Unit One Part 10:infrared spectroscopy andmass spectrometry my last lecture...yippee
10Unit OnePartinfrared spectroscopy?What isFunctional groups & IRMass spectrometry
determining structureUV uv-vis nmr irconjugation C–H functional groups
determining structureUV three common forms of spectroscopy, all used to uv-vis nmr ir look at molecules (and determine their structure)...conjugation C–H functional groups
determining structure UV three common forms of spectroscopy, all used to uv-vis nmr ir look at molecules (and determine their structure)...conjugationlast lecture...not did nmr C–H functional going to look at uv so that leaves ir and... groups
determining structure O CH3H3C N N H O N N CH3 Mass: caffeine 194.08 C8H10N4O2 we’ll also look at mass spectrometry (not a form of spectroscopy, but still useful!)
look at infraredspectroscopy ﬁrst what is infrared spectroscopy?
infrared spectroscopy molecule in excited energy state E2 energy energy basically we shine energy infrared radiation on a molecule and some of it is... energy molecule in energy state E1
infrared spectroscopy molecule in excited energy state E2 energy energy is absorbed, exciting the molecule...we measure what comes energy through and note what has been absorbed (unlike nmr when we measured what was emitted) energy molecule in energy state E1
what energy is absorbedby the molecule?
atom 1 atom 2two atoms connected by bonda bond can be thought to be the same as...
spring ...two masses / blobs joined by a spring...mass 1 mass 2
bonds bonds vibrate (stretching andcontracting) in the vibrate same way...
likesprings ...a spring would...
...this means we can use Hooke’s law on extension and elasticity...Hooke’s law
...of course, I hate maths so I’m not actually going to show you Hooke’s law but just tell you what it reveals...Hooke’s law
energy to vibrate bond depends on... spring the force needed to vibrate a bond / spring depends on the strength of the bond / spring and... mass 1 mass 2 strength
energy to vibrate bond depends on... spring ...the difference in mass of the two ends (or atoms)...this means... mass 1 mass 2difference in mass
strong (short) bond requires......so it is easy to seemultiple bonds in IR... more energy
light (hydrogen) atom vibrates faster ...and bonds to hydrogen (C–H, O–H etc) more energy
bonds only vibratewith fixed frequencies and... bonds vibrate with certain frequencies
...they will only absorb energy of the same frequency or wavelength... λ bond will only absorbenergy of same frequency
energy / lightlong wavelength short wavelength low frequency high frequency low energy high energy
energy / light ...so, how does all this mumbo-jumbo (I mean physics) effect IR spectroscopy?long wavelength short wavelength low frequency high frequency low energy high energy
infrared spectroscopy υ1 υ1 υ1 υ2 υ3 O υ2 EtO CH3 υ3 υ3 the cartoon version
infrared spectroscopy υ1 υ1 υ1 υ2 υ3 O υ2 EtO CH3 υ3 υ3 ...shine IR on molecule... certain bonds will absorb certain wavelengths of energy and thus by observing which wavelengths are missing we have a clue as to what bonds are in the molecule... the cartoon version
infrared spectroscopy high low energy energy υ1 υ1 υ1 υ2 υ3 O υ2 EtO CH3 υ3 υ3 high low the real version wave wavenumber number
infrared spectroscopy high low energy energy υ1 wavenumber is the υ 1 inverse (1/λ) of the wavelength in cm and measures υ1 υ2 υ3 O energy... υ2 EtO CH3 υ3 υ3 high low the real version wave wavenumber number
more on the theory of IR can be found at:www.massey.ac.nz/~gjrowlan ‘ in ‘Introduction to organicand bioorganic molecules and reactions’
interpreting IR spectra energy to cause vibration 4000 3000 2000 1500 1000 cm-1 O H C C C C C O change in scale N H C N C O C F C H C O C Cl bonds to triple double single hydrogen bonds bonds bonds regions of four the spectrum are important...
interpreting IR spectra energy to cause vibration 4000 3000 2000 1500 1000 cm-1 O H C C C C C O change in scale N H C N C O C F C H C O C Cl bonds to triple double single hydrogen bonds bonds bonds light atoms (H)
interpreting IR spectra energy to cause vibration 4000 3000 2000 1500 1000 cm-1 O H C C C C C O change in scale N H C N C O C F C H C O C Cl bonds to triple double single hydrogen bonds bonds bonds strong strong bonds bonds
interpreting IR spectra energy to cause vibration 4000 3000 2000 1500 1000 cm-1 O H C C C C C O change in scale N H C N C O C F C H C O C Cl bonds to triple double single hydrogen bonds region isn’t that actually fingerprint bonds bonds useful...it is unique to each molecule but almost impossible to interpret...good if you have a computer database I guess... 1500–400cm–1 ﬁngerprint region
interpreting IR spectra these two molecules have same functional group (ketone) and are almost identical in three regions... O O 1500–400cm–1 ﬁngerprint region
interpreting IR spectra only really differ in fingerprint... but I couldn’t tell you what bond stretching caused this peak! O O 1500–400cm–1 ﬁngerprint region
bad slide alert!
functional group absorptions
functional group absorptions can predict roughly where most functional groups will come...you’ll be given this in an exam if you need it...
examples of IR spectra H H NNH2 benzeneamine3480 aniline3395 two N–H stretches visible...but not for the reason you think (I’m not going to tell you why in case it confuses you!)
examples of IR spectra H3C H N N–H 3443 N-methylbenzenamine N-methylaniline
examples of IR spectra PhO–H3224 O O H H H H O O Ph Ph phenol - H-bonding
examples of IR spectra PhO–H3224 O O hydrogen bonding causes H H the peak to be very broad as the H H strength of H-bonds varies O O depending on factors like distance... Ph Ph phenol - H-bonding
examples of IR spectra O–H 3627 H O O H 2,6-di-tert-butyl-4-methylphenol
examples of IR spectra O–H 3627 H no H-bonding so O–H hasspecific strength bond and sharp O O peak (there is no H-bonding as the large tert-butyl groups H prevent the two molecules getting close to each other...) 2,6-di-tert-butyl-4-methylphenol
examples of IR spectra O C=C 1642 C=Ohex-5-en-2-one 1718
examples of IR spectra C=C C=O 1634 O 1674 note how putting the two groups in conjugation makes the bonds weaker (and hence have a lower wavenumber)(E)-hex-4-en-3-one
examples of IR spectra C=C C=O 1634 O 1674 why does conjugation make the bonds weaker? Think about the resonance forms...(E)-hex-4-en-3-one
examples of IR spectra O–H 3010 C=O 1712 O H O butanoic acid
examples of IR spectra N–H C=O 3356 1662 O 3184 1634 H N H butanamide
examples of IR spectra weaker C=O stretch in amide due to a resonance form involving the nitrogen lone pair...see IR can tell us a lot of useful information... N–H C=O 3356 1662 O 3184 1634 H N H butanamide
‘a mass spectroscopist issomeone who ﬁgures out whatsomething is by smashing it with ahammer & looking at the piecesJEOL (manufacturer) website
a mass spectrometerM + – M+ + – e 2e
a mass spectrometer basically, what you need to know is that a mass spectrometer fires electrons at your compound knocking one electron off the compound to form a radical cation that is then detected...M + – M+ + – e 2e
molecular mass molecular ion NH3 + – +NH + – e 3 2e because the mass of an electron is very very small...17.031 5.5 x 10–4 17.030mass-to-charge ratio (m/z)
molecular mass molecular ion NH3 + – +NH + – e the radical cation 3 2e (or molecular ion) effectively has the same mass as the original compound...17.031 5.5 x 10–4 17.030mass-to-charge ratio (m/z)
molecular mass molecular ion ...as a result mass spectrometry gives us the molecular mass...it also gives a lot more info but thats for another day (or course) NH3 + – +NH + – e 3 2e17.031 5.5 x 10–4 17.030mass-to-charge ratio (m/z)
cyclohexane +C6H12 + e– [C6H12 ] + 2e– m/z 84 C = 6x12 H = 12x1 a simple example m/z 85 showing the mass due to 13C of cyclohexane... isotope
cyclohexane +C6H12 + e– [C6H12 ] + 2e– m/z 84 C = 6x12 H = 12x1 ...all these other peaks are useful...but lets ignore m/z 85 due to 13C isotope
31 isotopes35Cl 37Cl chlorine exists as two isotopes (same element different mass due to number of neutrons)
31 isotopes35Cl 37Cl there is 3 times as much Cl mass 35 than Cl mass 37 (hence average is 35.5 as shown on most periodic tables)
isotopes C6H5Cl + e– [C6H5Cl]+ + 2e– the mass spectrum has two peaks...one for each isotope and the relative size of these peaks will m/z 112 be 3 : 1... due to 35Cl Cl m/z 114 due to 37ClC6H535Cl Mr = M+ = 112C6H537Cl Mr = M+2 = 114
11 isotopes79Br 81Br chlorine not the only element with isotopes...bromine exists as two isotopes in equal proportion so the spectrum of a bromide...
isotopes C6H5Br + e– [C6H5Br]+ + 2e– ...will have two peaks of equal intensity with a mass 2 apart m/z 156 due to 79Br m/z 158 due to Br 81BrC6H579Br Mr = M+ = 156C6H581Br Mr = M+2 = 158
what have....we learnt? • to determine molecular structure • thebasics of infrared spectroscopy • the basics of mass spectrometry