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DFT / GIAO simulation of cyclopentanedione

Discussion in 'Chemistry' started by Eric J, Oct 8, 2018.

  1. Eric J

    Eric J Guest

    I'm trying to simulate the 13C NMR spectrum of cyclopentane-1,3-dione ( CAS 3859-41-4 , SDBS spectrum No. 15258, canonical smiles C1CC(=O)CC1=O). There are three equivalence classes of carbons : The two directly attached to the oxygens (~197 ppm) (call this class 1), the one carbon between those two (~105 ppm) (call this class 2) and the "other two" (~31.3 ppm) (class 3).

    I've been using Gaussian16 via the guidance provided by Dean Tantillo's group at UC Davis in particular first performing geometry optimization at some level (say B3LYP/6-31+G(d,p)) and then calculating shieldings (mPW1PW91/6-311+G(2d,p) with a solvent (scrf=(solvent=chloroform,smd )). But I keep getting shielding values that are radically off.

    In particular, regardless of the settings I arrive at isotropic shielding values of ~-34, ~136, ~144, for classes (1, 2, 3). Now of course observed chemical shifts and shieldings are often related by a linear scaling, per the Tantillo group's work (and others), which means that it's very odd that classes 2 & 3 here (the non-Oxygen-attached carbons) have nearly identical shielding values.

    I am new to chemistry and DFT methods, so I worry I might be something wrong, but it might also be the case that this is just a "hard" molecule for DFT methods to get right. I wanted to know:


    1. Are there any higher levels of theory that I should try / better methods for shielding calculation that would be worth assessing?


    2. Is there any way of analyzing the output from Gaussian16 to get a hint that things might be going awry? The only warning I see in the output (which I can provide) is


    3. Is there something that just makes this a "hard" molecule to calculate? I get the impression that the contributions in liquid state from various rotamers, etc. is relatively minor as it seems quite rigid (although I have not quantified this).

    Thank you, any and all help is appreciated.

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