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Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid

Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid . Structure determination by Infrared Spectroscopy. Infrared Spectroscopy. Read Chapter 25, pp 833 to 867 Work problems on page 866-867. Typical Infrared Absorption Regions. O-H stretch.

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Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid

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  1. Experiment 56: Synthesis of an Ester from an Alcohol and Carboxylic Acid Structure determination by Infrared Spectroscopy WWU -- Chemistry

  2. Infrared Spectroscopy • Read Chapter 25, pp 833 to 867 • Work problems on page 866-867 WWU -- Chemistry

  3. Typical Infrared AbsorptionRegions WWU -- Chemistry

  4. O-H stretch • The OH peak is very strong and broad! • It usually appears near 3400 cm-1 WWU -- Chemistry

  5. C-H stretch • sp3 C-H stretch comes at 2950 cm-1 • sp2 C-H stretch comes at 3050 cm-1 • sp C-H stretch comes at 3300 cm-1 WWU -- Chemistry

  6. C=C stretch • C=C stretch occurs near 1650 cm-1 • It is often weak unless it is conjugated to a C=O WWU -- Chemistry

  7. Normal Base Values for C=O Stretching WWU -- Chemistry

  8. C-O stretch • The C-O band appears in the range of 1300 to 1000 cm-1 • Look for one or more strong bands appearing in this range! • Ethers, alcohols, esters and carboxylic acids have C-O bands WWU -- Chemistry

  9. C-H bendings • Bendings aren’t as important as stretches. • See page 848-850 • CH2 bending: 1450 cm-1 • CH3 bending: 1375 cm-1 WWU -- Chemistry

  10. Cyclohexanol bending O-H stretch C-O stretch sp3 C-H stretch WWU -- Chemistry

  11. Ethyl Butanoate C-O stretch sp3 C-H C=O stretch WWU -- Chemistry

  12. 4-Methyl-2-pentanoneC-H < 3000, C=O @ 1715 cm-1 C-H stretch C=O stretch WWU -- Chemistry

  13. 4-Methyl-3-penten-2-one C-H stretch C=O stretch C=C stretch WWU -- Chemistry

  14. Experiment 15Spearmint and Caraway • Experiment 15, pp124-131 • Essay: pp119-123 • Technique 25 (Infrared spectroscopy) • Technique 22 (Gas Chromatography, especially, Section 22.8 on page 806 and Fig 22.7) • Technique 23 (Polarimetry) WWU -- Chemistry

  15. Spearmint and Caraway Oils Limonene is present In both oils WWU -- Chemistry

  16. The experiment • Run the IR spectrum on the two enantiomers • Run a sample of each enantiomer through the gas chromatograph • Determine the optical rotations for each enantiomer • Calculate the specific rotation for each enantiomer WWU -- Chemistry

  17. Nucleophilic Substitution: Competing Nucleophiles Experiment 21 WWU -- Chemistry

  18. Reading Assignment • pp. 180-190 (Exp 21) and 510-514 (exp 58) • Chapter 10 in your lecture textbook • Review Techniques 1 through 6 • Technique 7 (Sections 7.2, 7.4, 7.5, 7.8) • Technique 12 (Sections 12.5, 12.9, 12.11) • Technique 22 • Technique 21 (omit this quarter) WWU -- Chemistry

  19. Changes to Exp 19 laboratory • We will be using 1-pentanol, 2-pentanol, 3-pentanol and 2-methyl-2-butanol instead of what is shown in the book. • The procedure stays the same; just change the alcohols. • Start 21A first, then do 21B, and then come back to finish 21A. • Part 21C: gas chromatography only • Signup for 1, 2, and 3-pentanol WWU -- Chemistry

  20. The SN2 Mechanism WWU -- Chemistry

  21. The SN1 Mechanism carbocation 1935: Hughes & Ingold WWU -- Chemistry

  22. Assisted SN2: Mechanism WWU -- Chemistry

  23. Assisted SN1: Mechanism WWU -- Chemistry

  24. WWU -- Chemistry

  25. Gas Chromatography Results: 1-Chloro pentane and 1-Bromopentane R-Br solvent R-Cl WWU -- Chemistry

  26. Gas Chromatography Results: 1-Chloropentane and 1-Bromopentane from 1-pentanol Ret. Time Width Peak Peak Result Time Offset Area Sep. 1/2 Status No. Name () (min) (min) (counts) Code (sec) Codes ---- ------------ ---------- ------- ------- ---------- ---- ----- ------ 1 9.9306 4.579 0.000 2113 BB 1.5 2 90.0694 5.332 0.000 19166 BB 1.4 ---- ------------ ========== ------- ======= ========== ---- ----- ------ Totals: 100.0000 0.000 21279 Total Identified Counts associated with halides: 21279 counts Round off the values to: 9.9 % 1-chloropentane 90.1 % 1-bromopentane Important! Other components may appear on the printout. If so, be sure to recalculate for only 1-chloro and 1-bromopentane! Make sure that the total equals 100% Assume that all response factors = 1.000 WWU -- Chemistry

  27. Do the results of this experiment with 1-pentanol fit with what you expected? WWU -- Chemistry

  28. Gas Chromatography Results: 2-Pentanol in the solvent nucleophile mixture 2-bromopentane 3-bromopentane solvent 2-chloropentane 3-chloropentane WWU -- Chemistry

  29. Gas Chromatography Results: 2-Pentanol in the solvent nucleophile mixture Ret. Time Width Peak Peak Result Time Offset Area Sep. 1/2 Status No. Name () (min) (min) (counts) Code (sec) Codes ---- ------------ ---------- ------- ------- ---------- ---- ----- ------ 1 21.7171 4.151 0.000 4100 BV 1.4 2 11.6505 4.216 0.000 2200 VB 1.5 3 46.7971 4.852 0.000 8835 BV 1.4 4 19.8353 4.932 0.000 3745 VB 1.5 ---- ------------ ========== ------- ======= ========== ---- ----- ------ Totals: 100.0000 0.000 18880 Total Identified Counts associated with halides: 18880 counts Important! Other components may appear on the printout. If so, be sure to recalculate for only the four halides! Make sure that the total equals 100% Round off the values to: 21.7 % 2-chloropentane 11.7 % 3-chloropentane 46.8 % 2-bromopentane 19.8 % 3-bromopentane WWU -- Chemistry

  30. Why are we obtaining mixtures of halides in this reaction? Time for chalk!! WWU -- Chemistry

  31. Gas Chromatography Results: 2-Methyl-2-butanol in the solvent nucleophile mixture 2-bromo-2-methylbutane 2-chloro-2-methylbutane solvent WWU -- Chemistry

  32. Gas Chromatography Results: 2-Methyl-2-butanol in the solvent nucleophile mixture Ret. Time Width Peak Peak Result Time Offset Area Sep. 1/2 Status No. Name () (min) (min) (counts) Code (sec) Codes ---- ------------ ---------- ------- ------- ---------- ---- ----- ------ 1 44.0704 3.844 0.000 5181 BB 1.4 2 55.9296 4.491 0.000 6575 BB 1.5 ---- ------------ ========== ------- ======= ========== ---- ----- ------ Totals: 100.0000 0.000 11756 Total Identified Counts for the two halides: 11756 counts Important! Other components may appear on the printout. If so, be sure to recalculate for only the two halides! Make sure that the total equals 100% Round off the values 44.1 % 2-chloro-2-methylbutane 55.9 % 2-bromo-2-methylbutane Assume that all response factors = 1.000 WWU -- Chemistry

  33. We expect that the reaction of 2-methyl-2-butanol with the solvent nucleophile mixture to be SN1. Why didn’t it come out as a 50-50 mixture? Life is never straightforward! WWU -- Chemistry

  34. Important notice about next week’s lab lecture Friday, May 19th Bond Hall 109 2:00 to 2:50 WWU -- Chemistry

  35. Important notice: Final Exam The exam will be held in SL 110 rather than SL 130 on Thursday, June 8th 3:30-5:30 WWU -- Chemistry

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