Dear all, My first question here, I hope that someone can help! In a paper describing a method to measure diffusion using multiple-quantum coherences (Zax, J.Chem.Phys, 1983, 78, 6333), I found the following two statements: 1 "(...) for N coupled protons the N-quantum transition is free of couplings." 2 "(...) except for double-quantum excitations, there is a loss of sensitivity because of inefficient transfer of magnetization." Regardless of the context of the article, can someone explain these two statements (for the second question, particularly the "except for double-quantum excitations"-part)?
asked
Daan |

Regarding 1: Consider a two proton J-coupled system. The states of the interest would be aa, ab, ba and bb. 1-quantum transitions are 1) aa->ab, 2) ba->bb, 3) aa->ba and 4) ab->bb. The presence of J-coupling modifies the energy level diagram in such a manner that the energy difference corresponding to the transition aa->ab is different from the energy difference corresponding to the transition ba->bb. This is the observable coupling/splitting of the .a->.b transition. However, there is only one 2-quantum transition in the system aa->bb. This would be true for any system with N-coupled protons, the N-quantum transition corresponds to the SINGLE transition between states a1a2...aN -> b1b2...bN and hence there is no observable 'splitting' (due to J-coupling) for this transition. Regarding 2, cannot say anything without knowledge of the contest. It may refer to the difficulty of converting N-quantum coherence to observable 1-quantum coherence, when N>2. Double quantum coherence can be converted to observable single quantum coherence with a single 90 pulse under favorable cases (e.g
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sekhar Talluri Thank you very much for your answer; completely clear. Regarding 2, you're probably right that these "difficulties" mean you can't do it with just a single pulse. - Daan (Sep 11 '12 at 00:34) |