[bioe419]

After synthesizing benzoin from benzaldehyde, I ran a UV-VIS on benzoin and obtained a spectrum with absorption bands at 210nm and 247nm. I know that the absorption band at 247nm is attributable to the presence of the phenyl ketone group C6H5-C=O, in which the carbonyl group is conjugated with the benzene ring.  However, I am unsure of what the band at 210nm represents.  My guess is it is from the presence of the carbonyl group, but I am unsure.  Also, I don't know how to tell if the absorption bands are n-pi* or pi-pi*.

Any help would be much appreciated.
Thanks.

[ARGOS++]

Dear BioE419;

Benzene and its derivates have two bands in the region of 200nm till 320nm, but they depend very much on the solvent you use.
The bands are called B-band (B for Benzene) and E-band (E for Ethane).

The B-band is of some “stable” location between 250nm till 285nm in Methanol and depends not so strong on the substituent.
The E-band in the same solvent depends much stronger on the substituent and starts for Benzene itself by 203nm, for Toluene (a UV/Vis-Analogon to one part of Benzoin) by 207nm, till finally 300nm.
From the conjugation of Benzene follows (for me) that in most such cases in the higher region a л – л* and in the lower region an σ – σ* of   -C-C- should be observed.
But you Spectrum must be more complicated, because it contains two different Benzenes and so the E-band for the Phenyl-methyl-Ketone arrives already at ~243nm, and that's why I’m missing your report about the B-band at ~275nm with a lower ε.

I’m sorry, but NIST contains no UV/Vis-Spectrum for it.

I hope it is anyway of some help to you.

Good Luck!
                    ARGOS⁺⁺