Nuclear magnetic resonance spectroscopy II. 13 C NMR. Reading: Pavia Chapter , 6.7, 6.11, 6.13

Transcription

1 Nuclear magnetic resonance spectroscopy II. 13 NMR Reading: Pavia hapter , 6.7, 6.11, 6.13

2 1. General - more/better/additional structural information for larger compounds -problems: a) isotopes and their natural abundance 1 H 99.95% % 1.1% nuclear spin I 0 ½ b) gyromagnetic ratio h E = hν = γ 2 B 1 H γ π 13 γ unit: radians/t for B T, ω( 1 H) 60 MHz ω( 13 ) 15 MHz much lower E needed for transition even less nuclei in excess in the lower level smaller signal - solutions: sum up more scans, use more sample, use a stronger field

3 2. hemical shift - reference signal is from 13 in TMS - δ range is 200 ppm: advantage: - signals are spread out more - more detailed information possible - order of deshielding follows that for 1 H: Table 6.1 you will need to provide this detail! N much detailed information provided: Fig. 6.2

4 3. Estimation of δ 13 - from increment systems - for alkanes δ 13 (ppm) = increments for α, β atoms and steric corrections see p. A-24 b δ 13 a = = 8.7 ppm (exp. 8.9 ppm) c d a α β γ δ 13 b = = 32.5 ppm α β γ -3.4 steric: 1, attached to 4 = 29.1 ppm (exp ppm) ok δ 13 c = = 43.5 ppm α β steric: 4, attached to 3 1 and 2 = 30.6 ppm (exp ppm) ok δ 13 d = = 44.1 ppm α β -7.5 steric: 2, attached to 4 (and 1 ) other substituents = 36.6 ppm (exp ppm) ok alkenes aromatic compounds ok

5 3. Estimation of δ 13 continued - from increment systems - for substituted alkanes - for alkenes - for substituted benzenes see Table A8.3, add to values from Table A8.2! (i.e., calculate from scratch!) see Tables A8.5 and A8.6 see Table A8.7 Work through these tables and examples on your own! You re (probably) ok if you can do the following two examples. Do not look at the answers before you have tried them!

6 Example 1 Are the assignments provided on the spectrum correct? (alculate/estimate δ 13!) Ph H 2

7 Example 1 answer Are the assignments provided on the spectrum correct? (alculate/estimate δ 13!) Ph d c b a H 2 δ 13 a = = 13.3 ppm (exp. 14 ppm) ok α β R δ 13 b = = 63.3 ppm (exp. 61 ppm) ok α α R Yes, all assignments are fine. δ 13 c : ester: ppm (exp. 171 ppm) ok δ 13 d = = 42.4 ppm (exp. 41 ppm) ok α R α Ph Ph: highest highest lowest lowest δ 13 1 = ? = ppm (exp. 136 ppm) δ 13 4 = ? = ppm (exp. 127 ppm) ipso H 3 para H 3 δ 13 2,6 = ? = ppm (exp. 130 ppm) δ 13 3,5 = ? = ppm (exp. 129 ppm) ortho H 3 meta H 3 all ok: H 3 might be the wrong choice, but it s equally wrong for all positions: error cancels

8 Example 2 For this 4 H 6 2, the 1 H NMR spectrum shows signals around 5 and 12 ppm. Which of the four possible isomers has the following spectrum? (Estimate δ 13!) Dl ppm

9 Example 2 answer For this 4 H 6 2, the 1 H NMR spectrum shows signals around 5 and 12 ppm. Which of the four possible isomers has the following spectrum? (Estimate δ 13!) I used Table A8.6; you can also use A8.4. Neither has steric corrections, yet I included the Z-correction for illustration: because of conjugation, steric hindrance is more than likely. n an exam, you would not have to consider this if I do not phrase the question accordingly. U = 2 H-= H H =-H difference is 26 ppm =-H Dl 3 H 3 an t decide: could be E or Z- crotonic acid: H E is a bit closer H ppm δ 13 (H): ppm ok δ 13 (H 3 ): = 17.2 ppm ok = H H H H = difference is 23.5 ppm ok = = = = = difference is 13.5 ppm too far off = = 37.3 too far off

10 4. Spin-spin coupling -as in 1 H NMR, but in principle two situations: - 13 coupling with 13 rare event: probability of 2 13 next to each other is very low - 13 coupling with 1 H heteronuclear coupling follows the n+1 rule 1 J coupling: H 3 13 signal split by 1 H into q H 2 H t d no proton: signal is not split, s (quaternary, R 2 =, R- N )

11 4. Spin-spin coupling continued - 13 spectra that show 1 J( 13-1 H) coupling are called proton-coupled or non-decoupled proton-coupled t t q s q?? 4 signals in here! What are they? not often recorded, because of frequent overlap between multiplets, especially if the molecule contains many (sp 3 ) atoms

12 4. Spin-spin coupling continued - 13 spectra that do not show 1 J( 13-1 H) coupling are called proton-decoupled only singlets are observed proton-coupled proton-decoupled H 2 - advantage: simpler spectra - disadvantage: important information on -H connectivity is lost

13 5. Integration - 13 spectra are not usually integrated 4 sp 2 carbon count: at least 5 sp 3 even more extreme example: - 2 ortho, 2 meta : spectrum shows 2 large peaks - 1 ipso, 1 para : spectrum shows 2 small peaks, but certainly not of the same area - make sure you always know the solvent peak, and do not count it as part of the molecule!

14 5. Integration continued - two problems are the reason: a) NE (nuclear verhauser enhancement) - in proton-decoupled spectra - intensity of a signal increases upon decoupling -signal for H 3 grows most -signal for H 2 grows less - =, ipso- or quaternary signal ends up being tiny - reason lies in the decoupling experiment Example application NE usage for peak assignment - NE also works through space = H ester amide chloride anhydride N H 3 syn anti H 3 Dl 3 2 sp 3 syn signal gets enhanced more: is closer to H at least 3

15 5. Integration continued - two problems are the reason: b) relaxation of a nucleus - FT-NMR records the FID signal follows the excited nucleus as it relaxes to the ground state de-excitation is fast for 1 H -variable for 13, depending on the environment 5 s 75 s 1 s 125 s - for a fast-relaxing nucleus, the FID is collected completely full signal is recorded - for a slow-relaxing nucleus, the FID is collected incompletely only partial signal is recorded Why don t we just wait longer between pulses???