What happens to the shielding of the proton when an alcohol is oxidized?

Question

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:

1. Removing nearby carbonyl groups
2. Adding electropositive atoms like silicon
3. Oxidation of primary alcohols with $\ce{KMnO4}$
4. 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.

Answer 1

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.