Four Questions for the chemistry and physics cognoscenti

# 1.  I went and bought “Transition Metals in the Synthesis of Complex Organic Molecules” by Hegedus and Soderberg on the advice of the chair of the chemistry department at my alma mater.  There on p. 3 is a diagram of the d electron configuration of the Transition Metals.  But there are only 8 columns.  What happened to Sc, Y, Lu, Zn, Cd, Hg?  They aren’t metals?   Is this because the d shell of these elements is empty (Sc Y, Lu) or full (Zn, Cd, Hg) ?

#2.  There’s a brand new book out on the same subject by Hartwig.  I picked it up at a local college library — certainly my best spent $35 of the year.  They have a fabulous shelf of just the new arrivals, which means that their faculty is doing my intellectual legwork.  The book is so new that it doesn’t have a review on Amazon.  Any thoughts on how good (or bad) it is?  It is 3 times the (physical) size of Hegedus. It’s probably more advanced than Hegedus, as Hartwig does not define neta(n) which Hegedus does on p. 4.

#3.  One can regard one resonance structure of the ylid  (Phenyl)3 P – CH2 as having a positive charge on the P and a negative charge on the C.  The other resonance structure has a double bond between the P and C.  If the latter, where do the ‘extra’ electrons go — presumably into a d orbital on the P.  But which one?  What is its spatial orientation?  Clayden is silent. 

#4.  Clayden admits that the details of the nuclear Overhauser effect are ‘beyond the scope of this book’  (p. 846). The Wikipedia entry is quite brief and unsatisfying.  It seems that pumping one nucleus to an excited state (and keeping it there) helps its nearby neighbors drop down from their excited state (enhancing the signal they emit, in volume, not in location).  How does this happen?  Given that the effect is quantum mechanical, there may not be an appropriate analogy at the macroscopic level.  Any suggestions of what to read?  

Finally, to James who posted a lot of answers to earlier questions.  Many thanks.  I should be able to interpolate them in the appropriate post this week, now that I’ve got things categorized.  Here are some more questions.

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  • Wavefunction  On June 29, 2010 at 4:29 pm

    NOE: The most accessible treatment I have read is from Lambert and Mazzola, “Introduction to NMR spectroscopy”. For the nuts and bolts, the standard work is THE NUCLEAR OVERHAUSER EFFECT

  • Yggdrasil  On June 29, 2010 at 9:55 pm

    Here’s a quick attempt at explaining how the NOE works. I believe the text I used when learning this is Derome’s Modern NMR techniques.

    First, some notation. We’re examining nucleus I and its coupling to nucleus S. In a NOESY experiment, you measure the NMR signal from nucleus I, then in a second experiment, you excite nucleus S then measure the NMR signal from nucleus I. If the signal from I is larger in the second experiment, you have a positive NOE. If it is smaller, the NOE is negative. If it is unchanged, the two nuclei are not coupled.

    Now, (assuming spin 1/2 nuclei) each of the two nuclei can exist in either a ground state or an excited state. Let’s denote these states as by lower case and upper case letters, respectively. In a system of two coupled spins, the system will have four states: is, Is, iS, and IS. At equilibrium, the ground state population will be greater than the excited state population by some amount (because it has a lower energy). The strength of the NMR signal from these nuclei are dependent on the population difference between the excited state and ground state. So, for nucleus I, whose signal we are measuring, the NMR signal is proportional to [P(is)-P(Is) + P(iS) – P(IS)]. (where P(is) denotes the population of state is).

    Exciting nucleus S prior to the experiment takes the system out of equilibrium: the is and Is states are underpopulated and the iS and IS states are overpopulated. If nuclei I and S are independent, this change in the distribution of nucleus S will not affect the distribution of nucleus I between the excited and ground states. The system will have only two single-quantum transitions available for relaxation iS–>is and IS–>Is, neither of which change change i spins into I spins or vice versa. However, if nuclei I and S are coupled, the system gains two new pathways for relaxation. The system can undergo “forbidden” transitions to help depopulate an overpopulated state (e.g. iS and IS) and repopulate an underpopulated state (e.g. is and Is). One such pathway is a double-quantum transition, IS–>is. This pathway changes I spins into i spins, decreasing the population of excited I nuclei, increasing the population of ground state i nuclei, and consequently increasing the NMR signal from nucleus I. Alternatively, the system can relax through a zero-quantum transition, iS–>Is, which changes i spins into I spins and decreases the NMR signal.

    Whether the double-quantum or zero-quantum pathway dominates depends on the size of the molecule. Small molecules (the kind organic chemists deal with) rotate very quickly in solution and therefore experience rapidly fluctuating EM fields from interactions with the environment. These high frequency fluctuations stimulate the higher energy double-quantum transition. Large molecules (such as proteins), on the other hand tumble very slowly in solution and therefore see very low frequency fluctuating EM fields. These low frequency fluctuations stimulate zero-quantum emission. Therefore, in general, organic chemists see positive NOEs while biochemists see negative NOEs.

    I’ll also note that the NOE is basically a dipole-induced dipole interaction (it displays a 1/r^6 distance dependence) and probably can be understood classically as well.

  • luysii  On June 30, 2010 at 12:10 pm

    Yggdrasil: Fabulous. That really helps.

    However it is my understanding that the signal we measure in NMR is the emitted by excited nuclei returning to the ground state. So although you say
    “the NMR signal is proportional to [P(is)-P(Is) + P(iS) – P(IS)]” , by these lights it should be NEGATIVELY proportional to [P(is)-P(Is) + P(iS) – P(IS)] assuming that capital letters mean the excited state and lower case letters mean the ground state. Correct?

  • Yggdrasil  On June 30, 2010 at 2:04 pm

    It is true that the NMR signal comes from excited nuclei returning to the ground state. So, it seems counterintuitive that the double-quantum transition that induces excited I nuclei to return to the ground state would increase the NMR signal. However, you must remember that after applying the pulse to saturate nucleus S but measuring the NMR signal, we apply a 180deg pulse to produce a population inversion: that is, after the 180deg pulse, the ground state and excited state populations get swapped. Thus, effects that increase the ground state population before we apply the 180deg pulse act to enhance the NMR signal.

  • Yggdrasil  On June 30, 2010 at 2:05 pm

    grammar correction: However, you must remember that after applying the pulse to saturate nucleus S but _before_ measuring the NMR signal…

  • luysii  On July 2, 2010 at 10:58 am

    Wavefunction: thanks for the reading tips. Another benefit of the annual fee of $35 at the local college library. I checked out Neuhaus and Williamson (the cheapest used on Amazon is #130+) and started in. It’s good and I learned something right off — spontaneous relaxation from the excited state is far to slow to account for the NMR spectra that we see.

    The mechanism of relaxation is a VERY big deal in medical MRI, and allows the gray matter and white matter of the brain to appear quite different on both longitudinal and transverse relaxation. I never delved into the physics back then (too busy) to understand what was going (or even what longitudinal and transverse relaxation really meant). Thanks again.

  • Yonemoto  On July 2, 2010 at 12:25 pm

    “NMR signal comes from excited nuclei returning to the ground state.”
    Not true at all.

    The NOE signal comes from excited nuclei returning to the ground state. But NMR signals, in general, are simply resonance measurements; the rotating nuclei are inducing an electrical current in a coil positioned parallel to the axis of rotation.

  • James  On July 2, 2010 at 10:52 pm

    Here’s my stab at things:
    1) good question. Scandium & yttrium are in the same column as the lanthanides and their chemistry is very similar – they like to be in the +3 oxidation state. Their chemistry is dominated by two things. First, they’re really good lewis acids. Second, when they actually possess a valence electron, they are fantastic reducing agents.That’s about it. From a transition metal chemistry standpoint, the mechanisms we write for transition metals (insertion, reductive elimination, etc) don’t really come into play for Sc/Y so they don’t get discussed along with the other transition metals.
    2) Haven’t read it, but I have no doubt the Hartwig book would be excellent. Always enjoy his talks, he goes into great detail on mechanism. As I said earlier I’d also recommend Christina White’s organometallic notes at UCIC – they are free, and fantastic.
    3) Crystal structures have been done on phosphorus ylides – I didn’t dig deeply, but in one example the P-C bond length was 1.72 A, which is intermediate between P=C (1.66) and P-C (1.80). The geometry about the P is nearly a perfect tetrahedron.

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