Question

$Ca{C_2}\xrightarrow{{{H_2}O}}X\xrightarrow{{(C{u_2}C{l_2} + N{H_4}Cl)}}Y$ ; Y is :
A. $CH \equiv CCu$
B. $CuC \equiv CCu$
C. $CH \equiv C - C \equiv CH$
D. $C{H_2} = CH - C \equiv CH$

Answer 1

Hint: When water reacts with calcium carbide the by product formed is calcium hydroxide .
Also X is a substance with 2 carbon atoms whereas Y is a substance with 4 carbon atoms .

Complete step by step answer:
 When calcium carbide reacts with water the following reaction takes place
$Ca{C_2} + 2{H_2}O \to HC \equiv CH + Ca{(OH)_2}$
As we can see from the equation a mixture of ethyne and calcium hydroxide is produced . It is a method used for the preparation of ethyne . Therefore the main product of this reaction is ethyne . Hence X is $HC \equiv CH$ .
Now , when ethyne is further subjected to react with $C{u_2}C{l_2} + N{H_4}Cl$ at first chloroprene is formed , then afterwards $HCl$ molecule is eliminated and we get but-1en-3-yne as the product , that is dimerisation of ethyne takes place .
$HC \equiv CH\xrightarrow{{(C{u_2}C{l_2} + N{H_4}Cl)}}C{H_2} = CH - C \equiv CH$
Hence Y is $C{H_2} = CH - C \equiv CH$.

So, the correct answer is Option D .

Additional Information:
The ethyne which is prepared by the above method contains impurities of hydrogen sulfide and phosphine due to the contamination of calcium sulphide and calcium phosphide in calcium carbide . Hydrogen sulphide is removed by bubbling the gas through an acidified solution of copper sulphate while phosphine is removed by passing the gas through a suspension of bleaching powder . Pure acetylene is finally collected over water .

Note:
The dimerisation of ethyne is only possible under suitable conditions , it can form polymer also with a large number of repeating units of ${(CH = C - CH = CH)_n}$ .