Ultraviolet Absorption of Pyrazine Vapor Due to n — π Transition

Abstract

The near ultraviolet absorption spectrum of pyrazine vapor was photographed and measured. A tentative vibrational analysis was made for some of the bands in the longest wavelength discrete band system (3200 A system), due to the n — π transition. The spectral band feature shows that the 3200 A system is a superposition of two absorption systems which belong to different electronic transitions. It appears that this conclusion is also supported by simple theoretical considerations.

Luminescence and radiationless transitions from single vibronic levels of the isolated pyrazine molecule in the S 1(n,π*) state

“…The strongest bands are accompanied at their high-frequency side by satellite bands corresponding probably to hot transitions. 14 The excitation (for the spectral, as well as for the yield measurements) was carried out at six different wavelengths corresponding to early members of both progressions: (0,0) = 30 880 cm-1 = 324 nm, 5~ = 31 245 cm-1 = 320 nm, 6a~= 31465 cm-1 : 318 nm, 5~ 6a~= 31800 cm-1 = 314. 3 nm, 6cfo = 32 040 cm-1 = 312.…”

On the Correlation between Ionization Potentials and Excitation Energies, part III: Pyrazine

“…Koopwzans' theorem [.i] to hold in (4), i t was implied in [4] cited above that, since E ( 2 .~~) -E (2ulu) = I P (n,) -113 pZa) = A r p 0, then also B (lII3u) -E = E (n;" f-w,) -E (z.? c ?<a) = A E M 0.…”

“…All these assignments were thought t o be reasona1)lc on the basis of (4) and (5), but they became somcwhat doubtful when photoelectron spxtroscopical investigations by Gleiter ~t al. [9] revealxl A I P = -1.72 eV4 A new approach to thc present problem is duc to Ilfoovnaw ct aE.…”

“…For the following discussion we need the difference ( A I P ) between the .n,-and the n,-IP in PY and DMPY as defined in (4). These values are given in Table 2.…”

“…I n earlier work [4] it was suggested that in view of the large spatial separation bctwLen the lone pair orbitals vzl and P Z~, assumed to be localized, the symmetry-adapted linear combinations 1 n, = (nl+n2), A, -symmetry (1) If2 should have roughly the same orbital energies E , i.e. : E (ns) M E (%a).…”