The reagents A,B and C respectively are: | vedclass

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(A) The starting material is $HOCH_{2}-CH=CH-CHO$.Step $A$: The product is $HOCH_{2}-CH=CH-CH_{2}OH$. This indicates the reduction of the aldehyde gro

Vedclass · Accepted Answer

$NaBH_{4}$,$PCC$,$H_{2} / Pd$

Vedclass · Answer

(A) The starting material is $HOCH_{2}-CH=CH-CHO$.Step $A$: The product is $HOCH_{2}-CH=CH-CH_{2}OH$. This indicates the reduction of the aldehyde group to a primary alcohol while leaving the $C=C$ double bond intact. $NaBH_{4}$ is a selective reducing agent that reduces aldehydes/ketones but not $C=C$ double bonds.Step $B$: The product is $HOCH_{2}-CH=CH-CHO$. This indicates that the aldehyde group remains unchanged. $PCC$ (Pyridinium chlorochromate) is an oxidizing agent used to oxidize alcohols to aldehydes,but it does not affect the aldehyde group itself.Step $C$: The product is $HOCH_{2}-CH_{2}-CH_{2}-CH_{2}OH$. This indicates the reduction of both the aldehyde group and the $C=C$ double bond. $H_{2} / Pd$ is a catalytic hydrogenation reagent that reduces both aldehydes and $C=C$ double bonds.Therefore,the reagents are $A = NaBH_{4}$,$B = PCC$,$C = H_{2} / Pd$.

The reagents $A$,$B$ and $C$ respectively are:

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