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4 a. The structure of free-base porphyrin $(\ce{H2P}$, chemical formula: $\ce{C20H14N4})$ is shown below. Given that this molecule belongs to the $D_\mathrm{2h}$ point group (character table given above) identify the symmetries (irreps) for all the vibrational modes of this planar polyatomic molecule;

Free base porphyrin

(i) Identify the number of vibrational modes in $\ce{H2P}$. [3 marks]
(ii) Work out the reducible representation for the vibrational modes. [6 marks]
(iii) Use the reduction formula to identify the (irreps) for all the vibrational modes. [6 marks]
(iv) Identify which of the vibrational modes are Raman-active and which are IR-active. [4 marks]

In particular I'm getting stuck with figuring out the unmoved atoms when applying symmetry operations. I know i need to calculate gamma vib in order to use the reduction formula but I cannot picture how to apply the symmetry operations to calculate unmoved atoms for each symmetry class (aside from E) because it is a more complicated structure than I'm used to dealing with (h20, benzene etc)

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    $\begingroup$ One thing to think about in the question: is the free base porphyrin really D2h symmetric? $\endgroup$ – Martin - マーチン May 13 at 22:48
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If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

enter image description here

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.

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