Aromatics: Long-Range Coupling H’s on aromatic rings may couple with non-neighboring protons due to long-range coupling. You will see this in lab! Why?

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Presentation transcript:

Aromatics: Long-Range Coupling H’s on aromatic rings may couple with non-neighboring protons due to long-range coupling. You will see this in lab! Why? Nuclei “communicate” via bonding electrons -  electrons that are in resonance will allow non-neighboring H’s to “communicate” and couple/split. This leads to complex splitting. Lecture 9

Nuclear Magnetic Resonance Use: To assist in the elucidation of a molecule’s structure Information Gained: Different chemical environments of nuclei being analyzed ( 1 H 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

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 1. Measure the length of the integral with a ruler 2. Establish a relative ratio of protons (divide each length by the lowest number)

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

Problem: LG 15.2 Draw the structure of a compound that fits each molecular formula and has a 1 H NMR spectrum showing a single peak (a singlet). (Hint: Consider HDI). Single peak must mean equivalent Hs! (Only one peak and no splitting.) (a) C 2 H 6 O (c) C 4 H 6 (f) C 3 H 6 O (g) C 4 H 9 Br

Problem: LG (a-d) Steps to solve problem: 1. Calculate HDI 2. Draw out possibilities of structure 3. Determine how many signals would be seen for each possibility & their splitting patterns 4. Work with the NMR data: (a) Count # of signals = the # of different types of H’s (b) Look at the splitting of each signal (c) Consider the ppm values for each signal - use correlation chart (d) Look at the integration values

LG 21.29(a) C 4 H 10 O 1.28 ppm (s, 9H) 4.5 ppm (s, 1H)

LG 21.29(b) C 3 H 7 Br 1.71 ppm (d, 6H) 4.32 ppm (m, 1H)

LG 21.29(c) C 4 H 9 Cl 1.04 ppm (d, 6H) 1.95 ppm (m, 1H) 3.35 ppm (d, 2H)

LG 21.29(d) C 8 H ppm (t, 3H) 2.68 ppm (q, 2H) 7.23 ppm (m, 5H)

LG The 1 H NMR spectrum of a compound C 3 H 3 Cl 5 shows peaks at 4.5 ppm (t, 1H) and 6.0 ppm (d, 2H). What is the compound’s structure?

IR Correlation Chart

1 H NMR Correlation Chart

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