Beckmann rearrangement v/s dehydration of Oximes In the above question, A is an oxime, which is then reacted with phosphorus pentoxide. Now, phosphorus pentoxide is both a dehydrating reagent and a reagent used for Beckmenn rearrangement, but which reaction will form the major product mostly, ie, which reaction will dominate? My teacher told to consider beckmenn rearrangement with Phosphorus pentoxide, but the answer of this question is A, that is, dehydration is favored . What I think from my experience is that aldoximes will favor dehydration, and ketoximes beckmenn rearrangement( dehydration obviously cannot  occur in them), but I am not sure. Please tell the conditions under which any reaction will dominate over other. Also what  would be the answer if phosphorus pentoxide is replaced by phosphorus pentachloride?
Do you think that the mechanism of &quot;Beckmann&quot; and &quot;dehydration&quot; are different?
Answer is No. They're essentially the same reaction. Aldoximes undergoes Beckmann rearrangement to form nitriles, or you can say &quot;they dehydrate&quot; (as hinted by Ben Norris in comments)
The mechanism for Beckmann rearrangement is as follows,

(Source: Wikipedia, Beckmann rearrangement)
Note: Just replace $\ce{R1}$ with $\ce{H}$ and then, in the intermediate (upper one) after second step, the arrow will be from $\ce{N-R1}$ bond to $\ce{N}$. That's dehydration!
Also, if you replace $\ce{P2O5}$ by $\ce{PCl5}$ or $\ce{H2SO4}$, the answer would be same for all.