Question

The reagent that can be used to distinguish acetophenone and benzophenone is _______________.
A. 2,4-dinitrophenylhydrazine
B. aqueous $NaHS{O_3}$
C. Benedict’s Reagent
D. ${I_2}$ and $NaOH$

Answer 1

In order to determine the reagent which is used to distinguish between acetophenone and benzophenone, we will treat them with a reagent which on reaction with acetophenone gives a yellow precipitate and on reaction with benzophenone it does not give any precipitate.

Complete step by step answer:
Acetophenone is a chemical compound having molecular formula ${C_6}{H_5}\mathop C\limits^{\mathop \parallel \limits^O } C{H_3}$ . It is a linear structure but if we talk about benzophenone it is a chemical compound having molecular formula ${({C_6}{H_5})_2}CO$ . It has two benzene rings.
In the above given tests, the test which is used to distinguish acetophenone and benzophenone is ${I_2}$ and $NaOH$ . On reacting with acetophenone, iodine molecules react with methyl groups and form methyl iodide. Due to the formation of methyl iodide it leads to the formation of yellow precipitate and it shows the compound is acetophenone. But on the other hand, if ${I_2}$ and $NaOH$ reacts with benzophenone then it will not give any precipitate. This reaction does not take place because benzophenone has a ring which does not lead any precipitate.
Hence, option D is correct.

Note:
2,4-dinitrophenylhydrazine and benedict’s reagent are also used to identify compounds but the difference is that, 2,4-dinitrophenylhydrazine is used to detect the carbonyl group attached to aldehyde or ketone. And benedict’s reagent is used in the place of fehling solution, more precisely benedict’s reagent is used to distinguish if the sugar is reducing or non reducing. Benedict’s reagent shows positive test for reducing sugar and negative test for non reducing sugar. It is a mixture of sodium bicarbonate, sodium citrate and copper(ll) sulphate pentahydrate.