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Lecture 10

Lecture 10. NMR: Integration & Coupling Constants Lab Guide Problems This week in lab: Ch 6: Procedure 1 & PreLab Due; Quiz 3 Due: Chapter 4 Final Report. Please do question #9 INSTEAD OF question #8!! Next week in lab: Ch 6: Procedure 2 Due: Chapter 5 Final Report. Splitting Summary.

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Lecture 10

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  1. Lecture 10 • NMR: Integration & Coupling Constants • Lab Guide Problems This week in lab: • Ch 6: Procedure 1 & PreLab Due; Quiz 3 • Due: Chapter 4 Final Report. Please do question #9 INSTEAD OF question #8!! Next week in lab: • Ch 6: Procedure 2 • Due: Chapter 5 Final Report

  2. Splitting Summary # of neighboring H’sMultiplet of Signal 0 Singlet (s) 1 Doublet (d) 2 Triplet (t) 3 Quartet (q) 4 Pentet (Quintet or Multiplet) (p, m) 5 Sextet (Multiplet) (m) 6 Septet (Multiplet) (m)

  3. Nuclear Magnetic Resonance Use: To assist in the elucidation of a molecule’s structure Information Gained: • Different chemical environments of nuclei being analyzed (1H nuclei): chemical shift • The number of nuclei with different chemical environments: number of signals in spectrum • The numbers of protons with the same chemical environment: integration • Determine how many protons are bonded to the same carbon: integration • Determine the number of protons that are adjacent to one another: splitting patterns • Determine which protons are adjacent to one another:coupling constants

  4. Integration • Area underneath signal; NMR machine will give integrals • First, gives the relative ratio of different types of protons in compound • Second, allows determination of actual ratio of different types of protons Measure the length of the integral with a ruler Establish a relative ratio of protons (divide each length by the lowest number)

  5. Coupling Constants (J) Protons that split each other’s peaks will have the same coupling constant or J value.

  6. Problem: LG 11.49 Draw the structure of a compound that fits each molecular formula and has a 1H NMR spectrum showing a single peak (a singlet). (Hint: Consider HDI). Single peak must mean equivalent H’s! (Only one peak and no splitting.) C2H6O (c) C4H6 (f) C3H6O (g) C4H9Br

  7. Problem: LG 11.53 (a-d) • Steps to solve problem: • Calculate HDI • Draw out possibilities of structure • Determine how many signals would be seen for each possibility & their splitting patterns 4. Work with the NMR data: • Count # of signals = the # of different types of H’s • Look at the splitting of each signal • Consider the ppm values for each signal - use correlation chart • Look at the integration values

  8. LG 11.53(a) C4H10O 1.28 ppm (s, 9H) 4.5 ppm (s, 1H) HDI = 0 Possible structures: 3. # of signals for each 4. NMR data

  9. LG 11.53(b) C3H7Br 1.71 ppm (d, 6H) 4.32 ppm (m, 1H) HDI = 0 (Replace Br with an H) Possible structures: 3. # of signals for each 4. NMR data

  10. LG 11.53(c) C4H9Cl 1.04 ppm (d, 6H) 1.95 ppm (m, 1H) 3.35 ppm (d, 2H) HDI = 0 (Replace Cl with an H) Possible structures: 3. # of signals for each 4. NMR data

  11. LG 11.53(d) C8H10 1.25 ppm (t, 3H) 2.68 ppm (q, 2H) 7.23 ppm (m, 5H) HDI = 4 Possible structures: 3. # of signals for each 4. NMR data

  12. LG 11.55 The 1H NMR spectrum of a compound C3H3Cl5 shows peaks at 4.5 ppm (t, 1H) and 6.0 ppm (d, 2H). What is the compound’s structure?

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