Question

$C{H_3}CHO + N{H_2}OH\mathop \to \limits^\Delta P\mathop \to \limits^{H \oplus } Q\mathop \to \limits^{B{r_2}/KOH} R(C{H_3}N{H_2})$ (as only product)
Following is correct:
(A) Oxime $P$ is syn form of geometrical isomer
(B) Oxime $P$ is anti form
(C) $Q$ is more basic than $R$
(D) $R$ is more basic than $Q$

Answer 1

In order to answer this question, as regarding the given reaction, to know which statement is correct, we can see that the given format of the reaction is matched with the Beckmann rearrangement, so we can go through it.

Complete answer:
In Beckmann rearrangement, the group that migrates is anti with respect to hydroxyl group.
In the Beckmann rearrangement $H$ migrates. It is anti with respect to hydroxyl groups.
So, oxime $P$ is an anti form.
The Beckmann rearrangement is a rearrangement of an oxime functional group of substituted amides, named after the German chemist Ernst Otto Beckmann (1853–1923). On haloimines and nitrones, the rearrangement has also been effective. Lactams are generated by cyclic oximes and haloimines. The Beckmann solution, which is made up of acetic acid, hydrochloric acid, and acetic anhydride, was commonly used to catalyse the reaction. Other acids have been used, including sulphuric acid, polyphosphoric acid, and hydrogen fluoride.
Hence, the correct option is (B).

Note:
Antidotes for nerve agents are oxime compounds. Acetylcholinesterase is phosphorylated and made inactive by a nerve agent. By binding to phosphorus, oxime compounds can reactivate acetylcholinesterase, forming an oxime-phosphonate, which then splits away from the acetylcholinesterase molecule.