Methyl alcohol (CH_3OH),ethyl alcohol (C_2H_5 | vedclass

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(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.$1$. Ethyl alcohol $(C_2H_5OH)$ contains the $CH

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Only ethyl alcohol and acetone

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(C) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.$1$. Ethyl alcohol $(C_2H_5OH)$ contains the $CH_3CH(OH)-$ group,so it gives a positive iodoform test: $C_2H_5OH + 4I_2 + 6NaOH \rightarrow CHI_3 + HCOONa + 5NaI + 5H_2O$.$2$. Acetone $(CH_3COCH_3)$ contains the $CH_3CO-$ group,so it gives a positive iodoform test: $CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$.$3$. Methyl alcohol $(CH_3OH)$ does not contain either of these groups,so it does not give the iodoform test.Therefore,only ethyl alcohol and acetone give the iodoform test.

Methyl alcohol $(CH_3OH)$,ethyl alcohol $(C_2H_5OH)$,and acetone $(CH_3COCH_3)$ were treated with iodine and sodium hydroxide solutions. Which substances will give the iodoform test?

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The above reduction can be best carried out by

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