Ketone reacts with NH_2OH to give oxime. The | vedclass

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(C) The simplest ketone is acetone $(CH_3COCH_3)$.When acetone reacts with $NH_2OH$,it forms acetone oxime: $(CH_3)_2C=N-OH$.This product does not exh

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$3$

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(C) The simplest ketone is acetone $(CH_3COCH_3)$.When acetone reacts with $NH_2OH$,it forms acetone oxime: $(CH_3)_2C=N-OH$.This product does not exhibit geometrical isomerism because the two groups attached to the carbon atom are identical ($CH_3$ groups).The next higher homologue of acetone is butan$-2-$one $(CH_3COCH_2CH_3)$.When butan$-2-$one reacts with $NH_2OH$,it forms butan$-2-$one oxime: $CH_3(C_2H_5)C=N-OH$.This product exhibits geometrical isomerism because the two groups attached to the carbon atom ($CH_3$ and $C_2H_5$) are different.Thus,it forms $2$ geometrical isomers ($syn$ and $anti$ forms).Total organic products formed = $1$ (from acetone) + $2$ (from butan$-2-$one) = $3$.

Ketone reacts with $NH_2OH$ to give oxime. The reaction is as follows:
$R_2C=O + H_2N-OH \rightarrow R_2C=N-OH + H_2O$
The total number of organic products (ketoxime) formed when the simplest ketone and its next higher homologue react with $NH_2OH$,is:

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Ketone reacts with $NH_2OH$ to give oxime. The reaction is as follows:$R_2C=O + H_2N-OH \rightarrow R_2C=N-OH + H_2O$The total number of organic products (ketoxime) formed when the simplest ketone and its next higher homologue react with $NH_2OH$,is: