Question

The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is:
A.2-butyne
B.2-butene
C.Acetylene
D.Ethene

Answer 1

We need to understand the structure of the reactant and then proceed with its reaction with silver powder. 1,1,1-trichloroethane is a 2 carbon compound with three chlorine atoms attached to the 1st carbon. Hence it is a haloalkane due to the presence of halogen Chlorine attached to an alkane which is ethane. We now study the reaction of haloalkanes with silver to predict the end product of the reaction.

Complete step by step answer:
We can write a chemical equation for when 1,1,1-trichloroethane reacts with silver powder the following takes place as,
$2C{H_3}CC{l_3}\xrightarrow[{Heat}]{{6Ag}}C{H_3}C \equiv CC{H_3} + 6AgCl$
We must remember that the 2 carbon atoms which have 3 chlorine atoms bonded to each of them form a triple bond by the removal of the 6 chlorine atoms by silver. This leads to the formation of a 4 carbon compound with a triple bond in between and the chlorine atoms are taken away by the silver to form a white precipitate of silver chloride.
Therefore, 2 molecules of 1,1,1-trichloroethane react with six molecules of silver to produce 2-butyne and silver chloride.
Hence, the correct option is option (A).

Note:
We must note that these reactions are basically used to detect the presence of a halogen and to specify which halogen is present in a given compound. For example, the white precipitate of AgCl dissolves in ammonia to give a colourless solution hence confirming the presence of chloride ions. Other halogens such as Bromine and Iodine provide a pale cream precipitate and pale yellow precipitate with silver respectively.