Question

${C_2}{H_5}Br\xrightarrow{{AgCN}}X\xrightarrow[{Zn - Hg/HCl}]{{\operatorname{Reduction}}}Y,$ Here Y is
(A) Ethyl methyl amine
(B) n-propylamine
(C) Isopropylamine
(D) Ethylamine

Answer 1

Hint:Alkyl halides undergoes nucleophilic substitution reactions in the presence of nucleophiles such as hydroxide, cyanide. Reagent -Zn - Hg/HCl is a strong reducing agent. It can be used to reduce one group into another in organic reactions.

Complete step-by-step answer:We know that, when Bromoethane is reacted as in the presence of silver cyanide, as a result we get the product as in the following reaction,
${C_2}{H_5}Br\xrightarrow{{AgCN}}{C_2}{H_5}NC + AgBr$

In the above equation we get that, when Bromoethane is reacted in the presence of Silver Cyanide, we get the product as in the form of Ethyl Cyanide.

As we get the first product(X), Now we have to find the second product(Y),
So, as completing the whole reaction as in the presence of Reduction(Zn-Hg/HCl) we get the following reaction,
${C_2}{H_5}NC\xrightarrow[{Zn - Hg/HCl}]{{\operatorname{Reduction}}}{C_2}{H_5}NHC{H_3}$

As a result, we get the second product as Ethyl methyl amine as writing the whole reaction we get the full data in the whole reaction as below,
${C_2}{H_5}Br\xrightarrow{{AgCN}}{C_2}{H_5}NC\xrightarrow[{Zn - Hg/HCl}]{{\operatorname{Reduction}}}{C_2}{H_5}NHC{H_3}$

Therefore, the correct answer of the reaction as in the product (Y) is Ethyl methyl amine.

Option ‘A’ is correct

Note: The alkyl halide ${C_2}{H_5}Br$ undergoes a nucleophilic substitution reaction in the presence of cyanide ion. The cyano group can be reduced to an amino group in the presence of reducing agents. This reaction is an example of preparation of amine from alkyl halides.