Magnetic Equivalence

Two or more protons may have identical chemical shifts (chemical equivalence) but may not have the same coupling constant (J) to another proton in the molecule. In this case they are chemically equivalent but not magnetically equivalent. In this case we usually label protons with a "prime": Ha and Ha/, to indicate their chemical equivalence but still distinguish them. For example, in a para-disubstituted benzene ring, X-p-C6H4-Y, we have chemical equivalence of the two protons ortho to group X (Ha and Ha/) and of the two protons ortho to group Y (Hb and Hb) because of the symmetry of the benzene ring. There is an ortho coupling (3Jhh) of 8-10 Hz between Ha and Hb, and between Ha/ and Hb/ (Fig. 2.27), and if the two systems were completely isolated from each other and identical, we could analyze this as an AB system. In fact, many such compounds show fairly clean AB systems in the 1H spectrum, but a closer look reveals that there are other lines, although the whole pattern is symmetrical. The reason for the additional lines is that there is a significant meta (4Jhh) coupling between Ha and Ha/ and between Hb and Hb/. Even though these pairs are chemically equivalent and should not split each other, their coupling complicates the overall

Jbb'

1-2 vHz Figure 2.27

Jbb'

1-2 vHz Figure 2.27

8-10 Hz

8-10 Hz

8-10 Hz Figure 2.28

pattern. This is because Ha and Ha/ are chemically equivalent but are not magnetically equivalent since we can find a third proton, Hb, that has a different coupling to Ha (810 Hz—ortho) than it does to Ha/ (~0 Hz—para). Instead of calling this an AB system (or two identical AB systems), we have to call it an AAfBB' system, which has a more complicated analysis. Computer programs can calculate the spectrum precisely, including the additional lines, given the values of 8a, 8b, Jab, Jaa/ and Jbb/.

A more dramatic example occurs for an ortho-disubstituted benzene with two identical substituents: X-o-C6H4-X (Fig. 2.28). In this case we can label the four adjacent protons on the benzene ring as Ha, Hb, Hb/ and Ha/ in that order. The two systems are very tightly connected because Hb and Hb/ have a large coupling (ortho or 3 JHH = 8-10 Hz), similar to Jab and Ja/b. This pattern is very distorted, and in many instances there is no recognizable underlying AB pattern.

Earlier on in this chapter, the X-CH2-CH2-Y system was mentioned and predicted to give a pair of leaning triplets (A2B2 pattern). In fact, it can be much more complex because this is actually an AA'BB' system. This might seem surprising, because there is no chiral center, and the molecule can be drawn with a mirror plane interchanging Ha and Ha/ (Fig. 2.29). But again the criterion for magnetic equivalence of Ha and Ha/ is not met: The coupling from Ha to Hb is not in general the same as the coupling between

Ha/ and Hb. This is because the dihedral angle (in the conformation where X and Y are anti) between Ha and Hb is 60° (gauche), and between Ha/ and Hb it is 180° (anti). This would lead us to expect a smaller coupling constant (~5 Hz) for Jab and a larger J value (810 Hz) for Ja/b. One might argue that wehavepicked an arbitrary conformation—what about the two gauche conformations? In one of these conformations, the criterion for magnetic equivalence is approximately satisfied (both dihedral angles are gauche). But if Ha and Ha/ are not magnetically equivalent in any one of the conformations, they cannot be considered magnetically equivalent, and we have to analyze the whole system as an AA'BB' system, not an A2B2 system. In fact, in many simple cases the X-CH2-CH2-Y system gives a very complex pair of resonances, symmetrical about the center ((5a + 5b)/2) but not resembling in any way a pair of triplets leaning toward each other. In some cases, the outer lines of the individual resonances are more intense than the inner lines, giving them a very odd appearance because we always see overlap in the inner parts of a first-order multiplet, not in the outer lines.

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