CHM1022 Week 3/Lecture 2 2023 NMR Spectroscopy PDF
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Monash University
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Summary
This document is a lecture about NMR spectroscopy, focusing on concepts like chemical shift, integration, and multiplicity. It covers the principles and practical applications of NMR analysis in organic chemistry for undergraduate students at Monash University.
Full Transcript
Week 3/Lecture 2: Chemical Detectives What does NMR spectroscopy tell us about….. connectivity? The multiplicity of a signal tells us about what functional group is attached. Chemical Shift of Hb Ha Ha “Hb feels” 2 magnetic fields and ap...
Week 3/Lecture 2: Chemical Detectives What does NMR spectroscopy tell us about….. connectivity? The multiplicity of a signal tells us about what functional group is attached. Chemical Shift of Hb Ha Ha “Hb feels” 2 magnetic fields and appears as “Ha feels” 3 magnetic fields and appears as three two lines: a doublet lines: a triplet What does NMR spectroscopy tell us about….. connectivity? The n+1 rule: A proton with n equivalent neighbours will be split into n+1 lines The integration of the lines is not always equal The separation of the lines is the coupling constant J Name of signal Pascal’s Triangle You can determine the integration of each line in a signal from first principles or Pascal’s triangle tells you about their height. Putting it all together. Chemical Shift: How electron withdrawing are the neighbouring functional groups? Integration: How many protons contribute to each signal? Multiplicity: How are the signals split? Cl H H H Cl Cl NMR summary Summary of 1H-NMR spectroscopy Chemical Shift: How electron withdrawing are the neighbouring functional groups? Integration: How many protons contribute to each signal? (Relates to symmetry) Multiplicity: How are the signals split? Summary of 13C-NMR spectroscopy Chemical Shift: How electron withdrawing are the neighbouring functional groups? Integration: N/A Splitting: N/A ACTIVITY 3: Unknown A Solve the structure of the following unknown compounds based on the IR, MS, 1H-NMR and 13C-NMR spectral information. Use the following “work flow” approach to solve these problems: 1. What is the molecular formula? 2. What is the Index of Hydrogen Deficiency? 3. Does the IR indicate any functional groups? 4. From the number of signals (chemical shift and integration) in the 1H-NMR and 13C-NMR what functional groups are present? 5. From the multiplicity in the 1H-NMR, how are the functional groups connected? 6. From the chemical shift in the 1H-NMR and 13C-NMR, how are the functional groups connected? CHECK. Does your structure match all the data? 15 mins. ACTIVITY 3: Unknown A Molecular formula: C4H8O2 15 mins. ACTIVITY 3: Discussion and feedback H(reference) for oxygen containing C4 is C4H10 2984 cm-1 presence of C−H bond 1741 cm-1 presence of C=O group IHD = H(reference)-H(unknown) 10-8 15 mins. = = 1 2 2 ACTIVITY 3: Unknown A 3H 3H 2H 15 mins. ACTIVITY 3: Discussion and feedback Integration 1 CH2; 2XCH3 Total = 8 ⇒ No symmetry Multiplicity H H H Multiplicity H 3H 3H H H H triplet (3 lines) ⇒ 2H H 2 neighbouring protons Quartet (4 lines) H ⇒ 3 neighbouring 1 CH2; 2XCH H 3 protons Total = 8 ⇒ No symmetryIntegration Chemical Shift Multiplicity ~ 4 ppm ⇒ attached to an singlet (1 lines) ⇒ H C O, N or halide 0 neighbouring protons ⇒O H H H H H Chemical Shift O O ~ 2 ppm H H C H ⇒ attached to a C=O group H H ACTIVITY 3: Discussion and feedback 1 13 O H-NMR Assignment C-NMR Assignment H 3C O CH3 d 1.2 3H, triplet, CH2CH3 d 15 CH2CH3 d 2.0 3H, singlet, O=C-CH3 d 20 CH3 Ethyl Acetate d 4.2 2H, quartet,CH2CH3 d 60 CH2 d 170 C=O IR MS 2984 cm-1 C−H bond C 4H 8O 2 1741 cm-1 C=O group IHD = 1 4 signals ⇒ no symmetry 3 below 100 ⇒ 3 sp3 C ~170 ppm ⇒ 1 C=O group ACTIVITY 3: Unknown B Molecular formula: C2H4Cl2 10 mins. ACTIVITY 3: Discussion and feedback Molecular formula: C2H4Cl2 H(reference) for C2 is C2H4 Not very informative IHD = 0 10 mins. ACTIVITY 3: Unknown B 3H 1H 10 mins. ACTIVITY 3: Discussion and feedback 1 3H H-NMR Assignment 1H d 2.0 3H, doublet, CHCH3 d 5.9 1H, quartet, CHCH3 H Cl CH3 Cl 13 C-NMR Assignment d 15 CH3 d 70 CH 10 mins. Summary Today we have: Introduced the origins of multiplicity in 1H-NMR Discussed how the integration, multiplicity, and chemical shift can be used to solve unknown structures Solved the structures for some more complex unknown molecules