Consider the following reaction scheme and ch | vedclass

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(C) $1$. Hydroboration-oxidation of styrene gives $P$ ($2$-phenylethanol,$C_6H_5CH_2CH_2OH$).$2$. Oxidation of $P$ with $CrO_3/H_2SO_4$ gives phenylac

Vedclass · Accepted Answer

$Q$: $2-$chloro$-2-$phenylacetic acid,$R$: $3-$phenylpropanoic acid,$S$: $1-$indanone

Vedclass · Answer

(C) $1$. Hydroboration-oxidation of styrene gives $P$ ($2$-phenylethanol,$C_6H_5CH_2CH_2OH$).$2$. Oxidation of $P$ with $CrO_3/H_2SO_4$ gives phenylacetic acid $(C_6H_5CH_2COOH)$.$3$. $HVZ$ reaction $(Cl_2/Red \ P)$ on phenylacetic acid gives $Q$ ($2$-chloro$-2-$phenylacetic acid,$C_6H_5CHClCOOH$).$4$. $P$ $(C_6H_5CH_2CH_2OH)$ reacts with $SOCl_2$ to give $C_6H_5CH_2CH_2Cl$,followed by $NaCN$ to give $C_6H_5CH_2CH_2CN$,and hydrolysis gives $R$ ($3$-phenylpropanoic acid,$C_6H_5CH_2CH_2COOH$).$5$. Intramolecular Friedel-Crafts acylation of $R$ using $conc. \ H_2SO_4$ gives $S$ ($1$-indanone).

Column-$I$ (Reactant)	Column-$II$ (Product)
$A$. $RCH_2OH \xrightarrow[(ii) H_3O^{+}]{(i) NaOH, KMnO_4} ?$	$i$. $C_6H_5COOH + C_2H_5OH$
$B$. $CH_3(CH_2)_8CH_2OH \xrightarrow{Jones reagent} ?$	$ii$. $RCOOH$
$C$. $RCONH_2 \xrightarrow{H_3O^{+}, \Delta} ?$	$iii$. $CH_3(CH_2)_8COOH$
$D$. $C_6H_5COOEt \xrightarrow{H_3O^{+}} ?$	$iv$. $RCOOH + NH_3$

Column-$I$ (Name)	Column-$II$ (Formula)	Column-$III$ (Example)
$(A)$ Ketone	$(i)$ $R-CO-OH$	$(p)$ $CH_3CHO$
$(B)$ Carboxylic acid	$(ii)$ $R-CO-H$	$(q)$ $CH_3COOH$
$(C)$ Amide	$(iii)$ $R-CO-R'$	$(r)$ $CH_3CONH_2$
$(D)$ Aldehyde	$(iv)$ $R-CO-NH_2$	$(s)$ $CH_3COC_2H_5$

Consider the following reaction scheme and choose the correct option for the major products $Q$,$R$ and $S$.

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