IAP5301 S petros copy Chemists use spectroscopy (Ir, UV-Vis, NMR) to determine molecular structure. spectroscopy: technique that measures the amount of radiation a substance absorbs at various wavelengths(based on the quantization of molecular energy levels Molecules exist in quantized energy states (rotational, vibrational, and electronic quantum states) Molecules absorb discrete amounts of energy(ae)and are excited to higher energy states We can use electromagnetic radiation to excite molecules h=6.62608x1034j △E=hv c=2998x10cm/s 入 length(cI Molecules have unique absorption spectra that depend on their structure Specific absorptions provide useful structural information. How do you know what type of electromagnetic radiation to use? EM Radiation Wavelength(cm Excitation Approximate AE cosmic ravs 10-12to10-10 amma ravs 10-10to103 10-8to10-6 ultraviolet 106to3:x105 electronic 70-300 kcal/mol visible 3.8x10-5 to 7.8x10-5 electronic 40-70 kcal/mol infrared 7.8x10-5 to 3x 10-2 vibrational 1-10 kcal /mol microwave 3x10-2to3x102 rotational 1 cal/mol 102to3x105 Radio frequency EM radiation is used in NMr spectroscopy
IAP 5.301 Spectroscopy Chemists use spectroscopy (IR, UV-Vis, NMR) to determine molecular structure. Spectroscopy: technique that measures the amount of radiation a substance absorbs at various wavelengths (based on the quantization of molecular energy levels) • Molecules exist in quantized energy states (rotational, vibrational, and electronic quantum states). • Molecules absorb discrete amounts of energy (∆E) and are excited to higher energy states. • We can use electromagnetic radiation to excite molecules. E1 E2 ∆E = hν (ν = c/λ) 10 cm/s ν ( ) λ h = 6.62608 x 10–34 J·s c = 2.998 x 10 = frequency Hz = wavelength (cm) • Molecules have unique absorption spectra that depend on their structure. • Specific absorptions provide useful structural information. How do you know what type of electromagnetic radiation to use? EM Radiation Wavelength (cm) 10 –10 10 –8 10–8 –6 10–6 –5 –5 –5 –5 –2 –2 2 2 5 Excitation –– –– –– –– Approximate ∆E –– –– –– / / –– cosmic rays gamma rays X-rays ultraviolet visible infrared microwave radio* –12 to 10 –10 to 10 to 10 to 3.8 x 10 3.8 x 10 to 7.8 x 10 7.8 x 10 to 3 x 10 3 x 10 to 3 x 10 3 x 10 to 3 x 10 electronic electronic vibrational rotational 70–300 kcal mol 40–70 kcal mol 1–10 kcal/mol ~ 1 cal/mol * Radio frequency EM radiation is used in NMR spectroscopy. 1
Infrared Spectroscopy (vibrational energy) Chemical bonds are not rigid; they are constantly vibrating Different vibrational modes are of higher energy than others, and complex organic molecules have a large number of vibrational modes(complicated! A nonlinear molecule containing n atoms has 3n-6 possible fundamental vibrational modes Luckily, different functional groups(parts of a molecule)exhibit unique chracteristic absorptions This allows us to look at an IR spectrum(% transmission versus wavenumber)and determine which functional groups are present based on the absorption band Wavenumber is the number of waves per centimeter(1/1) Characteristic IR Absorption Bands for Common Functional Groups Class Intensity Assi Alkanes C-H stretch Alkenes 3080-3140 C=C str s由 C-Hout-of-plane bend 3300 s 丝C-h 2100-214 C=C 600-700 =C-H bend 3400 bonded o-H stretch 1540-1870 C=O stretch Aromatic Hydrocarbons m Coplane bendi 1400-1600
Infrared Spectroscopy (vibrational energy) • Chemical bonds are not rigid; they are constantly vibrating. • Different vibrational modes are of higher energy than others, and complex organic molecules have a large number of vibrational modes (complicated!). • A nonlinear molecule containing n atoms has 3n–6 possible fundamental vibrational modes! • Luckily, different functional groups (parts of a molecule) exhibit unique chracteristic absorptions. • This allows us to look at an IR spectrum (% transmission versus wavenumber) and determine which functional groups are present based on the absorption bands. • Wavenumber is the number of waves per centimeter (1/λ). Characteristic IR Absorption Bands for Common Functional Groups Class Wavenumber, cm–1 Intensity Assignment Alkanes 2850–3000 s C–H stretch 1450–1470 s 1370–1380 s CH2 and CH3 bend 720–725 m Alkenes 3080–3140 1650 809–990 m =C–H stretch m C=C stretch s C–H out-of-plane bend Alkynes 3300 2100–2140 600–700 s ≡C–H stretch m C≡C stretch s ≡C–H bend Alcohols 3600 var free O–H stretch 3400 s bonded O–H stretch Ethers 1070–1150 s C–O stretch Carbonyls 1540–1870 s C=O stretch Aromatic Hydrocarbons 675–900 1000–1300 1400–1600 s C–H out-of-plane-bend m C–H in-plane bending m C–C stretching 2
Ultraviolet-Visible Spectroscopy(electronic energy) It is easy to think of atoms as tiny little solar systems with the electrons orbitting the nucleus in a similar fashion as planets orbiting the sun. This simplistic picture satisfies our intuition, but it does not correspond with what we know about the atom It is more effective to treat electrons as waves rather than particles (This may sound complicated but you' ll have a chance to explore this in great depth during your stay here at MIT.) What it boils down to is that electrons occupy discrete orbitals that are best described by mathematical functions called wavefunctions. Because these orbitals electrons from lower energy to higher energy orbitals excitation of ave quantized energy, we can use spectroscopy to investigate the These excitations usually involve h an electron moving from a low excitation energy bonding orbital to a high energy antibonding orbital E Bonding UV-Vis spectroscopy is most frequently used to detect the presence of a specific compound in solution or to measure concentrations very accurately. You will get a chance to do this later this month
Ultraviolet-Visible Spectroscopy (electronic energy) • It is easy to think of atoms as tiny little solar systems with the electrons orbitting the nucleus in a similar fashion as planets orbiting the sun. This simplistic picture satisfies our intuition, but it does not correspond with what we know about the atom. It is more effective to treat electrons as waves rather than particles! (This may sound complicated, but you'll have a chance to explore this in great depth during your stay here at MIT.) • What it boils down to is that electrons occupy discrete orbitals that are best described by mathematical functions called wavefunctions. Because these orbitals have quantized energy, we can use spectroscopy to investigate the excitation of electrons from lower energy to higher energy orbitals. i i hν E Ant bonding Bond ng excitation • These excitations usually involve an electron moving from a low energy bonding orbital to a high energy antibonding orbital. • UV-Vis spectroscopy is most frequently used to detect the presence of a specific compound in solution or to measure concentrations very accurately. You will get a chance to do this later this month. 3
Nuclear Magnetic Resonance(NMR) Spectroscopy NMr is the most common spectroscopic tool used by organic chemists nMR differs from other spectroscopy because the different energy states (nuclear spin states)exist only in the presence of a magnetic field The nuclei of most(but not all) atoms behave as if they are spinning on an axis(nuclear spin) Because nuclei are positively charged, these spinning nuclei create small magnetic moments In the absence of a magnetic field, these magnetic moments are oriented in a random fashion For some nuclei(those having nuclear spin =+1/2), the presence of an external magnetic field (h) results in the alignment of each nucleus either wit ith (a)or against(B)the magnetic field iC and are the three most common nuclei observed using NMR spectroscopy. In this class, we will focus on H(proton)NMR Nuclear Magnetic Moments H Nuclear Magnetic Moments o magnetic field (in a magnetic field) The energy difference between the a and B states(AE)is proportional to the strength of the AE is relatively small(-10cal/mol); radio waves are used to excite(flip) nuclei from a to B E △E={954x101[rH/2r △E=h y= magnetogyric ratio cific to nucle H=magnetic field strength (at the nuck (+1/2)
Nuclear Magnetic Resonance (NMR) Spectroscopy NMR is the most common spectroscopic tool used by organic chemists. • NMR differs from other spectroscopy because the different energy states (nuclear spin states) exist only in the presence of a magnetic field. • The nuclei of most (but not all) atoms behave as if they are spinning on an axis (nuclear spin). • Because nuclei are positively charged, these spinning nuclei create small magnetic moments. • In the absence of a magnetic field, these magnetic moments are oriented in a random fashion. • For some nuclei (those having nuclear spin = ±1/2), the presence of an external magnetic field (H) results in the alignment of each nucleus either with (α) or against (β) the magnetic field. • 1 H, 13C. and 19F are the three most common nuclei observed using NMR spectroscopy. In this class, we will focus on 1H (proton) NMR. α α α α α β β β β H ) Nuclear Magnetic Moments (no magnetic field) Nuclear Magnetic Moments (in a magnetic field • The energy difference between the α and β states (∆E) is proportional to the strength of the magnetic field (H) at the nucleus. • ∆E is relatively small (~10–2 cal/mol); radio waves are used to excite (flip) nuclei from α to β. β α ∆ ν ∆ [(γ·H) π]} γ ) ) β /2) α /2) ∆E E H E = h E = {9.54 x 10–11 /2 = magnetogyric ratio (specific to nucleus H = magnetic field strength (at the nucleus (–1 (+1 4
If the energy difference between the a and p spin states of a proton nucleus (AE) depends only on the H NMR magnetogyric ratio(y=2.6753 x 10 radians sec .gauss )and the magnetic field (H), then won,' t all of the protons in a molecule absorb radiation of the exact same energy? Luckily no! We wouldn' t be able to obtain any structural information from a H NMR spectrum if that were the case Remember, nuclei are surrounded by clouds of negatively charged electrons In the presence of an applied magnetic field(Ho), these electrons move in such a way that their motion induces their own small magnetic field (H) At the nucleus, the induced magnetic field(H)opposes the applied magnetic field(Ho). Therefore, the nucleus experiences a magnetic field (H) slightly less than the applied magnetic field (Ho) ced magne e H=H-H applied etic electrons fied(H)∵ This phenomenon is called diamagnetic shielding. Protons with a lot of electron density around them are well-shielded and experience a reduced magnetic field. As a result, there is a smaller energy difference between the two spin states, and lower energy radiation is required to flip the spin from a to p In other words, protons in different electronic environments experience different amounts of shielding and absorb radiation of different frequencies. These differences are referred to as chemical shifts
• If the energy difference between the α and β spin states of a proton nucleus (∆E) depends only on the 1 H NMR magnetogyric ratio (γ = 2.6753 x 104 radians·sec–1·gauss–1) and the magnetic field (H), then won't all of the protons in a molecule absorb radiation of the exact same energy? • Luckily, no! We wouldn't be able to obtain any structural information from a 1 H NMR spectrum if that were the case. • Remember, nuclei are surrounded by clouds of negatively charged electrons. In the presence of an applied magnetic field (H0), these electrons move in such a way that their motion induces their own small magnetic field (H'). • At the nucleus, the induced magnetic field (H') opposes the applied magnetic field (H0). Therefore, the nucleus experiences a magnetic field (H) slightly less than the applied magnetic field (H0). 0) 0 induced magnetic field (H') electrons applied magnetic field (H H = H – H' • This phenomenon is called diamagnetic shielding. Protons with a lot of electron density around them are well-shielded and experience a reduced magnetic field. As a result, there is a smaller energy difference between the two spin states, and lower energy radiation is required to flip the spin from α to β. • In other words, protons in different electronic environments experience different amounts of shielding and absorb radiation of different frequencies. These differences are referred to as chemical shifts. 5
That's all well and good, but what does it mean practically? downfield region upfield (deshielded, electron poor) (shielded, electron rich All chemical shifts(O)are reported relative to a standard peak. Tetramethylsilane(tms)is used as the standard because the protons are very electron rich, and the resulting absorption peak is upfield from most other protons. The chemical shift of the TMS peak is assigned a value of STMS=0.0 ppm The location of the absorbance peak on this scale gives the chemical shift (8)of that particular proton The chemical shift of a proton gives us information about its chemical environment, but there's more to be learned Let's look at a sample h nmr to get a better idea of the information we can take from it 6
6 That's all well and good, but what does it mean practically? The location of the absorbance peak on this scale gives the chemical shift (δ) of that particular proton. The chemical shift of a proton gives us information about its chemical environment, but there's more to be learned! Let's look at a sample 1 H NMR to get a better idea of the information we can take from it . . . δ δ downfield region (deshielded, electron poor) upfield region (shielded, electron rich) All chemical shifts ( ) are reported relative to a standard peak. Tetramethylsilane (TMS) is used as the standard because the protons are very electron rich, and the resulting absorption peak is upfield from most other protons. The chemical shift of the TMS peak is assigned a value of TMS = 0.0 ppm 9 8 7 6 5 4 3 2 1 0 -1 ppm
The H NMR spectrum of 1, 2, 2-trichloropropane has two peaks other than the TMS peak · How can you ou tell which peak corresponds to the two H, protons and which is from the three H2 protons? Here's a clue: Chlorine atoms are very electronegative and strongly electron withdrawing. Thi means that they suck electron density away from the atoms close to them. The two H, protons are attached to a carbon that is also attached to a chlorine atom. As a result, these protons have less electron density around them, are less effectively shielded, and have a chemical shift further downfield than the H, protons (H=A, H2=B) Integration In addition to chemical shift, the relative size of the peaks provides information When you take a H NMR, you will integrate the different peaks in your spectrum. The relative values of the integrals are proportional to the number of protons absorbing energy at that frequency For example(from above) The spectrum tells you that there is a 3: 2 ratio of different protons in 1, 2, 2-trichloropropane. Of course you know that there are three H, protons and two H, protons that fit this criteria 7
H1 H1 H2 H2 C C Cl C H2 Cl Cl • The 1 H NMR spectrum of 1,2,2-trichloropropane has two peaks other than the TMS peak. • How can you tell which peak corresponds to the two H1 protons and which is from the three H2 protons? Here's a clue: Chlorine atoms are very electronegative and strongly electron withdrawing. This means that they suck electron density away from the atoms close to them. • The two H1 protons are attached to a carbon that is also attached to a chlorine atom. As a result, these protons have less electron density around them , are less effectively shielded, and have a chemical shift further downfield than the H2 protons (H1 = A, H2 = B). Integration • In addition to chemical shift, the relative size of the peaks provides information. • When you take a 1 H NMR, you will integrate the different peaks in your spectrum. The relative values of the integrals are proportional to the number of protons absorbing energy at that frequency. For example (from above): B : A = 75 : 50 = 3 : 2 The spectrum tells you that there is a 3 : 2 ratio of different protons in 1,2,2-trichloropropane. Of course you know that there are three H2 protons and two H1 protons that fit this criteria. 7
Spin-Spin Splitting Unlike in the H NMR spectrum of 1, 2, 2-trichloropropane on the previous page, most peaks are not single lines(singlets). In fact, the multiplicity of a peak(singlet, doublet, triplet) provides even more information about the structure of a molecule. This is because protons on adjacent atoms communicate with eachother In simple systems, n adjacent protons cause splitting into n+ l peaks For example, look at the spectrum of 1-chloropropane There are three inequivalent types of protons in this molecule(Hy h2, H3) It is easy to assign chemical shifts to the appropriate protons because you already know that chlorine is an electron withdrawing group The protons closest to the chlorine will have the least electron density (deshielded, downfield), and the protons furthest from the chlorine will have the most electron density(shielded, upfield). Using this reasoning, you can assign A=Hlb=H2, C=H Now, let's look at the multiplicity of the three peaks
Spin-Spin Splitting • Unlike in the 1 H NMR spectrum of 1,2,2-trichloropropane on the previous page, most peaks are not single lines (singlets). • In fact, the multiplicity of a peak (singlet, doublet, triplet) provides even more information about the structure of a molecule. This is because protons on adjacent atoms communicate with eachother. In simple systems, n adjacent protons cause splitting into n + 1 peaks. • For example, look at the spectrum of 1-chloropropane: H1 H1 H3 H3 C C Cl C H3 H2 H2 • There are three inequivalent types of protons in this molecule (H1, H2, H3) . • It is easy to assign chemical shifts to the appropriate protons because you already know that chlorine is an electron withdrawing group. • The protons closest to the chlorine will have the least electron density (deshielded, downfield), and the protons furthest from the chlorine will have the most electron density (shielded, upfield). Using this reasoning, you can assign: A = H1, B = H2, C = H3 • Now, let's look at the multiplicity of the three peaks. . . 8
triplet H3 H3 sextet Peak A, corresponding to the two H protons, is a triplet(three lines). This is because there are two adjacent H, protons (2+1=3) Peak B, corresponding to the two H2 protons, is a sextet(six lines ). This is because there are five adjacent protons two h, and three h,(5+1=6) Peak C, corresponding to the three H3 protons, is a triplet(three lines). This is because there are two adjacent H2 protons (2+1= 3) Spin-spin splitting, also known as coupling, can get much more complicated than this, but we'll stick to simple systems for now It is very important to remember that spin-spin splitting and peak integrals are not related. The multiplicity of a peak does not have any correlation to the number of protons absorbing energy at that frequency! There is a useful table on p. 301 of Zubrick that will help you assign the peaks in the H NMR spectra you will be obtaining this month
Cl C C C H3 H1 H1 H2 H2 H3 H3 triplet sextet triplet • Peak A, corresponding to the two H1 protons, is a triplet (three lines). This is because there are two adjacent H2 protons (2 + 1 = 3). • Peak B, corresponding to the two H2 protons, is a sextet (six lines). This is because there are five adjacent protons, two H1 and three H3 (5 + 1 = 6). • Peak C, corresponding to the three H3 protons, is a triplet (three lines). This is because there are two adjacent H2 protons (2 + 1 = 3). • Spin-spin splitting, also known as coupling, can get much more complicated than this, but we'll stick to simple systems for now. • It is very important to remember that spin-spin splitting and peak integrals are not related. The multiplicity of a peak does not have any correlation to the number of protons absorbing energy at that frequency! • There is a useful table on p. 301 of Zubrick that will help you assign the peaks in the 1 H NMR spectra you will be obtaining this month. 9