How did you purify the sample?

In the aromatic region 7.6 is the proton between the formyl and methoxide groups. This is a singlet.

7.1 is also aromatic protons, but should be a doublet and count 2 protons, not 3, so something is wrong with the integration.

3.8 is the methoxide protons (3) and should be a singlet.

2.3 were the hydroxyl group was converted into an acetate ester (3 protons and a singlet)

~10-11 is were the formyl proton should be, but these often don't show up well (a singlet of 1 proton).

At 2.1, you have a singlet of 6 protons which matches the methyl protons on acetic acid or acetic anhydride. It is possible but unlikely that the formyl group was converted into a hemi-acetate (2 ester groups on the same carbon). I say unlikely because you normally need a strong electron withdrawing group (like CF3) also attached to the same carbon to stablize the hemi-acetate....

Go to www.sigmaaldrich.com/spectra...1609.PDF to see a FT-NMR of vanillin (the top is a carbon FT-NMR). Use that as a point of comparison.