An organic compound [A],molecular formula C_{ | vedclass

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(B, C) The hydrolysis of an ester $[A]$ $(C_{10}H_{20}O_2)$ yields a carboxylic acid $[B]$ and an alcohol $[C]$. Since oxidation of $[C]$ gives $[B]$,

Vedclass · Accepted Answer

$CH_3-CH_2-CH(CH_3)-O-CO-CH_2-CH(CH_3)-CH_2CH_3$

Vedclass · Answer

(B, C) The hydrolysis of an ester $[A]$ $(C_{10}H_{20}O_2)$ yields a carboxylic acid $[B]$ and an alcohol $[C]$. Since oxidation of $[C]$ gives $[B]$,$[C]$ must be a primary alcohol $(R-CH_2OH)$ and $[B]$ must be the corresponding carboxylic acid $(R-COOH)$. Thus,the ester $[A]$ must be formed from the same alkyl group $R$ in both the acid and alcohol parts,i.e.,$R-COO-CH_2-R$.$1$. Option $A$: $(CH_3)_3C-COOCH_2C(CH_3)_3$ has $10$ carbons. Hydrolysis gives $(CH_3)_3C-COOH$ and $(CH_3)_3C-CH_2OH$. Oxidation of the alcohol gives the acid. This is possible.$2$. Option $B$: $CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3$ has $8$ carbons $(C_8H_{16}O_2)$. This does not match the molecular formula $C_{10}H_{20}O_2$. This is not possible.$3$. Option $C$: This structure is not an ester but an anhydride. This is not possible.$4$. Option $D$: $CH_3-CH_2-CH(CH_3)-COO-CH_2-CH(CH_3)-CH_2CH_3$ has $10$ carbons. Hydrolysis gives $CH_3-CH_2-CH(CH_3)-COOH$ and $CH_3-CH_2-CH(CH_3)-CH_2OH$. Oxidation of the alcohol gives the acid. This is possible.Therefore,structures $B$ and $C$ are not possible.

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

An organic compound $[A]$,molecular formula $C_{10}H_{20}O_2$,was hydrolyzed with dilute sulphuric acid to give a carboxylic acid $[B]$ and an alcohol $[C]$. Oxidation of $[C]$ with $CrO_3-H_2SO_4$ produced $[B]$. Which of the following structures are not possible for $[A]$?

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