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Last updated date: 02nd Dec 2023
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What are X and Y in the following reaction sequence: ${C_2}{H_5}OH\xrightarrow{{C{l_2}}}\left( X \right)\xrightarrow{{C{l_2}}}Y$  $A.{C_2}{H_5}Cl,{\text{ }}C{H_3}CHO \\ B. C{H_3}CHO,{\text{ }}C{H_3}C{O_2}H \\ C. C{H_3}CHO,{\text{ }}CC{l_3}CHO \\ D. {C_2}{H_5}Cl,{\text{ }}CC{l_3}CHO \\$

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Hint :In this particular reaction ethanol is undergoing some reaction and the product formed is $X$ which is again undergoing chlorination and forming the product $Y$ . So for solving this question we have to look into a reaction that is completing within 2-3 steps and where ethanol is in the reactant.

For solving this question let us look into our reaction carefully. It is given that ${C_2}{H_5}OH$ is reacting with $C{l_2}$ and giving some product which we need to find and that product is again reacting with $C{l_2}$ . So in the first step, ${C_2}{H_5}OH$ will react with $C{l_2}$ and here oxidation reaction will take place where $C{l_2}$ will act as an oxidizing agent and will oxidize ${C_2}{H_5}OH$ to form the product acetaldehyde or ethanal having formula $C{H_3}CHO$ . Then in the next step, this ethanal $C{H_3}CHO$ is reacting with another $C{l_2}$ molecule but here chlorination will take place and the three hydrogens of $C{H_3}CHO$ will get replaced by three chlorines, and in the product instead of three hydrogens there will be three chlorines attached to the carbon along with $CHO$ , hence the product that will be formed here is $CC{l_3}CHO$
Hence the overall reaction is: ${C_2}{H_5}OH\xrightarrow{{C{l_2}}}C{H_3}CHO\xrightarrow{{C{l_2}}}CC{l_3}CHO$
Where X and Y are $C{H_3}CHO$ and $CC{l_3}CHO$ .
Therefore correct option is $C.\;\;\;\;\;C{H_3}CHO,{\text{ }}CC{l_3}CHO$
The reaction taking place here is called a synthesis of chloroform where these steps are the initial steps of the reaction. The chlorine which is reacting here with ${C_2}{H_5}OH$ and $C{H_3}CHO$ is coming from the bleaching powder. The complete reaction of synthesis of chloroform is shown
$CaC{l_2} + {H_2}O \to Ca(OH)2 + Cl2$
${C_2}{H_5}OH\xrightarrow{{C{l_2}}}C{H_3}CHO\xrightarrow{{C{l_2}}}CC{l_3}CHO + 3HCl$
$2CC{l_3}CHO + Ca{(OH)_2} \to CHC{l_3} + {(HCOO)_2}Ca$