Question

In this sequence, the gaseous product Y is:
$C{H_3}CO{C_2}{H_5}\xrightarrow{{NaOH/{I_2}}}\mathop X\limits_{yellowppt.} \xrightarrow{{Ag}}{Y_{(g)}}$
A.Ethyne
B.Ethane
C.Ethene
D.Propane

Answer 1

Hint: The haloform reaction is a chemical reaction where a haloform is produced by the exhaustive halogenation of a methyl ketone in the presence of a base. The reaction can be used to transform acetyl groups into carboxyl groups or to produce chloroform, bromoform, or iodoform .

Complete step by step answer:
When Iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed. It can be used to identify aldehydes or ketones. If an aldehyde gives a positive iodoform test, then it must be acetaldehyde.
$C{H_3}CO{C_2}{H_5}\xrightarrow{{NaOH/{I_2}}}C{H_3}CO{O^ - } + \mathop {CH{I_3}}\limits_{yellowppt.} $
So, X is $CH{I_3}$
Iodoform reacts with$Ag$ to form ACETYLENE gas.2 mole of iodoform reacts with 6 mole of Ag powder and forms 1 mole of acetylene gas and 6 mole of silver iodide.
$2CH{I_3} + 6Ag\xrightarrow{\Delta }{C_2}{H_2} + 6AgI$
Hence Y is${C_2}{H_2}$.
Hence the correct option is (A).

Note:
Acetylene (systematic name: ethyne) is the chemical compound with the formula ${C_2}{H_2}$ .It is a hydrocarbon and the simplest alkyne. This colorless gas is widely used as a fuel and a chemical building block. It is unstable in its pure form and thus is usually handled as a solution. As an alkyne, acetylene is unsaturated because its two carbon atoms are bonded together in a triple bond. The carbon–carbon triple bond places all four atoms in the same straight line.