Question

In order to get ethanethiol from bromoethane, the reagent used is
A.Sodium bisulphide
B.Sodium sulphide
C.Potassium thiocyanate
D.Potassium sulphide

Answer 1

Hint: We can understand that the reagent has to be bisulfide salt of alkali metal so that thiol substituted alkane and salt will be formed.

Complete step by step answer:
We have to know that the functional group thiol shows the presence of sulfur and hydrogen atoms $\left( {R - SH} \right).$ Here, R is the alkyl group (or) any organic substituent. It is given that one of the reagents is substituted alkane (or) haloalkane that is bromoethane.
We need to eliminate the bromo group and attach the thiol group with ethane. This can be done by using inorganic reagents of alkali metals (which belongs to group-1).
Now let us choose the most suitable reagent from the given options,
The formula of potassium thiocyanate is \[KSCN.\] If potassium thiocyanate reacts with bromoethane, the product formed will be ethyl thiocyanate and sodium bromide. The reaction is,
$KSCN + C{H_3}C{H_2}Br\xrightarrow{{}}C{H_3}C{H_2}SCN + KBr$
Option (D) is incorrect.
The formula of potassium sulphide is ${K_2}S.$ If potassium sulphide reacts with bromoethane, the product formed will be ethyl sulphide and sodium bromide. The reaction is,
${K_2}S + C{H_3}C{H_2}Br\xrightarrow{{}}C{H_3}C{H_2}S + KBr$
Option (C) is incorrect.
The formula of sodium sulphide is ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{\text{.}}$ If sodium sulphide reacts with bromoethane, the product formed will be ethyl sulphide and sodium bromide. The reaction is,
$N{a_2}S + C{H_3}C{H_2}Br\xrightarrow{{}}C{H_3}C{H_2}S + NaBr$
Option (B) is incorrect.
We can obtain ethanethiol from the reagent sodium bisulfide. The formula for sodium bisulfide is $NaSH$ the reaction is,
$NaSH + C{H_3}C{H_2}Br\xrightarrow{{}}C{H_3}C{H_2}SH + NaBr$
Hence, the correct option is option A.

Note:
We know that group-1 elements are alkali metals, and they form salts with non-metals (halogens). Therefore, we can use sodium or potassium salts to eliminate the bromine group present in haloalkane.