You have to distinguish two cases: a) two equal substituents, b) two different substituents.
Case a:
Just consider the symmetry of the products.
1. ortho-disubst. benzene: 1,2-X2-C6H4
1H NMR: non-first-order spectrum of type AA'BB' or AA'XX'(see NMR textbooks), i. e. two multiplets with different chemical shifts and integrals of 2 H each; examples: 1,2-dichlorobenzene or naphthalene.
13C NMR: three signals, 1 for C-1/C-2 (quaternary C), 1 for C-3/C-6 (CH), and 1 for C-4/C-5 (CH).
2. meta-disubst. benzene: 1,3-X2-C6H4
1H NMR: H-2: triplet with J = 1-2 Hz (integral: 1 H), H-4/H-6: doublet (d) of doublets with J = 7-8 and 1-2 Hz (integral: 2 H), H-5: triplet (t) with J = 7-8 Hz (integral: 1 H).
13C NMR: four signals, 1 for C-1/C-3 (quaternary C), 1 for C-2 (CH), 1 for C-4/C-6 (CH), and 1 for C-5 (CH).
3. para-disubst. benzene: 1,4-X2-C6H4
1H NMR: all 4 H have the same environment -> a single line.
13C NMR: two signals, 1 for C-1/C-4 (quaternary C), 1 for C-2/C-3/C-5/C-6 (CH).
Case b (assuming chemical shift differences much larger than coupling constants):
1. ortho-disubst. benzene: 1-X,2-Y-C6H4
1H NMR: H-3 and H-6: each gives a separate dd with J = 7-8 and 1-2 Hz,
H-4 and H-5: each gives a separate td (or ddd) with J = 7-8 (twice) and 1-2 Hz (once).
13C NMR: six signals, 2 for Cquat, 4 for CH.
2. meta-disubst. benzene: 1-X,3-Y-C6H4
1H NMR: H-2: dd (or t) with J = 1-2 Hz (twice), H-4: dt (or ddd) with J = 7-8 (once) and 1-2 Hz (twice), H-5: t (or dd) with J = 7-8 (twice), H-6: similar to H-4.
13C NMR: six signals, 2 for Cquat, 4 for CH.
3. para-disubst. benzene: 1-X,4-Y-C6H4
1H NMR: non-first-order spectrum of type AA'XX'(very different from case a1, see NMR textbooks), i. e. two multiplets with different chemical shifts and integrals of 2 H each; example: 4-nitroaniline.
13C NMR: four signals, 1 for C-1 (quaternary C), 1 for C-2/C-6 (CH), 1 for C-3/C-5 (CH) and 1 for C-4 (quaternary C).
In addition you can estimate the 1H and 13C chemical shifts by adding substituent increments to the shift of benzene (again: see textbooks).
I hope this helps and is not too confusing for a beginner.
M.Jameel its quit easy to differentiate aromatic proton . i will prefer Pavia spectroscopy book. No need to worry. As Mr lodger mentioned above is also easy. anyway i would like to add some additional tricks. Calculte the coupling constant if it is 2 to 3 Hz then the substituent will be at Para position. If the J value is 8 to 9 hz the the substituent will be at ortho position . If the j value is 0 then the substituent will lie at postion para.
You can use 13C NMR for distinguishing o-,m- and p-sub. groups considering benzene as a aromatic substrate.
1. For ortho, groups of an aromatic molecule will give 3 signals for Carbons
in 13C NMR
2. For meta sub. groups will give 4 signals for carbons in 13C NMR
3. For Para sub. groups will give 2 signals for carbons in 13C NMR.
Dear Selin Manoj Kumar
That is not correct, it goes like this:
J constants:
3J ortho = 7 -10 Hz
4J meta = 2 - 3 Hz
5J para = 0 - 1 Hz
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