IR spectroscopy - The University of York红外光谱-约克大学

IR spectroscopy - The University of York红外光谱-约克大学 A-level Spectroscopy An image of a human brain from a live patient recorded using magnetic resonance imaging - a 3D form of n.m.r. spectroscopy 1 Introduction Spectroscopy is a collective name for the various techniques that use the interaction between molecules and electromagnetic radiation to elucidate the structure of molecules. Spectroscopic methods are fundamental to the study of Chemistry, Molecular Biology, Medicine and Astrophysics. This booklet covers the following techniques:- A) Infrared Spectroscopy - from „Polymer Revolution? and „What?s in a Medicine?? , Use relevant given data to interpret (and make predictions of) infrared spectra for organic compounds containing a limited range of functional groups (hydroxyl, carbonyl and carboxylic acid groups). , Understand that every compound has a distinctive „fingerprint? in its infrared spectrum. , Use information given in the Data Sheet to interpret and predict infrared spectra for organic compounds, in terms of the functional group(s) present; understand that specific frequencies of infrared radiation make specific bonds vibrate more. B) Mass Spectrometry - from „What?s in a Medicine?? , Interpret and predict mass spectra: Identify the M+ peak and explain that it indicates the Mr (synoptic); explain how the molecular formula can be worked out from the high-resolution value of the M+ peak; recall that other peaks are due to positive ions from fragments and the mass differences between peaks indicate the loss of groups of atoms, suggest the origins of peaks e.g. peaks at masses of 15 and 77 are usually due to the presence of the methyl and phenyl positive ions; loss of a methyl group would be indicated by a mass difference of 15. C) Nuclear Magnetic Spectroscopy - from „Medicines by Design? , Describe and explain how proton nuclear magnetic resonance spectroscopy (NMR) can be used for the elucidation of molecular structure (including splitting patterns up to quartets – using the „n+1? rule; further explanation of splitting not required; , Explain how a combination of spectroscopic techniques [MS, IR(synoptic) and NMR] can be used to elucidate the structure of organic molecules. This work builds on AS topics of:- , Interaction of radiation with matter („Elements of Life? and „Atmosphere?); , Mass spectrometry („Elements of Life? 2 A) Infrared (i.r.) spectroscopy , Used to identify bonds / functional groups , Can only identify the exact molecule by comparison with library spectra Experiment 1 RSC Video , Infrared radiation is passed simultaneously through the sample and a reference cell. , The reference ensures that peaks due to water or carbon dioxide in the air can be cancelled out. , The frequencies of i.r. radiation absorbed are determined by passing through a rotating prism to focus one frequency at a time onto the detector. -1, The spectrum shows the ________________ (cm) on the x axis (which is 1/,) and the ____________________ on the y-axis. Calculations 1) c = ,, 2) Wavenumber = 1/,(cm) e.g. What wavenumber would appear on an i.r. spectrum if the frequency of 13radiation absorbed by a molecule was 2.5 x 10 Hz? 3 Theory , IR radiation corresponds to the energy required to make chemical bonds vibrate more / move to a higher vibrational energy level. , Therefore, energy of certain wavelengths is absorbed by molecules. , The actual energy depends on the mass of the atoms and the strength of the bond, so different bonds will absorb at different frequencies. c.f. spring oscillations m , Stronger bonds need more energy to make them vibrate, so absorb a higher frequency of i.r. radiation (higher wavenumber) e.g. hydrogen halides , Molecules with more than 2 atoms can vibrate in different ways e.g. sulphur dioxide , So these spectra will contain more absorptions 4 , Most organic molecules contain a number of types of bond, so characteristic absorptions will be seen for each bond. e.g. ethanol The following types of bond need to be recognised:- -1Bond Functional group Absorbance (cm) O – H Alcohols 3200 – 3600 / strong and broad* O – H Carboxylic acids 2500 – 3200 / medium and very broad* C=O Aldehydes / ketones / 1680 – 1750 / strong and sharp carboxylic acids/ esters C-O Alcohols / esters / ethers 1050 – 1300 / medium C-H Alkanes / alkenes etc 2850 – 3100 / medium *Broad due to Hydrogen Bonding between O-H groups i.r. bands.ppt 5 Examples of infrared spectra 1) ethanol (CHCHOH) 32 displayed formula i.r. spectrum Bond / (Functional group) -1Absorption / cm 6 2) ethanoic acid (CHCOOH) 3 displayed formula i.r. spectrum Bond / (Functional group) -1Absorption / cm 7 3) Ethyl Ethanoate (CHCOOCHCH) 332 H O H H C H O C C C H H H H i.r. spectrum Bond / (Functional group) -1Absorption / cm 1750 1250 3000 8 4) a) i.r. spectrum of an alcohol with molecular formula CHO. 38 NB: This Alcohol is oxidised to compound 4)b) when heated under distillation with acidified potassium dichromate and 4)c) when heated to reflux with acidified potassium dichromate. Clue? Bond / (Functional group) -1Absorption / cm Displayed Formula of 4a 9 4) b) i.r. spectrum of the compound with molecular formula CHO 36 obtained by distilling compound 4)a) with acidified potassium dichromate Bond / (Functional group) -1Absorption / cm Displayed Formula of compound 4b 10 4)c) i.r. spectrum of the compound with molecular formula CHO formed 362 when compound 4a is heated to reflux with acidified potassium dichromate Bond / (Functional group) -1Absorption / cm Displayed Formula of Compound 4c 11 5)a) i.r. spectrum of an isomer of 4a which forms the same product 5)b) whether it is heated to distil or reflux with acidified potassium dichromate Bond / (Functional group) -1Absorption / cm Displayed Formula of Compound 5a 12 5)b) i.r. spectrum of the product of the reaction of 5a with acidified potassium dichromate when heated to reflux or distillation. Bond / (Functional group) -1Absorption / cm Displayed Formula of Compound 5b 13 6) Salicylic Acid (2-hydroxybenzoic acid - (HOCHCOOH)) 64 displayed formula i.r. spectrum Bond / (Functional group) -1Absorption / cm 14 7) Aspirin (CHCOOCHCOOH) 364 i.r. spectrum Bond / (Functional group) -1Absorption / cm 2900 v. broad 1750 1700 1200 15 B) Mass Spectrometry +, Use M (molecular ion) to measure Mr , Use M+2 isotope peaks to identify Cl or Br , Use fragmentation pattern to confirm structure of molecule Experiment 2 – RSC video AS-level , Vaporisation of atoms or molecules; , Ionisation of atoms or molecules; , Acceleration of ions; , Time of Flight Measurement of ions (lighter ones quicker); , Detection of ions. 2K.E. = ?mv „Time of Flight? = 16 A2-level High Resolution Mass Spectrometry In high resolution mass spectrometry, the ions are focused into a much smaller range of kinetic energies before entering the „time of flight? measurement. Consequently:- , the mass to charge ratio of ions can be measured to a much higher level of precision; , using accurate relative isotopic masses we can distinguish between molecules with the same M; r , this allows us to determine the molecular formula. Isotope Accurate relative isotopic mass 1H 1.0078 12C 12.0000 14N 14.0031 16O 15.9949 Example CO and N have the same M at low resolution (28), but different at high 2r resolution. C 12.0000 N 14.0031 O 15.9949 N 14.0031 CO 27.9949 N 28.0062 2 Q Propane, carbon dioxide and ethanal all have a Mr of 44. A high resolution mass spectrum shows an accurate molecular ion peak at 44.0261, which molecule is it? 17 Fragmentation , The atoms or molecules are ionised by bombarding with high energy electrons:- -+-e.g. CHCOCH + e [CHCOCH] + 2 e 3333 +M , Usually, the resulting molecular ion has such high energy that it splits up into a smaller ion and an uncharged molecule (fragmentation) + +e.g. [CHCOCH] [CHCO] + CH 3333 + Mm/e = 58 43 ++or [CHCOCH] CHCO + [CH] 3333 58 15 NB The first fragmentation route is more likely because fragments containing +the [R-C=O] group (acylium cations) are particularly stable. , The following peaks are often seen in the fragmentation patterns of mass spectra – the highlighted peaks usually provide very useful clues in determining the structure of a molecule fragment m/e CH 15 3 CHCH or CHO 29 32 CHNH 30 22 CHOH 31 2 CHCO or CH 43 337 CONH 44 2 COOH 45 CH 77 65 CHCH 91 652 CHCO 105 65 18 Examples of fragmentation and the interpretation of mass spectra 1) Propanone (CHCOCH) 33 displayed formula mass spectrum m/z Formula m/z lost Group lost 58 43 15 19 2) Propanal (CHCHCHO) 32 displayed formula mass spectrum m/z Formula m/z lost Group lost 58 57 29 20 3) Methyl Benzoate (CHCOOCH) 653 O H O C C H H mass spectrum m/z Formula m/z lost Group lost 136 105 77 21 4) Ethyl Ethanoate (CHCOOCHCH) 332 H O H H C H O C C C H H H H mass spectrum m/z Formula m/z lost Group lost 88 73 43 29 15 22 5) Salicylic Acid (2-hydroxybenzoic acid - (HOCHCOOH)) 64 displayed formula mass spectrum m/z Formula m/z lost Group lost 138 * 120 92 * NB 3- or 4- hydroxybenzoic acid isomers cannot eliminate water –why not? 23 6) Aspirin (CHCOOCHCOOH) 364 mass spectrum m/z Formula m/z lost Group lost 180 138 120 43 24 7) ethanamide (CHCONH) 32 displayed formula mass spectrum m/z Formula m/z lost Group lost 59 44 43 25 8) paracetamol (4-hydroxyphenylethanamide) (HOCHNHCOCH) 643 displayed formula mass spectrum m/z Formula m/z lost Group lost 151 109 108 43 26 C) Nuclear Magnetic Resonance (n.m.r.) spectroscopy , The number of peaks – number of proton types , The chemical shift (δ) – what are the proton types , The integration – how many protons of each type Experiment 3 – RSC video , Sample is placed in a very strong magnetic field , A pulse of radiofrequency radiation is applied , Radiofrequency signal emitted from sample is detected Theory , Nuclei have a property called nuclear spin which generates a tiny magnetic field. The nuclei therefore behave like tiny bar magnets. , When such nuclei are placed in a large magnetic field they will become aligned with or against the direction of the external field. 27 , The nuclei lined up with the field are slightly more stable (lower energy) than those that oppose the external field. , The energy gap between these two states corresponds to radiofrequency radiation. , If the sample is irradiated with a pulse of radio waves, the nuclei in the lower energy state may be promoted to the higher energy state (the tiny bar magnets „flip? from being aligned with to against the external field). , The excited nuclei will then return to the ground state releasing fixed quanta of energy which will be detected. 28 , The energy gap depends on the chemical environment of the nuclei and can be used to deduce the exact structure of the molecule. , Ethanal has two proton types, so produces 2 signals in the n.m.r. spectrum. The important features of the spectrum are:- , The number of peaks – number of proton types , The integration – how many protons of each type , The chemical shift (δ) – what are the proton types The following table can be used to link the chemical shift to the proton type (chemical environment of H atom):- type of proton chemical shift δ / ppm RCH/ RCHR (alkane) 0.8 - 1.4 3 2 RCOCH(carbonyls, esters) 1.8 - 2.2 3 RCHHal 3.2 - 4.6 2 ROCH(esters, ethers) 3.2 - 3.5 3 ROH (alcohol) 1.0 - 6.0 RCHH (arenes) 6.0 - 9.0 64 RCHCH(methylarene) 2.2 - 2.4 643 RCONHR (amides) 7.0 - 10.0 RCHO (aldehydes) 9.7 - 9.8 RCOOH (carboxylic acids) 9.0 - 12.0 RCHOH (phenols) variable 64 RNHR (amines) variable [R represents an alkyl group] 29 Examples of the interpretation of n.m.r spectra 1) propanone (CHCOCH) 33 displayed formula nmr spectrum Proton integration inference δ / ppm inference Ha 30 2) ethanoic acid (CHCOOH) 3 nmr spectrum Proton integration inference δ / ppm inference Ha 11.4 Hb 2.1 31 3) propanal (CHCHCHO) 32 displayed formula nmr spectrum Proton integration inference δ / ppm inference Ha 9.7 Hb 2.4 Hc 1.1 32 High Resolution nmr Spectra , Most nmr spectra look more complicated than the first three examples. , The signal for each hydrogen atom may be split into a number of peaks. , The pattern of the splitting tells us how many hydrogen atoms are bonded to the adjacent carbon atom. The n+1 rule The no. peaks = the no. H atoms on the adjacent carbon + 1 e.g. ethanal O H a C H aC H b H a , The Ha protons have one adjacent H atom (Hb) , The signal will be split into ____ peaks – a doublet. , The Hb proton has three adjacent H atoms (Ha) , The signal will be split into ____ peaks – a quartet. High resolution spectra may be analysed as follows:- , The number of peaks – number of proton types , The integration – how many protons of each type , The chemical shift (δ) – what are the proton types , The splitting pattern – the number of H atoms on the adjacent C atom NB This level of analysis is now required for the A-level examinations, as the information from splitting patterns is extremely useful in working out the structure of complex molecules. 33 Why are the signals split by adjacent protons? , Each H nucleus generates its own tiny magnetic field, which may be aligned with or against the external magnetic field. This will affect the magnetic environment experienced by H nuclei bonded to adjacent C atoms. e.g. ethanal H O a C H aC H b H a , The Ha protons have one adjacent H atom (Hb) , Hb may be aligned with or against the field 1 : 1 , This means that there are two possible environments for the Ha protons, of equal probability. , The signal for Ha will be split into 2 peaks – a 1:1 doublet. 34 O H a C H aC H b H a , The situation is slightly more complicated for the Hb proton, which has three adjacent H atoms (Ha). , Each of the three Ha protons may be aligned with or against the field. , This means that there are four possible orientations of the Ha nuclei:- a. All nuclei are aligned with the field (1); b. Two nuclei with and one against the field (3); c. One nucleus with and two nuclei against the field (3); d. All nuclei aligned against the external magnetic field (1). 1 : 3 : 3 : 1 , This means that there are four possible environments for the Hb proton, with a relative probability of 1:3:3:1 , The signal will be split into 4 peaks – a 1:3:3:1 quartet. 35 4) Ethanal (CHCHO) 3 H O a C H C a H b H a High Resolution nmr Spectrum of ethanal 3 H a 1 H b 10 8 6 4 2 0 , / ppm Proton integration inference δ / ppm inference splitting inference Ha 2.1 Hb 9.7 36 5) ethyl ethanoate (CHCOOCHCH) 332 Ha O Hc Hb C Ha O C C C Hc Ha Hc Hb nmr spectrum Proton integration inference δ / ppm inference splitting inference Ha 2.1 Hb 4.1 Hc 1.2 37 Ha 6) propan-2-ol (CHCH(OH)CH) 33 O Hc Hc Hc C C C Hc Hc Hb Hc nmr spectrum Proton integration inference δ / ppm inference splitting inference Ha 2.1 singlet Hb 3.9 septet Hc 1.2 doublet 38 7) Salicylic Acid (2-hydroxybenzoic acid - (HOCHCOOH)) 64 displayed formula nmr spectrum Proton integration inference δ / ppm inference Ha 8.0 Hb 7.6 Hc 7.0 Where are the O-H groups? 39 8) Aspirin (CHCOOCHCOOH) 364 H H H H nmr spectrum Proton integration inference δ / ppm inference Ha 1 11.3 Hb 4 x 1 7 - 8 Hc 3 2.1 40 9) Mystery compound – “Why are there no aspirin in the jungle?” n.m.r. spectrum Proton integration inference δ / ppm inference Ha 1 9.7 Hb 1 9.1 Hc 2 7.4 Hd 2 6.7 He 3 2.0 Structure 41 Combined Spectral Techniques 1) Predict the ir, nmr and mass spectra of propanoic acid a) IR spectroscopy Bond / (Functional group) -1Absorption / cm b) Nmr spectroscopy Proton integration inference δ / ppm inference splitting inference Ha Hb Hc c) Mass Spectrometry m/z Formula m/z lost Group lost 42 2) Deduce the structure of the molecule from these spectra a. ir spectrum Bond / (Functional group) -1Absorption / cm 43 b) nmr spectrum Proton integration inference δ / ppm inference splitting inference Ha 2.2 singlet Hb 3.6 triplet overlapping quartet of Hc 1.5 triplets Hd 0.9 triplet 44 c) mass spectrum 31 60 m/z Formula m/z lost Group lost 60 31 Structure of Unknown Molecule Name 45