The iodoform test can differentiate between:A | vedclass

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(A) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.$A$. Methanol $(CH_3OH)$ does not give the test,

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$A$ & $E$ only

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(A) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.$A$. Methanol $(CH_3OH)$ does not give the test,while Ethanol $(CH_3CH_2OH)$ gives the test. Thus,they can be differentiated.$B$. Both $CH_3COOH$ and $CH_3CH_2COOH$ do not contain the required group,so they cannot be differentiated.$C$. Cyclohexene does not give the test,while cyclohexanone (if it had a methyl group at the alpha position) would,but simple cyclohexanone does not have the $CH_3CO-$ group. However,in many contexts,this is considered a differentiation between a ketone and an alkene,but strictly speaking,cyclohexanone does not give a positive iodoform test.$D$. Diethyl ether does not give the test,while Pentan$-3-$one $(CH_3CH_2COCH_2CH_3)$ does not contain the $CH_3CO-$ group and thus does not give the test.$E$. Anisole $(C_6H_5OCH_3)$ does not give the test,while acetone $(CH_3COCH_3)$ gives the test. Thus,they can be differentiated.Therefore,pairs $A$ and $E$ can be differentiated by the iodoform test.

The iodoform test can differentiate between:
$A$. Methanol and Ethanol
$B$. $CH_3COOH$ and $CH_3CH_2COOH$
$C$. Cyclohexene and cyclohexanone
$D$. Diethyl ether and Pentan$-3-$one
$E$. Anisole and acetone
Choose the correct answer from the options given below:

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The iodoform test can differentiate between:$A$. Methanol and Ethanol$B$. $CH_3COOH$ and $CH_3CH_2COOH$$C$. Cyclohexene and cyclohexanone$D$. Diethyl ether and Pentan$-3-$one$E$. Anisole and acetoneChoose the correct answer from the options given below: