Question

For the reaction $CaOC{l_2} + {H_2}O \to Ca{\left( {OH} \right)_2} + X:$
$X + CH{}_3CHO \to Y:$
$Y + Ca{\left( {OH} \right)_2} \to CHC{l_3}$
What is $Y$ ?

Answer 1

Chloroform which is also called by its IUPAC name as trichloromethane is an organic chemical compound with the chemical formula $CHC{l_3}$ . We shall perform the reactions in order to first find out X and then Y. The compound Y is an aldehyde with chlorine attached to the alpha carbon.

Complete Step by Step answer
The final product of the final chemical equation among the given chemical equations is $CHC{l_3}$ . We know that if you follow the IUPAC naming system rules to name the compound, we can call the compound, Trichloromethane. It is an organic compound that has 1 carbon atom and 3 substituent halogen atoms attached to it. It is known better among people when addressed by its common name chloroform.
Chloroform is an organic chemical compound that is used as an anaesthetic and anxiolytic drug that can be used as antidepressants. It has a number of other uses chemically, including the formation of Teflon.
We can see that from the problem we start with $CaOC{l_2}$ and end up in formation of chloroform.
In the first step of the reaction, we can see the compound $CaOC{l_2}$ which is known as bleaching powder commonly being hydrolysed.
We have the equation,
$CaOC{l_2} + {H_2}O\xrightarrow{{}}Ca{\left( {OH} \right)_2} + X$
Balancing the equation on both sides we get,
$CaOC{l_2} + {H_2}O\xrightarrow{{}}Ca{\left( {OH} \right)_2} + C{l_2}$
Thus, now we know that $X = C{l_2}$
Now, in the second equation we can see that the compound $X$ reacts with acetaldehyde $C{H_3}CHO$ . When acetaldehyde reacts with chlorine, we can observe that the 3 alpha hydrogen atoms of the acetaldehyde will be substituted by 3 chlorine atoms.
$3C{l_2} + CH{}_3CHO\xrightarrow{{}}CC{l_3}CHO + 3HCl$
Now we know that the compound $Y$ is Trichloroacetaldehyde, which is also called chloral $(CC{l_3}CHO)$ .
Now chloral reacts with calcium hydroxide in the final step to produce chloroform $\left( {CHC{l_3}} \right)$ .
$2CC{l_3}CHO + Ca{\left( {OH} \right)_2}\xrightarrow{{}}2CHC{l_3}$
Thus, the compound $Y$ is $CC{l_3}CHO$ which is one of the intermediate compounds in the formation of chloroform from bleaching powder.

Note
This method by which we manufacture chloroform is the laboratory process of manufacturing chloroform. The chloroform solution formed is purified further when passed through caustic soda solution, to obtain it at its purest form.