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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?

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    $\begingroup$ The aldoxime will tend to dehydrate with P2O5 or PCl5. No Beckmann. $\endgroup$
    – user55119
    Commented Jul 14, 2020 at 23:58
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    $\begingroup$ Or, A is the Beckmann product with aldoximes. $\endgroup$
    – Ben Norris
    Commented Jul 15, 2020 at 3:05
  • $\begingroup$ See beckmann fragmentation. $\endgroup$
    – Safdar Faisal
    Commented Jul 15, 2020 at 5:56

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Do you think that the mechanism of "Beckmann" and "dehydration" are different?

Answer is No. They're essentially the same reaction. Aldoximes undergoes Beckmann rearrangement to form nitriles, or you can say "they dehydrate" (as hinted by Ben Norris in comments)

The mechanism for Beckmann rearrangement is as follows,

aldoximes Beckmann rearrangement

(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.

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  • $\begingroup$ 1) The middle arrow in the intermediate is going in the wrong direction. 2) If you want phosphate to deprotonate intramolecularly, then draw the anti-oxime. 3) This is a dehydration, not a rearrangement. Atoms are not being reconnected as in the Beckmann rearrangement. $\endgroup$
    – user55119
    Commented Jul 15, 2020 at 15:52
  • $\begingroup$ @user55119: due to limited search via Google, I can't find a standard mechanism for dehydration of aldoximes. Hence, I've edited it with Beckmann rearrangement. $\endgroup$
    – Rahul Verma
    Commented Jul 15, 2020 at 17:04
  • $\begingroup$ What about Beckmann Fragmentation/abnormal Beckmann? $\endgroup$
    – Safdar Faisal
    Commented Jul 15, 2020 at 17:07
  • $\begingroup$ The top two arrows in the second structure need work. R1 with its electron pair migrates to nitrogen as water is lost. Then the electron pair on nitrogen can stabilize the carbocation. $\endgroup$
    – user55119
    Commented Jul 15, 2020 at 17:09
  • $\begingroup$ @user55119: "...Then the electron pair on nitrogen can stabilize the carbocation." - the intermediate which you're talking about is the lower one in square brackets. No? $\endgroup$
    – Rahul Verma
    Commented Jul 15, 2020 at 17:20