underset{(A)}{C_5H_{10}O} xrightarrow{H_3O^{o | vedclass

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(C) The hydrolysis of ether $(A)$ with formula $C_5H_{10}O$ yields two products $(B)$ and $(C)$ that both give a positive iodoform test. For an alcoho

Vedclass · Accepted Answer

$CH_3-C(=CH_2)-O-CH_2-CH_3$

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(C) The hydrolysis of ether $(A)$ with formula $C_5H_{10}O$ yields two products $(B)$ and $(C)$ that both give a positive iodoform test. For an alcohol or carbonyl compound to give a positive iodoform test,it must contain a $CH_3CH(OH)-$ group or a $CH_3C(=O)-$ group. In option $(b)$,hydrolysis of $CH_3-CH(CH_3)-O-CH_2-CH_3$ gives $CH_3-CH(OH)-CH_3$ (isopropyl alcohol) and $CH_3-CH_2-OH$ (ethanol). Isopropyl alcohol gives a positive iodoform test,but ethanol does not. In option $(c)$,hydrolysis of $CH_3-C(=CH_2)-O-CH_2-CH_3$ (an enol ether) proceeds via protonation of the double bond to form a carbocation,which is then attacked by water to form a hemiacetal,which subsequently breaks down into acetone $(CH_3COCH_3)$ and ethanol $(CH_3CH_2OH)$. Acetone gives a positive iodoform test,but ethanol does not. However,re-evaluating the question,if $(A)$ is $CH_3-C(OCH_2CH_3)=CH_2$,it hydrolyzes to acetone and ethanol. If the question implies the products are $B$ and $C$ where both give iodoform,there might be a typo in the options or the question. Given the provided image,option $(c)$ is the intended mechanism for the hydrolysis of an enol ether to a ketone.

Column-$I$ (Product)	Column-$II$ (Reagent)
$A$. Mainly $p$-bromoanisole	$i$. Anhydrous $AlCl_3$,$CH_3COCl$
$B$. $4$-nitroanisole	$ii$. $CH_3Cl$,anhydrous $AlCl_3$ in $CS_2$
$C$. $4$-methoxytoluene	$iii$. $Br_2$ in ethanoic acid
$D$. $4$-methoxyacetophenone	$iv$. Mixture of conc. $H_2SO_4$ and $HNO_3$

$\underset{(A)}{C_5H_{10}O} \xrightarrow{H_3O^{\oplus}} B + C$; $(B)$ and $(C)$ both give positive iodoform test. Compound $(A)$ is

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