Hi John,

As Kirk says the protons are not really equivalent - if you look at the spectrum from aldrich:

http://www.sigmaaldrich.com/spectra/fnmr/FNMR000250.PDF

and zoom in close to that peak, you can just about see that you have a triplet and a doublet, with the rightmost peak of the triplet very close to the leftmost peak of the doublet. Whilst looking at this I came across this from Ray Abraham at Modgraph:

http://www.modgraph.co.uk/Downloads/scs19.pdf

which indicates that the two most similar protons are 3 and 5 (where position 1 is the aldehyde, 2 is the methoxy etc), whereas your source has 3 and 4 as similar. I don't think 4 can have the same shift as 3 - the aldehyde is strongly ortho, para (pi) electron withdrawing, so would deshield proton 4, the methoxy is sigma-withdrawing (hence high shift of carbon 2) and ortho,para (pi) donating, so will have limited effect at proton 4. Thus you expect proton 6 to have the highest shift (ortho-withdrawing effect of aldehyde, no dotating effect from methoxy), then 4 (para-withdrawing effect of aldehyde, no donating effect from methoxy), 3 has sigma-withdrawing and pi-donating effects, and 5 not much of anything, so not possible to guess at those. A quick PERCH calculation agrees with this assignment.

For aromatics changing from CDCl3 to benzene-d6 can have a aprticularly pronounced effect on the shifts.

The four aromatic protons in this molecule are, in principle, non-equivalent. However, sometimes non-equivalent protons just happen to have very similar chemical shifts. The spectrum shown on the "sciencesoft" site is fairly low resolution. A higher resolution spectrum would probably show them as non-equivalent. Changing the solvent is another good trick when dealing with cases of accidental near equivalence of chemical shifts.