Chem 263 Sept. 8, 2009 NMR Continued (focusing on 1H NMR) If you apply a magnetic field (BO) to the nuclei there are two possible energy states, one with the spin aligned with the field and the other opposed. ΔE = hν (this is in the radio frequency range) !E which is in the radio frequency range Bo Absorption from the lower energy state to the higher one becomes possible, which allows a spectrum to be observed: Absorption Chemical Shift Chemical shift is the change in the frequency of absorption. The chemical shift is typically shown in units of parts per million (ppm) and is compared to a reference. For hydrogens this reference is usually tetramethylsilane (TMS). The hydrogen atoms in TMS are highly shielded and the signal for TMS is at 0 ppm. Deshielding occurs when electrons are pulled away from a hydrogen atom, which can be caused by attaching an electron withdrawing group, such as a carbonyl. When the electrons are pulled away, the nuclei is more exposed to the magnetic field resulting in a downfield chemical shift. CH3 H3C Si CH3 = TMS H CH3 R C N O H H H R H C C CH3 R C O CH2 10 9 8 7 6 5 4 3 2 1 0 (ppm) Downfield Upfield De-Shielded Shielded Intensity: 1 (2H’s) 3 (2CH3’s) Examples 3 H CH3 1 C C 2 H CH3 2-methylpropene 5 1.5 ppm The intensity depends on the number of hydrogen atoms, in this case 2:6 which is the same as 1:3 HC Cl HA HC HB C C HB HA 1-chloroethene aka chloroethylene 8 6 5 ppm vinyl chloride HB and HC are different due to the distance from the Cl; HB is trans to the Cl while HC is cis to the Cl Diastereotopic Protons 1 H3C HA C-2 is a stereogenic centre, there is no plane of symmetry in the molecule HB (S) HA and HB are in different chemical environments and have different C C chemical shifts; they are diastereotopic 2 3 H HO CH3 4 (S)-2-butanol butan-2-ol The methyl group is only one signal because the C-C bond can freely rotate and NMR cannot distinguish between the H’s. The barrier to bond rotation is only a few kilocalories per mole and about 15 to 20 kilocalories per mole of energy are available at room temperature. Hence rotation is very rapid. If you substitute Ha or Hb with deuterium (labeled as D), which is the isotope of hydrogen with one proton and one neutron, you get diastereomers; therefore, these two hydrogens are diastereotopic 1 1 H 3C HA H3C D 2 D HB 2 C C diastereomers C C 3 3 H H HO CH3 HO CH3 4 4 SS isomer SR isomer Newman Projection Round circle in structure below is back carbon (carbon 3 of 2-butanol) – carbon 2 is directly attached to it and is represented as the dot (point) where the bonds meet in front. HA (R) HB (S) H CH3 rotate H CH3 H3C HB (S) HA (R) CH3 OH OH Anti Staggered conformation Gauche conformation (most stable) When there is stereogenic center present, the two hydrogens, HA and HB, of the methylene (CH2) group cannot be brought into the same chemical environment. Bond rotation can bring HB to the position where HA was before, as represented by the above Newman projections of the two conformations, but the CH3’s are occupying different positions and therefore the two methylene H’s are not in the same chemical environment. Spin-Spin Splitting (Coupling) Ha Hb X Y X≠Y X Y HA is not a singlet due to the influence of HB on its spin and vice versa. This diagram shows the energy for HA, the spins for HB are shown in red only to show the effect on HA HA HA HB spin Bo JAB = coupling constant between atoms A and B, measured in Hertz (Hz) The HA coupling constant corresponds to the energy difference caused by the HB proton in alignment with the field and against the field. This energy difference is equal to the effect that the HA proton has on HB, so both of these coupling constants are equal. The coupled signal is centered around the same chemical shift value that the signal would have if there were no coupling. Limitations of Coupling 1. 2 to 3 bonds separating nuclei (typically not seen further apart than 3 bonds) 2. usually no coupling across O, N, S, C=O HA HB eg. C O C HA and HB would not couple due to the oxygen Note that there are exceptions to these rules Ha Hb Jab Cl Hb Jab Ha Cl Cl Hb Jab Jab 1,1,2-trichloroethane achiral (Not chiral) doublet triplet triplet doublet TMS Jab Jab Jab Ha Hb 0 (ppm) Pascal’s Triangle 1 1 1 1 2 1 1 3 31 1 4 6 4 1 above represents intensities of singlet, doublet (1 to 1), triplet (1 to 2 to 1), quartet ( 1 to 3 to 3 to 1), quintet (1 to 4 to 6 to 4 to 1) The degree of coupling (also referred to as multiplicity) can be described by the rule: 2nI + 1 where I = spin and n = the number of equivalent hydrogens that are near (2 or 3 bonds away) from the hydrogen that is being examined Since I = ½ for 1H, this rule can be simplified to the n+1 rule Consider examples below and the types hydrogens (a and b) (a) (b) (a) CH3 (b) O H3C O CH3 diethyl ether CH3 methyl tert-butyl ether or methyl 1,1-dimethylethyl ether (a) (b) (b) TMS TMS (a) 1 0 ppm 0 ppm 4 quartet triplet singlet singlet 2CH2’s 2 CH3’s 1CH3 3CH3’s Intensity would be 4:6 or 2:3 Intensity would be 3:9 or 1:3 No stereogenic centers present. No coupling across oxygen. Both molecules should have 2 signals. Both have ether functional groups. The intensity (of the entire coupled signal) is determined by the number of hydrogens that make the particular signal, for example, the quartet above is for the 2 CH2’s which is a total of 4 hydrogens, while the triplet is for the 2 CH3’s which is a total of 6 hydrogens.
Pages to are hidden for
"E which is in the radio frequency range"Please download to view full document