In the following reaction sequence,CH_3-CHO x | vedclass

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Question

(C) In the first step,acetaldehyde $(CH_3-CHO)$ reacts with chlorine $(Cl_2)$ and calcium hydroxide $(Ca(OH)_2)$ to undergo the haloform reaction,prod

Vedclass · Accepted Answer

$C_6H_5-NC$

Vedclass · Answer

(C) In the first step,acetaldehyde $(CH_3-CHO)$ reacts with chlorine $(Cl_2)$ and calcium hydroxide $(Ca(OH)_2)$ to undergo the haloform reaction,producing chloroform $(CHCl_3)$.$2CH_3-CHO + 3Cl_2 + Ca(OH)_2 \rightarrow CHCl_3 + (HCOO)_2Ca + 2H_2O$ (Note: The reaction produces $CHCl_3$ as the haloform).In the second step,chloroform $(CHCl_3)$ reacts with aniline $(C_6H_5-NH_2)$ and alcoholic $KOH$ (Carbylamine reaction) to form phenyl isocyanide $(C_6H_5-NC)$.$C_6H_5-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} C_6H_5-NC + 3KCl + 3H_2O$

List-$I$ (Reagents used)	List-$II$ (Compound with functional group detected)
$A.$ Alkaline solution of copper sulphate and sodium citrate	$I.$ $C_6H_5CH(OH)CH_3$
$B.$ Neutral $FeCl_3$ solution	$II.$ $CH_3CH(NH_2)CH_3$
$C.$ Alkaline chloroform solution	$III.$ $C_6H_5CH_2CHO$
$D.$ Potassium iodide and sodium hypochlorite	$IV.$ $m-C_2H_5C_6H_4OH$

In the following reaction sequence,$CH_3-CHO$ $\xrightarrow{Cl_2 / Ca(OH)_2} X$ $\xrightarrow{C_6H_5-NH_2 / \text{Alc. KOH}} Y$. $Y$ is:

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