Ph-CH(OH)-CH_3 xrightarrow{PCC} (A) xrightarr | vedclass

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(B) $PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that oxidizes secondary alcohols to ketones. Thus,$(A)$ is acetophenone $(Ph-C(=O)-CH_

Vedclass · Accepted Answer

$Ph-C(CH_3)=N-NH-C(=O)-NH_2$

Vedclass · Answer

(B) $PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that oxidizes secondary alcohols to ketones. Thus,$(A)$ is acetophenone $(Ph-C(=O)-CH_3)$. Acetophenone reacts with semicarbazide $(NH_2-NH-C(=O)-NH_2)$ to form acetophenone semicarbazone. In semicarbazide,the $-NH_2$ group attached to the carbonyl group is less nucleophilic due to resonance,so the terminal $-NH_2$ group of the hydrazine part reacts with the carbonyl group of the ketone. The product $(B)$ is $Ph-C(CH_3)=N-NH-C(=O)-NH_2$.

$Ph-CH(OH)-CH_3$ $\xrightarrow{PCC} (A)$ $\xrightarrow{NH_2-NH-C(=O)-NH_2} (B)$. Product $(B)$ is:

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