Question

Bottles containing C6H5I and C6H5CH2I lost their original labels. They were labelled A and B for testing. A and B were separately taken in a test tube and boiled with NaOH solution. The end solution in each tube was made acidic with dilute HNO3 and then some AgNO3 solution was added. Substance B gave a yellow precipitate. Which one of the following statements is true for this experiment? (a) A was C6H5I (b) A was C6H5CH2I (c) B was C6H5I (d) Addition of HNO3 was unnecessary

Answer 1

it is question of Haloalkanes and Haloarens.

Let's try to solve it.

first consider, C6H5I
C6H5I + NaOH ===> C6H5ONa

C6H5ONa + (HNO3/H+) ===> C6H5OH , benzenol which doesn't give a yellow precipitated compound when we treat benzenol with Silver nitrate, AgNO3

hence, A must be C6H5I

now took C6H5CH2I
C6H5CH2I + NaOH ===> C6H5CH2ONa

C6H5CH2ONa + (HNO3/H+) ===> C6H6CH2OH , it gives yellow precipitated compound when we treat it with silver nitrate, AgNO3. hence, B must be C6H5CH2I.

so, option (A) is correct.