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

List-$I$	List-$II$
$(a)$ $NH_2OH$	$(i)$ Cyanohydrin
$(b)$ $R-NH_2$	(ii) Oxime
$(c)$ $R-OH$	(iii) Schiff base
$(d)$ $H-C\equiv N$	(iv) Acetal

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

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The reagent$(s)$ which can be used to bring about the following transformation is(are):
$(A)$ $LiAlH_4$ in $(C_2H_5)_2O$
$(B)$ $BH_3$ in $THF$
$(C)$ $NaBH_4$ in $C_2H_5OH$
$(D)$ Raney $Ni / H_2$ in $THF$

What is Fehling solution?

Consider the above reaction,the products $X$ and $Y$ respectively are:

The product of reduction of benzaldehyde is

Match the reagents in List-$I$ with the products obtained from their reaction with carbonyl compounds.
List-$I$ List-$II$
$(a)$ $NH_2OH$ $(i)$ Cyanohydrin
$(b)$ $R-NH_2$ (ii) Oxime
$(c)$ $R-OH$ (iii) Schiff base
$(d)$ $H-C\equiv N$ (iv) Acetal

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