Ethene \(\frac{\mathrm{O}_{2}}{\mathrm{Ag} / \Delta} \mathrm{A}\) dil HCl/\Delta \(\mathrm{B}\). The product \(\mathrm{B}\) in the above reaction is (a) Ethanol (b) Ethane- 1,2 - diol (c) Ethanal (d) epoxyethane

Ethene is an unsaturated hydrocarbon that has a double bond between two carbon atoms. Ethene has a molecular formula of \(\mathrm{C}_2\mathrm{H}_4\).

As given in the exercise, Ethene reacts with Oxygen (\(\mathrm{O}_2\)) in the presence of a Silver (Ag) catalyst and heat (\(\Delta\)). This reaction involves the addition of an oxygen atom across the double bond of Ethene. It forms an intermediate molecule known as epoxyethane or ethylene oxide: $$\mathrm{C}_2\mathrm{H}_4\ +\ \mathrm{O}_2 \xrightarrow[\Delta]{\mathrm{Ag}} \mathrm{C}_2\mathrm{H}_4\mathrm{O}$$

In the next step, epoxyethane reacts with dilute HCl in the presence of heat (\(\Delta\)). In this reaction, a hydrochloric acid molecule (HCl) undergoes hydrolysis and forms product B. The reaction can be written as: $$\mathrm{C}_2\mathrm{H}_4\mathrm{O}\ +\ \mathrm{HCl} \xrightarrow[\Delta]{} \mathrm{B}$$

To determine the final product B, let's examine each of the given options: (a) Ethanol: \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\) – Ethanol is formed by adding one hydrogen and one hydroxide (\(\mathrm{OH}\)) group to Ethene, which is not the case here. So, this option is incorrect. (b) Ethane-1,2-diol: \(\mathrm{C}_2\mathrm{H}_4(\mathrm{OH})_2\) – In this option, there are two hydroxide groups added to the molecule, which matches the addition of oxygen and HCl in our reactions. So, this is the correct option. (c) Ethanal: \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}\) – This molecule is an aldehyde, which is not formed in the given reactions. So, this option is incorrect. (d) Epoxyethane: \(\mathrm{C}_2\mathrm{H}_4\mathrm{O}\) – This is the intermediate formed in the first step, not the final product B. So, this option is incorrect. Based on our analysis, the correct option is: **B) Ethane-1,2-diol**