NONFIRST ORDER SPECTRA CHCH 2 t d p

NON-FIRST ORDER SPECTRA >CH---CH 2 t d p. 173 is only true if D (in Hz) > 10 J This is easier to achieve with large fields: d 1. 5 1. 3 D = 0. 2 ppm J = 7 Hz (constant with field) At 60 MHz, 0. 2 ppm = 12 Hz, so D /J = 12/7 = ~2 At 250 MHz, 0. 2 ppm = 50 Hz, so D /J = 50/7 = ~7 At 600 MHz, 0. 2 ppm = 120 Hz, so D /J = 120/7 = ~17

If D <10 J, spectra are NOT as predictable p. 174 1 st order not 1 st order singlet when D =0

Brucine hydrate 60 MHz p. 175

p. 175 300 MHz CH 3 O are now readily resolved

p. 175 Higher field, more 1 st order 600 MHz 300 MHz

p. 176 Δ ~ 6 still triplets, but inside lines > outside lines

p. 176 triplets hardly recognizable, outside lines v small

p. 176 no longer recognizable

Spin system labels p. 177 1 st order not 1 st order Assumes ‘infinite’ resolution… actual lines will merge

NOMENCLATURE: Spin system notation p. 178 If 1 st order conditions, i. e. D > 10 J, use letters well apart in alphabet (these arbitrary but accepted) HF = AX; Cl. PF 2 = AX 2 F 2 P(CH 3) = AM 2 X 3 CH 3 CH 2 Br = A 2 X 3 Different nuclei by DEFINITION satisfy the 1 st order requirement! If not 1 st order, use letters close in alphabet Cl. CH 2 OH = A 2 B 2 X NOTE: the order is arbitrary, e. g. A 2 M 3 X above and does not imply ANY additional information

p. 179 CHEMICAL EQUIVALENCE: use same letter CH 3 CH 2 Br = A 2 X 3 the two A atoms are the same and the three X atoms are same as each other, BUT A = X In this example, the two F are CHEMICALLY EQUIVALENT and the two H are CHEMICALLY EQUIVALENT BUT…

The two H’s (or the two F’s) are p. 179 NOT MAGNETICALLY EQUIVALENT 3 J = 2 J Because: the coupling constant of H is not the same TO EACH F The F’s are MAGNETICALLY INEQUIVALENT

same argument for the H’s p. 179 Fx does not have same J to HA as to HA’ 2 J 3 J so we call this an AA’XX’ system

TO TEST MAGNETIC EQUIVALENCE: p. 179 A) hydrogens: Pick a different COUPLED SPIN ACTIVE NUCLEUS compare J of this atom to each of the atoms you wish to test JFH is not same as JFH’ so H and H’ are MAGNETICALLY INEQUIVALENT if you get a “YES” (equivalent), test ALL nuclei !!!

p. 180 Test: HA and HA’ X HA HA’ HB HB’ Pick different spin nucleus, HC is JCA’ = JCA YES, so test again!!! HC NOW PICK HB: JBA is not same as JBA’ so HA and HA’ are NOT equivalent once you have one NO, no need to test again Test HB and HB’ NOT equivalent using HA or HA’ = AA’BB’C system

p. 180 d 6. 8 Pick one H d 7. 8 are J’s same to both H NO, so H are not equiv Likewise pick H and test H Chem shifts: Dd ~ 1 ppm (300 Hz), J=8, so (D /J) >10 so again NO AA’XX’

p. 181 Symmetry and its effects on NMR spectra What is symmetry? “A property of a physical system that allows the system to remain unchanged by a specific physical or mathematical transformation, such as translation or rotation” Or more plainly for a molecule: “A transformation of a molecule that duplicates the type and spatial arrangement of ALL atoms in space”

p. 181 Transformations that meet this criterion are called ‘symmetry operations’ and the features of the molecule that they operate on are known as ‘symmetry elements’ Symmetry Element Symmetry Operation Symbol 1. Identity ‘do nothing’ E 2. Inversion centre invert the molecule through a point i 3. Mirror plane reflect through a plane σ 4. Proper axis of rotation rotate about an axis by (360/n)° Cn 5. Improper axis of rotation rotate about an axis by (360/n)° and reflect in a plane perpendicular to the axis Sn All molecules possess the symmetry element E and many we commonly encounter also possess rotation axes (Cn) or mirror planes (σ); fewer contain inversion centres (i) or improper rotation axes (Sn).

p. 182 Example 1 BF 3 is a trigonal planar molecule with several symmetry elements including the BF 3 plane itself (not shown): The C 3 axis makes all 3 of the 19 F signal is observed in the 19 F NMR. nuclei equivalent and only one

p. 182 Example 2 Naphthalene is a flat aromatic molecule with rotation axes (shown), mirror planes that include the C 2 axes (not shown for clarity) and an inversion centre marked with a star (*) : These symmetry elements result in only 3 unique types of 1 H nuclei. 13 C nuclei and 2

p. 182 Example 3 Water has the familiar bent structure and possesses a C 2 axis as shown and two mirror planes: the plane of the page and one perpendicular to the page that bisects the H-O-H angle (not shown):

p. 182 Example 4 Methane is tetrahedral and has mirror planes (σ), proper (Cn) and improper (Sn) axes of rotation, but NO inversion centre. Symmetry Elements: E, 6 σ, 4 C 3, 3 C 2 , 3 S 4 E, 3 σ, C 3 E, σ E

p. 182 The molecule on the far right is recognizable as a CHIRAL molecule because it cannot be superimposed on its mirror image. If the molecule ITSELF had a mirror plane as a symmetry element then this would not be true. We therefore have a simple test for chirality based on symmetry elements: Chiral molecules cannot possess mirror planes, inversion centres or improper axes of rotation as symmetry elements!

p. 183 However, chiral molecules CAN possess proper axes of rotation (Cn) as illustrated below:

p. 183 How does symmetry affect NMR spectra? Basically, the presence of a symmetry element in a molecule reduces the number of resonances because carrying out the symmetry operations renders some atoms equivalent to others. Pay close attention to ANY reduction in the number of observed resonances. Symmetry can be an enormous help in solving structures if you understand the relationship between the symmetry elements present and the number of resonances. Make symmetry your friend!

p. 183 Beware of CHIRAL compounds!!! A-H C-D B-H’ C-H B-D A-H’ so H and H’ are never the same, they have different chemical shifts, so are an AB or AX and couple to each other, JAX ~ 15 Hz, they are called pro-chiral or diastereotopic hydrogens

A X D /J > 10 AB ‘quartet’ D /J < 10

p. 184 If molecule frozen 3 J values depend on dihedral angles and are different so would be AA’A’’XX’ But at room temp, there is free rotation, so all J values average to 7 Hz, so A 3 X 2 2

Inorganic systems, need to know geometry! axial F’s different from equatorial F’s If frozen, AX 2 Y 3 but in fact fluxional, so AX 5 Lower T 31 P is triplet of quartets sextet p. 184 -5

ASIDE: Interconversion of 5 -coordinate structures Berry Pseudo-rotation

In general, you need to know geometries to predict the spectra BF 3 CH 4 Pt. Cl 4 (square planar) PF 5 p. 184 SF 6

PF 3(CH 3)2 however has three isomers, which all have different spectra P = q 7 P = dtqq F = d 7 F = dt 7 H = ddt F = dtqq F = ddqq H = ddtq p. 185

SPIN DILUTE SYSTEMS p. 186 Pt has one isotope that has I = 1/2 , 195 Pt = 34% abundant so ~2/3 Pt has no spin, 1/3 has spin and so couples

p. 187 -8 1 4 1 3

ASSIGNMENT 7