What happens to the shielding of the proton when an alcohol is oxidized? The full question:


  For H-NMR, the magnetic field experienced by a hydrogen atom can be
  externally applied or internally induced or some combination of the
  two. When the applied and induced fields act in concert, a proton in
  these fields is said to be deshielded. Which of the following
  processes will result in the greater deshielding of the proton? Please
  choose from one of the following options:
  
  
  Removing nearby carbonyl groups
  Adding electropositive atoms like silicon
  Oxidation of primary alcohols with $\ce{KMnO4}$ 
  Reducing terminal alkyne groups
  


This is what the question is asking me, but I'm confused because if you oxidize a primary alcohol, this gives an aldehyde or a carboxylic acid (depending on reaction conditions), and the proton in question goes into forming water. So if an H disassociates, does this equate to a total deshielding, or how is this explained with respect to proton NMR? 
I realize that water does show up on NMR spectra, but I'm unclear what I am being asked. It also obviously depends if this is primary, secondary, or tertiary alcohol.

The answer given is (3) Oxidation of primary alcohols with $\ce{KMnO4}$. I don't understand why.
The alcohol proton becomes an acid proton. Potassium permanganate oxidizes primary alcohols to carboxylic acids. In the process, the hydroxyl proton goes from 2-5 ppm to >10 ppm (generally). Therefore, since it is shifted to a higher ppm, it is being more deshielded.