Spectroscopy of Organic Compounds

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Presentation transcript:

Spectroscopy of Organic Compounds Chemistry 344 Spectroscopy of Organic Compounds

Modern Spectroscopic Methods Revolutionized the study of Organic Chemistry Determine the exact structure of small to medium size molecules in a few minutes. Nuclear Magnetic Resonance (NMR) and Infrared Spectroscopy (IR) are particularly powerful techniques which we will focus on and use in this course.

Interaction of Light and Matter The Physical Basis of Spectroscopy Quantum properties of light (photons) Quantum properties of matter (quantized energy states). Photons of light act as our “quantum probes” at the molecular level giving us back precise information about the energy levels within molecules.

The Electromagnetic Spectrum Continuous Covers a wide range of wavelengths of “light” from radio waves to gamma rays. Wavelengths (l) range from more than ten meters to less than 10-12 meter.

The Electromagnetic Spectrum

Relationship Between Wavelength, Frequency and Energy Speed of light (c) is the same for all wavelengths. Frequency (n), the number of wavelengths per second, is inversely proportional to wavelength: n = c/l Energy of a photon is directly proportional to frequency and inversely proportional to wavelength: E = hn = hc/l (where h = Plank’s constant)

Wavelength/Spectroscopy Relationships Spectral Region Photon Energy Molecular Energy Changes UV-Visible ~ 100 kcal/mole Electronic Infrared (IR) ~ 10 kcal/mole Bond vibrations Radio < 0.1 kcal/mol Nuclear Spin states in a magnetic field

Spin of Atomic Nuclei Spin 1/2 atoms: mass number is odd. examples: 1H and 13C. Spin 1 atoms: mass number is even. examples: 2H and 14N. Spin 0 atoms: mass number is even. examples: 12C and 16O.

Magnetic Properties of the Proton Related to Spin

Energy States of Protons in a Magnetic Field = l a b s o r e d i g h t A p M n c F H x

Nuclear Magnetic Resonance (NMR) Nuclear – spin ½ nuclei (e.g. protons) behave as tiny bar magnets. Magnetic – a strong magnetic field causes a small energy difference between + ½ and – ½ spin states. Resonance – photons of radio waves can match the exact energy difference between the + ½ and – ½ spin states resulting in absorption of photons as the protons change spin states.

The NMR Experiment The sample, dissolved in a suitable NMR solvent (e.g. CDCl3 or CCl4) is placed in the strong magnetic field of the NMR. The sample is bombarded with a series of radio frequency (Rf) pulses and absorption of the radio waves is monitored. The data is collected and manipulated on a computer to obtain an NMR spectrum.

The NMR Spectrometer

The NMR Spectrometer

The NMR Spectrum The vertical axis shows the intensity of Rf absorption. The horizontal axis shows relative energy at which the absorption occurs (in units of parts per million = ppm) Tetramethylsilane (TMS) in included as a standard zero point reference (0.00 ppm) The area under any peak corresponds to the number of hydrogens represented by that peak.

The NMR Spectrum

Chemical Shift (d) The chemical shift (d) in units of ppm is defined as: d = distance from TMS (in hz) radio frequency (in Mhz) A standard notation is used to summarize NMR spectral data. For example p-xylene: d 2.3 (6H, singlet) d 7.0 (4H, singlet) Hydrogens in identical chemical environments (equivalent hydrogens) have identical chemical shifts.

Shielding – The Reason for Chemical Shift Differences Circulation of electrons within molecular orbitals results in local magnetic fields that oppose the applied magnetic field. The greater this “shielding” effect, the greater the applied field needed to achieve resonance, and the further to the right (“upfield”) the NMR signal.

Structure Effects on Shielding Electron donating groups increase the electron density around nearby hydrogen atoms resulting in increased shielding, shifting peaks to the right. Electron withdrawing groups decrease the electron density around nearby hydrogen atoms resulting in decreased shielding, (deshielding) shifting peaks to the left.

Structure Effects on Shielding The Deshielding effect of an electronegative substituent can be seen in the NMR spectrum of 1-Bromobutane Br – CH2-CH2-CH2-CH3 d (ppm): 3.4 1.8 1.5 0.9 No. of H’s: 2 2 2 3

Some Specific Structural Effects on NMR Chemical Shift Type of Hydrogen d (ppm Alkyl (C – H) 0.8 – 1.7 Alkyl Halide (RCH2X) 3 - 4 Alkene (R2C=CH2) 4 - 6 Aromatic (e.g. benzene) 6 - 8 Carboxylic Acid (RCOOH) 10 - 12

Spin-Spin Splitting Non-equivalent hydrogens will almost always have different chemical shifts. When non-equivalent hydrogens are on adjacent carbon atoms spin-spin splitting will occur due to the hydrogens on one carbon feeling the magnetic field from hydrogens on the adjacent carbon. The magnitude of the splitting between two hydrogens (measured in Hz) is the coupling constant, J.

Spin-Spin Splitting Origin of the Doublet

Spin-Spin Splitting Origin of the Triplet

Spin-Spin Splitting Origin of the Quartet

Pascal’s Triangle Pattern Relative Peak Height Singlet 1 Doublet 1 : 1 Triplet 1 : 2 : 1 Quartet 1 : 3 : 3 : 1 Quintet 1 :4 : 6 : 4 : 1

The n + 1 Rule If Ha is a set of equivalent hydrogens and Hx is an adjacent set of equivalent hydrogens which are not equivalent to Ha: The NMR signal of Ha will be split into n+1 peaks by Hx. (where n = # of hydrogens in the Hx set.) The NMR signal of Hx will be split into n+1 peaks by Ha. (where n = # of hydrogens in the Ha set.)

1H NMR Spectrum of Bromoethane

1H NMR Spectrum of 1-Nitropropane

Unknown A (Figure 9.20 Solomons 7th ed.) Formula: C3H7I IHD = 0 No IR data provided 1HNMR d: 1.90 (d, 6H), 4.33 (sept., 1H)

Unknown B (Figure 9.20 Solomons 7th ed.) Formula C2H4Cl2 IHD = 0 No IR data given 1HNMR d: 2.03 (d, 3H), 4.32 (quartet, 1H)

Unknown C (Figure 9.20 Solomons 7th ed.) Formula: C3H6Cl2 IHD = 0 No IR data given 1HNMR d: 2.20 (pent., 2H), 3.62 (t, 4H)

Infrared Spectroscopy Energy of photons in the IR region corresponds to differences in vibrational energy levels within molecules. Vibrational energy levels are dependent on bond types and bond strengths. IR is highly useful to determine if certain types of bonds (functional groups) are present in the molecule.

IR Spectrum of Ethanol

IR Correlation Table

Key Functional Groups by Region of the IR Spectrum

IR Spectrum of Benzaldehyde

IR Spectrum of Cyclohexanone

IR Spectrum of Propanoic Acid

NMR: Exceptions to the n+1 Rule The n+1 rule does not apply when a set of equivalent H’s is split by two or more other non-equivalent sets with different coupling constants. The n+1 rule does not apply to second order spectra in which the chemical shift difference between two sets of H’s is not much larger than the coupling constant.

NMR: Some Specific Functional Group Characteristics O-H and N-H will often show broad peaks with no resolved splitting, and the chemical shift can vary greatly. Aldehyde C-H is strongly deshielded. (d = 9-10 ppm) and coupling to alkyl H’s on adjacent carbon is small. Carboxylic Acid O-H is very strongly deshielded. (d = 10-12 ppm)

NMR: Some Specific Functional Group Characteristics Ortho splitting on aromatic rings is often resolved, but meta and para splitting is not. Cis and trans H’s on alkenes usually show strong coupling, but geminal H’s on alkenes show little or no resolved coupling.

NMR: Mixtures of Compounds Unlike most textbook examples, “real world” NMR samples commonly contain mixtures of compounds. It is very common to see extra signals due to impurities. The impurities can often be identified. In some cases it is useful (or necessary) to analyze mixtures that contain comparable amounts of two or more compounds.

NMR: Mixtures of Compounds Synthesis of Ethyl Acetate by Fischer Esterification Results in an Equilibrium Mixture:

Unknown A (Figure 14.27 Solomons 7th ed.) Formula = C9H12 IHD = 4 IR shows no medium or strong bands above 1650 cm-1 except C-H stretching bands around 3,000 cm-1 1HNMR d: 1.26 (d, 6H), 2.90 (sept., 1H), 7.1-7.5 (m, 5H)

Unknown B (Figure 14.27 Solomons 7th ed.) Formula = C8H11N IHD = 4 IR shows two medium peaks between 3300 and 3500 cm-1 . No other medium or strong bands above 1650 cm-1 except C-H stretching bands around 3,000 cm-1 1HNMR d: 1.4 (d, 3H), 1.7 (s, br, 2H), 4.1(quart., 1H), 7.2-7.4 (m, 5H)

Unknown C (Figure 14.27 Solomons 7th ed.) Formula = C9 H10 IHD = 5 IR shows no medium or strong bands above 1650 cm-1 except C-H stretching bands around 3,000 cm-1 1H NMR d: 2.05 (pent., 2H), 2.90 (trip., 4H), 7.1-7.3 (m, 4H)

Unknown H (Figure 9.48 Solomons 7th ed.) Formula = C3H4Br2 IHD = 1 No IR data given 1HNMR d: 4.20 (2H), 5.63 (1H), 6.03 (1H)

Unknown Y (Figure 14.34 Solomons 7th ed.) Formula = C9H12O IHD = 4 IR shows a strong, broad, absorbance centered at 3400 cm-1 1HNMR d: 0.85 (t, 3H), 1.75 (m, 2H), 4.38 (s, br, 1H), 4.52 (t, 1H), 7.2-7.4 (m, 5H)