Lab+Lecture+9+-+Methyl+Diantilis_Formal+Lab+Report_Annotated_2023Oct24

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Nov 24, 2024

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1 CHEM 12AL Organic Chemistry Lab
2 Synthesis of Methyl Diantilis 1) Two step synthesis - Everyone performs the first step of the reaction - Split into 4 groups of 4 and carry out second step to produce different products 2) Combine all techniques up to this point - Set up reactions - Purify intermediates and final products - Characterize products by 1 H NM and 13 C NMR 3) Formal Lab report is written based on Methyl Diantilis Synthesis - Detailed formal lab report guidelines available on bCourses
3 Synthesis of methyl diantilis Two steps: i) aldehyde reduction, ii) etherification NaBH 4 reduction mechanism H- 0 - H - N Hy - 0 - Hz I is oft -> Ot ON . not a 1 , 'Yout 3 equiv of hydride remaining (H0)
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4 Amberlyst-15 (H + ) catalyzed etherification mechanism Synthesis of methyl diantilis Two steps: i) aldehyde reduction, ii) etherification Sheterogeneous H * /Acid Catalyst Ju T i , it " In t - - " os wo , an I -> dH · benzylic cation R -OH · resonance stabilized F · Sul - H I hi It en 1 - I · similar stability tab" catius In -Ott · compare purity
5 Due on Monday, Dec 4 (30pts) Typed: 12pt font, double spaced Page count: 3-5 pages including figures Use ChemDraw to draw molecules and reaction schemes Formal Lab Report is a separate assignment to the experimental lab notebook. * Can not be used as your dropped lab score* Synthesis of methyl diantilis Formal Lab Report Formatting Guidelines Use ChemDraw to draw molecules and reaction schemes Formal Lab Report is a separate assignment to the experimental lab notebook.
6 Synthesis of methyl diantilis Formal Lab Report Lab Report Outline 1) Introduction 2) Results and Discussion 3) Conclusion 4) Experimental 5) References 6) Supplemental Information
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7 Synthesis of methyl diantilis Formal Lab Report Lab Report Outline Introduction [5 pt] Introduce the synthesis of methyl diantilis in general (why it’s important, etc). In this introduction, include at least two reference to a peer-reviewed journal article. Indicate the references using superscripted numbers which correspond to a numbered list of properly formatted citations at the end of the report. Draw a reaction scheme for the specific reactions that you carried out. Describe this reaction scheme in 2-3 sentences. Results, Discussion, and Conclusions [10 pt] Write a summary of your results. Write a cohesive discussion regarding the experimental details for the reaction that you carried out. Your discussion should include: An analysis of all NMR spectra of your group A rational arrow-pushing mechanism for both reactions A discussion of your yields with all four alcohols Discuss the composition of each product. In this discussion, include information gained from TLC as well as 1 H NMR and 13 C NMR. Include the purity calculations and a qualitative comparison of your group’s products with the products from other groups in your lab section. Describe the experimental factors that you believe had the biggest effect in causing any variability between different groups in your laboratory section. Write a concise conclusion that summarizes the results of your experiment.
8 Synthesis of methyl diantilis Formal Lab Report Lab Report Outline Experimental [6 pt] Write a concise description of the steps that were taken to carry out this reaction in the lab. This should be written in the third person passive tense and should constitute an overview of the procedure without including all the details. Report the 1 H NMR and 13 C NMR data for your intermediate and products. References [2 pt] Formatted according to the American Chemical Society style Supporting Information: [7 pt] Attach copies of each NMR spectrum, annotated as typical for this course.
9 NMR Spectroscopy 1 H NMR of Aromatic Molecules - Circulating pi electrons create ring currents - Induced magnetic field increases B effective near H atoms Increase in B effective deshields H atoms and shifts downfield benzene ethylene Induced magnetic field B 0 Induced magnetic field = 6-8 ppm = 5-6 ppm
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10 NMR Spectroscopy 1 H NMR of Aromatic Molecules 2-Isopropylphenol 7.00 6.75 7.25 7.50 1.0 1.0 1.0 1.0 7.35 7.21 7.01 6.80 Ch - He Haz CHf H ! Ad · 4 sets of aromatic 'It's f · complex splitting - Of He
11 NMR Spectroscopy 1 H NMR of Aromatic Molecules 7.00 6.75 7.25 7.50 1.0 1.0 1.0 1.0 7.35 7.21 7.01 6.80 How do we assign the aromatic protons given the information that we have available: i) Chemical shift values , ii) Peak Areas , and iii) Peak Splitting ? Origins of aromatic 1 H Chemical Shift Values Consider ring substituent resonance forms it * 3 Elite j I - =I E - OH is pi-donating I I - I -> lone pairs on oxygen donate intoring's shield ortholpara positions - -oH is withdrawing -> oxygen is more dectronegative than carbons is an Eltria the o-csigma bond , hart to predict mary : substituents affect shielding/deshielding of aromatict's in sometimes wats .
12 NMR Spectroscopy 1 H NMR of Aromatic Molecules 7.00 6.75 7.25 7.50 1.0 1.0 1.0 1.0 7.35 7.21 7.01 6.80 Origins of aromatic 1 H Chemical Shift Values
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13 NMR Spectroscopy 1 H NMR of Aromatic Molecules Origins of aromatic 1 H Peak Splitting 3-bund coupling Hist 1X I I ing strengths -H 1 Ortho - H - - H or tho coupling : 6-10 Hz (normal 3-bond coupling) pare # meta coupling : 1-3Hz (4-bond coupling) para para coupling : 0-1H1 /S-bond coupling ( · Conjugation enables long-range coupling · S-bond coupling is uncommon in nonvaromatic systems
14 NMR Spectroscopy 1 H NMR of Aromatic Molecules 7.00 6.75 7.25 7.50 Origins of aromatic 1 H Peak Splitting fanity t 3-bond coupling a Stab , "Ybc , "Tod Ab-lit I - · Ha or Hd e ef 4-bund coupling HborHe - "Fac , " Abd Hb: He have more coupling Partiers
15 NMR Spectroscopy Aromatic 1 H Splitting trees 3 J a b = 8 Hz 4 J a c = 2 Hz 3 J d c = 8 Hz 4 J d b = 1 Hz 1 box = 1 Hz Ha Ald -- -- N - Ha-Hb /Ortho) N Hd-H , (orthol - 8 - - r IT Ha-He (metc) l Hd-Hb (meta) 1 : 1 : I : 1 1 : 1 iii) I I us it te - doublets of doublets doublets of doublets (dd] (dd]
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16 NMR Spectroscopy Aromatic 1 H Splitting trees 3 J b a = 8 Hz 3 J b c = 7 Hz 4 J b d = 1 Hz 3 J c d = 8 Hz 3 J c b = 7 Hz 4 J c a = 2 Hz 1 box = 1 Hz Hb Ha I N - = Hb- Ha (orthol F -> Ha-Hd (ortha) V 8 8 --- -- I - - 8 y 8 S /Hb-Hc(orthol - - TH - Hb lot # T ST Ab-Hd (metal 11 is is ~ MHc-Ha 1 : 1 2 : 2 1 : 1 : I 2 : 2 1 : I expected Mee le triple of doublets triplet of doublets ( + d) ( +d)
17 NMR Spectroscopy 1 H NMR of Aromatic Molecules 7.00 6.75 7.25 7.50 1.0 1.0 1.0 1.0 7.35 7.21 7.01 6.80 C - He n CH Ad Hc Hb Ha - dd dd I d Id HA - OH He