Question

a) An organic compound (A) of molecular formula ${{C}_{2}}{{H}_{6}}O$ liberates hydrogen with metallic sodium. Compound (A) on heating with con.${{H}_{2}}S{{O}_{4}}$ at 440K gives an alkene (B). Compound (B) when oxidized by Baeyer’s reagent gives compound (C) of molecular formula ${{C}_{2}}{{H}_{6}}{{O}_{2}}$. Identify A, B and C and explain the above reactions.
b) The metal (A) is extracted from its sulphide ore. On treatment with dilute nitric acid metal (A) gives a compound (B), which is also known as lunar caustic.(B) on heating at 723K gives (C) and ${{O}_{2}}$. Identify A, B and C and explain the above reactions.

Answer 1

Hint: In the first question it’s given that the compound A reacts with metallic sodium to give hydrogen gas. From this we can assume that A is a primary alcohol and from the given molecular formula we can find A and the following B and C also. In the second question, the ore mentioned is argentite and its sulphide ore of A. By knowing this we can find compound A and subsequently we can find B and C.

Complete step by step answer:
  a) The given question can be simplified as follows
\[A\xrightarrow[Con.{{H}_{2}}S{{O}_{4}}]{440K}B\xrightarrow[\operatorname{Re}agent]{Baeyer's}C\]
(i) It’s given that the organic compound A liberates hydrogen gas when it is treated with metallic sodium. Therefore it must be a primary alcohol and the molecular formula of the compound is given as ${{C}_{2}}{{H}_{6}}O$ .Thus the compound A is ethanol ${{C}_{2}}{{H}_{5}}OH$ and the above reaction can be written as follows
\[2{{C}_{2}}{{H}_{5}}OH+2Na\to 2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\]

(ii) The compound A that is ethanol on treatment with concentrated sulphuric acid at 440 K gives ethylene which has the molecular formula ${{C}_{2}}{{H}_{4}}$ as the product (B). The above reaction can be written as follows
\[{{C}_{2}}{{H}_{5}}OH\xrightarrow[{{H}_{2}}S{{O}_{4}}]{Conc.}C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}O\]

(iii) The compound B that is ethylene on treatment with Baeyer’s reagent gives ethylene glycol as the product (C). Baeyer’s reagent is cold dilute$KMn{{O}_{4}}$. The above reaction can be written as follows
\[C{{H}_{2}}=C{{H}_{2}}\xrightarrow{KMn{{O}_{4}}}OH-C{{H}_{2}}-C{{H}_{2}}-OH\]
 Therefore A is ethanol, B is ethylene and C is ethylene glycol.

b) (i) The metal which is extracted from sulphide ore is Silver. It’s extracted from the sulphide ore argentite $A{{g}_{2}}S$.

(ii) The metal A that is silver will react with dilute nitric acid to give silver nitrate (B) which is also known as lunar caustic. The above reaction can be written as follows
\[3Ag+4HN{{O}_{3}}\to 3AgN{{O}_{3}}+NO+2{{H}_{2}}O\]

(iii) The compound B that is silver nitrate on heating at 723 K gives silver nitrite (C) and oxygen. The reaction is written below
\[2AgN{{O}_{3}}\xrightarrow{723K}2AgN{{O}_{2}}+{{O}_{2}}\]
 Therefore A is silver, B is silver nitrate and C is silver nitrite.

Note: Do not confuse silver nitrate with silver nitrite. Silver nitrate has the molecular formula $AgN{{O}_{3}}$ and is prepared by the reaction between silver (Ag) and nitric acid whereas silver nitrite has the molecular formula $AgN{{O}_{2}}$ and it’s formed by heating silver nitrate heating at 723K.