$C H_{3} C H O$ and $C_{6} H_{5} C H_{2} C H O$ can be distinguished chemically by:

Benedict test

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Iodoform test

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Tollen's reagent test

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Fehling solution test

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Iodoform test
$C H_{3} C H O$ and $C_{6} H_{5} C H_{2} C H O$ react with Tollen's reagent, Fehlings solution and Benedict's solution. But $C H_{3} C H O$ reacts with $N a O H$ and $I_{2}$ to give yellow crystals of Iodoform while $C_{6} H_{5} C H_{2} C H O$ doesn't react with it. So, iodoform test is used.
$C H_{3} C H O + 3 I_{2} + 4 N a O H ⟶ C H I_{3} + H C O O N a + 3 N a I + 3 H_{2} O$
$C_{6} H_{5} C H_{2} C H O + I_{2} + 4 N a O H ⟶$ No reaction.

Suggest Corrections

Similar questions

$C H_{3} C H O$ and $C_{6} H_{5} C H_{2} C H O$ can be distinguished by:

n

C H_{3} C H O

C H_{3} C O C H_{3}

Join BYJU'S Learning Program