Aldehydes and Ketones reacts with NH_2OH to form Aldoximes and Ketoximes respectively.Configuration of these can be determined by Beckmann rearrangement as that group migrates which is anti w.r.t-OH It is interesting to note that the migration of group is completely retentive and no loss of optical activity is seen. CH_3CHO+NH_2OHoverset(Delta)toPoverset(H^(o+))toQoverset(Br_2//KOH)toR(CH_3NH_2) (as only product) Following is correct

Aldehydes and Ketones reacts with NH_2OH to form Aldoximes and Ketoxim

1.	Aldehydes and Ketones reacts with NH_2OH to form Aldoximes and Ketoximes respectively.Configuration of these can be determined by Beckmann rearrangement as that group migrates which is anti w.r.t-OH It is interesting to note that the migration of group is completely retentive and no loss of optical activity is seen. CH_3CHO+NH_2OHoverset(Delta)toPoverset(H^(o+))toQoverset(Br_2//KOH)toR(CH_3NH_2) (as only product) Following is correct
Answer» <P>Oxime P is syn form of geometrical isomers Oxime P is anti form Q is more basic than R Q is `H-undersetunderset(O)(\|\|)C-NH-CH_3` Solution :`CH_3-undersetunderset(O)(\|\|)C-H+H_2N-OHtounderset((P))(CH_3-CH=N-OH("anti"))tounderset((Q))(CH_3-undersetunderset(O)(\|\|)C-NH_2)OVERSET("Heffmann Bromamide")toCH_3-NH_2`