact with crystal PSI can be controlled

Transmission vs. ATR ATR Advantages High Quality Spectrum for qualitative analysis Minimal sample preparation Non destructive Time efficient Spectra not affected by sample thickness Radiation penetrates only a few micrometers Highly reproducible results Wide variety of sample types Threads, yarns, fabrics, fibers, pastes, powders, suspensions, polymers, rubbers

ATR forensic applications Drug analysis Fiber analysis Paint chip analysis Ink analysis Paper analysis Biological analysis

Spectra Comparison Resulting peaks from ATR are very similar in intensity and wavelength to transmittance technique Koulis, Cynthia, et. al. Comparison of Transmission and Internal Reflection Infrared Spectra of Cocaine. Journal of Forensic Sciences, 2001.

ATR Peak Shift Small variations in peak intensity and position occur: Carbonyl band Absorption of cocaine shows ATR peak shift Koulis, Cynthia, et. al. Comparison of Transmission and Internal Reflection Infrared Spectra of Cocaine. Journal of Forensic Sciences, 2001.

C-H Stretch occurs around 3000 cm-1 C-H Stretch occurs around 3000 cm-1. In alkanes (except strained ring compounds), sp3 C-H absorption always occurs at frequencies less than 3000 cm-1 (3000-2840 cm-l). Methylene groups have a characteristic bending absorption of approximately 1465 cm-1. Methyl groups have a characteristic bending absorption of approximately 1375 cm-1. The bending (rocking) motion associated with four or more CH2 groups in an open chain occurs at about 720 cm-1 (called a long-chain band).

=C-H Stretch for sp2 C-H occurs at values between 3100-3010 cm-1. =C-H Out-of-plane (oop) bending occurs in the range 1000-650 cm-1. Stretch occurs at 1660-1600 cm-1; often conjugation moves C=C stretch to lower frequencies and increases the intensity. Symmetrically substituted bonds (e.g., 2,3-dimethyl-2-butene) do not absorb in the infrared (no dipole change). Symmetrically disubstituted (trans) double bonds are often vanishingly weak in absorption; cis are stronger.

Alkene isomers =CH bend =CH bend

Aromatics C–H stretch from 3100-3000 cm-1 overtones, weak, from 2000-1665 cm-1 C–C stretch (in-ring) from 1600-1585 cm-1 C–C stretch (in-ring) from 1500-1400 cm-1 C–H "oop" from 900-675 cm-1

Alkynes

Alcohols O–H stretch, hydrogen bonded 3500-3200 cm-1 C–O stretch 1260-1050 cm-1 (s) 

http://www2. chemistry. msu http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

equatorial C-O absorption at 1068 is relatively stronger than the axial C-O absorption at 970 http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

Carbonyls Conjugation with a double bond or benzene ring lowers the stretching frequency. http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

Carbonyls Changing an alkyl substituent of a ketone for an electron releasing or withdrawing group. Stronger CO Weaker CO Incorporation of the carbonyl group in a small ring (5, 4 or 3-membered), raises the stretching frequency. http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

Carbonyls C-CO-C bond angle Incorporation of the carbonyl group in a small ring (5, 4 or 3-membered), raises the stretching frequency & makes CO stronger. 120° CO sigma bond becomes more s-orbital rich 108° 90° http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

1715 cm1 in six-membered ring and acyclic ketones C=O in Ketones 1715 cm1 in six-membered ring and acyclic ketones 1750 cm1 in 5-membered ring ketones 1690 cm1 in ketones next to a double bond or an aromatic ring 1735 cm1 in saturated esters 1715 cm1 in esters next to aromatic ring or a double bond C=O in Esters http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

Aldehydes H–C=O stretch 2830-2695 cm-1 C=O stretch: aliphatic aldehydes 1740-1720 cm-1 alpha, beta-unsaturated aldehydes 1710-1685 cm-1 

Esters C=O stretch aliphatic from 1750-1735 cm-1 a,b-unsaturated from 1730-1715 cm-1 C–O stretch from 1300-1000 cm-1 

Esters

Amides

Carboxylic acids O–H stretch from 3300-2500 cm-1 C=O stretch from 1760-1690 cm-1 C–O stretch from 1320-1210 cm-1 O–H bend from 1440-1395 and 950-910 cm-1 

Carboxylate

Acid Anhydrides Two C=O and two C-O stre http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

Acid Anhydrides Two C=O and two C-O stre http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed

Amines C-N stretching Aromatic amines 1200 to 1350 cm-1 http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/InfraRed C-N stretching Aromatic amines 1200 to 1350 cm-1 Aliphatic amines 1000 to 1250 cm-1

Nitro compounds N–O asymmetric stretch from 1550-1475 cm-1 N–O symmetric stretch from 1360-1290 cm-1 

Halogenated organics C–H wag (-CH2X) from 1300-1150 cm-1 C–X stretches (general) from 850-515 cm-1 C–Cl stretch 850-550 cm-1 C–Br stretch 690-515 cm-1

Infrared-spectroscopical Polymer Identification 1790-1720 very strong 1610-1590, 1600-1580 and 1510-1490 Modif. Epoxies Polycarbon- ates Alkyd-, Polyesters, Cellulose ether, PVC(plasticized) Polyvinyl acetate, PVC-copolymers Cellulose ester Polyurethane Acrylics, Polyester Phenol derivatives, Polystyrenes, Arylsilicones, Aryl-alkyl Silicone Copolymers Polyamides, amines Nitrocellulose cellophane Cellophan, Alkylcellulose, PVA, PEO PAN, PVC, Polyvinylidene chloride POM Alkylsilicone, aliphatic hy drocarbons, Polytetra fluorethylene Thiokol 1450 -1410 sharp 1680 - 1630 strong 1550 - 1530 1610 –1590, 1600 – 1580 and 1510 - 1490 3500 - 3200 1100 - 1000 1450 - 1410 840 - 820 All numbers have the meaning of wave numbers and are given in cm-1 yes no Infrared-spectroscopical Polymer Identification

1610-1590 1600-1580 cm-1 1510-1490 820-840 cm-1 1790-1720 cm-1 modified epoxides, polycarbonate, Alkyd resins, polyester, cellulose ester, cellulose ether, PVC (plast), PVAc, PVC-copolym., PU, acrylics modified epoxides, polycarbonate, Alkyd resins, polyester, cellulose ester, cellulose ether, PVC (plast) modified epoxies, polycarbonate polycarbonate

Determination of a Layer Thickness Intensity, arbitrary units wave length 1/d= 2/n (1/l1 -1/l2) n = number of minima between two maxima l1 and l2

Figure 16-1 Infrared absorption spectrum of a thin polystyrene film recorded with a modern infrared spectrophotometer. Note that the abscissa scale changes at 2000 cm-1. Usually wavenumber is used: directly proportional to energy;