Spin-Spin Coupling. J trans > J cis > J gem. Structure Evaluation. cis. J trans = Hz J cis = 9-12 Hz J gem = 1-3 Hz. H a. H b gem.

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1 cis R gem J trans = z J cis = 9-12 z J gem = 1-3 z trans J trans > J cis > J gem C J ab = 4-10 z

2 vinyl acetate C 3 C 3 C 3 C 3

3 vinyl acetate C 3 z : : :

4 vinyl acetate C 3 J ac = 6.2 z J bc = 14.0 z J ac = 6.2 z J bc = 14.0 z J ba = 1.5 z J ab = 1.5 z

5 crotonic acid C 3 5 C 2 C 2 C 2 3 C C 2 C 3 3 C A B C

6 C 3 5 C 2 C 2 C 3 3 C C 2 3 C C 2 A B C crotonic acid z J ac = 7.0 z J ab = 1.7 z J = 7.0, 1.7 z doublet of doublets (dd)

7 C 3 5 C 2 C 2 C 3 3 C C 2 3 C C 2 A B C crotonic acid z J = 15.5 z J = 1.6 z J = 1.6 z J = 1.6 z J = 15.5, 1.6 z doublet of quartets (dq)

8 C 3 5 C 2 C 2 C 3 3 C C 2 3 C C 2 A B C crotonic acid 15.5 z J = 15.5, 6.9 z

9 C 3 5 C 2 C 2 C 3 3 C C 2 3 C C 2 A B C crotonic acid J = 15.5 z J = 6.9 z J = 6.9 z J = 6.9 z

10 cis R gem J trans = z J cis = 9-12 z J gem = 1-3 z trans J trans > J cis > J gem C J ab = 4-10 z

11 crotonic acid J = 15.5, 6.9 z J = 15.5, 1.6 z J = 7.0, 1.7 z C 2 C 2 3 C C 2 C 3 3 C A B C J gem = 1-3 z J allyl = 1-3 J trans = J allyl = 1-3 J cis = 9-12 J allyl = 1-3 crotonic acid: J = 15.5, 7.0, 1.6 z

12 Analysis of Coupling Constants First rder Spectra Some helpful constraints: For every signal split into a multiplet, the component J-value(s) must match some other multiplet in the spectrum The distance (z) between the two outermost peaks in a multiplet is equal to the sum of each of the coupling constants The smallest J-value is typically given by the difference between the first and second peaks in the multiplet First order multiplets are symmetrically distributed about the center

13 Analysis of Coupling Constants First rder Spectra Start with the simplest multiplet ( ): is an apparent doublet of doublets (dd) Step 1: The distance between the first two lines always represents the smallest J value If the ratio of these two lines (integral) is 1:1, this J is unique; if it is 1:2, 1:3, etc. there are two or more identical smallest J s - 1:1 comes from coupling to one nucleus (e.g. doublet) - 1:2 first J comes from coupling of 2 nuclei with same coupling constant à part of a 1:2:1 pattern - 1:3 first J comes from coupling of 3 nuclei with same coupling constant à part of a 1:3:3:1 pattern Label this J small (z) = 1.80 z

14 Analysis of Coupling Constants First rder Spectra Step 2: (most difficult step for complex multiplets) Find the full set of pairs within the multiplet that are separated by J small Each pair will have a reflected partner through the center of the multiplet For pairs where one of the lines has a relative intensity >1, that line will be part of more than one pair = 1.80 z = 5.25 z = 1.64 z While these two may not seem equal, they must be matched if 1 st order

15 Analysis of Coupling Constants First rder Spectra Step 3: Find the centers of each of the pairs generated in step 2 (average) These will collectively represent a new pattern (as if the J small was selectively decoupled) As with step 1, the spacing between the first two lines of this multiplet represent the next smallest J Label this J as med-small, etc. as necessary = 6.97

16 Analysis of Coupling Constants First rder Spectra Step 4: Step 5: Find the midpoint(s) of this new pair(s), and repeat step 3 Repeat steps 2 and 3 as necessary until all J-values have been found Remember, it must be internally consistent and all the J values must add up to the difference between the outer peaks of the multiplet. CECK: The two J s we determined are 1.80 (1.64) and ( )/ = 8.69 The difference between the outermost peaks of the multiplet: = 8.69

17 Analysis of Coupling Constants First rder Spectra This process effectively generates a tree-diagram! J med = 7.0 z J small = 1.7 z (average of 1.80 and 1.64) This proton is coupled to two other protons, with coupling constants of 1.7 and 7.0; (1, dd, J = 7.0, 1.7 z)

18 Repeat the analysis with the next most complex multiplet Look for possible J-values of 1.72 and 6.97 Analyzing the proton; apparent doublet of quartets (dq): Analysis of Coupling Constants First rder Spectra Step 1&2 Step 3& z 1.64 z 1.65 z 1.64 z 1.80 z 1.64 z z 1.64 z 1.72 z 1.72 z Step z z Check: = = 20.51

19 Analysis of Coupling Constants First rder Spectra Tree diagram for : J large = 15.5 z J small = 1.64 z J small = 1.64 z J small = 1.64 z

20 Analysis of Coupling Constants First rder Spectra Tree diagram for : J med = 15.4 z J small = 6.9 z J small = 6.9 z J small = 6.9 z

21 d 4-allyloxyanisole C 3 Ar

22 d 4-allyloxyanisole C 3 expect ( )? Ar

23 d expect ( ): J ad = 4-10 z (7) Tree Diagrams C 3 J ab = 0-3 z (2) J ac = 0-3 z (2)!!

24 d C 3 4-allyloxyanisole doublet of triplets d 4.49 (1, dt, J = 5.2, 1.5 z)

25 d 4-allyloxyanisole C 3 expect ( )? Ar

26 d expect ( ): J bd = 9-12 z (10) Tree Diagrams C 3 J bc = 1-3 z (2) J ba = 0-3 z (2)!!

27 d C 3 4-allyloxyanisole doublet of quartets J = 10.3, 1.5 z

28 d 4-allyloxyanisole C 3 expect ( )? Ar

29 d expect ( ): J cd = z (16) Tree Diagrams C 3 J cb = 0-3 z (2) J ca = 1-3 z (2)!!

30 d C 3 4-allyloxyanisole doublet of quartets J = 17.3, 1.5 z

31 d 4-allyloxyanisole C 3 expect ( d )? Ar

32 d expect ( d ): J da = 4-10 z (5) - 2 Tree Diagrams C 3 J db = 9-12 z (10) J dc = z (17)!!

33 d C 3 4-allyloxyanisole d J dc 17 z J db 10 z J da1 5 z J da2 5 z!

34 d C 3 4-allyloxyanisole d doublet of doublet of triplets? J = 17.3, 10.7, 5.2 z

35 Aromatic Systems coupling constants J ab = 7-10 z J ab = 2-3 z J ab = 0-1 z ortho meta para

36 Aromatic Systems substituent effects on chemical shift C 3 C C 2 C 3 para ortho meta meta ortho & para meta ortho & para

37 Aromatic Systems N 2 N 2 N 2 N 2 d d N 2 N

38 Aromatic Systems C 3 N 2 N 2 solvent

39 ther Resources for Coupling Constant Analysis 1 st order multiplet evaluation: Mann J. Chem. Ed. 1995, 72, 614. oye J. rg. Chem. 1994, 59, pattern calculator: